Contents
1
The Real Number System
1.1 Definition and Properties of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Bounded Sets, Suprema and Infima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Natural Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Sequences
2.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Bounded and Monotonic Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Limits and Continuity of Functions
3.1 Limits of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Limits at Infinity and Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Limit Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4 Continuity of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5 The Intermediate Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Differentiation
4.1 Revision – Self Study . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 The Chain Rule and Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Series
5.1 Definitions and Basic Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Convergence Tests for Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Riemann Integration
6.1 Suprema and Infima (self study) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Riemann Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.3 Refinements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4 The Riemann Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.5 Properties of the Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Introduction to Metric Spaces
7.1 Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 Complete Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3 Banach’s Fixed Point Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.4 Existence of solutions of differential equations . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.5 Power Series and the Exponential Function in 𝑀𝑛 (𝔽 ) . . . . . . . . . . . . . . . . . . . . . . . .
7.6 Solutions of Linear Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Chapter 1
The Real Number System
1.1
Definition and Properties of Real Numbers
We will give an axiomatic definition of the real numbers. We will define the set of real numbers, denoted by ℝ, as
a set with two operations + and ⋅, called addition and multiplication, as well as an ordering <, which satisfy the
laws of addition (A), the laws of multiplication (M), the distributive law (D), the order laws (O) and the Dedekind
Completeness Axiom (C). These laws and axioms will be given and discussed below.
Additional Note. An alternative way to define the real numbers is to use Peano’s axiom to define the natural
numbers, and then construct the integers and rational numbers in turn. Then the axioms (A), (M), (D) and (O)
become properties of rational numbers, so that we call them laws.
Addition and multiplication are maps which assign to every two elements in 𝑎, 𝑏 ∈ ℝ an element in ℝ which is
denoted by 𝑎+𝑏 and 𝑎⋅𝑏 (in general, written 𝑎𝑏), respectively. We require that these operations satisfy the following
axioms.
A. Axioms of addition
(A1)
Associative Law: 𝑎 + (𝑏 + 𝑐) = (𝑎 + 𝑏) + 𝑐 for all 𝑎, 𝑏, 𝑐 ∈ ℝ.
(A2)
Commutative Law: 𝑎 + 𝑏 = 𝑏 + 𝑎 for all 𝑎, 𝑏 ∈ ℝ.
(A3)
Zero: There is a real number 0 such that 𝑎 + 0 = 𝑎 for all 𝑎 ∈ ℝ.
(A4)
Additive inverse: For each 𝑎 ∈ ℝ there is −𝑎 ∈ ℝ such that 𝑎 + (−𝑎) = 0.
Notation. For 𝑎, 𝑏 ∈ ℝ one writes 𝑎 − 𝑏 ∶= 𝑎 + (−𝑏).
Additional Notes. 1. Any set with the operation + satisfying (A1), (A3), (A4) is called a group. If also (A2) is
satisfied, the group is called an Abelian (or commutative) group.
2. You will encounter detailed discussions of (Abelian) groups in algebra courses.
3. The additive inverse of a number is also called the negation of a number.
Theorem 1.1 (Basic group properties).
(a) The number 0 is unique.
(b) For all 𝑎 ∈ ℝ, the number −𝑎 is unique.
(c) For all 𝑎, 𝑏 ∈ ℝ, the equation 𝑎 + 𝑥 = 𝑏 has a unique solution. This solution is 𝑥 = 𝑏 − 𝑎.
(d) ∀𝑎 ∈ ℝ, −(−𝑎) = 𝑎.
(e) ∀𝑎, 𝑏 ∈ ℝ, −(𝑎 + 𝑏) = −𝑎 − 𝑏.
(f) −0 = 0.
Proof. (a) Let 0, 0′ ∈ ℝ such that 𝑎 + 0 = 𝑎 and 𝑎 + 0′ = 𝑎 for all 𝑎 ∈ ℝ. We must show that 0 = 0′ :
0 = 0 + 0′
= 0′ + 0
by (A2)
= 0′
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(b) Let 𝑎 ∈ ℝ and 𝑎′ , 𝑎′′ ∈ ℝ such that 𝑎 + 𝑎′ = 0 and 𝑎 + 𝑎′′ = 0. We must show that 𝑎′ = 𝑎′′ :
𝑎′ = 𝑎′ + 0
by (A3)
′
′′
= 𝑎 + (𝑎 + 𝑎 )
= (𝑎′ + 𝑎) + 𝑎′′
by (A1)
′
by (A2)
′′
= (𝑎 + 𝑎 ) + 𝑎
′′
=0+𝑎
= 𝑎′′ + 0
by (A2)
= 𝑎′′
by (A3)
(c) First we show that 𝑥 = 𝑏 − 𝑎 is a solution. So let 𝑥 = 𝑏 − 𝑎.
Then
𝑎 + 𝑥 = 𝑎 + (𝑏 − 𝑎) = 𝑎 + (𝑏 + (−𝑎))
= 𝑎 + ((−𝑎) + 𝑏)
by (A2)
= (𝑎 + (−𝑎)) + 𝑏
by (A1)
=0+𝑏
by (A4)
=𝑏+0
by (A2)
=𝑏
by (A3)
To show that the solution is unique let 𝑥 ∈ ℝ such that 𝑎 + 𝑥 = 𝑏.
Then
𝑥 = 𝑥 + 0 = 𝑥 + (𝑎 + (−𝑎))
by (A3), (A4)
= (𝑥 + 𝑎) + (−𝑎)
by (A1)
= (𝑎 + 𝑥) + (−𝑎)
by (A2)
=𝑏−𝑎
∵𝑎+𝑥=𝑏
This shows that the solution is unique.
(d) Note that
−𝑎 + (−(−𝑎)) = 0 by (A4).
On the other hand
−𝑎 + 𝑎 = 𝑎 + (−𝑎) = 0 by (A2), (A4).
By part (b), it follows that
−(−𝑎) = 𝑎.
Alternatively:
−(−𝑎) = −(−𝑎) + 0 = −(−𝑎) + (𝑎 + (−𝑎))
by (A3), (A4)
= (𝑎 + (−𝑎)) + (−(−𝑎))
by (A2)
= 𝑎 + ((−𝑎) + (−(−𝑎)))
by (A1)
=𝑎+0
=𝑎
by (A4)
by (A3)
(e)
(𝑎 + 𝑏) + (−𝑎 − 𝑏) = (𝑎 + 𝑏) + ((−𝑎) + (−𝑏))
= ((𝑏 + 𝑎) + (−𝑎)) + (−𝑏)
by (A1), (A2)
= (𝑏 + (𝑎 + (−𝑎))) + (−𝑏)
by (A1)
= (𝑏 + 0) + (−𝑏)
by (A4)
= 𝑏 + (−𝑏)
by (A3)
=0
by (A4)
By part (b), −(𝑎 + 𝑏) = −𝑎 − 𝑏.
(f) 0 + 0 = 0 by (A3), so that −0 = 0 by (A4) and (b).
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M. Axioms of multiplication
(M1)
Associative Law: 𝑎(𝑏𝑐) = (𝑎𝑏)𝑐 for all 𝑎, 𝑏, 𝑐 ∈ ℝ.
(M2)
Commutative Law: 𝑎𝑏 = 𝑏𝑎 for all 𝑎, 𝑏 ∈ ℝ.
(M3)
One: There is a real number 1 such that 1 ≠ 0 and 𝑎 ⋅ 1 = 𝑎 for all 𝑎 ∈ ℝ.
(M4)
Multiplicative inverse: For each 𝑎 ∈ ℝ with 𝑎 ≠ 0 there is 𝑎−1 ∈ ℝ such that 𝑎𝑎−1 = 1.
D. The distributive law axiom
(D)
Distributive Law: 𝑎(𝑏 + 𝑐) = 𝑎𝑏 + 𝑎𝑐 for all 𝑎, 𝑏, 𝑐 ∈ ℝ.
Additional Notes. 1. Any set with the operations +, ⋅ satisfying the axioms (A1)–(A4), (M1)–(M4) and (D) is
called a field.
2. You will encounter detailed discussions of fields in algebra courses.
3. The multiplicative inverse of a number is also briefly called the inverse of a number.
4. The set of nonzero real numbers, ℝ ⧵ {0}, is an Abelian group with respect to multiplication.
Theorem 1.2 (Basic field properties: Distributive laws).
(a) ∀𝑎, 𝑏, 𝑐 ∈ ℝ, (𝑎 + 𝑏)𝑐 = 𝑎𝑐 + 𝑏𝑐.
(b) ∀𝑎 ∈ ℝ, 𝑎 ⋅ 0 = 0.
(c) ∀𝑎, 𝑏 ∈ ℝ, 𝑎𝑏 = 0 ⇔ 𝑎 = 0 or 𝑏 = 0.
(d) ∀𝑎, 𝑏 ∈ ℝ, (−𝑎)𝑏 = −(𝑎𝑏).
(e) ∀𝑎 ∈ ℝ, (−1)𝑎 = −𝑎.
(f) ∀𝑎, 𝑏 ∈ ℝ, (−𝑎)(−𝑏) = 𝑎𝑏.
Proof. (a) By (M2) and (D),
(𝑎 + 𝑏)𝑐 = 𝑐(𝑎 + 𝑏) = 𝑐𝑎 + 𝑐𝑏 = 𝑎𝑐 + 𝑏𝑐.
(b) By (A3) and (D),
𝑎 ⋅ 0 = 𝑎(0 + 0) = 𝑎 ⋅ 0 + 𝑎 ⋅ 0,
and by (A3), (A2), 𝑎 ⋅ 0 = 0 + 𝑎 ⋅ 0. Then Theorem 1.1,(c) gives 𝑎 ⋅ 0 = 0.
(c) If 𝑏 = 0, then 𝑎𝑏 = 𝑎 ⋅ 0 = 0 by (b).
If 𝑎 = 0, then 𝑎𝑏 = 𝑏𝑎 = 𝑏 ⋅ 0 = 0 by (M2) and (b).
Now assume that 𝑎𝑏 = 0. If 𝑏 = 0, the property “𝑎 = 0 or 𝑏 = 0” follows. So now assume 𝑏 ≠ 0. Then
𝑎 = 𝑎 ⋅ 1 = 𝑎(𝑏𝑏−1 ) = (𝑎𝑏)𝑏−1 = 0 ⋅ 𝑏−1 = 0
by (M3), (M4), (M1), (M2) and (b).
(d) Using field laws, we get
𝑎𝑏 + (−𝑎)𝑏 = (𝑎 + (−𝑎))𝑏 = 0 ⋅ 𝑏 = 0,
and from Theorem 1.1(b), (−𝑎)𝑏 = −𝑎𝑏.
(e) is a special case of (d).
(f) From (d) and other laws and rules (which one’s?) we find
(−𝑎)(−𝑏) = −[𝑎(−𝑏)] = −[(−𝑏)𝑎] = −[−𝑏𝑎] = 𝑏𝑎 = 𝑎𝑏.
Theorem 1.3 (Basic field properties: multiplication).
(a) The number 1 is unique.
(b) For all 𝑎 ∈ ℝ with 𝑎 ≠ 0, the number 𝑎−1 is unique.
(c) For all 𝑎, 𝑏 ∈ ℝ with 𝑎 ≠ 0, the equation 𝑎𝑥 = 𝑏 has a unique solution. This solution is 𝑥 = 𝑎−1 𝑏.
(d) ∀𝑎 ∈ ℝ ⧵ {0}, (𝑎−1 )−1 = 𝑎.
(e) ∀𝑎, 𝑏 ∈ ℝ ⧵ {0}, (𝑎𝑏)−1 = 𝑎−1 𝑏−1 .
(f) ∀𝑎 ∈ ℝ ⧵ {0}, (−𝑎)−1 = −𝑎−1 .
(g) 1−1 = 1.
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Proof. See tutorials. The proofs are similar to those of Theorem 1.1.
𝑏
1
Notation. 𝑎−1 𝑏 is also written as , In particular, 𝑎−1 = .
𝑎
𝑎
Tutorial 1.1.1. 1. Prove Theorem 1.3.
Next we give the axioms for the set of positive real numbers. It is convenient to use the notation 𝑎 > 0 for positive
numbers 𝑎.
O. The order axioms
(O1) Trichotomy: For each 𝑎 ∈ ℝ, exactly one of the following statements is true:
𝑎 > 0 or 𝑎 = 0 or −𝑎 > 0.
(O2)
If 𝑎 > 0 and 𝑏 > 0, then 𝑎 + 𝑏 > 0.
(O3)
If 𝑎 > 0 and 𝑏 > 0, then 𝑎𝑏 > 0.
The definition of positivity of real numbers gives rise to an order relation for real numbers:
Definition. Let 𝑎, 𝑏 ∈ ℝ. Then 𝑎 is called larger than 𝑏, written 𝑎 > 𝑏, if 𝑎 − 𝑏 > 0.
Notes. 1. Since 𝑎 − 0 = 𝑎, the notation 𝑎 > 0 is consistent.
2. It is convenient to introduce the following notations:
𝑎 ≥ 𝑏 ⇔ 𝑎 > 𝑏 or 𝑎 = 𝑏,
𝑎 < 𝑏 ⇔ 𝑏 > 𝑎,
𝑎 ≤ 𝑏 ⇔ 𝑎 < 𝑏 or 𝑎 = 𝑏.
3. We will define general powers later. Below we use the notation 𝑎2 = 𝑎 ⋅ 𝑎.
Theorem 1.4 (Basic order properties). Let 𝑎, 𝑏, 𝑐, 𝑑 ∈ ℝ. Then
(a) 𝑎 < 0 ⇔ −𝑎 > 0.
(b) 𝑎 < 𝑏 and 𝑏 < 𝑐 ⇒ 𝑎 < 𝑐.
(c) 𝑎 < 𝑏 ⇒ 𝑎 + 𝑐 < 𝑏 + 𝑐.
(d) 𝑎 < 𝑏 and 𝑐 < 𝑑 ⇒ 𝑎 + 𝑐 < 𝑏 + 𝑑.
(e) 𝑎 < 𝑏 and 𝑐 > 0 ⇒ 𝑐𝑎 < 𝑐𝑏.
(f) 0 ≤ 𝑎 < 𝑏 and 0 ≤ 𝑐 < 𝑑 ⇒ 𝑎𝑐 < 𝑏𝑑.
(g) 𝑎 < 𝑏 and 𝑐 < 0 ⇒ 𝑐𝑎 > 𝑐𝑏.
(h) 𝑎 ≠ 0 ⇒ 𝑎2 > 0.
(i) 𝑎 > 0 ⇒ 𝑎−1 > 0 and 𝑎 < 0 ⇒ 𝑎−1 < 0.
(j) 0 < 𝑎 < 𝑏 ⇒ 𝑏−1 < 𝑎−1 .
(k) 1 > 0.
Proof. (a)
𝑎<0⇔0>𝑎
by definition of <
⇔0−𝑎>0
by definition of 0 > 𝑎
⇔ −𝑎 > 0
∵ 0 − 𝑎 = −𝑎 + 0 = −𝑎
(b)
𝑎 < 𝑏 and 𝑏 < 𝑐 ⇒ 𝑏 − 𝑎 > 0 and 𝑐 − 𝑏 > 0
by definition
⇒ (𝑏 − 𝑎) + (𝑐 − 𝑏) > 0
by (O2)
⇒𝑐−𝑎>0
⇒𝑎<𝑐
by (A1)-(A4)
by definition
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(h) Since 𝑎 ≠ 0, either 𝑎 > 0 or 𝑎 < 0.
If 𝑎 > 0, then 𝑎2 = 𝑎𝑎 > 0 by (O3).
If 𝑎 < 0, then −𝑎 > 0 by (a) and 𝑎2 = (−𝑎)(−𝑎) by Theorem 1.2, (f).
Hence 𝑎2 = (−𝑎)(−𝑎) > 0 by (O3).
(i) Since 𝑎 ≠ 0, 𝑎−1 exists with 𝑎𝑎−1 = 1 by (M4). Then 𝑎−1 ≠ 0 by Theorem 1.2, (c). Hence (𝑎−1 )2 > 0 by (h).
Thus, if 𝑎 > 0,
𝑎−1 = 𝑎(𝑎−1 )2 > 0 by (O3).
Similarly, use (g) if 𝑎 < 0.
(j) By (i), 𝑎−1 > 0 and 𝑏−1 > 0. Hence 𝑎−1 𝑏−1 > 0 by (O3). Then
𝑏−1 = 𝑎(𝑎−1 𝑏−1 ) < 𝑏(𝑎−1 𝑏−1 )
−1
−1
= (𝑏𝑏 )𝑎
by (e)
−1
=𝑎 .
(k) 1 = 1 ⋅ 1 = 12 > 0 by (h). (c)–(g): See tutorials.
Note. There is still one axiom missing, the axiom of Dedekind completeness. However, we will postpone the
formulation of this axiom to the next section since we need some further definitions.
Tutorial 1.1.2. 1. Prove Theorem 1.4, (c)–(g).
2. The absolute value function. Define the following function on ℝ:
|𝑥| =
{
𝑥
if 𝑥 ≥ 0,
−𝑥
if 𝑥 < 0.
Prove the following statements for 𝑥, 𝑦 ∈ ℝ:
(a) |𝑥| ≥ 0,
(b) |𝑥𝑦| = |𝑥| |𝑦|,
(c) |𝑦| < 𝑥 ⇔ −𝑥 < 𝑦 < 𝑥,
(d) |𝑥 + 𝑦| ≤ |𝑥| + |𝑦|.
3. Let 𝑥, 𝑦, 𝑧 ∈ ℝ. Which of the following statements are true and which are false?
(a) 𝑥 ≤ 𝑦 ⇒ 𝑥𝑧 ≤ 𝑦𝑧,
(b) 0 < 𝑥 ≤ 𝑦 ⇒ 1𝑦 ≤ 𝑥1 ,
(c) 𝑥 < 𝑦 < 0 ⇒ 1𝑦 < 𝑥1 ,
(d) 𝑥2 < 1 ⇒ 𝑥 < 1,
(e) 𝑥2 < 1 ⇒ −1 < 𝑥 < 1,
(f) 𝑥2 > 1 ⇒ 𝑥 > 1.
4. In each of the following questions fill in the
𝑎
(a) 𝑎 ≥ 3 ⇒ 𝑎−2
,
7
7
3
(b) 𝑎 ≥ 1 ⇒ 𝑎+1
3
,
𝑎
(c) 𝑎 > 1 ⇒ 𝑎9
10
,
𝑎−1
(d) 𝑎 > 1 ⇒ 𝑎12
1
,
𝑎
(e) 𝑎 ≥ 2 ⇒ 𝑎21−1
1
,
𝑎
(f) 𝑎 > 3 ⇒ −3
𝑎
−2
.
𝑎−1
with < or >.
5. Let 𝑥 ≥ 0 and 𝑦 ≥ 0. Show that 𝑥 < 𝑦 ⇔ 𝑥2 < 𝑦2 .
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Bounded Sets, Suprema and Infima
Definition 1.1. Let 𝑆 be a nonempty subset of ℝ. Then
1. If there is a number 𝐴 ∈ ℝ such that 𝑥 ≤ 𝐴 for all 𝑥 ∈ 𝑆, then 𝐴 is said to be an upper bound of 𝑆, and 𝑆 is
said to be bounded above.
2. If there is a number 𝑎 ∈ ℝ such that 𝑥 ≥ 𝑎 for all 𝑥 ∈ 𝑆, then 𝑎 is said to be a lower bound of 𝑆, and 𝑆 is said
to be bounded below.
3. If 𝑆 is bounded above and bounded below, then 𝑆 is called bounded.
Note. Recall the definition of intervals. [Note for lecturers: We do not need to write this down. Students have seen
this in first year. Just refer students to the notes in case they have forgotten the definitions.] Let 𝑎, 𝑏. Then
(𝑎, 𝑏) = {𝑥 ∈ ℝ ∶ 𝑎 < 𝑥 < 𝑏},
[𝑎, 𝑏] = {𝑥 ∈ ℝ ∶ 𝑎 ≤ 𝑥 ≤ 𝑏},
(𝑎, 𝑏] = {𝑥 ∈ ℝ ∶ 𝑎 < 𝑥 ≤ 𝑏},
[𝑎, 𝑏) = {𝑥 ∈ ℝ ∶ 𝑎 ≤ 𝑥 < 𝑏},
(−∞, 𝑏) = {𝑥 ∈ ℝ ∶ 𝑥 < 𝑏},
(−∞, 𝑏] = {𝑥 ∈ ℝ ∶ 𝑥 ≤ 𝑏},
(𝑎, ∞) = {𝑥 ∈ ℝ ∶ 𝑎 < 𝑥},
[𝑎, ∞) = {𝑥 ∈ ℝ ∶ 𝑎 ≤ 𝑥}.
Example 1.2.1. Let 𝑆 = (−∞, 1). Does 𝑆 have upper and lower bounds?
Solution. Since 𝑥 < 1 for all 𝑥 ∈ 𝑆, 1 is an upper bound of 𝑆, and 𝑆 is therefore bounded above. But also 2 and
7.3 are upper bounds of 𝑆, for example.
Assume that 𝑆 has a lower bound 𝑚. Then 𝑚 ≤ 0 since 0 ∈ 𝑆, and 𝑚 − 1 < 𝑚 ≤ 0 < 1 shows that 𝑚 − 1 ∈ 𝑆. So
𝑚 ≤ 𝑚 − 1 since 𝑚 is a lower bound of 𝑆. But this is false, so that 𝑚 cannot be a lower bound of 𝑆. Therefore 𝑆
has no lower bound, and 𝑆 is therefore not bounded below.
Definition 1.2. Let 𝑆 be a nonempty subset of ℝ.
1. If 𝑆 has an upper bound 𝑀 which is an element of 𝑆, then 𝑀 is called the greatest element or maximum of
𝑆, and we write 𝑀 = max 𝑆.
2. If 𝑆 has a lower bound 𝑚 which is an element of 𝑆, then 𝑚 is called the least element or minimum of 𝑆, and
we write 𝑚 = min 𝑆.
From the definition, we immediately obtain
Proposition 1.5. Let 𝑆 be a nonempty subset of ℝ. Then
1. 𝑀 = max 𝑆 ⇔ 𝑀 ∈ 𝑆 and 𝑥 ≤ 𝑀 for all 𝑥 ∈ 𝑆,
2. 𝑚 = min 𝑆 ⇔ 𝑚 ∈ 𝑆 and 𝑥 ≥ 𝑚 for all 𝑥 ∈ 𝑆.
We have already implicitly used in our notation that maximum and minimum are unique if they exist. The next
result states this formally.
Proposition 1.6. Let 𝑆 be a nonempty subset of ℝ. If maximum or minimum of 𝑆 exist, then they are unique.
Proof. Assume that 𝑆 has maxima 𝑀1 and 𝑀2 . We must show 𝑀1 = 𝑀2 . Since 𝑀1 ∈ 𝑆 and 𝑀2 is an upper
bound of 𝑆, we have 𝑀1 ≤ 𝑀2 . Since 𝑀2 ∈ 𝑆 and 𝑀1 is an upper bound of 𝑆, we have 𝑀2 ≤ 𝑀1 . Hence
𝑀1 = 𝑀2 .
A similar proof holds for the minimum.
Example 1.2.2. Let 𝑎, 𝑏 ∈ ℝ, 𝑎 < 𝑏, and 𝑆 = [𝑎, 𝑏). Then min 𝑆 = 𝑎, but 𝑆 has no maximum.
Solution. Clearly, 𝑎 ∈ 𝑆 and 𝑎 ≤ 𝑥 for all 𝑥 ∈ 𝑆, so that 𝑎 = min 𝑆.
𝑀 +𝑏
Assume 𝑆 has a maximum 𝑀. Since 𝑀 ∈ 𝑆, 𝑎 ≤ 𝑀 < 𝑏. Put 𝑐 =
. Then
2
𝑎≤𝑀 =
(1 + 1)𝑀
𝑀 +𝑀
𝑀 +𝑏 𝑏+𝑏
2𝑀
=
=
<
<
= 𝑏.
2
2
2
2
2
Therefore
𝑎 ≤ 𝑀 < 𝑐 < 𝑏,
which shows 𝑐 ∈ 𝑆 and 𝑀 < 𝑐, contradicting that 𝑀 = max 𝑆 is an upper bound of 𝑆.
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Definition 1.3. Let 𝑆 be a nonempty subset of ℝ.
A real number 𝑀 is said to be the supremum or least upper bound of 𝑆, if
(a) 𝑀 is an upper bound of 𝑆, and
(b) if 𝐿 is any upper bound of 𝑆, then 𝑀 ≤ 𝐿.
The supremum of 𝑆 is denoted by sup 𝑆.
Definition 1.4. Let 𝑆 be a nonempty subset of ℝ.
A real number 𝑚 is said to be the infimum or greatest lower bound of 𝑆, if
(a) 𝑚 is a lower bound of 𝑆, and
(b) if 𝑙 is any lower bound of 𝑆, then 𝑚 ≥ 𝑙.
The infimum of 𝑆 is denoted by inf 𝑆.
Note. Let 𝑆 be a nonempty subset of ℝ.
1. By definition, if 𝑆 has a supremum, then sup 𝑆 is the minimum of the (nonempty) set of the upper bounds of
𝑆. Hence sup 𝑆 is unique by Proposition 1.6.
2. Similarly, inf 𝑆 is unique if it exists.
Proposition 1.7. Let 𝑆 be a nonempty subset of ℝ.
1. If max 𝑆 exists, then sup 𝑆 exists, and sup 𝑆 = max 𝑆.
2. If min 𝑆 exists, then inf 𝑆 exists, and inf 𝑆 = min 𝑆.
Proof. 1. max 𝑆 is an upper bound of 𝑆 by definition. Since max 𝑆 ∈ 𝑆, max 𝑆 ≤ 𝐿 for any upper bound 𝐿 of
𝑆. Hence max 𝑆 = sup 𝑆.
The proof of 2. is similar.
Example 1.2.3. Let 𝑎 < 𝑏 and put 𝑆 = [𝑎, 𝑏). Find inf 𝑆 and sup 𝑆 if they exist.
Solution. By Proposition 1.7 and Example 1.2.2, inf 𝑆 = min 𝑆 = 𝑎 exists.
From 𝑥 < 𝑏 for all 𝑥 ∈ 𝑆 we have that 𝑏 is an upper bound of 𝑆. If there were an an upper bound 𝐿 of 𝑆 with
𝐿 < 𝑏, it would follow as in the solution to Example 1.2.2 that there is 𝑐 ∈ 𝑆 with 𝐿 < 𝑐, which contradicts the
fact that 𝐿 is an upper bound of 𝑆.
Hence sup 𝑆 = 𝑏.
Note. It is crucial for Analysis that sufficiently may subsets have a supremum and/or infimum. For example, with
the currently formulated axioms, we cannot decide if the sets
𝑆1 = {𝑥 ∈ ℝ ∶ 𝑥 > 0 and 𝑥2 < 2} or 𝑆2 = {𝑥 ∈ ℝ ∶ 𝑥 > 0 and 𝑥2 ≤ 2}
have suprema.
√
√
Note that the naïve approach putting 𝑆2 = (0, 2] is invalid as we do not know if 2, i. e., a real number 𝑥 > 0
with 𝑥2 = 2, exists.
Thus we have the last axiom of the real number system.
C. The Dedekind completeness axiom
(C)
Every nonempty subset of ℝ which is bounded above has a supremum.
Additional Note. It can be shown that the real number system is uniquely determined by the axioms (A1)-(A4),
(M1)-(M4), (D), (O1)-(O3), (C).
Theorem 1.8 (Positive square root). Let 𝑎 ≥ 0. Then there is a unique 𝑥 ≥ 0 such that 𝑥2 = 𝑎. We write
√
1
𝑥 = 𝑎 = 𝑎2 .
√
Proof. If 𝑎 = 0, we have 02 = 0 and 𝑥2 > 0 if 𝑥 > 0, so that 0 = 0 is the unique number 𝑥 ≥ 0 such that 𝑥2 = 0.
Now let 𝑎 > 0. Note that 𝑥 ≥ 0 and 𝑥2 = 𝑎 > 0 gives 𝑥 > 0. For the uniqueness proof let 𝑥1 > 0 and 𝑥2 > 0 such
that 𝑥21 = 𝑎 and 𝑥22 = 𝑎. Then
0 = 𝑎 − 𝑎 = 𝑥21 − 𝑥22 = (𝑥1 + 𝑥2 )(𝑥1 − 𝑥2 ).
From 𝑥1 + 𝑥2 > 0 it follows that 𝑥1 − 𝑥2 = 0, i. e., the uniqueness 𝑥1 = 𝑥2 .
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For the existence of the square root define
𝑆𝑎 = {𝑥 ∈ ℝ ∶ 0 < 𝑥, 𝑥2 < 𝑎}.
First we are going to show that 𝑆𝑎 ≠ ∅ and that 𝑆𝑎 is bounded.
If 𝑎 < 1, then 𝑎2 < 𝑎 ⋅ 1 = 𝑎, so that 𝑎 ∈ 𝑆𝑎 .
If 𝑎 ≥ 1, then ( 12 )2 < 12 = 1 ≤ 𝑎, so that 21 ∈ 𝑆𝑎 .
For 𝑥 > 𝑎 + 1 we have
𝑥2 > (𝑎 + 1)2 = 𝑎2 + 2𝑎 + 1 > 2𝑎 > 𝑎,
so that any 𝑥 > 𝑎 + 1 does not belong to 𝑆𝑎 . Thus 𝑥 ≤ 𝑎 + 1 for all 𝑥 ∈ 𝑆𝑎 . We have shown that 𝑆𝑎 is bounded
above with upper bound 𝑎 + 1.
By the Dedekind completeness axiom there exists 𝑀𝑎 = sup 𝑆𝑎 . Note that 𝑀𝑎 > 0 since 𝑆𝑎 has positive elements.
To complete the proof we will show that 𝑀𝑎2 = 𝑎.
By proof by contradiction, assume that 𝑀𝑎2 ≠ 𝑎.
Case I: 𝑀𝑎2 < 𝑎. Put
}
{
𝑎 − 𝑀𝑎2
, 𝑀𝑎 .
𝜀 = min
4𝑀𝑎
Then 𝜀 > 0 and
(𝑀𝑎 + 𝜀)2 − 𝑎 = 𝑀𝑎2 + 2𝑀𝑎 𝜀 + 𝜀2 − 𝑎
= 𝑀𝑎2 − 𝑎 + (2𝑀𝑎 + 𝜀)𝜀
≤ 𝑀𝑎2 − 𝑎 + 3𝑀𝑎 𝜀
≤ 𝑀𝑎2 − 𝑎 + 3𝑀𝑎
𝑎 − 𝑀𝑎2
4𝑀𝑎
1
= (𝑀𝑎2 − 𝑎) < 0.
4
Thus
(𝑀𝑎 + 𝜀)2 < 𝑎,
giving 𝑀𝑎 + 𝜀 ∈ 𝑆𝑎 , contradicting the fact that 𝑀𝑎 is an upper bound of 𝑆𝑎 .
Case II: 𝑀𝑎2 > 𝑎. Put
𝑀2 − 𝑎
𝜀= 𝑎
.
2𝑀𝑎
Then 0 < 𝜀 < 12 𝑀𝑎 and
(𝑀𝑎 − 𝜀)2 − 𝑎 = 𝑀𝑎2 − 2𝑀𝑎 𝜀 + 𝜀2 − 𝑎
> 𝑀𝑎2 − 𝑎 − 2𝑀𝑎 𝜀
= 𝑀𝑎2 − 𝑎 − 2𝑀𝑎
𝑀𝑎2 − 𝑎
2𝑀𝑎
= 0.
Hence for all 𝑥 ≥ 𝑀𝑎 − 𝜀 > 12 𝑀𝑎 > 0,
𝑥2 ≥ (𝑀𝑎 − 𝜀)2 > 𝑎
so that any 𝑥 ≥ 𝑀𝑎 − 𝜀 does not belong to 𝑆𝑎 . Hence 𝑀𝑎 − 𝜀 is an upper bound of 𝑆𝑎 , contradicting the fact that
𝑀𝑎 is the least upper bound of 𝑆𝑎 .
So 𝑀𝑎2 ≠ 𝑎 is impossible, and 𝑀𝑎2 = 𝑎 follows.
The completeness axiom can be thought of as ensuring that there are no ’gaps’ on the real line.
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Theorem 1.9 (Characterizations of the supremum). Let 𝑆 be a nonempty subset of ℝ. Let 𝑀 ∈ ℝ. The following
are equivalent:
(a) 𝑀 = sup 𝑆;
(b) 𝑀 is an upper bound of 𝑆, and for each 𝜀 > 0, there is 𝑠 ∈ 𝑆 such that 𝑀 − 𝜀 < 𝑠;
(c) 𝑀 is an upper bound of 𝑆, and for each 𝑥 < 𝑀, there exists 𝑠 ∈ 𝑆 such that 𝑥 < 𝑠.
Proof. (a) ⇒ (b): Since (a) holds, 𝑀 is the least upper bound and thus an upper bound.
Let 𝜀 > 0. Then 𝑀 − 𝜀 is not an upper bound of 𝑆 since 𝑀 is the least upper bound of 𝑆. Hence 𝑥 ≤ 𝑀 − 𝜀 does
not hold for all 𝑥 ∈ 𝑆, and therefore there must be 𝑠 ∈ 𝑆 such that 𝑀 − 𝜀 < 𝑠.
(b) ⇒ (c): Let 𝑥 < 𝑀. Then 𝜀 = 𝑀 − 𝑥 > 0, and by assumption there is 𝑠 ∈ 𝑆 with 𝑀 − 𝜀 < 𝑠. Then
𝑥 = 𝑀 − 𝜀 < 𝑠.
(c) ⇒ (a): By assumption, 𝑀 is an upper bound of 𝑆. Suppose that 𝑀 is not the least upper bound. Then there is
an upper bound 𝐿 of 𝑆 with 𝐿 < 𝑀. By assumption (c), there is 𝑠 ∈ 𝑆 such that 𝐿 < 𝑠, contradicting that 𝐿 is an
upper bound of 𝑆. So 𝑀 must be indeed the least upper bound of 𝑆.
There is an apparent asymmetry in the Dedekind completion. However, there is a version for infima, which is
obtained by reflection. To this end, if 𝑆 is a (nonempty) subset of ℝ set
−𝑆 = {−𝑥 ∶ 𝑥 ∈ 𝑆}.
Note that because of −(−𝑥) = 𝑥 this can also can be written as
−𝑆 = {𝑥 ∈ ℝ ∶ −𝑥 ∈ 𝑆}.
Since 𝑥 ≤ 𝑦 ⇔ −𝑥 ≥ −𝑦 it is easy to see that the following properties hold:
Proposition 1.10. Let 𝑆 be be nonempty subset of ℝ. Then
(a) −𝑆 is bounded below if and only if 𝑆 is bounded above, inf (−𝑆) = − sup 𝑆, and if max 𝑆 exists, then min(−𝑆)
exists and min(−𝑆) = − max 𝑆.
(b) −𝑆 is bounded above if and only if 𝑆 is bounded below, sup(−𝑆) = − inf 𝑆, and if min 𝑆 exists, then max(−𝑆)
exists and max(−𝑆) = − min 𝑆.
(c) −𝑆 is bounded if and only if 𝑆 is bounded.
Thus the Dedekind completeness axiom immediately gives
Theorem 1.11. Every nonempty subset of ℝ which is bounded below has an infimum.
Theorem 1.12 (Characterizations of the infimum). Let 𝑆 be a nonempty subset of ℝ. Let 𝑚 ∈ ℝ. The following
are equivalent:
(a) 𝑚 = inf 𝑆;
(b) 𝑚 is a lower bound of 𝑆, and for each 𝜀 > 0, there is 𝑠 ∈ 𝑆 such that 𝑠 < 𝑚 + 𝜀;
(c) 𝑚 is a lower bound of 𝑆, and for each 𝑥 > 𝑚, there exists 𝑠 ∈ 𝑆 such that 𝑠 < 𝑥.
Theorem 1.13 (Dedekind cut). Let 𝐴 and 𝐵 be nonempty subsets of ℝ such that
(i) 𝐴 ∩ 𝐵 = ∅,
(ii) 𝐴 ∪ 𝐵 = ℝ,
(iii) ∀ 𝑎 ∈ 𝐴 ∀ 𝑏 ∈ 𝐵, 𝑎 ≤ 𝑏.
Then there is 𝑐 ∈ ℝ such that 𝑎 ≤ 𝑐 ≤ 𝑏 for all 𝑎 ∈ 𝐴 and 𝑏 ∈ 𝐵.
Proof. 𝐴 is nonempty and bounded above (any 𝑏 ∈ 𝐵 is an upper bound of 𝐴), so 𝑐 = sup 𝐴 exists by the Dedekind
completeness axiom, and 𝑐 ≥ 𝑎 for all 𝑎 ∈ 𝐴 by definition of upper bound. (iii) says that each 𝑏 ∈ 𝐵 is an upper
bound of 𝐴, and hence 𝑐 ≤ 𝑏 since 𝑐 is the least upper bound of 𝐴.
Note. The above theorem says that Dedekind completeness implies the Dedekind cut property. Conversely, if the
ordered field axioms and the Dedekind cut property are satisfied, then the Dedekind completeness axiom holds
(see tutorials).
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Tutorial 1.2.1. 1. In each of the following cases, state if the given set is bounded above or not. If a set is bounded
above, give two different upper bounds for the set, give the supremum of the set and state if the set has a maximum
or not.
(a) (−3, 2)
(b) (1, ∞)
(c) [10, 11]
(d) {5, 4}
(e) {10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, −1, −2, −5}
(f) (−∞, 2]
(g) {𝑥 ∈ ℝ ∶ 𝑥2 < 3}
(h) {𝑥 ∈ ℝ ∶ 𝑥2 ≤ 3}
2. For each of the sets in Q. 1, state if the given set is bounded below or not. If a set is bounded below, give two
different lower bounds for the set, give the infimum of the set and state if the set has a minumum or not.
3. Let 𝑆 be a nonempty subset of ℝ.
(a) If 𝑎 is the greatest element of 𝑆, then what is sup 𝑆?
(b) If sup 𝑆 = 𝑎, then what are the upper bounds of 𝑆? .
(c) If sup 𝑆 = 𝑎, does 𝑆 have a maximum?
(d) If sup 𝑆 = 𝑎 and 𝑎 ∈ 𝑆, does 𝑆 have a maximum?
4. Prove Proposition 1.10.
5. Prove Theorem 1.11.
6. Prove Theorem 1.12.
7. Let 𝑆 and 𝑇 be non-empty subsets of ℝ which are bounded above. Use Theorem 1.9 to prove that sup(𝑆 + 𝑇 ) =
sup 𝑆 + sup 𝑇 .
Step 1: Let 𝐾 = sup 𝑆 and 𝑀 = sup 𝑇 . Show that 𝐾 + 𝑀 is an upper bound of 𝑆 + 𝑇 .
Step 2: Let 𝜀 > 0. Then 2𝜀 > 0 as well. Since sup 𝑆 = 𝐾, by Theorem 1.9 there is 𝑠 ∈ 𝑆 such that 𝐾 − 2𝜀 < 𝑠. Also,
since sup 𝑇 = 𝑀, there is 𝑡 ∈ 𝑇 such that 𝑀 − 2𝜀 < 𝑡. Show that there exists 𝑢 ∈ 𝑆 + 𝑇 such that 𝐾 + 𝑀 − 𝜀 < 𝑢.
Step 3: Conclude, using Theorem 1.9, that sup(𝑆 + 𝑇 ) = 𝐾 + 𝑀 = sup 𝑆 + sup 𝑇 .
8. Some simple questions. Decide which of the following statements are true and which are false.
(b) 3 ∈ (0, 3)
(c) 17 ∈ [0, 17]
(a) 12 ∈ {0, 1}
(d) 17 ∈ (−3, 18)
(e) 17 ∈ [16, 18]
(f) 2 ∈ {1, 3, 5, 7}
(g) 2.5 ∈ {𝑥 ∈ ℝ ∶ 𝑥2 ≥ 4}
(h) −1 ∈ {𝑥 ∈ ℝ ∶ 2𝑥 + 7 < 5}
9∗ . Assume that the Dedekind cut property, Theorem 1.13 as well as the ordered field axioms are satisfied. Show
that the Dedekind completeness holds.
1.3
Natural Numbers
Naïvely, the natural numbers are 0, 1, 2, 3, … , where the dots stand for the numbers which are obtained by adding
one to the previous number. However, our axioms of the real number system do not tell us that this is possible.
Hence we need another axiom to characterize the natural numbers.
P. Peano’s axioms of the natural numbers
There is a unique subset ℕ of ℝ, called the set of natural numbers, satisfying the following properties:
(P1)
0 ∈ ℕ,
(P2)
𝑎 ∈ ℕ ⇒ 𝑎 + 1 ∈ ℕ,
(P3)
∀ 𝑎 ∈ ℕ, 𝑎 + 1 ≠ 0,
(P4)
If 𝑆 ⊂ ℕ such that 0 ∈ 𝑆 and 𝑎 + 1 ∈ 𝑆 for all 𝑎 ∈ 𝑆, then 𝑆 = ℕ.
Additional Notes. 1. Here we define natural numbers to start at 0. This is common if one uses the axiomatic
definition of the natural numbers. You may start the natural numbers with 1, if you are consistent with this.
2. If one defines the natural numbers without reference to real number, there is no arithmetic on ℕ needed. Then
the number 𝑎 + 1 is called the successor of 𝑎, and axioms (P2) and (P3) can be phrased as follows: Every natural
number has a successor, and 0 is not the successor of a natural number.
3. The natural numbers are used to “count”. In everyday life, counting refers to concrete objects (‘There are 220
students in the class’, ‘I have got no pen with me’, ‘I ate three rolls this morning’). However, they can also be
defined as cardinalities of abstract (finite) sets. Here the cardinality #𝑆 of a (finite) set 𝑆 is the number of elements
in the set. The empty set ∅ has no elements, so #∅ = 0 serves as the definition of the number 0. To get the next
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number, consider the set which consists of the empty set. It has exactly one element, namely the empty set. So
#{∅} = 1. Next #{∅, {∅}} = 2, and so on.
Recall that the integers ℤ and the rational numbers ℚ are defined by
ℤ = {𝑥 ∈ ℝ ∶ 𝑥 ∈ ℕ or − 𝑥 ∈ ℕ},
{
}
𝑝
ℚ = 𝑥 ∈ ℝ ∶ 𝑥 = , 𝑝, 𝑞 ∈ ℤ, 𝑞 ≠ 0 .
𝑞
As a consequence of (P4) one has
The Principle of Mathematical Induction
Suppose we have a statement 𝐴(𝑛) associated with each 𝑛 ∈ ℕ. If
(i) the statement 𝐴(0) is true, and
(ii) whenever the statement 𝐴(𝑛) is true, also the statement 𝐴(𝑛 + 1) is true,
then the statement 𝐴(𝑛) is true for all 𝑛 ∈ ℕ.
Note. The statement 𝐴(0) is called the induction base. The induction base does not need to be taken at 0, one can
take any integer which is suitable for the statements to be proved.
Theorem 1.14. 1. All nonzero natural numbers 𝑛 satisfy 𝑛 ≥ 1.
2. ∀ 𝑛 ∈ ℕ ⧵ {0} ∃ 𝑚 ∈ ℕ, 𝑛 = 𝑚 + 1.
3. ∀ 𝑚, 𝑛 ∈ ℕ, 𝑚 + 𝑛 ∈ ℕ.
4. for all 𝑚, 𝑛 ∈ ℕ with 𝑚 ≥ 𝑛 there is 𝑘 ∈ ℕ such that 𝑚 = 𝑛 + 𝑘.
Proof. A selection of these proofs may be presented in class.
1. For each 𝑛 ∈ ℕ our statement 𝐴(𝑛) is: if 𝑛 ≠ 0, then 𝑛 ≥ 1.
We have to show that this statement is true for all 𝑛 ∈ ℕ.
Induction base: 𝐴(0) is true since the assumption 0 ≠ 0 does not hold.
Induction step: Now let 𝑛 ∈ ℕ and assume that 𝐴(𝑛) holds. We must show that 𝐴(𝑛 + 1) holds.
If 𝑛 = 0 we prove 𝐴(1) directly: 0 + 1 = 1 ≥ 1.
Now let 𝑛 ≠ 0. Then 𝑛 ≥ 1 since 𝐴(𝑛) is true. Since 1 > 0 by Theorem 1.4, (k), we therefore have 𝑛+1 ≥ 1+1 ≥ 1.
Hence, by the principle of mathematical induction, 𝐴(𝑛) is true for all 𝑛 ∈ ℕ.
2. Let 𝑆 = {0} ∪ {𝑛 ∈ ℕ ∶ ∃ 𝑚 ∈ ℕ, 𝑛 = 𝑚 + 1}.
Then 𝑆 ⊂ ℕ, 0 ∈ 𝑆, and if 𝑛 ∈ 𝑆 then 𝑛 + 1 ∈ 𝑆 since 𝑛 ∈ ℕ.
Hence 𝑆 = ℕ by (P4).
3. For 𝑛 ∈ ℕ let 𝐴(𝑛) be the statement:
For all 𝑚 ∈ ℕ, 𝑚 + 𝑛 ∈ ℕ.
𝐴(0) is trivially true.
Now assume that 𝐴(𝑛) is true. Then, for 𝑚 ∈ ℕ,
𝑚 + (𝑛 + 1) = (𝑚 + 1) + 𝑛 ∈ ℕ
by induction hypothesis. Hence 𝐴(𝑛 + 1) holds.
By the principle of mathematical induction, 𝐴(𝑛) holds for all 𝑛 ∈ ℕ. The proof is complete.
4. For 𝑛 ∈ ℕ let 𝐴(𝑛) be the statement:
For all 𝑚 ∈ ℕ with 𝑚 ≥ 𝑛 there exists 𝑘 ∈ ℕ such that 𝑚 = 𝑛 + 𝑘.
𝐴(0) is true with 𝑘 = 𝑚 for all 𝑚.
Now assume 𝐴(𝑛) is true and let 𝑚 ∈ ℕ with 𝑚 ≥ 𝑛 + 1. By induction hypothesis, since 𝑚 ≥ 𝑛 + 1 ≥ 𝑛 there is
𝑘 ∈ ℕ such that 𝑚 = 𝑛 + 𝑘.
If 𝑘 = 0, then 𝑚 = 𝑛 < 𝑛 + 1, which is impossible.
Hence 𝑘 ≠ 0, and so 𝑘 = 𝑙 + 1 for some 𝑙 ∈ ℕ by part 2. It follows that
𝑚 = 𝑛 + 𝑙 + 1 = (𝑛 + 1) + 𝑙,
and therefore 𝐴(𝑛 + 1) is true.
By the principle of mathematical induction, 𝐴(𝑛) hold for all 𝑛 ∈ ℕ. The proof is complete.
Finally, we state and prove two important principles which are consequences of the Dedekind completeness and
Peano’s axioms.
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Theorem 1.15 (Well-ordering principle). Every nonempty subset of ℕ has a smallest element.
Proof. Let 𝑆 ⊂ ℕ, 𝑆 ≠ ∅. Assume 𝑆 has no minimum. Let
𝑇 = {𝑛 ∈ ℕ ∶ ∀ 𝑚 ∈ 𝑆, 𝑛 < 𝑚}.
Note that 𝑛 ∈ 𝑇 and 𝑚 ∈ 𝑆 gives 𝑛 < 𝑚, so that 𝑛 ≠ 𝑚. Hence 𝑇 ∩ 𝑆 = ∅.
Assume that 0 ∈ 𝑆. Since 0 ≤ 𝑛 for all 𝑛 ∈ ℕ by Theorem 1.14, 1, and since 𝑆 ⊂ ℕ, we have 0 ≤ 𝑚 for all 𝑚 ∈ 𝑆.
But 𝑆 has no minimum, and therefore 0 ∉ 𝑆. Thus 0 < 1 ≤ 𝑚 for all 𝑚 ∈ 𝑆.
Hence 0 ∈ 𝑇 .
Now let 𝑛 ∈ 𝑇 . We want to show that 𝑛 + 1 ∈ 𝑇 .
By definition of 𝑇 , we have 𝑛 < 𝑚 for all 𝑚 ∈ 𝑆. By Theorem 1.14, 4, For each such 𝑚 there is 𝑘 ∈ ℕ such that
𝑚 = 𝑛 + 𝑘. Since 𝑛 < 𝑚, it follows that 𝑘 ≠ 0, and hence 𝑘 ≥ 1 by Theorem 1.14, 1. Thus
𝑚=𝑛+𝑘≥𝑛+1
for all 𝑚 ∈ 𝑆. Since we assume that 𝑆 has no minimum, it follows that 𝑛 + 1 ∉ 𝑆 and hence 𝑛 + 1 ≠ 𝑚 for all
𝑚 ∈ 𝑆. Altogether 𝑛 + 1 < 𝑚 for all 𝑚 ∈ 𝑆. This shows that 𝑛 + 1 ∈ 𝑇 .
By (P4), 𝑇 = ℕ. Hence
𝑆 = ℕ ∩ 𝑆 = 𝑇 ∩ 𝑆 = ∅,
a contradiction.
Note. One can show that if 𝑆 ⊂ ℤ, 𝑆 ≠ ∅, and 𝑆 is bounded below, then 𝑆 has a minimum.
Theorem 1.16 (The Archimedean principle). For each 𝑥 ∈ ℝ there is 𝑛 ∈ ℕ such that 𝑛 > 𝑥.
Proof. Assume the Archimedian principle is false. Then there is 𝑥 ∈ ℝ such that 𝑛 ≤ 𝑥 for all 𝑛 ∈ ℕ. That means,
ℕ is bounded above and therefore has a supremum 𝑀.
By Theorem 1.9 there is 𝑛 ∈ ℕ such that 𝑀 − 1 < 𝑛. Then
𝑛 + 1 > (𝑀 − 1) + 1 = 𝑀
and 𝑛 + 1 ∈ ℕ contradict the fact that 𝑀 is an upper bound of ℕ. Hence the Archimedian principle must be
true.
Definition 1.5. A subset 𝑆 of ℝ is said to be dense in ℝ if for all 𝑥, 𝑦 ∈ ℝ with 𝑥 < 𝑦 there is 𝑠 ∈ 𝑆 such that
𝑥 < 𝑠 < 𝑦.
Real numbers which are not rational numbers are called irrational numbers.
√
Note. 1. 2 is irrational (see tutorials).
2. ℚ is an ordered field, i. e., ℚ satisfies all axioms of ℝ except the Dedekind completeness axiom (see tutorials).
3. If 𝑎 is rational and 𝑏 is irrational, then 𝑎 + 𝑏 is irrational (see tutorials).
Theorem 1.17. The set of rational numbers as well as the set of irrational numbers are dense in ℝ.
Proof. Let 𝑥, 𝑦 ∈ ℝ, 𝑥 < 𝑦. By the Archimedian principle, there is a natural number 𝑛0 such that
𝑛0 >
2
> 0.
𝑦−𝑥
Let
𝑆 = {𝑛 ∈ ℤ ∶ 𝑛 > 𝑛0 𝑥}.
By the Archimedean principle, 𝑆 ≠ ∅. Also, 𝑆 is bounded below. So 𝑆 has a minimum 𝑚. Then 𝑚 − 1 ∉ 𝑆, so
that 𝑚 − 1 ≤ 𝑛0 𝑥.
√
From 1 < 2 < 4 it follows that 1 < 2 < 2. Then
√
𝑛0 𝑥
𝑚 − 1 + 2 𝑚 + 1 𝑛0 𝑥 + 2
𝑦−𝑥
𝑚
2
𝑥=
<
<
<
≤
=𝑥+
<𝑥+2
= 𝑦.
𝑛0
𝑛0
𝑛0
𝑛0
𝑛0
𝑛0
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Then
𝑚
𝑢=
∈ ℚ,
𝑛0
𝑚−1+
𝑣=
𝑛0
√
2
13
∈ℝ⧵ℚ
and, by above,
𝑥 < 𝑢 < 𝑣 < 𝑦,
so that we have have a rational and an irrational number between each pair of real numbers 𝑥, 𝑦 with 𝑥 < 𝑦. Hence
both ℚ and ℝ ⧵ ℚ are dense in ℝ.
√
√
Example 1.3.1. Let 𝑆 = {𝑥 ∈ ℚ ∶ 𝑥 < 2} and 𝑇 = {𝑥 ∈ ℚ ∶ 𝑥 ≤ 2}. Then 𝑆 = 𝑇 , and 𝑆 and 𝑇 have no
supremum in ℚ.
Example 1.3.2 (Bernoulli’s inequality). For all 𝑥 ∈ ℝ with 𝑥 ≥ −1 and all 𝑛 ∈ ℕ, (1 + 𝑥)𝑛 ≥ 1 + 𝑛𝑥.
Solution. For 𝑛 = 0, (1 + 𝑥)𝑛 = (1 + 𝑥)0 = 1 = 1 + 0 ⋅ 𝑥 = 1 + 𝑛𝑥, so that the inequality holds for 𝑛 = 0.
Let 𝑛 ∈ ℕ and assume the Bernoulli inequality holds for this 𝑛. Then
(1 + 𝑥)𝑛+1 = (1 + 𝑥)(1 + 𝑥)𝑛 ≥ (1 + 𝑥)(1 + 𝑛𝑥) = 1 + 𝑥 + 𝑛𝑥 + 𝑛𝑥2 = 1 + (𝑛 + 1)𝑥 + 𝑛𝑥2 ≥ 1 + (𝑛 + 1)𝑥.
The first inequality follows from 1 + 𝑥 ≥ 0 and the induction hypothesis, the last from 𝑛𝑥2 ≥ 0. By the principle
of induction, the Bernoulli inequality holds for all 𝑛 ∈ ℕ.
Note. We will make use of more rules of real numbers without proof. See the tutorials for the power rules, for
example.
Tutorial 1.3.1. 1. Show that if 𝑆 ⊂ ℤ, 𝑆 ≠ ∅, and 𝑆 is bounded below, then 𝑆 has a minimum.
√
√
2. Show that 2 is irrational. Hint. Assume that 2 = 𝑞𝑝 with positive integers 𝑝 and 𝑞. You may assume that 𝑝
and 𝑞 do not have common factors, i. e., there are no positive integers 𝑘, 𝑝1 , 𝑞1 with 𝑘 ≠ 1 such that 𝑝 = 𝑝1 𝑘 and
𝑞 = 𝑞1 𝑘.
3. Show that the rational numbers satisfy the axioms (A1)–(A4), (M1)–(M4), (D), and (O1)–(O3).
4. Let 𝑎, 𝑏 ∈ ℚ with 𝑏 ≠ 0 and 𝑟 ∈ ℝ ⧵ ℚ. Show that 𝑎 + 𝑏𝑟 ∈ ℝ ⧵ ℚ.
5. For 𝑥 ∈ ℝ and 𝑛 ∈ ℕ, 𝑥𝑛 is defined inductively by
(i) 𝑥0 = 1,
(ii) 𝑥𝑛+1 = 𝑥𝑥𝑛 for 𝑛 ∈ ℕ.
Show that (a) 𝑥𝑛 𝑥𝑚 = 𝑥𝑛+𝑚 , (b) (𝑥𝑛 )𝑚 = 𝑥𝑛𝑚 , (c) 𝑥𝑛 𝑦𝑛 = (𝑥𝑦)𝑛 .
Chapter 2
Sequences
Students have seen sequences and learnt rules for limits and how to apply these rules to find limits. In this chapter,
students will learn the proper definitions of limits and the proofs of the rules for limits.
2.1
Sequences
Definition 2.1. A (real) sequence is an ordered list of infinitely many real numbers
𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 , … .
We usually denote a sequence by (𝑎𝑛 ) = 𝑎1 , 𝑎2 , 𝑎3 , … .
The number 𝑎𝑛 is called the 𝑛-th term of the sequence. The subscript 𝑛 is called the index.
Note. The index of the sequence does not have to start at 1, and therefore one also writes (𝑎𝑛 )∞
𝑛=𝑛0 , where 𝑛0 can
∞
∞
∞
be any integer, e. g., (2𝑛 − 3)𝑛=1 , (2𝑛 − 3)𝑛=0 , (2𝑛 − 3)𝑛=−5 .
Recall the following note and sketch from First Year Calculus. [Lecturers: do not write this down in class.]
Note. (i) A sequence can be identified with a function on the (positive) integers:
𝑎𝑛 = 𝑓 (𝑛).
(ii) A sequence can be plotted as points of the real axis or as the graph of the function, see (i).
Below is a plot for the sequence
(
𝑛
𝑛+1
)∞
.
𝑛=1
𝑎1
𝑎2
1
2
0
𝑎3
1
𝑎𝑛
1
𝑛
1 2 3 4 5 6 7 8
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15
In first year calculus students have seen the following informal definition. [Lecturers: You do not need to write
this down. In any case, emphasize that this notion has to be formalized.]
Informal Definition. A sequence {𝑎𝑛 } is said to converge if there is a number 𝐿 such that 𝑎𝑛 is as close to 𝐿 as we
like for all sufficiently large 𝑛.
We have to formalize “as close to” and “for all sufficiently large”: “as close to” means that the distance does not
exceed any given (small) positive number. The distance between two numbers 𝑥 and 𝑦 is 𝑦 − 𝑥 if 𝑥 < 𝑦 and 𝑥 − 𝑦
if 𝑥 > 𝑦. Hence the distance between 𝑥 and 𝑦 is |𝑦 − 𝑥|.
Therefore we have the following formal definition:
Definition 2.2. Let (𝑎𝑛 ) be a sequence and 𝐿 ∈ ℝ.
1. The statement ‘𝑎𝑛 tends to 𝐿 as 𝑛 tends to infinity’, which we write as ’𝑎𝑛 → 𝐿 as 𝑛 → ∞’, is defined by:
∀ 𝜀 > 0 ∃ 𝐾 ∈ ℝ ∀𝑛 ∈ ℕ, 𝑛 ≥ 𝐾, |𝑎𝑛 − 𝐿| < 𝜀.
2. If 𝑎𝑛 → 𝐿 as 𝑛 → ∞, we say that (𝑎𝑛 ) converges to 𝐿, and we also write lim 𝑎𝑛 = 𝐿.
𝑛→∞
3. The sequence (𝑎𝑛 ) is said to be convergent if it converges to some real number. Otherwise, the sequence (𝑎𝑛 ) is
said to be divergent.
Note. 1. It is useful to note that |𝑎𝑛 − 𝐿| < 𝜀 is equivalent to 𝐿 − 𝜀 < 𝑎𝑛 < 𝐿 + 𝜀.
2. The number 𝐾 depends on 𝜀. We may write 𝐾𝜀 to emphasize the dependence on 𝜀.
3. If one wants to prove convergence of a sequence from first principles, then one has to ‘guess’ a limit 𝐿 and then
prove that it is indeed the limit.
4. The ‘first few terms’ do not matter for convergence and the limit. That is, the sequence (𝑎𝑛 )∞
𝑛=𝑛0 converges if
converges,
and
their
limits
coincide.
and only if the sequence (𝑎𝑛 )∞
𝑛=𝑚0
5. By the Archimedean principle, there are only finitely many natural numbers such that 𝑛 < 𝐾, and one may also
assume 𝐾 ∈ ℕ, without loss of generality.
)
(
𝑛
converges and find its limit.
Example 2.1.1. Prove that the sequence (𝑎𝑛 ) = 𝑛+1
Solution. We first have to guess the limit. From the rules used in first year, we guess that the limit should be 𝐿 = 1.
Now let 𝜀 > 0. We must find 𝐾𝜀 such that
𝑛 ≥ 𝐾𝜀 ⇒ |𝑎𝑛 − 1| < 𝜀.
For this we first simplify:
| 𝑛
|
1
|𝑎𝑛 − 1| = ||
− 1|| =
.
𝑛
+
1
𝑛
+
1
|
|
1
< 𝜀, which can be written as 𝑛 + 1 > 1𝜀 or 𝑛 > 1𝜀 − 1. Hence if we take
𝑛+1
𝑛
𝐾𝜀 = 1𝜀 − 1, then |𝑎𝑛 − 1| < 𝜀 for all 𝑛 > 𝐾𝜀 . Since 𝜀 > 0 was arbitrary, we have shown that 𝑛+1
→ 1 as 𝑛 → ∞.
Hence |𝑎𝑛 − 1| < 𝜀 provided that
Lemma 2.1. If 𝐿, 𝑀 ∈ ℝ such that |𝐿 − 𝑀| < 𝜀 for all 𝜀 > 0, then 𝐿 = 𝑀.
Proof. Assume 𝐿 ≠ 𝑀. Then either 𝐿 < 𝑀 or 𝐿 > 𝑀. But
𝐿 < 𝑀 ⇒ |𝐿 − 𝑀| = 𝑀 − 𝐿 > 0,
𝑀 < 𝐿 ⇒ |𝐿 − 𝑀| = 𝐿 − 𝑀 > 0,
so that |𝐿 − 𝑀| > 0. Then
0<
|𝐿 − 𝑀|
< |𝐿 − 𝑀|,
2
which contradicts |𝐿 − 𝑀| < 𝜀 for 𝜀 = |𝐿−𝑀|
. Hence the assumption 𝐿 ≠ 𝑀 must be false, and 𝐿 = 𝑀
2
follows.
Theorem 2.2. If the sequence (𝑎𝑛 ) converges, then its limit is unique.
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Proof. Assume that (𝑎𝑛 ) converges to 𝐿 and 𝑀. Let 𝜀 > 0. Then there are numbers 𝑘𝐿 and 𝑘𝑀 such that
(i) |𝑎𝑛 − 𝐿| < 2𝜀 if 𝑛 ≥ 𝑘𝐿 ,
(ii) |𝑎𝑛 − 𝑀| < 2𝜀 if 𝑛 ≥ 𝑘𝑀 .
Put 𝐾 = max{𝑘𝐿 , 𝑘𝑀 }. For positive integers 𝑛 ≥ 𝐾 we have 𝑛 ≥ 𝑘𝐿 and 𝑛 ≥ 𝑘𝑀 and therefore
|𝐿 − 𝑀| = |(𝐿 − 𝑎𝑛 ) + (𝑎𝑛 − 𝑀)| ≤ |𝑎𝑛 − 𝐿| + |𝑎𝑛 − 𝑀| <
𝜀 𝜀
+ = 𝜀.
2 2
By Lemma 2.1, 𝐿 = 𝑀.
We are now going to prove the rules you have learnt in first year.
Theorem 2.3 (Limit Laws). Let 𝑐 ∈ ℝ and suppose that lim 𝑎𝑛 = 𝐿 and lim 𝑏𝑛 = 𝑀 both exist. Then
𝑛→∞
𝑛→∞
(a) lim 𝑐 = 𝑐.
𝑛→∞
(b) lim (𝑎𝑛 + 𝑏𝑛 ) = lim 𝑎𝑛 + lim 𝑏𝑛 = 𝐿 + 𝑀.
𝑛→∞
𝑛→∞
𝑛→∞
(c) lim (𝑐𝑎𝑛 ) = 𝑐 lim 𝑎𝑛 = 𝑐𝐿.
𝑛→∞
(𝑛→∞ ) (
)
(d) lim (𝑎𝑛 𝑏𝑛 ) = lim 𝑎𝑛
lim 𝑏𝑛 = 𝐿𝑀.
𝑛→∞
𝑛→∞
𝑛→∞
lim 𝑎𝑛
𝑎𝑛
𝑛→∞
𝐿
=
=
.
𝑛→∞ 𝑏𝑛
lim 𝑏𝑛
𝑀
𝑛→∞
𝑎
(f) If 𝐿 ≠ 0 and 𝑀 = 0, then lim 𝑛 does not exist.
𝑛→∞ 𝑏𝑛
(
)𝑘
(g) If 𝑘 ∈ ℤ+ , lim 𝑎𝑘𝑛 = lim 𝑎𝑛 = 𝐿𝑘 .
𝑛→∞
𝑛→∞
(e) If 𝑀 ≠ 0, lim
√
√
√
𝑘
(h) If 𝑘 ∈ ℤ+ , lim 𝑘 𝑎𝑛 = 𝑘 lim 𝑎𝑛 = 𝐿. If 𝑘 is even, we assume 𝑎𝑛 ≥ 0 and 𝐿 ≥ 0.
𝑛→∞
𝑛→∞
(i) If lim |𝑎𝑛 | = 0, then lim 𝑎𝑛 = 0.
𝑛→∞
𝑛→∞
Proof. (a) For all 𝜀 > 0 and all indices 𝑛, with 𝑎𝑛 = 𝑐, |𝑎𝑛 − 𝑐| = |𝑐 − 𝑐| = 0 < 𝜀.
(b) Let 𝜀 > 0. Then there are numbers 𝑘𝐿 and 𝑘𝑀 such that
(i) |𝑎𝑛 − 𝐿| < 2𝜀 if 𝑛 ≥ 𝑘𝐿 ,
(ii) |𝑏𝑛 − 𝑀| < 2𝜀 if 𝑛 ≥ 𝑘𝑀 .
Put 𝐾 = max{𝑘𝐿 , 𝑘𝑀 }. For positive integers 𝑛 ≥ 𝐾 we have 𝑛 ≥ 𝑘𝐿 and 𝑛 ≥ 𝑘𝑀 and therefore
|(𝑎𝑛 + 𝑏𝑛 ) − (𝐿 + 𝑀)| = |(𝑎𝑛 − 𝐿) + (𝑏𝑛 − 𝑀)| ≤ |𝑎𝑛 − 𝐿| + |𝑏𝑛 − 𝑀| <
𝜀 𝜀
+ = 𝜀.
2 2
Hence (𝑎𝑛 + 𝑏𝑛 ) converges to 𝐿 + 𝑀.
(c) Let 𝜀 > 0. Then there is a number 𝐾 such that
𝜀
|𝑎𝑛 − 𝐿| < 1+|𝑐|
if 𝑛 ≥ 𝐾.
For positive integers 𝑛 ≥ 𝐾 we have
|𝑐𝑎𝑛 − 𝑐𝐿| = |𝑐(𝑎𝑛 − 𝐿)| = |𝑐| |𝑎𝑛 − 𝐿| < |𝑐|
𝜀
< 𝜀.
1 + |𝑐|
(d) We consider 2 cases: one special case, to which then the general case is reduced.
Case I: 𝐿 = 𝑀 = 0. Let 𝜀 > 0 and put 𝜀′ = min{1, 𝜀}. Then 0 < 𝜀′ ≤ 𝜀, and there are numbers 𝑘𝐿 and 𝑘𝑀 such
that
(i) |𝑎𝑛 − 0| < 𝜀′ if 𝑛 ≥ 𝑘𝐿 ,
(ii) |𝑏𝑛 − 0| < 𝜀′ if 𝑛 ≥ 𝑘𝑀 .
Put 𝐾 = max{𝑘𝐿 , 𝑘𝑀 }. For positive integers 𝑛 ≥ 𝐾 we have 𝑛 ≥ 𝑘𝐿 and 𝑛 ≥ 𝑘𝑀 and therefore
2
|𝑎𝑛 𝑏𝑛 | = |𝑎𝑛 | |𝑏𝑛 | < 𝜀′ ≤ 𝜀′ ≤ 𝜀.
Case II: 𝐿 and 𝑀 are arbitrary. Then
𝑎𝑛 𝑏𝑛 = (𝑎𝑛 − 𝐿)(𝑏𝑛 − 𝑀) + 𝐿(𝑏𝑛 − 𝑀) + 𝑎𝑛 𝑀.
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Introductory Analysis 2017 Lecture Manual
By (a), (b), (c), (𝑎𝑛 − 𝐿) → 0 and (𝑏𝑛 − 𝑀) → 0 as 𝑛 → ∞, and by (b), (c), and Case I, it follows that (𝑎𝑛 𝑏𝑛 )
converges with
lim 𝑎𝑛 𝑏𝑛 = 0 + 𝐿 ⋅ 0 + 𝐿𝑀 = 𝐿𝑀.
𝑛→∞
(e) First consider 𝑎𝑛 = 1. Then
|1
|
| − 1 | = |𝑀 − 𝑏𝑛 | .
|𝑏
|
|𝑏𝑛 𝑀|
| 𝑛 𝑀|
We must assure that 𝑏𝑛 stays away from 0, and hence we estimate
|𝑀| = |𝑏𝑛 + 𝑀 − 𝑏𝑛 | ≤ |𝑏𝑛 | + |𝑏𝑛 − 𝑀|,
which gives
|𝑏𝑛 | ≥ |𝑀| − |𝑏𝑛 − 𝑀|.
Since 𝑏𝑛 → 𝑀 ≠ 0, there is 𝑘1 such that |𝑏𝑛 − 𝑀| < |𝑀|
for 𝑛 ≥ 𝑘1 , and hence
2
|𝑀| |𝑀|
=
.
2
2
|𝑏𝑛 | ≥ |𝑀| −
If we therefore let 𝜀 > 0 and choose 𝐾 such that
|𝑏𝑛 − 𝑀| < min
{
|𝑀| 𝑀 2
,
𝜀
2
2
}
for 𝑛 ≥ 𝐾, it follows that
2
|1
|
| − 1 | = |𝑀 − 𝑏𝑛 | ≤ |𝑀 − 𝑏𝑛 | 2 < |𝑀| 𝜀 2 = 𝜀.
|𝑏
|
|𝑏𝑛 | |𝑀|
2
|𝑀|2
|𝑀|2
| 𝑛 𝑀|
The general case now follows with (d):
𝑎𝑛
1
1
= lim 𝑎𝑛 ⋅ lim
=𝐿⋅
.
𝑛→∞ 𝑏𝑛
𝑛→∞
𝑛→∞ 𝑏𝑛
𝑀
lim
𝑎
(f) Assume that 𝑃 = lim 𝑏𝑛 exists. Then, by (d),
𝑛→∞ 𝑛
(
𝐿 = lim 𝑎𝑛 = lim
𝑛→∞
𝑛→∞
𝑎𝑛
𝑏
𝑏𝑛 𝑛
)
𝑎𝑛
⋅ lim 𝑏 = 𝑃 ⋅ 𝑀 = 𝑃 ⋅ 0 = 0,
𝑛→∞ 𝑏𝑛 𝑛→∞ 𝑛
= lim
which contradicts 𝐿 ≠ 0.
(g) Follows from (d) by induction.
(h) We will only prove the case 𝑘 = 2 as we have not yet established the existence of the 𝑘-th root.
√
√
√
If 𝐿 = 0, let 𝜀 > 0 and choose 𝐾 such that 𝑎𝑛 < 𝜀2 for 𝑛 ≥ 𝐾. Then 𝑎𝑛 < 𝜀 for these 𝑛, and lim 𝑎𝑛 = 0 = 𝐿.
𝑛→∞
If 𝐿 > 0, then
|
|
√
| 𝑎 −𝐿 |
√
|𝑎 − 𝐿|
1
| 𝑎𝑛 − 𝐿| = || √ 𝑛 √ || = √ 𝑛 √ ≤ √ |𝑎𝑛 − 𝐿|,
| 𝑎𝑛 + 𝐿 |
𝑎𝑛 + 𝐿
𝐿
|
|
√
√
√
and choosing 𝐾 such that |𝑎𝑛 − 𝐿| < 𝐿𝜀 for 𝑛 ≥ 𝐾, it follows that | 𝑎𝑛 − 𝐿| < 𝜀 for 𝑛 ≥ 𝐾.
(i) is trivial.
Theorem 2.4 (Sandwich Theorem). If 𝑎𝑛 ≤ 𝑏𝑛 ≤ 𝑐𝑛 and lim 𝑎𝑛 = lim 𝑐𝑛 = 𝐿, then lim 𝑏𝑛 = 𝐿.
𝑛→∞
𝑛→∞
𝑛→∞
Proof. Let 𝜀 > 0 and choose 𝑘1 and 𝑘2 such that |𝑎𝑛 − 𝐿| < 𝜀 if 𝑛 ≥ 𝑘1 and |𝑐𝑛 − 𝐿| < 𝜀 if 𝑛 ≥ 𝑘2 . In particular,
for 𝑛 ≥ 𝐾 = max{𝑘1 , 𝑘2 },
𝐿 − 𝜀 < 𝑎𝑛 < 𝐿 + 𝜀, 𝐿 − 𝜀 < 𝑐𝑛 < 𝐿 + 𝜀
gives
𝐿 − 𝜀 < 𝑎𝑛 ≤ 𝑏𝑛 ≤ 𝑐𝑛 < 𝐿 + 𝜀.
Hence |𝑏𝑛 − 𝐿| < 𝜀 if 𝑛 ≥ 𝐾.
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Introductory Analysis 2017 Lecture Manual
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Theorem 2.5. Let 𝑥 ∈ ℝ. Then 𝑥𝑛 converges if and only if −1 < 𝑥 ≤ 1.
Proof. For 𝑥 = 1, 1𝑛 = 1, so that lim 1𝑛 = 1 by Theorem 2.3 (a).
𝑛→∞
Let 0 < 𝑥 < 1 and let 𝜀 > 0. Then 0 < 1 < 𝑥1 and therefore
𝑦 ∶=
1
− 1 > 0.
𝑥
Then
( )𝑛
1
1
=
= (1 + 𝑦)𝑛 ≥ 1 + 𝑛𝑦
𝑛
𝑥
𝑥
1−𝜀
. Then, for 𝑛 > 𝐾,
by Bernoulli’s inequality. Put 𝐾 =
𝑦𝜀
0 < 𝑥𝑛 ≤
1
1
<
= 𝜀.
1 + 𝑛𝑦 1 + 1−𝜀
𝜀
Hence 𝑥𝑛 → 0 as 𝑛 → ∞. Now let 𝑥 > 1. If 𝑥𝑛 would converge to some 𝐿 as 𝑛 → ∞, then
1 = lim 1𝑛 = lim
𝑛→∞
( )𝑛
1
⋅ lim 𝑥𝑛 = 0 ⋅ 𝐿 = 0,
𝑛→∞
𝑥
which is impossible.
The remaining cases are left as an exercise.
Theorem 2.6. If 𝑟 > 0, then
lim
1
𝑛→∞ 𝑛𝑟
= 0.
Proof. We are only going to prove the case that 𝑟 is a positive integer.
Let 𝑟 = 1 and choose 𝜀 > 0. Let 𝐾 = 2𝜀 . Then, for 𝑛 ≥ 𝐾,
0<
1
1
𝜀
≤
= < 𝜀,
𝑛 𝐾
2
| |
whence | 1𝑛 | < 𝜀.
| |
Now assume the statement holds for an integer 𝑟 > 0. Then
lim
1
𝑛→∞ 𝑛𝑟+1
= lim
1
𝑛→∞ 𝑛
⋅ lim
1
𝑛→∞ 𝑛𝑟
= 0 ⋅ 0 = 0.
So the result follows for all positive integers by induction.
(
)
𝑛
Tutorial 2.1.1. 1. (a) Prove, using the definition of convergence, that the sequence 𝑛+1
does not converge to 2.
𝑛
(b) Prove, using the definition of convergence, that the sequence ((−1) ) does not converge to any 𝐿.
2. (a) Prove that if lim 𝑎𝑛 = 𝐿, then lim |𝑎𝑛 | = |𝐿|. (Hint: use (and prove) the inequality ||𝑥| − |𝑦|| ≤ |𝑥 − 𝑦|.)
𝑛→∞
𝑛→∞
(b) Give an example to show that the converse to part (a) is not true.
3. Contractive maps. Suppose that for some 𝑐 ∈ ℝ with 0 < 𝑐 < 1, we have |𝑎𝑛+1 − 𝐿| ≤ 𝑐|𝑎𝑛 − 𝐿| for all 𝑛 ∈ ℕ.
(a) Use induction to prove that |𝑎𝑛 − 𝐿| ≤ 𝑐 𝑛 |𝑎0 − 𝐿| for all 𝑛 ∈ ℕ.
(b) Use the Sandwich Theorem and the fact that lim 𝑐 𝑛 = 0 to prove that lim 𝑎𝑛 = 𝐿.
𝑛→∞
𝑛→∞
√
4. Recursive algorithm for finding 𝑎. Let 𝑎 > 1 and define
(
)
1
𝑎
𝑎0 = 𝑎 and 𝑎𝑛 =
𝑎𝑛−1 +
for 𝑛 ≥ 1.
2
𝑎𝑛−1
√
√
(a) Prove that 0 < 𝑎𝑛 − 𝑎 = 2𝑎1 (𝑎𝑛−1 − 𝑎)2 for 𝑛 ≥ 1.
𝑛−1
√
√
(b) Use (a) to prove that 0 ≤ 𝑎𝑛 − 𝑎 ≤ 21 (𝑎𝑛−1 − 𝑎) for 𝑛 ≥ 1.
√
(c) Deduce that lim 𝑎𝑛 = 𝑎.
𝑛→∞
√
(d) Apply four steps of the recursive algorithm with 𝑎 = 3 to approximate 3.
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Introductory Analysis 2017 Lecture Manual
19
Bounded and Monotonic Sequences
Definition 2.3. 1. A sequence (𝑎𝑛 ) is said to be bounded above if there is a number 𝑀 ∈ ℝ such that 𝑎𝑛 ≤ 𝑀
for all indices 𝑛.
2. A sequence (𝑎𝑛 ) is said to be bounded below if there is a number 𝑚 ∈ ℝ such that 𝑎𝑛 ≥ 𝑚 for all indices 𝑛.
3. A sequence (𝑎𝑛 ) is said to be bounded if it is a bounded above and bounded below.
Note. A sequence (𝑎𝑛 )∞
𝑛=𝑛0 is bounded (above, below) if and only if the set {𝑎𝑛 ∶ 𝑛 ∈ ℤ, 𝑛 ≥ 𝑛0 } is bounded
(above, below).
Definition 2.4. A sequence {𝑎𝑛 }∞
is called
𝑛=1
increasing if 𝑎𝑛 ≤ 𝑎𝑛+1 for all indices 𝑛,
strictly increasing if 𝑎𝑛 < 𝑎𝑛+1 for all indices 𝑛,
decreasing if 𝑎𝑛 ≥ 𝑎𝑛+1 for all indices 𝑛,
strictly decreasing if 𝑎𝑛 > 𝑎𝑛+1 for all indices 𝑛,
monotonic if it is either increasing or decreasing,
strictly monotonic if it is either strictly increasing or strictly decreasing.
Theorem 2.7. Every convergent sequence is bounded.
Proof. Here we use the fact, which is easy to prove (by induction), that a set of the form {𝑎𝑛 ∶ 𝑛 ∈ ℤ, 𝑛0 ≤ 𝑛 ≤ 𝑛1 }
is bounded (and indeed has minimum and maximum). We also use that the union of two bounded sets is bounded.
Since (𝑎𝑛 ) converges, there are 𝐿 and 𝑘 such that |𝑎𝑛 − 𝐿| < 1 for all 𝑛 ≥ 𝑘. Hence {𝑎𝑛 ∶ 𝑛 < 𝑘} and {𝑎𝑛 ∶ 𝑛 ≥
𝑘} ⊂ (𝐿 − 1, 𝐿 + 1) are bounded. So the sequence is bounded by the above note.
Although each unbounded sequence diverges, there are some divergent sequences which have a limiting behaviour
which we want to exploit. Recall that we use the notation ‘𝑛 → ∞’ in the definition of the limit. This extends to
‘𝑎𝑛 → ∞’, and we have therefore the
Definition 2.5. 1. We say that 𝑎𝑛 tends to infinity as 𝑛 tends to infinity and write 𝑎𝑛 → ∞ as 𝑛 → ∞ if for every
𝐴 ∈ ℝ there is 𝐾 ∈ ℝ such that 𝑎𝑛 > 𝐴 for all 𝑛 ≥ 𝐾.
2. We say that 𝑎𝑛 tends to minus infinity as 𝑛 tends to infinity and write 𝑎𝑛 → −∞ as 𝑛 → ∞ if for every 𝐴 ∈ ℝ
there is 𝐾 ∈ ℝ such that 𝑎𝑛 < 𝐴 for all 𝑛 ≥ 𝐾.
Notes. 1. For the definition of 𝑎𝑛 → ∞(−∞) as 𝑛 → ∞, we may restrict 𝐴 to 𝐴 > 𝐴0 (𝐴 < 𝐴0 ) for suitable
numbers 𝐴0 .
2. For 𝑎𝑛 → ∞ as 𝑛 → ∞ we also write lim 𝑎𝑛 = ∞ and say that (𝑎𝑛 ) diverges to ∞.
𝑛→∞
3. For 𝑎𝑛 → −∞ as 𝑛 → ∞ we also write lim 𝑎𝑛 = −∞ and say that (𝑎𝑛 ) diverges to −∞.
𝑛→∞
Example 2.2.1. 1. Let 𝑎𝑛 = (−1)𝑛 . Then (𝑎𝑛 ) is bounded but not convergent.
2. Let 𝑎𝑛 = 𝑛. Then (𝑎𝑛 ) is unbounded (by the Archimedean principle). Hence (𝑎𝑛 ) diverges.
Example 2.2.2. Prove that 𝑛2 − 𝑛3 + 10 → −∞ as 𝑛 → ∞.
Solution. Let 𝐴 < 0 and put 𝐾 = −𝐴 + 11. Then 𝐾 > 11, and for 𝑛 ≥ 𝐾 we have
𝑛2 − 𝑛3 + 10 = −𝑛2 (𝑛 − 1) + 10 ≤ −10𝑛2 + 10 ≤ −10𝐾 2 + 10
= −10(𝐾 − 1)(𝐾 + 1) ≤ −100(𝐾 + 1) ≤ −𝐾 = 𝐴 − 11 ≤ 𝐴.
Theorem 2.8. 1. Let (𝑎𝑛 ) be a sequence with 𝑎𝑛 > 0 for all indices 𝑛. Then
1
lim 𝑎 = ∞ ⇔ lim
= 0.
𝑛→∞ 𝑛
𝑛→∞ 𝑎𝑛
2. Let (𝑎𝑛 ) be a sequence with 𝑎𝑛 < 0 for all indices 𝑛. Then
1
lim 𝑎 = −∞ ⇔ lim
= 0.
𝑛→∞ 𝑛
𝑛→∞ 𝑎𝑛
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Introductory Analysis 2017 Lecture Manual
Proof. 1. Assume that lim 𝑎𝑛 = ∞. To show that lim
0 < 𝐴 < 𝑎𝑛 for all 𝑛 ≥ 𝐾. Then
0<
for these 𝑛, which shows that lim
1
𝑛→∞ 𝑎𝑛
Conversely, assume that lim
1
𝑛→∞ 𝑎𝑛
Hence there is 𝐾 ∈ ℝ such that
The proof of part 2 is similar.
1
𝑛→∞ 𝑎𝑛
𝑛→∞
= 0 let 𝜀 > 0. Put 𝐴 =
MATH2022
1
. Then there is 𝐾 such that
𝜀
1
1
<
=𝜀
𝑎𝑛
𝐴
= 0.
= 0. Let 𝐴 ∈ ℝ and put 𝐴0 = max{𝐴, 1}. Then 𝐴0 > 0 and put 𝜀 =
|1|
1
1
= || || < 𝜀 for all 𝑛 ≥ 𝐾. This give 𝑎𝑛 > = 𝐴0 ≥ 𝐴 for all 𝑛 ≥ 𝐾.
𝑎𝑛 | 𝑎𝑛 |
𝜀
1
> 0.
𝐴0
Apart from this rule, there are more rules for infinite limits. Some of these rules are listed below. [Note to lecturers:
do not write this down, just refer the students to the notes.]
Theorem 2.9. Consider 𝑘 ∈ ℝ and sequences with the following properties as 𝑛 → ∞:
𝑎𝑛 → ∞, 𝑏𝑛 → ∞, 𝑐𝑛 → 𝑐 ∈ ℝ, 𝑑𝑛 → −∞.
Then, as 𝑛 → ∞,
⎧∞
if k>0,
⎪
(a) 𝑘𝑎𝑛 → ⎨ −∞ if k<0,
⎪0
if k=0.
⎩
(b) 𝑎𝑛 + 𝑏𝑛 → ∞,
(c) 𝑎𝑛 + 𝑐𝑛 → ∞,
(d) −𝑑𝑛 → ∞,
{
∞
if c>0,
(e) 𝑎𝑛 𝑐𝑛 →
−∞ if c<0.
(f) 𝑎𝑛 𝑏𝑛 → ∞.
Proof. See tutorials.
It is convenient to extend the notions of supremum and infimum to unbounded set.
Definition 2.6. Let 𝐴 ⊂ ℝ be nonempty. If 𝐴 is not bounded above, we write sup 𝐴 = ∞, and if 𝐴 is not bounded
below, we write inf 𝐴 = −∞.
We know that not every sequence converges, and it may be hard to decide if a sequence converges. However, the
situation is different with monotonic sequences:
Theorem 2.10. 1. Let (𝑎𝑛 ) be an increasing sequence. If (𝑎𝑛 ) is bounded, then (𝑎𝑛 ) converges, and
lim 𝑎𝑛 = sup{𝑎𝑛 ∶ 𝑛 ∈ ℕ}. If (𝑎𝑛 ) is not bounded, then (𝑎𝑛 ) diverges to ∞.
𝑛→∞
2. Let (𝑎𝑛 ) be a decreasing sequence. If (𝑎𝑛 ) is bounded, then (𝑎𝑛 ) converges, and
lim 𝑎𝑛 = inf {𝑎𝑛 ∶ 𝑛 ∈ ℕ}. If (𝑎𝑛 ) is not bounded, then (𝑎𝑛 ) diverges to −∞.
𝑛→∞
Proof. 1. Assume (𝑎𝑛 ) is bounded. Put 𝐿 = sup{𝑎𝑛 ∶ 𝑛 ∈ ℕ} and let 𝜀 > 0.
We must show that there is 𝐾 ∈ ℕ such that 𝑛 ≥ 𝐾 gives 𝐿 − 𝜀 < 𝑎𝑛 < 𝐿 + 𝜀.
Indeed by the definition of the supremum we have 𝑎𝑛 ≤ 𝐿 < 𝐿 + 𝜀 for all 𝑛 and by Theorem 1.9 there is 𝐾 such
that 𝐿 − 𝜀 < 𝑎𝐾 . Then we have for all 𝑛 ≥ 𝐾 that
𝐿 − 𝜀 < 𝑎𝐾 ≤ 𝑎𝑛 .
Hence 𝐿 − 𝜀 < 𝑎𝑛 < 𝐿 + 𝜀 for all 𝑛 ≥ 𝐾, which proves 𝑎𝑛 → 𝐿 as 𝑛 → ∞.
Now assume that (𝑎𝑛 ) is not bounded. Since 𝑎0 ≤ 𝑎𝑛 for all 𝑛, (𝑎𝑛 ) is bounded below. Hence (𝑎𝑛 ) is not bounded
above. Let 𝐴 ∈ ℝ. Then there is an index 𝐾 such that 𝑎𝐾 > 𝐴, and thus 𝑎𝑛 ≥ 𝑎𝐾 > 𝐴 for all 𝑛 ≥ 𝐾.
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Introductory Analysis 2017 Lecture Manual
21
(
)𝑛
(
)𝑛+1
Example 2.2.3. Let 𝑎𝑛 = 1 + 1𝑛 and 𝑏𝑛 = 1 + 1𝑛
. Then 𝑎𝑛 < 𝑏𝑛 for all 𝑛 ∈ ℕ, (𝑎𝑛 ) is an increasing
sequence, (𝑏𝑛 ) is a decreasing sequence, both sequences converge to the same number, and the limit is denoted by
𝑒, Euler’s number.
Proof. The inequality 𝑎𝑛 < 𝑏𝑛 is obvious from
(
)
1
𝑏𝑛 = 𝑎𝑛 1 +
.
𝑛
Using Bernoulli’s inequality, see Example 1.3.2, we calculate
(
)
𝑛 + 2 𝑛+1 (
)𝑛+1
𝑎𝑛+1
𝑛(𝑛 + 2)
𝑛+1
𝑛+1
= (
=
)
𝑎𝑛
𝑛
𝑛+1 𝑛
(𝑛 + 1)2
𝑛
)𝑛+1
(
𝑛+1
1
= 1−
𝑛
(𝑛 + 1)2
)
(
𝑛+1
𝑛+1
𝑛 𝑛+1
≥ 1−
=
=1
𝑛
𝑛+1 𝑛
(𝑛 + 1)2
and
(
)
𝑛 + 1 𝑛+1 (
)𝑛+2
𝑏𝑛
(𝑛 + 1)2
𝑛
=
=( 𝑛 )
𝑏𝑛+1
𝑛(𝑛 + 2)
𝑛+1
𝑛 + 2 𝑛+2
𝑛+1
)𝑛+2
(
1
𝑛
= 1+
𝑛(𝑛 + 2)
𝑛+1
(
)
𝑛+2
𝑛+1 𝑛
𝑛
≥ 1+
=
= 1.
𝑛(𝑛 + 2) 𝑛 + 1
𝑛 𝑛+1
Hence the sequence (𝑎𝑛 ) is increasing and bounded (by 𝑏1 ). By Theorem 2.10, (𝑎𝑛 ) has a limit 𝛼. Similarly, (𝑏𝑛 )
has a limit 𝛽. Finally
𝛽 = lim 𝑏𝑛 = lim
𝑛→∞
((
𝑛→∞
1+
) )
(
)
1
1
𝑎𝑛 = lim 1 +
lim 𝑎 = 𝛼.
𝑛→∞
𝑛
𝑛 𝑛→∞ 𝑛
Definition 2.7. 1. With every sequence (𝑎𝑛 ) which is bounded above, we can associate the sequence of numbers
sup{𝑎𝑘 ∶ 𝑘 ≥ 𝑛}.
Since {𝑎𝑘 ∶ 𝑘 ≥ 𝑛} ⊃ {𝑎𝑘 ∶ 𝑘 ≥ 𝑛 + 1}, this sequence is decreasing, and we denote its limit by
lim sup 𝑎𝑛 = lim sup{𝑎𝑘 ∶ 𝑘 ≥ 𝑛}.
𝑛→∞
𝑛→∞
If (𝑎𝑛 ) is not bounded above, we put lim sup 𝑎𝑛 = ∞.
𝑛→∞
2. Similarly, if the sequence (𝑎𝑛 ) is bounded below, the sequence of numbers
inf {𝑎𝑘 ∶ 𝑘 ≥ 𝑛}
is increasing, and we denote its limit by
lim inf 𝑎𝑛 = lim inf {𝑎𝑘 ∶ 𝑘 ≥ 𝑛}.
𝑛→∞
𝑛→∞
If (𝑎𝑛 ) is not bounded below, we put lim inf 𝑎𝑛 = −∞.
𝑛→∞
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Note. 1. Since inf 𝑆 ≤ sup 𝑆 for every nonempty set 𝑆, it follows that lim inf 𝑎𝑛 ≤ lim sup 𝑎𝑛 , where we write
𝑛→∞
𝑛→∞
−∞ < 𝑥 and 𝑥 < ∞ for each 𝑥 ∈ ℝ.
2. The sequence (𝑎𝑛 ) is bounded if and only if both lim inf 𝑎𝑛 and lim sup 𝑎𝑛 are real numbers (also called finite).
𝑛→∞
𝑛→∞
We have now the following characterization of the limit of a sequence and its existence:
Theorem 2.11. 1. The sequence (𝑎𝑛 ) converges if and only if lim inf 𝑎𝑛 and lim sup 𝑎𝑛 are finite and equal, and
𝑛→∞
𝑛→∞
then
lim inf 𝑎𝑛 = lim 𝑎𝑛 = lim sup 𝑎𝑛 .
𝑛→∞
𝑛→∞
𝑛→∞
2. lim 𝑎𝑛 = ∞ ⇔ lim inf 𝑎𝑛 = ∞ and lim sup 𝑎𝑛 = ∞.
𝑛→∞
𝑛→∞
𝑛→∞
3. lim 𝑎𝑛 = −∞ ⇔ lim inf 𝑎𝑛 = −∞ and lim sup 𝑎𝑛 = −∞.
𝑛→∞
𝑛→∞
𝑛→∞
Proof. 1. For a sequence (𝑎𝑛 ), denote 𝑏𝑛 = inf {𝑎𝑘 ∶ 𝑘 ≥ 𝑛} and 𝑐𝑛 = sup{𝑎𝑘 ∶ 𝑘 ≥ 𝑛}. Note that 𝑏𝑛 ≤ 𝑎𝑛 ≤ 𝑐𝑛 .
1. Assume that (𝑎𝑛 ) converges to 𝐿. Let 𝜀 > 0. Then there is 𝐾 ∈ ℕ such that for all 𝑛 ≥ 𝐾, 𝐿 − 3𝜀 < 𝑎𝑛 < 𝐿 + 3𝜀 .
Hence
𝜀
𝜀
𝐿 − ≤ 𝑏𝑛 ≤ 𝑎𝑛 ≤ 𝑐𝑛 ≤ 𝐿 + .
3
3
Since (𝑏𝑛 ) is increasing and (𝑐𝑛 ) is decreasing, we have
𝐿−
𝜀
𝜀
≤ lim inf 𝑎𝑛 ≤ lim sup 𝑎𝑛 ≤ 𝐿 + .
𝑛→∞
3
3
𝑛→∞
Hence
0 ≤ lim sup 𝑎𝑛 − lim inf 𝑎𝑛 ≤
𝑛→∞
𝑛→∞
2𝜀
< 𝜀.
3
From Lemma 2.1 we obtain that lim inf 𝑎𝑛 = lim sup 𝑎𝑛 .
𝑛→∞
𝑛→∞
Conversely, if lim inf 𝑎𝑛 = lim sup 𝑎𝑛 , then it follows from 𝑏𝑛 ≤ 𝑎𝑛 ≤ 𝑐𝑛 and the Sandwich Theorem that (𝑎𝑛 )
𝑛→∞
𝑛→∞
converges and that
lim inf 𝑎𝑛 = lim 𝑎𝑛 = lim sup 𝑎𝑛 .
𝑛→∞
𝑛→∞
𝑛→∞
2., 3. Exercise.
Another important concept is that of a Cauchy sequence:
Definition 2.8 (Cauchy sequence). A seqence (𝑎𝑛 ) is called a Cauchy sequence if for all 𝜀 > 0 there is 𝐾 ∈ ℝ
such that for all 𝑛, 𝑚 ∈ ℕ with 𝑛, 𝑚 ≥ 𝐾, |𝑎𝑛 − 𝑎𝑚 | < 𝜀.
Theorem 2.12. A sequence (𝑎𝑛 ) converges if and only if it is a Cauchy sequence.
Proof. Let (𝑎𝑛 ) be a convergent sequence with limit 𝐿. Let 𝜀 > 0 and let 𝐾 such that |𝑎𝑛 − 𝐿| < 2𝜀 for 𝑛 ≥ 𝐾.
Then it follows for 𝑛, 𝑚 ≥ 𝐾 that
|𝑎𝑛 − 𝑎𝑚 | = |(𝑎𝑛 − 𝐿) − (𝑎𝑚 − 𝐿)| ≤ |𝑎𝑛 − 𝐿| + |𝑎𝑚 − 𝐿| <
𝜀 𝜀
+ = 𝜀.
2 2
Conversely, assume that (𝑎𝑛 ) is a Cauchy sequence. Let 𝜀 > 0 and choose 𝐾 such that |𝑎𝑛 − 𝑎𝑚 | < 3𝜀 for all
𝑚, 𝑛 ≥ 𝐾. Then
𝜀
𝜀
< 𝑎𝑛 < 𝑎𝑚 +
3
3
𝜀
𝜀
⇒ 𝑎𝑚 − ≤ lim inf 𝑎𝑛 ≤ lim sup 𝑎𝑛 ≤ 𝑎𝑚 +
𝑛→∞
3
3
𝑛→∞
(
) (
)
𝜀
𝜀
2𝜀
⇒ 0 ≤ lim sup 𝑎𝑛 − lim inf 𝑎𝑛 ≤ 𝑎𝑚 +
− 𝑎𝑚 −
=
< 𝜀.
𝑛→∞
3
3
3
𝑛→∞
𝑎𝑚 −
By Lemma 2.1, lim inf 𝑎𝑛 = lim sup 𝑎𝑛 , and an application of Theorem 2.11, part 1, completes the proof.
𝑛→∞
𝑛→∞
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Introductory Analysis 2017 Lecture Manual
Tutorial 2.2.1. 1. Prove Theorem 2.9.
2. Use suitable rules or first principles to find
(b) lim (𝑛 − 1𝑛 )
(a) lim (𝑛2 + 2𝑛 − 10)
𝑛→∞
𝑛→∞
𝑛3 − 3𝑛2
𝑛→∞ 𝑛 + 1
(c) lim
3. Prove that if lim |𝑎𝑛 | = ∞, then (𝑎𝑛 ) diverges.
𝑛→∞
4. Prove that if 𝑝 ∈ ℕ, 𝑝 > 0, then 𝑛𝑝 → ∞ as 𝑛 → ∞.
5. Define a sequence as follows:
𝑎0 = 0, 𝑎1 =
1
1
, 𝑎
= (1 + 𝑎𝑛 + 𝑎3𝑛−1 ) for 𝑛 ≥ 2.
2 𝑛+1 3
(a) Use induction to show that 0 ≤ 𝑎𝑛 < 23 for all 𝑛 ∈ ℕ.
(b) Use induction to show that 𝑎𝑛 ≤ 𝑎𝑛+1 for all 𝑛 ∈ ℕ.
(c) Explain why we may conclude that (𝑎𝑛 ) converges.
(d) Using the fact that lim 𝑎𝑛 = lim 𝑎𝑛−1 = lim 𝑎𝑛+1 , find lim 𝑎𝑛 .
𝑛→∞
𝑛→∞
𝑛→∞
𝑛→∞
6. Let lim 𝑎𝑛 = ∞, lim 𝑏𝑛 = ∞, lim 𝑐𝑛 = 0. Show, by giving examples, that no general conclusion can be made
𝑛→∞
𝑛→∞
𝑛→∞
about the behaviour of the following sequences:
𝑎
𝑎
(a) 𝑎𝑛 − 𝑏𝑛 , (b) 𝑎𝑛 𝑐𝑛 , (c) 𝑛 , (d) 𝑛 .
𝑏𝑛
𝑐𝑛
7. Let (𝑎𝑛 ) and (𝑏𝑛 ) be sequences such that 𝑎𝑛 ≤ 𝑏𝑛 for all 𝑛 ∈ ℕ. Show that lim inf 𝑎𝑛 ≤ lim inf 𝑏𝑛 and lim sup 𝑎𝑛 ≤
𝑛→∞
lim sup 𝑏𝑛 .
𝑛→∞
(
)
𝑥 𝑛
8. (a) Show that exp(𝑥) = lim 1 +
exists for all 𝑥 ∈ ℝ and that exp(1) = 𝑒.
𝑛→∞
𝑛
Hint. Adapt the proof of Example 2.2.3.
(b) Use Bernoulli’s inequality to prove that
(
lim
𝑛→∞
1 + 𝑥+𝑦
𝑛
1 + 𝑥+𝑦
+ 𝑥𝑦
𝑛
𝑛2
)𝑛
=1
for all 𝑥, 𝑦 ∈ ℝ.
(c) Use (b) to show that exp(𝑥 + 𝑦) = exp(𝑥) exp(𝑦) for all 𝑥, 𝑦 ∈ ℝ.
(d) Show that exp(𝑥) ≥ 1 + 𝑥 for all 𝑥 > 0.
(e) Show that exp(𝑥) > 0 for all 𝑥 ∈ ℝ.
(f) Show that exp is strictly increasing.
(g) Show that exp(𝑛) = 𝑒𝑛 for all 𝑛 ∈ ℤ.
𝑛→∞
𝑛→∞
Chapter 3
Limits and Continuity of Functions
The single most important concept in all of analysis is that of a limit. Every single notion of analysis is encapsulated
in one sense or another to that of a limit. In the previous chapter, you learnt about limits of sequences. Here this
notion is extended to limits of functions, which leads to the notion of continuity. You have learnt an intuitive
version in Calculus I. Here you will learn a precise definition and you will learn how to prove the results you have
learnt in Calculus I.
In this course all functions will have domains and ranges which are subsets of ℝ unless otherwise stated. Such
functions are called real functions.
3.1
Limits of Functions
Definition 3.1. Let 𝑎 ∈ ℝ. An interval of the form (𝑐, 𝑑) with 𝑐 < 𝑎 < 𝑑 is called a neighbourhood of 𝑎, and the
set (𝑐, 𝑑) ⧵ {𝑎} is called a deleted neighbourhood of 𝑎.
In the definition of the limit of sequences you have seen formal definitions for ‘𝑛 tends to ∞’ and ‘𝑎𝑛 tends to 𝐿’,
and the latter one has an obvious generalization to ‘𝑥 tends to 𝑎’ and ‘𝑓 (𝑥) tends to 𝐿’, which you will encounter
in the next definition.
Definition 3.2 (Limit of a function). Let 𝑓 be a real function, 𝑎, 𝐿 ∈ ℝ and assume that the domain of 𝑓 contains
a deleted neighbourhood of 𝑎, that is, 𝑓 (𝑥) is defined for all 𝑥 in a deleted neighbourhood of 𝑓 .
Then ‘𝑓 (𝑥) → 𝐿 as 𝑥 → 𝑎’ is defined to mean:
∀ 𝜀 > 0 ∃ 𝛿 > 0 ∀ 𝑥 ∈ (𝑎 − 𝛿, 𝑎 + 𝛿) ⧵ {𝑎}, 𝑓 (𝑥) ∈ (𝐿 − 𝜀, 𝐿 + 𝜀),
i. e., ∀ 𝜀 > 0 ∃ 𝛿 > 0 (0 < |𝑥 − 𝑎| < 𝛿 ⇒ |𝑓 (𝑥) − 𝐿| < 𝜀).
If 𝑓 (𝑥) → 𝐿 as 𝑥 → 𝑎, then we write lim 𝑓 (𝑥) = 𝐿.
𝑥→𝑎
Note. 1. 𝐿 in the definition above is also called the limit of the function at 𝑎. For the definition of the limit of a
function at 𝑎, the number 𝑓 (𝑎) is never used, and indeed, the function need not be defined at 𝑎.
2. Observe that the number 𝛿 will depend on 𝜀, and also on 𝑎, if we vary 𝑎.
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25
Exercise. Use the notation in the above definition to label the following graph:
Example 3.1.1. Prove, from first principles, that 𝑥2 → 4 as 𝑥 → 2.
Solution. Let 𝜀 > 0. We must show that there is 𝛿 > 0 such that 0 < |𝑥 − 2| < 𝛿 implies |𝑥2 − 4| < 𝜀. Since we
want to use |𝑥 − 2| < 𝛿 to get |𝑥2 − 4| < 𝜀, we try to factor out |𝑥 − 2| from |𝑥2 − 4|. Thus
|𝑥2 − 4| = |𝑥 − 2| |𝑥 + 2| ≤ |𝑥 − 2| (|𝑥 − 2| + 4).
It is often convenient to make an initial restriction on 𝛿, like 𝛿 ≤ 1. Then we can continue the above estimate to
obtain from |𝑥 − 𝑎| < 𝛿 that
|𝑥2 − 4| ≤ |𝑥 − 2| (1 + 4) = 5|𝑥 − 2|.
If we now put
}
{
𝜀
,
𝛿 = min 1,
5
then we can conclude (note that we already used 𝛿 ≤ 1) for 0 < |𝑥 − 2| < 𝛿 that
𝜀
|𝑥2 − 4| ≤ 5|𝑥 − 2| < 5 = 𝜀.
5
To justify the notation lim 𝑓 (𝑥) = 𝐿 we have to show
𝑥→𝑎
Theorem 3.1. If 𝑓 (𝑥) → 𝐿 as 𝑥 → 𝑎, then 𝐿 is unique.
Proof. Let 𝑓 (𝑥) → 𝐿 and 𝑓 (𝑥) → 𝑀 as 𝑥 → 𝑎. We must show that 𝐿 = 𝑀.
So let 𝜀 > 0. Then there are 𝛿1 > 0 and 𝛿2 > 0 such that
𝜀
𝜀
0 < |𝑥 − 𝑎| < 𝛿1 ⇒ |𝑓 (𝑥) − 𝐿| <
and 0 < |𝑥 − 𝑎| < 𝛿2 ⇒ |𝑓 (𝑥) − 𝑀| < .
2
2
For 𝛿 = min{𝛿1 , 𝛿2 } and 0 < |𝑥 − 𝑎| < 𝛿 (we indeed only need one such 𝑥 here) it follows that
𝜀 𝜀
|𝐿 − 𝑀| = |(𝐿 − 𝑓 (𝑥)) − (𝑓 (𝑥) − 𝑀)| ≤ |𝑓 (𝑥) − 𝐿| + |𝑓 (𝑥) − 𝑀| < + = 𝜀.
2 2
Hence 𝐿 = 𝑀 by Lemma 2.1.
Definition 3.3 (One sided limits). 1. Let 𝑓 be a real function, 𝑎, 𝐿 ∈ ℝ and assume that the domain of 𝑓 contains
an interval (𝑎, 𝑑) with 𝑑 > 𝑎, that is, 𝑓 (𝑥) is defined for all 𝑥 in (𝑎, 𝑑).
Then ‘𝑓 (𝑥) → 𝐿 as 𝑥 → 𝑎+ ’ is defined to mean:
∀ 𝜀 > 0 ∃ 𝛿 > 0 (𝑎 < 𝑥 < 𝑎 + 𝛿 ⇒ |𝑓 (𝑥) − 𝐿| < 𝜀).
If 𝑓 (𝑥) → 𝐿 as 𝑥 → 𝑎+ , then we write lim+ 𝑓 (𝑥) = 𝐿.
𝑥→𝑎
2. Let 𝑓 be a real function, 𝑎, 𝐿 ∈ ℝ and assume that the domain of 𝑓 contains an interval (𝑐, 𝑎) with 𝑐 < 𝑎, that
is, 𝑓 (𝑥) is defined for all 𝑥 in (𝑐, 𝑎).
Then ‘𝑓 (𝑥) → 𝐿 as 𝑥 → 𝑎− ’ is defined to mean:
∀ 𝜀 > 0 ∃ 𝛿 > 0 (𝑎 − 𝛿 < 𝑥 < 𝑎 ⇒ |𝑓 (𝑥) − 𝐿| < 𝜀).
If 𝑓 (𝑥) → 𝐿 as 𝑥 → 𝑎− , then we write lim− 𝑓 (𝑥) = 𝐿.
𝑥→𝑎
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Theorem 3.2. If 𝑓 (𝑥) is defined in a deleted neighbourhood of 𝑎, then
lim 𝑓 (𝑥) = 𝐿 ⇔ lim− 𝑓 (𝑥) = 𝐿 = lim+ 𝑓 (𝑥).
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
Proof. ⇒: Assume that lim 𝑓 (𝑥) = 𝐿 and let 𝜀 > 0. Then
𝑥→𝑎
∃ 𝛿 > 0 (𝑥 ∈ (𝑎 − 𝛿, 𝑎 + 𝛿), 𝑥 ≠ 𝑎 ⇒ |𝑓 (𝑥) − 𝐿| < 𝜀)
gives
∃ 𝛿 > 0 (𝑥 ∈ (𝑎 − 𝛿, 𝑎) ⇒ |𝑓 (𝑥) − 𝐿| < 𝜀) and
∃ 𝛿 > 0 (𝑥 ∈ (𝑎, 𝑎 + 𝛿) ⇒ |𝑓 (𝑥) − 𝐿| < 𝜀).
Hence lim− 𝑓 (𝑥) = 𝐿 and lim+ 𝑓 (𝑥) = 𝐿.
𝑥→𝑎
𝑥→𝑎
⇐: Assume that lim− 𝑓 (𝑥) = 𝐿 = lim+ 𝑓 (𝑥) and let 𝜀 > 0. Then there are 𝛿− > 0 and 𝛿+ > 0 such that
𝑥→𝑎
𝑥→𝑎
𝑥 ∈ (𝑎 − 𝛿− , 𝑎) ⇒ |𝑓 (𝑥) − 𝐿| < 𝜀 and
𝑥 ∈ (𝑎, 𝑎 + 𝛿+ ) ⇒ |𝑓 (𝑥) − 𝐿| < 𝜀.
Let 𝛿 = min{𝛿− , 𝛿+ }. Then
𝑥 ∈ (𝑎 − 𝛿, 𝑎 + 𝛿), 𝑥 ≠ 𝑎 ⇒ 𝑥 ∈ (𝑎 − 𝛿− , 𝑎) or 𝑥 ∈ (𝑎, 𝑎 + 𝛿+ )
⇒ |𝑓 (𝑥) − 𝐿| < 𝜀.
This proves that lim 𝑓 (𝑥) = 𝐿.
𝑥→𝑎
𝑥
for 𝑥 ∈ ℝ ⧵ {0}. Then 𝑓 (𝑥) = 1 if 𝑥 > 0 and 𝑓 (𝑥) = −1 if 𝑥 < 0. It is easy
|𝑥|
to prove from the definition that lim+ 𝑓 (𝑥) = 1 and lim− 𝑓 (𝑥) = −1. Now it follows from Theorem 3.2 that the
Example 3.1.2. Let 𝑓 (𝑥) =
𝑥→0
𝑥→0
function 𝑓 does not have a limit as 𝑥 tends to 𝑎.
Tutorial 3.1.1. 1. Prove from the definitions that
√
1
2
1
(a) 2 + 𝑥 + 𝑥2 → 8 as 𝑥 → 2, (b)
→ − as 𝑥 → , (c) lim− 1 − 𝑥 = 0.
𝑥→1
𝑥−2
3
2
2. By negating the definition of limit of a function show that the statement 𝑓 (𝑥) →
̸ 𝐿 as 𝑥 → 𝑎 is equivalent to the
following:
∃ 𝜀 > 0 ∀ 𝛿 > 0 ∃ 𝑥 with 0 < |𝑥 − 𝑎| < 𝛿 such that |𝑓 (𝑥) − 𝐿| ≥ 𝜀.
1
3. For 𝑥 ∈ ℝ ⧵ {0} let 𝑓 (𝑥) = sin . Prove that 𝑓 does not tend to any limit as 𝑥 → 0.
𝑥
4. Let 𝑓 (𝑥) = 𝑥 − ⌊𝑥⌋. For each integer 𝑛, find lim− 𝑓 (𝑥) and lim+ 𝑓 (𝑥) if they exist.
𝑥→𝑛
3.2
𝑥→𝑛
Limits at Infinity and Infinite Limits
In 𝑥 → 𝑎 the inequalities 0 < |𝑥 − 𝑎| < 𝛿 occur. For 𝑥 → ∞, this has to be replaced by 𝑥 > 𝐾. Hence we have the
following
Definition 3.4. 1. Let 𝑓 be a real function defined on a set containing an interval of the form (𝑐, ∞). Then
‘𝑓 (𝑥) → 𝐿 as 𝑥 → ∞’ is defined to mean:
∀ 𝜀 > 0 ∃ 𝐾(> 0) (𝑥 > 𝐾 ⇒ |𝑓 (𝑥) − 𝐿| < 𝜀).
If 𝑓 (𝑥) → 𝐿 as 𝑥 → ∞, then we write lim 𝑓 (𝑥) = 𝐿.
𝑥→∞
2. Let 𝑓 be a real function defined on a set containing an interval of the form (−∞, 𝑐).
Then ‘𝑓 (𝑥) → 𝐿 as 𝑥 → −∞’ is defined to mean:
∀ 𝜀 > 0 ∃ 𝐾(< 0) (𝑥 < 𝐾 ⇒ |𝑓 (𝑥) − 𝐿| < 𝜀).
If 𝑓 (𝑥) → 𝐿 as 𝑥 → −∞, then we write lim 𝑓 (𝑥) = 𝐿.
𝑥→−∞
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Example 3.2.1. Let 𝑓 (𝑥) =
1
. Find lim 𝑓 (𝑥) and lim 𝑓 (𝑥).
𝑥→∞
𝑥→−∞
𝑥
Solution. Guess that both limits are 0. Let 𝜀 > 0 and put 𝐾 =
1
. Then 𝑥 > 𝐾 gives
𝜀
|1| 1
| | = < 1 = 𝜀.
|𝑥| 𝑥 𝐾
| |
So lim 𝑓 (𝑥) = 0.
𝑥→∞
1
Now let 𝜀 > 0 and put 𝐾 = − . Then 𝑥 < 𝐾 gives
𝜀
|1|
| | = − 1 < 1 = 𝜀.
|𝑥|
𝑥 −𝐾
| |
So lim 𝑓 (𝑥) = 0.
𝑥→−∞
For infinite limits we have similar definitions:
Definition 3.5. Let 𝑓 be a real function whose domain includes a deleted neighbourhood of the number 𝑎.
1. ‘𝑓 (𝑥) → ∞ as 𝑥 → 𝑎’ is defined to mean:
∀ 𝐾(> 0) ∃ 𝛿 > 0 (0 < |𝑥 − 𝑎| < 𝛿 ⇒ 𝑓 (𝑥) > 𝐾).
If 𝑓 (𝑥) → ∞ as 𝑥 → 𝑎, then we write lim 𝑓 (𝑥) = ∞.
𝑥→𝑎
2. ‘𝑓 (𝑥) → −∞ as 𝑥 → 𝑎’ is defined to mean:
∀ 𝐾(< 0) ∃ 𝛿 > 0 (0 < |𝑥 − 𝑎| < 𝛿 ⇒ 𝑓 (𝑥) < 𝐾).
If 𝑓 (𝑥) → −∞ as 𝑥 → 𝑎, then we write lim 𝑓 (𝑥) = −∞.
𝑥→𝑎
Example 3.2.2. Prove that
2
→ ∞ as 𝑥 → 0.
𝑥2
Solution. The following sketch illustrates the result.
40
35
30
25
20
15
10
5
−2
−1
0
1
2
We may choose 𝐾 to be sufficiently large. So choose 𝐾 > 1 and let 𝛿 = 𝐾1 . Then, for 0 < |𝑥| < 𝛿, and because of
𝛿 < 1 we have
2
1 2
2
2
=
≥
> = 2𝐾 > 𝐾.
|𝑥| |𝑥| |𝑥| 𝛿
𝑥2
Definition 3.6. Let 𝑓 be a real function defined on a set containing an interval of the form (𝑐, ∞).
Then ‘𝑓 (𝑥) → ∞ as 𝑥 → ∞’ is defined to mean:
∀ 𝐴(> 0) ∃ 𝐾(> 0) (𝑥 > 𝐾 ⇒ 𝑓 (𝑥) > 𝐴).
If 𝑓 (𝑥) → ∞ as 𝑥 → ∞, then we write lim 𝑓 (𝑥) = ∞.
𝑥→∞
Similar definition hold with all other combinations of limits involving ∞ and −∞, including one-sided limits.
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𝑥
sin 𝑥
Example 3.2.3. Prove that 𝑓 (𝑥) =
+ 10
→ ∞ as 𝑥 → ∞. A section of the function’s graph is plotted
10
𝑥
below.
8
6
4
2
20
40
60
80
Additional notes. 1. You have learnt different notations for limits depending on whether real numbers, ∞ or
−∞ are considered. However, using the notion of (deleted) neighbourhood, one can give just one definition which
applies to each case. To define (deleted) neighbourhoods of ∞ and −∞, we first observe that ∞ and −∞ are just
symbols for our purposes, and thus never belong to sets of numbers which we consider. Therefore, no distinction
is made here between neighbourhoods and deleted neighbourhoods of ∞ and −∞, respectively.
A (deleted) neighbourhood of ∞ is any set of the form (𝑐, ∞) with 𝑐 ∈ ℝ.
A (deleted) neighbourhood of −∞ is any set of the form (−∞, 𝑑) with 𝑑 ∈ ℝ.
Now let 𝑎, 𝐿 ∈ ℝ ∪ {∞, −∞} and 𝑓 be a function defined in a deleted neighbourhood of 𝑎. Then 𝑓 (𝑥) → 𝐿 as
𝑥 → 𝑎 if and only if for each neighbourhood 𝑈 of 𝐿 there is a deleted neighbourhood 𝑉 of 𝑎 such that 𝑥 ∈ 𝑉
implies 𝑓 (𝑥) ∈ 𝑈 .
If we define one-sided deleted neighbourhoods of 𝑎 ∈ ℝ to be the sets (𝑐, 𝑎) and (𝑎, 𝑑), respectively, then the above
formulation also covers one-sided limits.
2. What we call neighbourhood here should actually be called basic neighbourhood. In general topology, a neighborhood is any subset (of a given set) which contains a basic neighbourhood. But since we do not need general
neighbourhoods, we conveniently dropped the word ’basic’.
1
. Find lim− 𝑓 (𝑥), lim+ 𝑓 (𝑥), lim 𝑓 (𝑥).
𝑥→0
𝑥→0
𝑥→0
𝑥
1
1
1
Solution. Let 𝐾 > 0 and put 𝛿 = . Then, for 0 < 𝑥 < 𝛿, > = 𝐾. Thus lim+ 𝑓 (𝑥) = ∞.
𝑥→0
𝐾
𝑥 𝛿
1
1
1
Now let 𝐾 < 0 and put 𝛿 = − > 0. Then, for −𝛿 < 𝑥 < 0, <
= 𝐾. Thus lim− 𝑓 (𝑥) = −∞.
𝑥→0
𝐾
𝑥 −𝛿
Since lim− 𝑓 (𝑥) ≠ lim+ 𝑓 (𝑥), lim 𝑓 (𝑥) does not exist.
Example 3.2.4. Let 𝑓 (𝑥) =
𝑥→0
𝑥→0
𝑥→0
Tutorial 3.2.1. 1. Prove from the definitions that
1 − 3𝑥
3
1
= − , (b)
→ ∞ as 𝑥 → 1+ ,
(a) lim
𝑥→−∞ 2𝑥 − 1
2
𝑥−1
(c)
1
→ −∞ as 𝑥 → 1− .
𝑥−1
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Limit Laws
Rather than calculating limits from the definition, in general one will use limit laws. In this section we state and
prove some of these laws.
Theorem 3.3 (Limit Laws). Let 𝑎, 𝑐 ∈ ℝ and suppose that the real functions 𝑓 and 𝑔 are defined in a deleted
neighbourhood of 𝑎 and that lim 𝑓 (𝑥) = 𝐿 ∈ ℝ and lim 𝑔(𝑥) = 𝑀 ∈ ℝ both exist. Then
𝑥→𝑎
𝑥→𝑎
(a) lim 𝑐 = 𝑐.
𝑥→𝑎
(b) lim [𝑓 (𝑥) + 𝑔(𝑥)] = lim 𝑓 (𝑥) + lim 𝑔(𝑥) = 𝐿 + 𝑀.
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
(c) lim [𝑓 (𝑥) − 𝑔(𝑥)] = lim 𝑓 (𝑥) − lim 𝑔(𝑥) = 𝐿 − 𝑀.
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
(d) lim [𝑐𝑓 (𝑥)] = 𝑐 lim 𝑓 (𝑥) = 𝑐𝐿.
𝑥→𝑎
𝑥→𝑎
[
][
]
(e) lim [𝑓 (𝑥)𝑔(𝑥)] = lim 𝑓 (𝑥) lim 𝑔(𝑥) = 𝐿𝑀.
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
lim 𝑓 (𝑥)
𝑓 (𝑥) 𝑥→𝑎
𝐿
=
=
.
𝑥→𝑎 𝑔(𝑥)
lim 𝑔(𝑥)
𝑀
(f) If 𝑀 ≠ 0, lim
𝑥→𝑎
𝑓 (𝑥)
does not exist.
𝑥→𝑎 𝑔(𝑥)
[
]𝑛
(h) If 𝑛 ∈ ℕ, lim [𝑓 (𝑥)𝑛 ] = lim 𝑓 (𝑥) = 𝐿𝑛 .
(g) If 𝐿 ≠ 0 and 𝑀 = 0, lim
𝑥→𝑎
𝑥→𝑎
(i) lim 𝑥 = 𝑎.
𝑥→𝑎
(j) If 𝑛 ∈ ℕ, lim 𝑥𝑛 = 𝑎𝑛 .
𝑥→𝑎
√
√
(k) If 𝑛 ∈ ℕ, lim 𝑛 𝑥 = 𝑛 𝑎. If 𝑛 is even, we assume that 𝑎 > 0.
𝑥→𝑎
√
√
√
𝑛
(l) If 𝑛 ∈ ℕ, lim 𝑛 𝑓 (𝑥) = 𝑛 lim 𝑓 (𝑥) = 𝐿.
𝑥→𝑎
𝑥→𝑎
(m) If lim |𝑓 (𝑥)| = 0, then lim 𝑓 (𝑥) = 0.
𝑥→𝑎
𝑥→𝑎
Proof. The proofs are similar to those in Theorem 2.3 and we will only prove (b), (e), (f).
(b) Let 𝜀 > 0. Then there are numbers 𝛿1 > 0 and 𝛿2 > 0 such that
(i) |𝑓 (𝑥) − 𝐿| < 2𝜀 if 0 < |𝑥 − 𝑎| < 𝛿1 ,
(ii) |𝑔(𝑥) − 𝑀| < 2𝜀 if 0 < |𝑥 − 𝑎| < 𝛿2 .
Put 𝛿 = min{𝛿1 , 𝛿2 }. For 0 < |𝑥 − 𝑎| < 𝛿 we have 0 < |𝑥 − 𝑎| < 𝛿1 and 0 < |𝑥 − 𝑎| < 𝛿2 and therefore
|(𝑓 (𝑥) + 𝑔(𝑥)) − (𝐿 + 𝑀)| = |(𝑓 (𝑥) − 𝐿) + (𝑔(𝑥) − 𝑀)| ≤ |𝑓 (𝑥) − 𝐿| + |𝑔(𝑥) − 𝑀| <
𝜀 𝜀
+ = 𝜀.
2 2
Hence 𝑓 (𝑥) + 𝑔(𝑥) converges to 𝐿 + 𝑀 as 𝑥 → 𝑎.
(e) We consider 2 cases: one special case, to which then the general case is reduced.
Case I: 𝐿 = 𝑀 = 0. Let 𝜀 > 0 and put 𝜀′ = min{1, 𝜀}. Then 0 < 𝜀′ ≤ 𝜀, and there are numbers 𝛿1 and 𝛿2 such
that
(i) |𝑓 (𝑥)| < 𝜀′ if 0 < |𝑥 − 𝑎| < 𝛿1 ,
(ii) |𝑔(𝑥)| < 𝜀′ if 0 < |𝑥 − 𝑎| < 𝛿2 .
Put 𝛿 = min{𝛿1 , 𝛿2 }. For 0 < |𝑥 − 𝑎| < 𝛿 we have
2
|𝑓 (𝑥)𝑔(𝑥)| = |𝑓 (𝑥)| |𝑔(𝑥)| < 𝜀′ ≤ 𝜀′ ≤ 𝜀.
Case II: 𝐿 and 𝑀 are arbitrary. Then
𝑓 (𝑥)𝑔(𝑥) = (𝑓 (𝑥) − 𝐿)(𝑔(𝑥) − 𝑀) + 𝐿(𝑔(𝑥) − 𝑀) + 𝑓 (𝑥)𝑀.
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By (a), (b), (c), (𝑓 (𝑥) − 𝑀) → 0 and (𝑔(𝑥) − 𝑀) → 0 as 𝑥 → 𝑎, and by (b), (c), (d), and Case I, it follows that
lim 𝑓 (𝑥)𝑔(𝑥) = 0 + 𝐿 ⋅ 0 + 𝐿𝑀 = 𝐿𝑀.
𝑥→𝑎
(f) First consider 𝑓 = 1. Since 𝑀 ≠ 0 and 𝑔(𝑥) → 𝑀 as 𝑥 → 𝑎, there is 𝛿0 > 0 such that |𝑔(𝑥) − 𝑀| <
0 < |𝑥 − 𝑎| < 𝛿0 . Then, for 0 < |𝑥 − 𝑎| < 𝛿0 ,
|𝑀|
for
2
|𝑀| |𝑀|
=
2
2
| 1
| |𝑀 − 𝑔(𝑥)| 2|𝑔(𝑥) − 𝑀|
1
|=
⇒ ||
−
≤
.
|
|𝑔(𝑥)𝑀|
|𝑀|2
| 𝑔(𝑥) 𝑀 |
|𝑔(𝑥)| ≥ |𝑀| − |𝑔(𝑥) − 𝑀| ≥ |𝑀| −
Now let 𝜀 > 0 and 𝛿1 > 0 such that 0 < |𝑥 − 𝑎| < 𝛿1 implies |𝑔(𝑥) − 𝑀| ≤
follows for 0 < |𝑥 − 𝑎| < 𝛿 that
|𝑀|2
𝜀. Put 𝛿 = min{𝛿0 , 𝛿1 }. It
2
| 1
|𝑀|2
1 || 2|𝑔(𝑥) − 𝑀|
1
|
<2
= 𝜀.
𝜀
| 𝑔(𝑥) − 𝑀 | ≤
2
2
2
|𝑀|
|𝑀|
|
|
The general case now follows with (d):
𝑓 (𝑥)
1
1
= lim 𝑓 (𝑥) ⋅ lim
=𝐿⋅
.
𝑥→𝑎 𝑔(𝑥)
𝑥→𝑎
𝑥→𝑎 𝑔(𝑥)
𝑀
lim
Recall that a polynomial function is of the form
𝑓 (𝑥) = 𝑏𝑛 𝑥𝑛 + 𝑏𝑛−1 𝑥𝑛−1 + ⋯ + 𝑏2 𝑥2 + 𝑏1 𝑥 + 𝑏0
with 𝑏𝑖 ∈ ℝ for 𝑖 = 1, 2, … , 𝑛 and 𝑛 any non-negative integer. A rational function is of the form 𝑓 (𝑥) =
𝑝(𝑥)
with
𝑞(𝑥)
𝑝(𝑥) and 𝑞(𝑥) polynomials. We then have the following as a consequence of Theorem 2.3, (b),(d),(i),(j).
Corollary 3.4. If 𝑓 is a polynomial or a rational function and 𝑎 is in the domain of 𝑓 , then lim 𝑓 (𝑥) = 𝑓 (𝑎).
𝑥→𝑎
Corollary 3.5. All the limit rules in Theorem 3.3 remain true if 𝑥 → 𝑎 is replaced by any of the following: 𝑥 → 𝑎+ ,
𝑥 → 𝑎− , 𝑥 → ∞, 𝑥 → −∞.
Proof. For 𝑥 → 𝑎+ and 𝑥 → 𝑎− one just has to replace 0 < |𝑥 − 𝑎| < 𝛿 with 0 < 𝑥 − 𝑎 < 𝛿 and −𝛿 < −(𝑥 − 𝑎) < 0,
respectively, in the proof of each of the statements. For 𝑥 → ∞ and 𝑥 → −∞, the proofs are very similar to those
for sequences.
Similar rules hold if the functions have infinite limits. We state some of the results for 𝑥 → 𝑎, observing that there
are obvious extensions as in Corollary 3.5.
Theorem 3.6. Assume that lim 𝑓 (𝑥) = ∞, lim 𝑔(𝑥) = ∞ and lim ℎ(𝑥) = 𝑐 ∈ ℝ. Then
𝑥→𝑎
(a) 𝑓 (𝑥) + 𝑔(𝑥) → ∞ as 𝑥 → 𝑎,
(b) 𝑓 (𝑥) + ℎ(𝑥) → ∞ as 𝑥 → 𝑎,
(c) 𝑓 (𝑥)𝑔(𝑥) → ∞
{ as 𝑥 → 𝑎,
∞
if c>0
(d) 𝑓 (𝑥)ℎ(𝑥) →
−∞ if c<0
𝑥→𝑎
𝑥→𝑎
as 𝑥 → 𝑎.
Proof. We prove (c) and leave the other parts as exercises.
Let 𝐴 > 1 (we can take any 𝐴 ∈ ℝ, but the restriction 𝐴 > 1 is convenient here). Then there are 𝛿1 > 0 and 𝛿2 > 0
such that
(i) 𝑓 (𝑥) > 𝐴 if 0 < |𝑥 − 𝑎| < 𝛿1 ,
(ii) 𝑔(𝑥) > 𝐴 if 0 < |𝑥 − 𝑎| < 𝛿2 .
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With 𝛿 = min{𝛿1 , 𝛿2 } it follows for 0 < |𝑥 − 𝑎| < 𝛿 that
𝑓 (𝑥)𝑔(𝑥) > 𝐴 ⋅ 𝐴 > 𝐴.
Theorem 3.7 (Sandwich Theorem). Let 𝑎 ∈ ℝ ∪ {∞, −∞} and assume that 𝑓 , 𝑔 and ℎ are real functions defined
in a deleted neighbourhood of 𝑎. If 𝑓 (𝑥) ≤ 𝑔(𝑥) ≤ ℎ(𝑥) for 𝑥 in a deleted neighbourhood of 𝑎 and
lim 𝑓 (𝑥) = 𝐿 = lim ℎ(𝑥), then lim 𝑔(𝑥) = 𝐿.
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
Proof. Note that 𝐿 ∈ ℝ ∪ {∞, −∞}. We will prove this theorem in the case 𝑎 ∈ ℝ and 𝐿 ∈ ℝ. The other cases
are left as an exercise.
Let 𝜀 > 0. Then there are 𝛿1 and 𝛿2 such that
(i) |𝑓 (𝑥) − 𝐿| < 𝜀 if 0 < |𝑥 − 𝑎| < 𝛿1 ,
(ii) |ℎ(𝑥) − 𝐿| < 𝜀 if 0 < |𝑥 − 𝑎| < 𝛿2 .
Put 𝛿 = min{𝛿1 , 𝛿2 }. Then, for 0 < |𝑥 − 𝑎| < 𝛿,
𝐿 − 𝜀 < 𝑓 (𝑥) < 𝐿 + 𝜀,
𝐿 − 𝜀 < ℎ(𝑥) < 𝐿 + 𝜀
gives
𝐿 − 𝜀 < 𝑓 (𝑥) ≤ 𝑔(𝑥) ≤ ℎ(𝑥) < 𝐿 + 𝜀.
Hence |𝑔(𝑥) − 𝐿| < 𝜀 if 0 < |𝑥 − 𝑎| < 𝜀.
Theorem 3.8. Let 𝑓 be defined on an interval (𝑎, 𝑏), where 𝑎 = −∞ and 𝑏 = ∞ are allowed. With the convenient
notation 𝑎+ = −∞ if 𝑎 = −∞ and 𝑏− = ∞ if 𝑏 = ∞, we obtain
(a) if 𝑓 is increasing, then
lim− 𝑓 (𝑥) = sup{𝑓 (𝑥) ∶ 𝑥 ∈ (𝑎, 𝑏)} and lim+ 𝑓 (𝑥) = inf {𝑓 (𝑥) ∶ 𝑥 ∈ (𝑎, 𝑏)};
𝑥→𝑏
𝑥→𝑎
(b) if 𝑓 is decreasing, then
lim− 𝑓 (𝑥) = inf {𝑓 (𝑥) ∶ 𝑥 ∈ (𝑎, 𝑏)} and lim+ 𝑓 (𝑥) = sup{𝑓 (𝑥) ∶ 𝑥 ∈ (𝑎, 𝑏)}.
𝑥→𝑏
𝑥→𝑎
Proof. Since all four cases have similar proofs, we only prove (a) in the case 𝑏 ∈ ℝ.
Let 𝐿 = sup{𝑓 (𝑥) ∶ 𝑥 ∈ (𝑎, 𝑏)}.
Case I: 𝐿 ∈ ℝ.
Let 𝜀 > 0. By Theorem 1.9 there is 𝑐 ∈ (𝑎, 𝑏) such that 𝐿 − 𝜀 < 𝑓 (𝑐). Put 𝛿 = 𝑏 − 𝑐 > 0. Now let 𝑏 − 𝛿 < 𝑥 < 𝑏,
i. e., 𝑥 ∈ (𝑐, 𝑏). Then 𝑐 < 𝑥 gives 𝑓 (𝑐) ≤ 𝑓 (𝑥) since 𝑓 is increasing and 𝑓 (𝑥) ≤ 𝐿 for all 𝑥 ∈ (𝑐, 𝑏) ⊂ (𝑎, 𝑏) by
definition of the supremum, so that
𝐿 − 𝜀 < 𝑓 (𝑐) ≤ 𝑓 (𝑥) ≤ 𝐿 < 𝐿 + 𝜀
for these 𝑥. By definition, this mean 𝑓 (𝑥) → 𝐿 as 𝑥 → 𝑏− .
Case II: 𝐿 = ∞. In this case, {𝑓 (𝑥) ∶ 𝑥 ∈ (𝑎, 𝑏)} is not bounded above. Therefore, for each 𝐴 ∈ ℝ there is
𝑐 ∈ (𝑎, 𝑏) such that 𝑓 (𝑐) > 𝐴. Since 𝑓 is increasing, it follows for all 𝑥 ∈ (𝑐, 𝑏) that 𝐴 < 𝑓 (𝑐) ≤ 𝑓 (𝑥). Therefore
𝑓 (𝑥) → ∞ as 𝑥 → 𝑏− .
Tutorial 3.3.1. 1. Let 𝑛 be a positive integer. Prove that
(a) lim 𝑥𝑛 = ∞,
𝑥→∞
{
∞
if n is even,
𝑛
(b) lim 𝑥 =
𝑥→−∞
−∞ if n is odd,
(c) lim+ 𝑥−𝑛 = ∞,
𝑥→0
{
∞
if n is even,
−𝑛
(d) lim− 𝑥 =
𝑥→0
−∞ if n is odd.
2. (a) Let 𝑓 , 𝑔 be defined in a deleted neighbourhood of 𝑎 and assume that 𝑓 (𝑥) < 𝑔(𝑥) for all 𝑥 in a deleted
neighbourhood of 𝑎. Show that if lim 𝑓 (𝑥) = 𝐿 and lim 𝑔(𝑥) = 𝑀 exist, then 𝐿 ≤ 𝑀.
𝑥→𝑎
𝑥→𝑎
(b) Give examples for 𝐿 < 𝑀 and for 𝐿 = 𝑀 in (a).
(c) Formulate and prove the result corresponding to (a) for one-sided limits.
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3. Using rules for limits, determine the behaviour of 𝑓 (𝑥) as 𝑥 tends to the given limit:
4𝑥
(a) 𝑓 (𝑥) =
as 𝑥 → 3− ,
3−𝑥
(𝑥 − 4)(𝑥 − 1)
(b) 𝑓 (𝑥) =
as 𝑥 → 2+ ,
𝑥−2
2𝑥 + 1
(c) 𝑓 (𝑥) =
as 𝑥 → 0+ .
𝑥2 − 𝑥
3.4
Continuity of Functions
Definition 3.7 (Continuity of a function at a point). Let 𝑓 be a real function, 𝑎 ∈ ℝ and assume that the domain
of 𝑓 contains a neighbourhood of 𝑎, that is, 𝑓 (𝑥) is defined for all 𝑥 in a neighbourhood of 𝑎.
We say that 𝑓 is continuous at 𝑎 if
∀ 𝜀 > 0 ∃ 𝛿 > 0 ∀ 𝑥 ∈ (𝑎 − 𝛿, 𝑎 + 𝛿), 𝑓 (𝑥) ∈ (𝑓 (𝑎) − 𝜀, 𝑓 (𝑎) + 𝜀),
i. e., ∀ 𝜀 > 0 ∃ 𝛿 > 0 (|𝑥 − 𝑎| < 𝛿 ⇒ |𝑓 (𝑥) − 𝑓 (𝑎)| < 𝜀).
Note that the sketch on page 25 also illustrates the definition of continuity.
We realize that the definition of the existence of a limit of a function at a point and continuity at that point are very
similar, but that there are subtle (and important) differences:
For limits, 𝑓 does not need to be defined at 𝑎, and even if 𝑓 (𝑎) exists, this value is not used at all when finding the
limit of the function 𝑓 at 𝑎.
We conclude
𝑓 is continuous at 𝑎 ⇔ ∀ 𝜀 > 0 ∃ 𝛿 > 0 (|𝑥 − 𝑎| < 𝛿 ⇒ |𝑓 (𝑥) − 𝑓 (𝑎)| < 𝜀) (by definition)
⇔ ∀ 𝜀 > 0 ∃ 𝛿 > 0 (0 < |𝑥 − 𝑎| < 𝛿 ⇒ |𝑓 (𝑥) − 𝑓 (𝑎)| < 𝜀)
(∵ 𝑥 = 𝑎 ⇒ 𝑓 (𝑥) = 𝑓 (𝑎))
⇔ lim 𝑓 (𝑥) = 𝑓 (𝑎).
𝑥→𝑎
Hence we have shown
Theorem 3.9. 𝑓 is continuous at 𝑎 if and only if the following three conditions are satisfied:
1. 𝑓 (𝑎) is defined, i. e., 𝑎 is in the domain of 𝑓 ,
2. lim 𝑓 (𝑥) exists, i.e., lim− 𝑓 (𝑥) = lim+ 𝑓 (𝑥), and
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
3. 𝑓 (𝑎) = lim 𝑓 (𝑥).
𝑥→𝑎
Example 3.4.1. 1. Let
{
𝑓 (𝑥) =
𝑥2
if 𝑥 ≠ 2,
2
if 𝑥 = 2.
Then lim 𝑓 (𝑥) = 4 exists, but this limit is different from 𝑓 (2) = 2. Hence 𝑓 is not continuous at 2.
𝑥→2
sin 𝑥
for 𝑥 ≠ 0 while 𝑓 is not defined at 𝑥 = 0. Then lim 𝑓 (𝑥) = 1 exists, but 𝑓 is not defined at 0.
𝑥→0
𝑥
Hence 𝑓 is not continuous at 0.
3. Let
{
sin 𝑥
if 𝑥 ≠ 0,
𝑓 (𝑥) =
𝑥
1
if 𝑥 = 0.
2. Let 𝑓 (𝑥) =
Then lim 𝑓 (𝑥) = 1 exists, and 𝑓 (0) = 1. Hence 𝑓 is continuous at 0.
𝑥→0
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Theorem 3.10. If 𝑓 and 𝑔 are continuous at 𝑎 and if 𝑐 ∈ ℝ, then
1. the sum 𝑓 + 𝑔,
2. the difference 𝑓 − 𝑔,
3. the product 𝑓 𝑔,
𝑓
4. the quotient if 𝑔(𝑎) ≠ 0 and
𝑔
5. the scalar multiple 𝑐𝑓
are functions that are also continuous at 𝑎.
Proof. The statements follow immediate from the limit laws, Theorem 3.3, and Theorem 3.9.
For example, for 3. we have
lim 𝑓 (𝑥) = 𝑓 (𝑎) and lim 𝑔(𝑥) = 𝑔(𝑎),
𝑥→𝑎
𝑥→𝑎
and then Theorem 3.3 gives
lim (𝑓 𝑔)(𝑥) = lim 𝑓 (𝑥)𝑔(𝑥) = lim 𝑓 (𝑥) lim 𝑔(𝑥) = 𝑓 (𝑎)𝑔(𝑎) = (𝑓 𝑔)(𝑎).
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
Then Theorem 3.9 says that 𝑓 𝑔 is continuous at 𝑎.
Recall that the composite 𝑔◦𝑓 of two functions 𝑓 and 𝑔 is defined by (𝑔◦𝑓 )(𝑥) = 𝑔(𝑓 (𝑥)).
Theorem 3.11. If 𝑓 is continuous at 𝑎 and 𝑔 is continuous at 𝑓 (𝑎), then 𝑔◦𝑓 is continuous at 𝑎.
Proof. Let 𝜀 > 0. Since 𝑔 is continuous at 𝑓 (𝑎), there is 𝜂 > 0 such that
|𝑦 − 𝑓 (𝑎)| < 𝜂 ⇒ |𝑔(𝑦) − 𝑔(𝑓 (𝑎))| < 𝜀.
(1)
Since 𝑓 is continuous at 𝑎, there is 𝛿 > 0 such that
|𝑥 − 𝑎| < 𝛿 ⇒ |𝑓 (𝑥) − 𝑓 (𝑎)| < 𝜂.
Putting 𝑦 = 𝑓 (𝑥) in (1) it follows from (1) and (2) that
|𝑥 − 𝑎| < 𝛿 ⇒ |𝑓 (𝑥) − 𝑓 (𝑎)| < 𝜂 ⇒ |𝑔(𝑓 (𝑥)) − 𝑔(𝑓 (𝑎))| < 𝜀
that is,
|𝑥 − 𝑎| < 𝛿 ⇒ |(𝑔◦𝑓 )(𝑥) − (𝑔◦𝑓 )(𝑎)| < 𝜀.
Hence 𝑔◦𝑓 is continuous at 𝑎.
Definition 3.8. 1. A function 𝑓 is continuous from the right at 𝑎 if lim+ 𝑓 (𝑥) = 𝑓 (𝑎).
𝑥→𝑎
2. A function 𝑓 is continuous from the left at 𝑎 if lim− 𝑓 (𝑥) = 𝑓 (𝑎).
𝑥→𝑎
Example 3.4.2. 1. Let
⎧ |𝑥| + 𝑥
⎪
𝑓 (𝑥) = ⎨ 2𝑥
⎪0
⎩
if 𝑥 ≠ 0,
if 𝑥 = 0.
Determine the right and left continuity of 𝑓 at 𝑥 = 0.
Solution. 𝑓 (0) = 0 whilst
lim 𝑓 (𝑥) = lim−
|𝑥| + 𝑥
−𝑥 + 𝑥
= lim−
= lim− 0 = 0
𝑥→0
𝑥→0
2𝑥
2𝑥
lim 𝑓 (𝑥) = lim+
|𝑥| + 𝑥
𝑥+𝑥
= lim+
= lim− 1 = 1.
𝑥→0
𝑥→0
2𝑥
2𝑥
𝑥→0−
𝑥→0
and
𝑥→0+
𝑥→0
(2)
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Since 𝑓 (0) = 0 = lim− 𝑓 (𝑥), 𝑓 is continuous from the left at 𝑥 = 0. Since 𝑓 (0) = 0 ≠ 1 = lim+ 𝑓 (𝑥), 𝑓 is not
𝑥→0
𝑥→0
continuous from the right at 𝑥 = 0.
Note. 1. It is easy to show that 𝑓 is continuous at 𝑎 if and only if 𝑓 is continuous from the right and continuous
from the left at 𝑎.
2. If 𝑎 ∈ dom(𝑓 ) and if there is 𝜀 > 0 such that dom(𝑓 ) ∩ (𝑎 − 𝜀, 𝑎 + 𝜀) = (𝑎 − 𝜀, 𝑎], then we say that 𝑓 is continuous
at 𝑎 if lim− = 𝑓 (𝑎).
𝑥→𝑎
3. If 𝑎 ∈ dom(𝑓 ) and if there is 𝜀 > 0 such that dom(𝑓 ) ∩ (𝑎 − 𝜀, 𝑎 + 𝜀) = [𝑎, 𝑎 + 𝜀), then we say that 𝑓 is continuous
at 𝑎 if lim+ = 𝑓 (𝑎).
𝑥→𝑎
4. The convention in 2 and 3 is consistent with what you will learn in General Topology about continuity. Just
note that the condition |𝑓 (𝑥) − 𝑓 (𝑎)| < 𝜀 has to be checked for all 𝑥 ∈ dom(𝑓 ) which satisfy |𝑥 − 𝑎| < 𝛿.
Lemma 3.12. If 𝑓 (𝑥) → 𝑏 as 𝑥 → 𝑎 (𝑎+ , 𝑎− ) and 𝑔 is continuous at 𝑏, then 𝑔(𝑓 (𝑥)) → 𝑔(𝑏) as 𝑥 → 𝑎 (𝑎+ , 𝑎− ),
which can be written, e. g., as
(
)
lim 𝑔(𝑓 (𝑥)) = 𝑔 lim 𝑓 (𝑥) .
𝑥→𝑎
Proof. The function
{
𝑓̃(𝑥) =
𝑥→𝑎
𝑓 (𝑥)
if 𝑥 ∈ dom(𝑓 ), 𝑥 ≠ 𝑎,
𝑏
if 𝑥 = 𝑎,
is continuous (from the right, from the left) at 𝑎. Hence the result follows from Theorem 3.11.
Definition 3.9. A function is continuous on a set 𝑋 ⊂ ℝ if 𝑓 is continuous at each 𝑥 ∈ 𝑋. Here continuity is
understood in the sense of the above note with 𝑋 = dom(𝑓 ).
√
Example 3.4.3. Show that 𝑓 (𝑥) = 𝑥2 − 4 is continuous.
Solution. The domain of 𝑓 is {𝑥 ∈ ℝ ∶ |𝑥| ≥ 2} = (−∞, −2] ∪ [2, ∞). By Theorem 3.10, the function 𝑥 → 𝑥2 − 4
is continuous on ℝ, and by Theorem 3.3 (k), the square root is continuous at each positive number. So also the
composite function 𝑓 is continuous at 𝑎. Also, the proof of Theorem 3.3 (k) can be easily adapted to show that the
square root is continuous from the right at 0. Then it easily follows that 𝑓 is continuous (from the right) at 2 and
continuous (from the left) at −2.
Theorem 3.13. The following functions are continuous on their domains.
1. Polynomials 𝑝(𝑥) = 𝑎𝑛 𝑥𝑛 + 𝑎𝑛−1 𝑥𝑛−1 + ⋯ + 𝑎2 𝑥2 + 𝑎1 𝑥 + 𝑎0 , 𝑎𝑖 ∈ ℝ, 𝑛 ∈ ℕ.
2. Rational functions
𝑝(𝑥)
, 𝑝 and 𝑞 ≠ 0 polynomials.
𝑞(𝑥)
3. Sums, differences, products and quotients of continuous functions.
4. Root functions.
5. The trigonometric functions sin 𝑥, cos 𝑥, tan 𝑥, cosec 𝑥, sec 𝑥 and cot 𝑥.
6. The exponential function exp(𝑥).
7. The absolute value function |𝑥|.
Proof. 1, 2 and 3 easily follow from previous theorems on limits and continuity, as does 7. However, 7 can be
easily proved dircetly: For each 𝜀 > 0 let 𝛿 = 𝜀. Then, for |𝑥 − 𝑎| < 𝛿 we have
| |𝑥| − |𝑎| | ≤ |𝑥 − 𝑎| < 𝛿 = 𝜀.
The continuity of sin and cos follows from the sum of angles formulae and from the limits proved in Calculus I
(the proofs used the Sandwich Theorem, which now has been proved). The continuity of the other trigonometric
functions then follows from part 3.
Finally, the continuity of exp is a tutorial problem.
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Theorem 3.14. Let 𝑎 ∈ ℝ and let 𝑓 be a real function which is defined in a neighbourhood of 𝑎. Then 𝑓 is
continuous at 𝑎 if and only if for each sequence (𝑥𝑛 ) in dom(𝑓 ) with lim 𝑥𝑛 = 𝑎 the sequence 𝑓 (𝑥𝑛 ) satisfies
𝑛→∞
lim 𝑓 (𝑥𝑛 ) = 𝑓 (𝑎).
𝑛→∞
Proof. ⇒∶ Let (𝑥𝑛 ) be a sequence in dom(𝑓 ) with lim 𝑥𝑛 = 𝑎. We must show that lim 𝑓 (𝑥𝑛 ) = 𝑓 (𝑎). Hence let
𝑛→∞
𝑛→∞
𝜀 > 0. Since 𝑓 is continuous at 𝑎, there is 𝛿 > 0 such that
|𝑥 − 𝑎| < 𝛿 ⇒ |𝑓 (𝑥) − 𝑓 (𝑎)| < 𝜀.
(1)
Since lim 𝑥𝑛 = 𝑎, there is 𝐾 ∈ ℝ such that for 𝑛 > 𝐾, |𝑥𝑛 − 𝑎| < 𝛿. But then, by (1), |𝑓 (𝑥𝑛 ) − 𝑓 (𝑎)| < 𝜀 for
𝑛→∞
𝑛 > 𝐾.
⇐∶ (indirect proof) Assume that 𝑓 is not continuous at 𝑎. Then
∃ 𝜀 > 0 ∀ 𝛿 > 0 ∃ 𝑥 ∈ dom(𝑓 ), |𝑥 − 𝑎| < 𝛿 and |𝑓 (𝑥) − 𝑓 (𝑎)| ≥ 𝜀.
In particular, for 𝛿 = 1𝑛 , 𝑛 = 1, 2, … we find 𝑥𝑛 ∈ dom(𝑓 ) such that |𝑥𝑛 − 𝑎| <
then lim 𝑥𝑛 = 𝑎, whereas (𝑓 (𝑥𝑛 )) does not converge to 𝑓 (𝑎).
1
and |𝑓 (𝑥𝑛 ) − 𝑓 (𝑎)| ≥ 𝜀. But
𝑛
𝑛→∞
Tutorial 3.4.1. 1. Consider the function
{
𝑓 (𝑥) =
⌊𝑥⌋
𝑥
−1
if 𝑥 ≠ 0,
if 𝑥 = 0.
Investigate continuity from the left and the right at 𝑥 = 0, 𝑥 = 𝜋 and 𝑥 = 1.
{
0 if 𝑥 ≠ 0,
2. Let 𝑓 (𝑥) = 𝑥 sin 𝑥1 for 𝑥 ≠ 0 and 𝑔(𝑥) =
1 if 𝑥 = 0.
Show that 𝑓 (𝑥) → 0 as 𝑥 → 0 and that 𝑔(𝑥) → 0 as 𝑥 → 0, but that 𝑔(𝑓 (𝑥)) does not have a limit as 𝑥 → 0.
Explain this behaviour.
3. Find the values of 𝑎 and 𝑏 which make the function
⎧𝑥 − 1
⎪
𝑓 (𝑥) = ⎨ 𝑎𝑥2 + 𝑐
⎪𝑥 + 1
⎩
if 𝑥 ≤ −2,
if −2 < 𝑥 < 1,
if 𝑥 ≥ 1,
continuous at 𝑥 = −2 and 𝑥 = 1.
4. Prove that if lim− 𝑓 (𝑥) exists, then lim+ 𝑓 (−𝑥) = lim− 𝑓 (𝑥).
𝑥→0
𝑥→0
𝑥→0
5. Prove that exp is continuous. You may use the following steps.
(a) The inequality exp(𝑥) ≥ 1 + 𝑥 is true for all 𝑥 ∈ ℝ.
(b) lim− exp(𝑥) = 1.
𝑥→0
(c) lim+ exp(𝑥) = 1. Hint. Use tutorial problem 4.
𝑥→0
3.5
The Intermediate Value Theorem
The following important theorem on continuous functions tells us that the graph of a continuous function cannot
jump from one side of a horizontal line 𝑦 = 𝑘 to the other without intersecting the line at least once.
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𝑓 (𝑏)
𝑦=𝑘
𝑓 (𝑎)
𝑎
𝑐
𝑏
Theorem 3.15 (Intermediate Value Theorem (IVT)). Suppose that 𝑓 is continuous on the closed interval [𝑎, 𝑏]
with 𝑓 (𝑎) ≠ 𝑓 (𝑏). Then for any number 𝑘 between 𝑓 (𝑎) and 𝑓 (𝑏) there exists a number 𝑐 in the open interval (𝑎, 𝑏)
such that 𝑓 (𝑐) = 𝑘.
Proof. Let
𝑔(𝑥) = 𝑓 (𝑥) − 𝑘,
(𝑥 ∈ [𝑎, 𝑏]).
Then 𝑔 is continuous, and 𝑔(𝑎) and 𝑔(𝑏) have opposite signs: 𝑔(𝑎)𝑔(𝑏) < 0.
Let [𝑎0 , 𝑏0 ] = [𝑎, 𝑏] and use bisection to define intervals [𝑎𝑛 , 𝑏𝑛 ] as follows: If [𝑎𝑛 , 𝑏𝑛 ] with 𝑔(𝑎𝑛 )𝑔(𝑏𝑛 ) < 0 has
been found, let 𝑑 be the midpoint of the interval [𝑎𝑛 , 𝑏𝑛 ]. If 𝑔(𝑑) = 0, the result follows with 𝑐 = 𝑑. If 𝑔(𝑑)
has the same sign as 𝑔(𝑏𝑛 ), then 𝑔(𝑎𝑛 ) and 𝑔(𝑑) have opposite signs, and putting 𝑎𝑛+1 = 𝑎𝑛 , 𝑏𝑛+1 = 𝑑, we have
𝑔(𝑎𝑛+1 )𝑔(𝑏𝑛+1 ) < 0. Otherwise, if 𝑔(𝑑) has the opposite sign to 𝑔(𝑏𝑛 ), we put 𝑎𝑛+1 = 𝑑, 𝑏𝑛+1 = 𝑏𝑛 and get again
𝑔(𝑎𝑛+1 )𝑔(𝑏𝑛+1 ) < 0.
If this procedure does not stop, we obtain an increasing sequence (𝑎𝑛 ) and a decreasing sequence (𝑏𝑛 ), both of
which converge by Theorem 2.10. We observe that
1
𝑏𝑛 = 𝑎𝑛 + (𝑏𝑛−1 − 𝑎𝑛−1 ) = 𝑎𝑛 + 2−𝑛 (𝑏 − 𝑎).
2
Then
𝑐 ∶= lim 𝑏𝑛 = lim 𝑎𝑛 + lim 2−𝑛 (𝑏 − 𝑎) = lim 𝑎𝑛 .
𝑛→∞
𝑛→∞
𝑛→∞
𝑛→∞
Since 𝑎 ≤ 𝑐 ≤ 𝑏 and 𝑔 is continuous at 𝑐, it follows in view of Theorem 3.14 that
𝑔(𝑐)2 = lim 𝑔(𝑎𝑛 ) lim 𝑔(𝑏𝑛 ) = lim 𝑔(𝑎𝑛 )𝑔(𝑏𝑛 ) ≤ 0.
𝑛→∞
𝑛→∞
𝑛→∞
Therefore 𝑔(𝑐) = 0, which gives 𝑓 (𝑐) = 𝑘.
Since 𝑓 (𝑎) ≠ 𝑘 and 𝑓 (𝑏) ≠ 𝑘, it follows that 𝑐 ≠ 𝑎 and 𝑐 ≠ 𝑏, so that 𝑎 < 𝑐 < 𝑏.
Note. You have seen the definition of interval in first year and you will recall that that the definition required several
cases, depending on whether the endpoints belong to the interval or not and whether the interval is bounded (above,
below), namely (𝑎, 𝑏), [𝑎, 𝑏), (𝑎, 𝑏], [𝑎, 𝑏], (−∞, 𝑏), (−∞, 𝑏], (𝑎, ∞), [𝑎, ∞), (−∞, ∞) where 𝑎, 𝑏 ∈ ℝ and 𝑎 < 𝑏.
However, intervals can be characterized by one common property. For this we need the following notion: A subset
𝑆 of ℝ is called a singleton if the set 𝑆 has exactly one element.
Definition 3.10. 1. A set 𝑆 ⊂ ℝ is called an interval if
(i) 𝑆 ≠ ∅,
(ii) 𝑆 is not a singleton,
(iii) if 𝑥, 𝑦 ∈ 𝑆, 𝑥 < 𝑦, then each 𝑧 ∈ ℝ with 𝑥 < 𝑧 < 𝑦 satisfies 𝑧 ∈ 𝑆.
2. An interval of the form [𝑎, 𝑏] with 𝑎 < 𝑏 is called a closed bounded interval.
Note. A subset 𝑆 of ℝ is an interval if and only if it contains at least two elements and if all real numbers between
any two elements in 𝑆 also belong to 𝑆.
Definition 3.11. For a function 𝑓 ∶ 𝑋 → 𝑌 and 𝐴 ⊂ 𝑋, the set
𝑓 (𝐴) = {𝑦 ∈ 𝑌 ∶ ∃ 𝑥 ∈ 𝐴 ∩ dom(𝑓 ), 𝑓 (𝑥) = 𝑦} = {𝑓 (𝑥) ∶ 𝑥 ∈ 𝐴 ∩ dom(𝑓 )}
is called the image of 𝐴 under 𝑓 .
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Corollary 3.16. Let 𝐼 be an interval and let 𝑓 be a continuous real function on 𝐼. Then 𝑓 (𝐼) is either an interval
or a singleton.
Proof. Since 𝐼 ≠ ∅, there is 𝑥 ∈ 𝐼 and so 𝑓 (𝑥) ∈ 𝑓 (𝐼). Hence 𝑓 (𝐼) ≠ ∅. Hence we have to show that if 𝑓 (𝐼) is
not a singleton, then it is an interval, that is, we must show that for any 𝑥, 𝑦 ∈ 𝐼 with 𝑓 (𝑥) < 𝑓 (𝑦) and any 𝑘 ∈ ℝ
with 𝑓 (𝑥) < 𝑘 < 𝑓 (𝑦) there is 𝑐 ∈ 𝐼 such that 𝑓 (𝑐) = 𝑘.
Indeed, from 𝑓 (𝑥) ≠ 𝑓 (𝑦), we have 𝑥 ≠ 𝑦. If 𝑥 < 𝑦, then 𝑓 is continuous on [𝑥, 𝑦], and by the intermediate value
theorem, there is 𝑐 ∈ (𝑥, 𝑦) with 𝑓 (𝑐) = 𝑘. A similar argument holds for 𝑥 > 𝑦.
Example 3.5.1. Let 𝑓 (𝑥) = 𝑥2 . Then 𝑓 ((−1, 2)) = [0, 4). Notice that 𝐼 = (−1, 2) is an open interval, while 𝑓 (𝐼)
is not.
Theorem 3.17. Let 𝑓 be a real function which is continuous on [𝑎, 𝑏], where 𝑎 < 𝑏. Then 𝑓 is bounded on [𝑎, 𝑏],
i. e., 𝑓 ([𝑎, 𝑏]) is bounded.
Proof. Assume that 𝑓 ([𝑎, 𝑏]) is unbounded. Let 𝑑 be the midpoint of the interval [𝑎, 𝑏]. Then at least one of the sets
𝑓 ([𝑎, 𝑑]), 𝑓 ([𝑑, 𝑏]) would be unbounded, because otherwise 𝑓 ([𝑎, 𝑏]) = 𝑓 ([𝑎, 𝑑]) ∪ 𝑓 ([𝑑, 𝑏]) would be bounded.
By induction, we find subintervals [𝑎𝑛 , 𝑏𝑛 ] of [𝑎, 𝑏] such that (𝑎𝑛 ) is increasing, (𝑏𝑛 ) is decreasing, 𝑓 ([𝑎𝑛 , 𝑏𝑛 ]) is
unbounded, and
1
𝑏𝑛 = 𝑎𝑛 + (𝑏𝑛−1 − 𝑎𝑛−1 ) = 2−𝑛 (𝑏 − 𝑎),
2
see the proof of Theorem 3.15, and we infer that both sequences converge with
𝑐 ∶= lim 𝑏𝑛 = lim 𝑎𝑛 ∈ [𝑎, 𝑏].
𝑛→∞
𝑛→∞
Since 𝑓 is continuous at 𝑐, there is 𝛿 > 0 such that
𝑓 ((𝑐 − 𝛿, 𝑐 + 𝛿) ∩ [𝑎, 𝑏]) ⊂ (𝑓 (𝑐) − 1, 𝑓 (𝑐) + 1).
Since 𝑐 = lim 𝑏𝑛 , there is 𝐾 ∈ ℕ such that |𝑏𝑛 −𝑐| < 𝛿 for 𝑛 > 𝐾. Similarly, there is 𝑀 ∈ ℕ such that |𝑎𝑛 −𝑐| < 𝛿
𝑛→∞
for 𝑛 > 𝑀. Let 𝑛 > max{𝐾, 𝑀}. Then
𝑐 − 𝛿 < 𝑎𝑛 ≤ 𝑐 ≤ 𝑏𝑛 < 𝑐 + 𝛿,
and
𝑓 ([𝑎𝑛 , 𝑏𝑛 ]) ⊂ 𝑓 ((𝑐 − 𝛿, 𝑐 + 𝛿) ∩ [𝑎, 𝑏]) ⊂ (𝑓 (𝑐) − 1, 𝑓 (𝑐) + 1)
would give the contradiction that 𝑓 ([𝑎𝑛 , 𝑏𝑛 ]) would have to be bounded as well as unbounded.
Theorem 3.18. A continuous function on a closed bounded interval achieves its supremum and infimum.
Proof. Let [𝑎, 𝑏] be a closed bounded interval and 𝑓 be a continuous function on [𝑎, 𝑏]. We must show that there
are 𝑥1 , 𝑥2 ∈ [𝑎, 𝑏] such that 𝑓 (𝑥1 ) = inf 𝑓 ([𝑎, 𝑏]) and 𝑓 (𝑥2 ) = sup 𝑓 ([𝑎, 𝑏]).
We are going to show the latter; the proof of the first statement is similar.
By Theorem 3.17, 𝑆 = 𝑓 ([𝑎, 𝑏]) is bounded. Let 𝑀 = sup 𝑆. By proof of contradiction, assume that 𝑓 (𝑥) ≠ 𝑀
for all 𝑥 ∈ [𝑎, 𝑏]. Define
1
𝑔(𝑥) =
, 𝑥 ∈ [𝑎, 𝑏].
𝑀 − 𝑓 (𝑥)
By Theorem 3.13, 𝑔 is continuous on [𝑎, 𝑏], and by Theorem 3.17, 𝑔([𝑎, 𝑏]) is bounded. So there is 𝐾 ∈ ℝ such
that 0 < 𝑔(𝑥) ≤ 𝐾 for all 𝑥 ∈ [𝑎, 𝑏]. Hence for all 𝑥 ∈ [𝑎, 𝑏]:
1
1
1
≤
= 𝑀 − 𝑓 (𝑥) ⇒ 𝑓 (𝑥) ≤ 𝑀 − ,
𝐾
𝑔(𝑥)
𝐾
1
so that the number 𝑀 −
< 𝑀 would be an upper bound of 𝑓 ([𝑎, 𝑏]). This contradicts the fact that 𝑀 =
𝐾
sup 𝑓 ([𝑎, 𝑏]).
Corollary 3.19. If 𝑓 is continuous on [𝑎, 𝑏], 𝑎 < 𝑏, then either 𝑓 ([𝑎, 𝑏]) is a singleton or 𝑓 ([𝑎, 𝑏]) = [𝑐, 𝑑] with
𝑐 < 𝑑.
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Proof. From Corollary 3.16 and Theorems 3.17 and 3.18 it follows that 𝑓 ([𝑎, 𝑏]) is either a singleton or a bounded
interval which contains both its infimum and supremum. But such an interval is of the form [𝑐, 𝑑] with 𝑐 < 𝑑.
Theorem 3.20. Let 𝐼 be an interval and 𝑓 ∶ 𝐼 → ℝ be a stricly monotonic continuous function. Then 𝑓 (𝐼) is an
interval, and the inverse function 𝑓 −1 ∶ 𝑓 (𝐼) → ℝ is continuous.
Proof. By Corollary 3.16, 𝑓 (𝐼) is an interval. Assume that 𝑓 is strictly increasing. Then also 𝑓 −1 is strictly
increasing. Let 𝑏 ∈ 𝑓 (𝐼), i. e., 𝑏 = 𝑓 (𝑎) for some 𝑎 ∈ 𝐼. If 𝑎 is not the left endpoint of 𝐼, then 𝑏 is not the left
endpoint of 𝑓 (𝐼), and for 𝑦 ∈ 𝑓 (𝐼) with 𝑦 < 𝑏 = 𝑓 (𝑎) we have 𝑓 −1 (𝑦) < 𝑓 −1 (𝑓 (𝑎)) = 𝑎. Therefore, 𝑓 −1 is
bounded above and increasing on 𝑓 (𝐼) ∩ (−∞, 𝑏), and thus
𝛼 = lim− 𝑓 −1 (𝑦)
𝑦→𝑏
exists and 𝛼 ≤ 𝑓 −1 (𝑏) = 𝑎, see Theorem 3.8. Let 𝑥0 ∈ 𝐼 ∩ (−∞, 𝑎). Then 𝑓 (𝑥0 ) < 𝑓 (𝑎) = 𝑏 and hence
𝑥0 = 𝑓 −1 (𝑓 (𝑥0 )) ≤ 𝛼 ≤ 𝑎.
Since 𝑥0 and 𝑎 belong to the interval 𝐼, also 𝛼 ∈ 𝐼. Since 𝑓 is continuous, it follows from Lemma 3.12 that
(
)
−1
𝑓 (𝛼) = 𝑓 lim− 𝑓 (𝑦) = lim− 𝑓 (𝑓 −1 (𝑦)) = lim− 𝑦 = 𝑏
𝑦→𝑏
𝑦→𝑏
𝑦→𝑏
which gives
lim 𝑓 −1 (𝑦) = 𝛼 = 𝑓 −1 (𝑏).
𝑦→𝑏−
Therefore 𝑓 −1 is continuous from the left. Similarly, one can show that 𝑓 −1 is continuous from the right. Therefore
𝑓 −1 is continuous. The case 𝑓 strictly decreasing is similar.
Tutorial 3.5.1. 1. Prove that a subset 𝑆 of ℝ is an interval if and only if 𝑆 has the form (𝑎, 𝑏), [𝑎, 𝑏), (𝑎, 𝑏], [𝑎, 𝑏],
(−∞, 𝑏), (−∞, 𝑏], (𝑎, ∞), [𝑎, ∞) or (−∞, ∞) where 𝑎, 𝑏 ∈ ℝ and 𝑎 < 𝑏.
Hint. Consider inf 𝑆 and sup 𝑆.
2. Let 𝑓 be a real function and let ∅ ≠ 𝐴 ⊂ 𝐵 ⊂ dom(𝑓 ) such that for each 𝑥 ∈ 𝐴 there is 𝜀 > 0 such that
(𝑥 − 𝜀, 𝑥] ⊂ 𝐴 or [𝑥, 𝑥 + 𝜀) ⊂ 𝐴 or (𝑥 − 𝜀, 𝑥 + 𝜀) ⊂ 𝐴, and the same property for 𝐵. Show that if 𝑓 is continuous
on 𝐵, then 𝑓 is also continuous on 𝐴.
3. A fixed point theorem. Let 𝑎 < 𝑏 and let 𝑓 be a continuous function on [𝑎, 𝑏] such that 𝑓 ([𝑎, 𝑏]) ⊂ [𝑎, 𝑏].
Show that there is 𝑥 ∈ [𝑎, 𝑏] such that 𝑓 (𝑥) = 𝑥.
4. Let 𝐼 be an interval and 𝑓 be a continuous function on 𝐼 such that 𝑓 (𝐼) is unbounded. What can you say about
𝑓 (𝐼)? Find examples which illustrate your answer.
5. Let 𝑓 ∶ [0, 1] → ℝ be a continuous function which only assumes rational values. Show that 𝑓 is constant.
6. Find a continuous function 𝑓 ∶ [−1, 1] → ℝ which is one-to-one when restricted to rational numbers in [−1, 1]
but which is not one-to-one on the whole interval [−1, 1].
Chapter 4
Differentiation
4.1
Revision – Self Study
[Note to lecturers: None of these results have to be presented formally. But, time permitting, some of these definitions and results may be discussed in class; foremost, of course, the definition of the derivative itself.]
In this section we recall definitions and results which have been stated and proved in first year calculus. These
results must be known but will not be examined directly.
In this section let 𝐴 ⊂ ℝ such that for each 𝑎 ∈ 𝐴 there is 𝜀 > 0 such that (𝑎 − 𝜀, 𝑎] ⊂ 𝐴, [𝑎, 𝑎 + 𝜀) ⊂ 𝐴 or
(𝑎 − 𝜀, 𝑎 + 𝜀) ⊂ 𝐴.
Definition 4.1. Let 𝑓 ∶ 𝐴 → ℝ and 𝑎 ∈ 𝐴.
1. 𝑓 is called differentiable at 𝑎 if the limit
𝑓 ′ (𝑥) = lim
𝑥→𝑎
𝑓 (𝑥) − 𝑓 (𝑎)
𝑥−𝑎
exists. The number 𝑓 ′ (𝑥) is called the derivative of 𝑓 .
2. 𝑓 is called differentiable on 𝐴 if 𝑓 is differentiable at each 𝑎 ∈ 𝐴.
Note. We also use the notations
𝑑𝑓
𝑑
or
𝑓 for 𝑓 ′ .
𝑑𝑥
𝑑𝑥
Theorem 4.1. Let 𝑓 ∶ 𝐴 → ℝ and 𝑎 ∈ 𝐴. If 𝑓 is differentiable at 𝑎, then 𝑓 is continuous at 𝑎.
Proof. From
𝑓 (𝑥) = 𝑓 (𝑎) +
𝑓 (𝑥) − 𝑓 (𝑎)
(𝑥 − 𝑎)
𝑥−𝑎
it follows that
lim 𝑓 (𝑥) = 𝑓 (𝑎) + lim
𝑥→𝑎
𝑥→𝑎
𝑓 (𝑥) − 𝑓 (𝑎)
lim (𝑥 − 𝑎) = 𝑓 (𝑎) + [𝑓 ′ (𝑎)](0) = 𝑓 (𝑎).
𝑥→𝑎
𝑥−𝑎
Theorem 4.2 (Rules for the derivative). Let 𝑓 , 𝑔 ∶ 𝐴 → ℝ, 𝑎 ∈ 𝐴, 𝑓 and 𝑔 differentiable at 𝑎, and 𝑐 ∈ ℝ. Then
1. Linearity of the derivative: 𝑓 + 𝑔 and 𝑐𝑓 are differentiable at 𝑎, and
(𝑓 + 𝑔)′ (𝑎) = 𝑓 ′ (𝑎) + 𝑔 ′ (𝑎) and (𝑐𝑓 )′ (𝑎) = 𝑐𝑓 ′ (𝑎).
2. Product Rule: 𝑓 𝑔 is differentiable at 𝑎, and
(𝑓 𝑔)′ (𝑎) = 𝑓 ′ (𝑎)𝑔(𝑎) + 𝑓 (𝑎)𝑔 ′ (𝑎).
3. Quotient Rule: If
𝑓
𝑓
is defined at 𝑎, i. e., 𝑔(𝑎) ≠ 0, then is differentiable at 𝑎, and
𝑔
𝑔
( )′
𝑓
𝑓 ′ (𝑎)𝑔(𝑎) − 𝑓 (𝑎)𝑔 ′ (𝑎)
(𝑎) =
.
𝑔
[𝑔(𝑎)]2
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Proof. 1. From first principles and rules of limits, we have
(𝑓 + 𝑔)(𝑥) − (𝑓 + 𝑔)(𝑎)
𝑥−𝑎
[𝑓 (𝑥) + 𝑔(𝑥)] − [𝑓 (𝑎) + 𝑔(𝑎)]
= lim
𝑥→𝑎
𝑥−𝑎
𝑔(𝑥) − 𝑔(𝑎)
𝑓 (𝑥) − 𝑓 (𝑎)
+ lim
= lim
𝑥→𝑎
𝑥→𝑎
𝑥−𝑎
𝑥−𝑎
= 𝑓 ′ (𝑎) + 𝑔 ′ (𝑎)
(𝑓 + 𝑔)′ (𝑎) = lim
𝑥→𝑎
and
(𝑐𝑓 )′ (𝑎) = lim
𝑥→𝑎
(𝑐𝑓 )(𝑥) − (𝑐𝑓 )(𝑎)
𝑐𝑓 (𝑥) − 𝑐𝑓 (𝑎)
𝑓 (𝑥) − 𝑓 (𝑎)
= lim
= 𝑐 lim
= 𝑐𝑓 ′ (𝑎).
𝑥→𝑎
𝑥→𝑎
𝑥−𝑎
𝑥−𝑎
𝑥−𝑎
2. Observing Theorem 4.1 we have
(𝑓 𝑔)(𝑥) − (𝑓 𝑔)(𝑎)
𝑥−𝑎
𝑓 (𝑥)𝑔(𝑥) − 𝑓 (𝑎)𝑔(𝑥) + 𝑓 (𝑎)𝑔(𝑥) − 𝑓 (𝑎)𝑔(𝑎)
= lim
𝑥→𝑎
𝑥−𝑎
[𝑓 (𝑥) − 𝑓 (𝑎)]𝑔(𝑥)
𝑓 (𝑎)[𝑔(𝑥) − 𝑔(𝑎)]
+ lim
= lim
𝑥→𝑎
𝑥→𝑎
𝑥−𝑎
𝑥−𝑎
𝑔(𝑥) − 𝑔(𝑎)
𝑓 (𝑥) − 𝑓 (𝑎)
lim 𝑔(𝑥) + 𝑓 (𝑎) lim
= lim
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
𝑥−𝑎
𝑥−𝑎
= 𝑓 ′ (𝑎)𝑔(𝑎) + 𝑓 (𝑎)𝑔 ′ (𝑎).
(𝑓 𝑔)′ (𝑎) = lim
𝑥→𝑎
3. We first consider the case 𝑓 = 1:
( )
( )
1
1
(𝑥) −
(𝑎)
( )′
𝑔
𝑔
1
(𝑎) = lim
𝑥→𝑎
𝑔
𝑥−𝑎
1
1
−
𝑔(𝑥) 𝑔(𝑎)
= lim
𝑥→𝑎
𝑥−𝑎
𝑔(𝑎) − 𝑔(𝑥)
= lim
𝑥→𝑎 𝑔(𝑥)𝑔(𝑎)(𝑥 − 𝑎)
𝑔(𝑎) − 𝑔(𝑥)
1
= lim
𝑥→𝑎
𝑥 − 𝑎 𝑔(𝑎) lim 𝑔(𝑥)
𝑥→𝑎
𝑔 ′ (𝑎)
=−
.
[𝑔(𝑎)]2
To prove the statement for general 𝑓 , we use the product rule:
( )′
( )′
( )
( )′
(
)
𝑓
𝑓 ′ (𝑎)
𝑔 ′ (𝑎)
1
1
1
′
(𝑎) = 𝑓
(𝑎) = 𝑓 (𝑎)
(𝑎) + 𝑓 (𝑎)
(𝑎) =
+ 𝑓 (𝑎) −
𝑔
𝑔
𝑔
𝑔
𝑔(𝑎)
[𝑔(𝑎)]2
′
′
𝑓 (𝑎)𝑔(𝑎) − 𝑓 (𝑎)𝑔 (𝑎)
=
.
[𝑔(𝑎)]2
𝑑𝑐
Example 4.1.1. 1. If 𝑐 ∈ ℝ, then 𝑑𝑥
= 0.
𝑑
2. For 𝑛 ∈ ℕ∗ we have
𝑥𝑛 = 𝑛𝑥𝑛−1 .
𝑑𝑥
Proof. 1. follows immediately from the definition.
2. We present an alternate proof to that given in first year: proof by induction. Again, for 𝑛 = 1 the statement is
straightforward:
𝑦−𝑥
𝑑
𝑥 = lim
= 1 = (1)𝑥1−1 .
𝑦→𝑥 𝑦 − 𝑥
𝑑𝑥
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Now assume the statement is true for 𝑛 ∈ ℕ∗ . Then, making use of the Product Rule,
(
)
𝑑 𝑛+1
𝑑
𝑑 𝑛
𝑑
𝑥
=
𝑥 (𝑥) + 𝑥𝑛 𝑥 = 𝑛𝑥𝑛−1 (𝑥) + 𝑥𝑛 (1) = (𝑛 + 1)𝑥𝑛 = (𝑛 + 1)𝑥(𝑛+1)−1 .
((𝑥𝑛 )(𝑥)) =
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑 𝑛
𝑥 = 𝑛𝑥𝑛−1 for negative integers 𝑛 with the aid of the
𝑑𝑥
quotient rule. Also, the derivatives of the trigonometric functions sin, cos, tan have been derived in first year
calculus and should be known.
As in first year calculus, we now obtain the derivative
The proof of the differentiability of 𝑒𝑥 in first year calculus was incomplete; it will be given in the next section.
Also, the differentiability of the inverse trigonometric functions, ln 𝑥 and 𝑥𝑟 for 𝑟 ∈ ℝ ⧵ ℕ will be postponed to the
next section.
Theorem 4.3 (Fermat’s Theorem). Let 𝐼 be an interval, 𝑓 ∶ 𝐼 → ℝ, and let 𝑐 be in the interior of 𝐼. If 𝑓 has a
local maximum or local minimum at 𝑐 and 𝑓 is differentiable at 𝑐, then 𝑓 ′ (𝑐) = 0.
Proof. Since 𝑐 is an interior point of 𝐼, there is 𝜀0 > 0 such that (𝑐 − 𝜀0 , 𝑐 + 𝜀0 ) ⊂ 𝐼. Assume that 𝑓 has a local
maximum at 𝑐. Then there is 𝜀 ∈ (0, 𝜀0 ) such that 𝑓 (𝑥) ≤ 𝑓 (𝑐) for all 𝑥 ∈ (𝑐 − 𝜀, 𝑐 + 𝜀). Therefore
{
𝑓 (𝑥) − 𝑓 (𝑐) ≥ 0 for 𝑥 ∈ (𝑐 − 𝜀, 𝑐),
𝑥−𝑐
≤ 0 for 𝑥 ∈ (𝑐, 𝑐 + 𝜀).
Hence, since 𝑓 is differentiable at 𝑐,
𝑓 ′ (𝑐) = lim−
𝑥→𝑐
𝑓 (𝑥) − 𝑓 (𝑐)
≥ lim− 0 = 0
𝑥→𝑐
𝑥−𝑐
𝑓 ′ (𝑐) = lim+
𝑓 (𝑥) − 𝑓 (𝑐)
≤ lim+ 0 = 0.
𝑥→𝑐
𝑥−𝑐
and
𝑥→𝑐
Therefore 0 ≤ 𝑓 ′ (𝑐) ≤ 0, which proves 𝑓 ′ (𝑐) = 0.
The case of a local minimum at 𝑐 can be dealt with in a similar way, or we may use the fact that then −𝑓 has a
local maximum at 𝑐 and therefore −𝑓 ′ (𝑐) = 0 by the first part of the proof.
Theorem 4.4 (Rolle’s Theorem). Let 𝑎 < 𝑏 be real numbers and 𝑓 ∶ [𝑎, 𝑏] → ℝ be a function having the following
properties:
1. 𝑓 is continuous on the closed interval [𝑎, 𝑏],
2. 𝑓 is differentiable on the open interval (𝑎, 𝑏),
3. 𝑓 (𝑎) = 𝑓 (𝑏).
Then there is 𝑐 ∈ (𝑎, 𝑏) such that 𝑓 ′ (𝑐) = 0.
Proof. If 𝑓 is constant, then 𝑓 ′ = 0, and the statement is true for any 𝑐 ∈ (𝑎, 𝑏). If 𝑓 is not constant, then there is
𝑥0 ∈ [𝑎, 𝑏] such that 𝑓 (𝑥0 ) ≠ 𝑓 (𝑎). We may assume 𝑓 (𝑥0 ) > 𝑓 (𝑎). Otherwise, 𝑓 (𝑥0 ) < 𝑓 (𝑎), and we can consider
−𝑓 . Since 𝑓 is continuous on [𝑎, 𝑏] by property 1, 𝑓 has a maximum on [𝑎, 𝑏], see Corollary 3.19. That is, there
is some 𝑐 ∈ [𝑎, 𝑏] such that 𝑓 (𝑐) ≥ 𝑓 (𝑥) for all 𝑥 ∈ [𝑎, 𝑏]. In particular, 𝑓 (𝑐) ≥ 𝑓 (𝑥0 ) > 𝑓 (𝑎). Since 𝑓 (𝑎) = 𝑓 (𝑏),
we have 𝑐 ≠ 𝑎 and 𝑐 ≠ 𝑏, and therefore 𝑐 ∈ (𝑎, 𝑏). Hence 𝑓 ′ (𝑐) = 0 by Fermat’s Theorem.
Theorem 4.5 (First Mean Value Theorem). Let 𝑎 < 𝑏 be real numbers and 𝑓 ∶ [𝑎, 𝑏] → ℝ be a continuous
function which is differentiable on (𝑎, 𝑏).
Then there is 𝑐 ∈ (𝑎, 𝑏) such that
𝑓 (𝑏) − 𝑓 (𝑎)
= 𝑓 ′ (𝑐).
𝑏−𝑎
Proof. The function
𝑓 (𝑏) − 𝑓 (𝑎)
(𝑥 − 𝑎)
𝑏−𝑎
is continuous on [𝑎, 𝑏], differentiable on (𝑎, 𝑏), and 𝑔(𝑎) = 𝑓 (𝑎) = 𝑔(𝑏). Hence 𝑔 satisfies the assumptions of
Rolle’s theorem. Therefore, there is 𝑐 ∈ (𝑎, 𝑏) such that 𝑔 ′ (𝑐) = 0. But
𝑔(𝑥) = 𝑓 (𝑥) −
𝑔 ′ (𝑥) = 𝑓 ′ (𝑥) −
𝑓 (𝑏) − 𝑓 (𝑎)
,
𝑏−𝑎
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and substituting 𝑐 completes the proof.
Tutorial 4.1.1. Let 𝑎 < 𝑏 be real numbers, 𝑓 , 𝑔 ∶ [𝑎, 𝑏] → ℝ be continuous and differentiable on (𝑎, 𝑏).
1. Show that 𝑔 is injective on [𝑎, 𝑏] if 𝑔 ′ (𝑥) ≠ 0 for all 𝑥 ∈ (𝑎, 𝑏).
2. Prove the Second Mean Value Theorem: If 𝑔 ′ (𝑥) ≠ 0 for all 𝑐 ∈ (𝑎, 𝑏), then there is 𝑐 ∈ (𝑎, 𝑏) such that
𝑓 (𝑏) − 𝑓 (𝑎) 𝑓 ′ (𝑐)
= ′ .
𝑔(𝑏) − 𝑔(𝑎)
𝑔 (𝑐)
Hint. Consider the function
ℎ(𝑥) = 𝑓 (𝑥) −
𝑓 (𝑏) − 𝑓 (𝑎)
(𝑔(𝑥) − (𝑔(𝑎)).
𝑔(𝑏) − 𝑔(𝑎)
3. Show that the statement of the Second Mean Value Theorem remains correct if one replaces the condition
that 𝑔 ′ (𝑥) ≠ 0 for all 𝑥 ∈ (𝑎, 𝑏) with the weaker condition that 𝑔(𝑎) ≠ 𝑔(𝑏) and that there is no 𝑥 ∈ (𝑎, 𝑏) with
𝑔 ′ (𝑥) = 𝑓 ′ (𝑥) = 0.
4. Prove the following one-sided version of l’Hôpital’s Rule: If 𝑓 (𝑎) = 𝑔(𝑎) = 0, 𝑔 ′ (𝑥) ≠ 0 for 𝑥 near 𝑎 and
𝑓 ′ (𝑥)
exists, then
lim+ ′
𝑥→𝑎 𝑔 (𝑥)
𝑓 (𝑥)
𝑓 ′ (𝑥)
lim+
= lim+ ′ .
𝑥→𝑎 𝑔(𝑥)
𝑥→𝑎 𝑔 (𝑥)
*Tutorial 4.1.2. Let −∞ ≤ 𝑎 < 𝑏 ≤ ∞ and let 𝑓 , 𝑔 ∶ (𝑎, 𝑏) → ℝ be differentiable on (𝑎, 𝑏) such that 𝑔 ′ (𝑥) ≠ 0 for
all 𝑥 ∈ (𝑎, 𝑏). Prove the following one-sided versions of l’Hôpital’s Rule.
𝑓 ′ (𝑥)
1. If lim+ 𝑓 (𝑥) = lim+ 𝑔(𝑥) = 0 and lim+ ′
exists as a proper or improper limit, then
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎 𝑔 (𝑥)
lim+
𝑥→𝑎
2. If lim+ 𝑓 (𝑥) = lim+ 𝑔(𝑥) = ∞ and lim+
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
𝑓 ′ (𝑥)
exists as a proper or improper limit, then
𝑔 ′ (𝑥)
lim+
𝑥→𝑎
4.2
𝑓 (𝑥)
𝑓 ′ (𝑥)
= lim+ ′ .
𝑔(𝑥) 𝑥→𝑎 𝑔 (𝑥)
𝑓 (𝑥)
𝑓 ′ (𝑥)
= lim+ ′ .
𝑔(𝑥) 𝑥→𝑎 𝑔 (𝑥)
The Chain Rule and Inverse Functions
Proposition 4.6. Let 𝐼 be an interval, 𝑓 ∶ 𝐼 → ℝ and 𝑎 ∈ 𝐼. The following are equivalent:
(i) 𝑓 is differentiable at 𝑎.
(ii) There is 𝛾 ∈ ℝ such that the function
{
𝑓 (𝑥) − 𝑓 (𝑎)
if 𝑥 ∈ 𝐼 ⧵ {𝑎},
𝑓 ∗ (𝑥) =
𝑥−𝑎
𝛾
if 𝑥 = 𝑎,
is continuous at 𝑎.
(iii) There is a function 𝑓̃ ∶ 𝐼 → ℝ such that 𝑓̃ is continuous at 𝑎 and
𝑓 (𝑥) = 𝑓 (𝑎) + 𝑓̃(𝑥)(𝑥 − 𝑎),
𝑥 ∈ 𝐼.
(iv) There are 𝛾 ∈ ℝ and a function 𝜀 ∶ 𝐼 → ℝ such that 𝜀(𝑎) = 0, 𝜀 is continuous at 𝑎, and
𝑓 (𝑥) = 𝑓 (𝑎) + 𝛾(𝑥 − 𝑎) + 𝜀(𝑥)(𝑥 − 𝑎),
𝑥 ∈ 𝐼.
If (ii) or (iv) hold, then the number 𝛾 is unique and 𝛾 = 𝑓 ′ (𝑎). If (iii) holds, then 𝑓̃(𝑎) = 𝑓 ′ (𝑎).
Proof. (i)⇒(ii): If 𝑓 is differentiable at 𝑎, then
lim 𝑓 ∗ (𝑥) = 𝑓 ′ (𝑥),
𝑥→𝑎
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43
so that 𝑓 ∗ is continuous at 𝑎 for 𝛾 = 𝑓 ′ (𝑎).
(ii)⇒(iii): We simply put 𝑓̃ = 𝑓 ∗ . In particular, 𝑓̃(𝑎) = 𝛾.
(iii)⇒(iv): We have
𝑓 (𝑥) = 𝑓 (𝑎) + 𝑓̃(𝑎)(𝑥 − 𝑎) + (𝑓̃(𝑥) − 𝑓̃(𝑎))(𝑥 − 𝑎).
Then (iv) holds with 𝜀(𝑥) = 𝑓̃(𝑥) − 𝑓̃(𝑎) since 𝜀(𝑎) = 0 and the continuity of 𝑓̃ at 𝑎 implies the continuity of 𝜀 at
𝑎.
(iv)⇒(i): For 𝑥 ∈ 𝐼 ⧵ {𝑎} we have
𝑓 (𝑥) − 𝑓 (𝑎)
= 𝛾 + 𝜀(𝑥) → 𝛾 as 𝑥 → 𝑎.
𝑥−𝑎
Hence 𝑓 is differentiable at 𝑎, and 𝛾 = 𝑓 ′ (𝑎).
Theorem 4.7 (Chain Rule). Let 𝐼 and 𝐽 be intervals, 𝑔 ∶ 𝐽 → ℝ and 𝑓 ∶ 𝐼 → ℝ with 𝑓 (𝐼) ⊂ 𝐽 , and let 𝑎 ∈ 𝐼.
Assume that 𝑓 is differentiable at 𝑎 and that 𝑔 is differentiable at 𝑓 (𝑎). Then 𝑔◦𝑓 is differentiable at 𝑎 and
(𝑔◦𝑓 )′ (𝑎) = 𝑔 ′ (𝑓 (𝑎))𝑓 ′ (𝑎).
Proof. Using the property (iii) of Proposition 4.6 we obtain a function 𝑓̃ on 𝐼 which is continuous at 𝑎 with
𝑓 (𝑥) = 𝑓 (𝑎) + 𝑓̃(𝑥)(𝑥 − 𝑎)
and a function 𝑔̃ on 𝐽 which is continuous at 𝑓 (𝑎) with
𝑔(𝑦) = 𝑔(𝑓 (𝑎)) + 𝑔(𝑦)(𝑦
̃
− 𝑓 (𝑎)).
Then we can write
(𝑔◦𝑓 )(𝑥) = 𝑔(𝑓 (𝑥)) = 𝑔(𝑓 (𝑎)) + 𝑔(𝑓
̃ (𝑥))(𝑓 (𝑥) − 𝑓 (𝑎))
̃
= (𝑔◦𝑓 )(𝑎) + 𝑔(𝑓
̃ (𝑥))𝑓 (𝑥)(𝑥 − 𝑎).
Since 𝑓 and 𝑓̃ are continuous at 𝑎 and since 𝑔 is continuous at 𝑓 (𝑎), (𝑔◦𝑓
̃ )𝑓̃ is continuous at 𝑎, see Theorem 3.11,
and by Proposition 4.6, (𝑔◦𝑓 )′ (𝑎) = 𝑔(𝑓
̃ (𝑎))𝑓̃(𝑎) = 𝑔 ′ (𝑓 (𝑎))𝑓 ′ (𝑎).
Theorem 4.8. Let 𝐼 be an interval, let 𝑓 ∶ 𝐼 → ℝ be continuous and strictly increasing or decreasing, and
𝑏 ∈ 𝑓 (𝐼). Assume that 𝑓 is differentiable at 𝑎 = 𝑓 −1 (𝑏) with 𝑓 ′ (𝑎) ≠ 0. Then 𝑓 −1 is differentiable at 𝑏 = 𝑓 (𝑎)
and
1
1
(𝑓 −1 )′ (𝑏) =
or, equivalently, (𝑓 −1 )′ (𝑓 (𝑎)) = ′ .
′
−1
𝑓 (𝑎)
𝑓 (𝑓 (𝑏))
Proof. By Proposition 4.6 there is a function 𝑓̃ on 𝐼 which is continuous at 𝑎 such that
𝑓 (𝑥) − 𝑓 (𝑎) = 𝑓̃(𝑥)(𝑥 − 𝑎),
(∗)
and 𝑓̃(𝑎) = 𝑓 ′ (𝑎). By Theorem 3.20, 𝑓 −1 is continuous, and therefore 𝑓̃◦𝑓 −1 is continuous at 𝑓 (𝑎). Since
𝑓 ′ (𝑎) ≠ 0, 𝑓̃(𝑎) ≠ 0, and since 𝑓 is injective, (∗) gives that 𝑓̃(𝑥) ≠ 0 for 𝑥 ∈ 𝐼 ⧵ {𝑎}. Therefore the function
1
is well-defined on 𝑓 (𝐼) and continuous at 𝑓 (𝑎). Setting 𝑥 = 𝑓 −1 (𝑦) for 𝑦 ∈ 𝑓 (𝐼) we get
̃
𝑓 ◦𝑓 −1
𝑦 − 𝑓 (𝑎) = 𝑓̃(𝑓 −1 (𝑦))(𝑓 −1 (𝑦) − 𝑎),
or, equivalently,
𝑓 −1 (𝑦) − 𝑓 −1 (𝑓 (𝑎)) =
1
(𝑦 − 𝑓 (𝑎)).
𝑓̃(𝑓 −1 (𝑦))
By Proposition 4.6, 𝑓 −1 is differentiable at 𝑓 (𝑎), and
(𝑓 −1 )′ (𝑓 (𝑎)) =
1
1
= ′ .
−1
̃
𝑓
(𝑎)
𝑓 (𝑓 (𝑓 (𝑎)))
This result now allows us to find the derivatives of arcsin and arctan.
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Example 4.2.1. Show that arcsin ∶ [−1, 1] → [− 𝜋2 , 𝜋2 ] is continuous on [−1, 1] and differentiable on (−1, 1), and
𝑑
find
arcsin.
𝑑𝑥
Solution. sin ∶ [− 𝜋2 , 𝜋2 ] → [−1, 1] is strictly increasing and continuous. By Theorem 3.20, arcsin is continuous
on [−1, 1]. Since
(
)
𝜋 𝜋
𝑑
sin 𝑥 = cos 𝑥 ≠ 0 for 𝑥 ∈ − ,
,
𝑑𝑥
2 2
that is, if sin 𝑥 ∈ (−1, 1), it follows by Theorem 4.8 that arcsin is differentiable on (−1, 1) and
𝑑
1
1
1
arcsin 𝑥 =
=√
=√
𝑑𝑥
cos(arcsin 𝑥)
1 − 𝑥2
1 − sin2 (arcsin 𝑥)
since cos 𝑡 > 0 for 𝑡 ∈ (− 𝜋2 , 𝜋2 ).
Finally, in this section we shall show that 𝑒𝑥 = exp(𝑥) is differentiable and find its derivative. This will then allow
us to find the derivative of ln 𝑥 as an inverse function.
Theorem 4.9 (Derivative of 𝑒𝑥 ).
1
1. 𝑒𝑥 ≥ 1 + 𝑥 for 𝑥 ∈ ℝ and 𝑒𝑥 ≤
for 𝑥 < 1.
1−𝑥
𝑑 𝑥
2. 𝑒𝑥 is differentiable and
𝑒 = 𝑒𝑥 .
𝑑𝑥
Proof. 1. Let 𝑥 ∈ ℝ and 𝑛 ∈ ℕ with 𝑛 > |𝑥|. Then 𝑥𝑛 > −1, and with the aid of Bernoulli’s inequality we calculate
(
)𝑛
𝑥
≥1+𝑛 =1+𝑥
𝑛
(
)
𝑥 𝑛
𝑥
⇒ 𝑒 = lim 1 +
≥ 1 + 𝑥.
𝑛→∞
𝑛
1+
𝑥
𝑛
Thus we have shown that 𝑒𝑥 ≥ 1 + 𝑥 for all 𝑥 ∈ ℝ. Replacing 𝑥 with −𝑥 gives 𝑒−𝑥 ≥ 1 − 𝑥. Taking inverses for
1
1
𝑥 < 1, i. e., 1 − 𝑥 > 0, leads to 𝑒𝑥 = −𝑥 ≤
for 𝑥 < 1.
𝑒
1−𝑥
2. The above estimates give for ℎ < 1 that
⎧ 1 ≤ 𝑒ℎ − 1 ≤ 1
⎪
ℎ
1
ℎ
1−ℎ
−1=
⇒⎨
ℎ ≤ 𝑒ℎ − 1 ≤
ℎ−1
1−ℎ
1−ℎ
𝑒
1
⎪
⎩1−ℎ ≤ ℎ ≤1
if h>0,
if h<0.
Application of the Sandwich Theorem leads to
𝑒ℎ − 1
= 1.
ℎ→0
ℎ
lim
Therefore
𝑒𝑥+ℎ − 𝑒𝑥
𝑒𝑥 𝑒ℎ − 𝑒𝑥
𝑒ℎ − 1
𝑑 𝑥
𝑒 = lim
= lim
= 𝑒𝑥 lim
= 𝑒𝑥 .
ℎ→0
ℎ→0
ℎ→0
𝑑𝑥
ℎ
ℎ
ℎ
Tutorial 4.2.1. 1. Find the derivatives of
(a) arctan 𝑥, (b) arccos 𝑥, (c) ln 𝑥 and ln |𝑥|,
(d) ln |𝑓 (𝑥)|, where 𝑓 ∶ 𝐼 → ℝ ⧵ {0} is differentiable on 𝐼.
1
𝑝
2. Let 𝑥 > 0 and let 𝑟 = with 𝑝 ∈ ℤ and 𝑞 ∈ ℕ∗ . Define 𝑥𝑟 = (𝑥𝑝 ) 𝑞 .
𝑞
Show that 𝑥𝑟 = exp(𝑟 ln 𝑥).
3. In view of tutorial problem 2 above, we define 𝑥𝑟 = exp(𝑟 ln 𝑥) for all 𝑥 > 0 and 𝑟 ∈ ℝ.
(a) Show that lim exp(𝑥) = 0.
𝑥→−∞
(b) Show that lim
1
𝑥→∞ 𝑥𝑟
= 0 for all 𝑟 > 0.
Chapter 5
Series
5.1
Definitions and Basic Examples
Given a sequence {𝑎𝑛 }∞
of real numbers, the symbol
𝑛=1
∞
∑
𝑎𝑛
𝑛=1
is called a series (of real numbers).
Definition 5.1. Let
1. The number 𝑠𝑛 =
2. The series
∞
∑
∞
∑
𝑎𝑛 be a series.
𝑛=1
𝑛
∑
𝑎𝑖 is called the 𝑛-th partial sum of the series.
𝑖=1
𝑎𝑛 is said to converge if the sequence (𝑠𝑛 ) converges.
𝑛=1
In this case, the number 𝑠 = lim 𝑠𝑛 is called the sum of the series and we write
𝑛→∞
∞
∑
𝑎𝑛 = 𝑠
𝑛=1
A series which does not converge is said to diverge or be divergent.
Note. 1. Observe that
∞
∑
𝑎𝑛 denotes a series (convergent or divergent) as well as its sum if it converges.
𝑛=1
2. A series does not have to start at 𝑛 = 1. The change of notation is obvious for other starting indices.
Example 5.1.1 (See Calculus I). [Note to lecturers: Students have seen this in Calculus I, so one does not need to
write down much.] Consider the geometric series
∞
∑
𝑎𝑟𝑛 ,
𝑎 ≠ 0, 𝑟 ∈ ℝ.
𝑛=0
Recall the partial sums for 𝑟 ≠ 1:
𝑠𝑛 = 𝑎 + 𝑎𝑟 + 𝑎𝑟2 + ⋯ + 𝑎𝑟𝑛 ,
𝑟𝑠𝑛 =
𝑎𝑟 + 𝑎𝑟2 + ⋯ + 𝑎𝑟𝑛 + 𝑎𝑟𝑛+1 .
Subtracting these equations gives
𝑠𝑛 − 𝑟𝑠𝑛 = 𝑎 − 𝑎𝑟𝑛+1 ,
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and so
𝑎(1 − 𝑟𝑛+1 )
.
1−𝑟
By Theorem 2.5, (𝑠𝑛 ) and thus the geometric series converges if |𝑟| < 1, with
𝑠𝑛 =
∞
∑
𝑎𝑟𝑛 = lim 𝑠𝑛 =
𝑛→∞
𝑛=0
𝑎
,
1−𝑟
and diverges if |𝑟| > 1 or 𝑟 = −1. Finally, for 𝑟 = 1, 𝑠𝑛 = 𝑎(𝑛 + 1), and thus 𝑠𝑛 → ∞ as 𝑛 → ∞.
Thus we have shown
Theorem 5.1. The geometric series
∞
∑
𝑎𝑟𝑛 ,
𝑎 ≠ 0,
𝑎𝑟𝑛 =
𝑎
.
1−𝑟
𝑛=0
is convergent if |𝑟| < 1, with sum
∞
∑
𝑛=0
If |𝑟| ≥ 1, the geometric series diverges.
Example 5.1.2. Is the series
∞
∑
22𝑛 31−𝑛 convergent or divergent?
𝑛=1
Solution. The 𝑛-th term of the series is obviously of the form 𝑎𝑟𝑛 :
𝑎𝑛 = 22𝑛 31−𝑛 =
So 𝑟 =
( )𝑛
4
3 ⋅ 4𝑛
.
=
3
3𝑛
3
4
> 1, and the series diverges.
3
Example 5.1.3. Write the number 5.4417 = 5.4417417417 … as a ratio of integers.
Solution. Write
5.4417 = 5.4 +
417 417
417
+ 7 +
+ ….
4
10
10
1010
417
−3 . Hence the series converges,
Starting with the second term we have a geometric series with 𝑎 = 10
−4 and 𝑟 = 10
and
5.4417 = 5.4 +
=
417
10−4
1 − 10−3
= 5.4 +
417
539877 179959
54
+
=
=
.
10 9990
9990
3330
The following theorem has been proved in Calculus I.
Theorem 5.2. If the series
∞
∑
𝑛=1
417
104 − 10
𝑎𝑛 converges, then lim 𝑎𝑛 = 0.
𝑛→∞
Proof. Let
𝑠𝑛 = 𝑎1 + 𝑎2 + … 𝑎𝑛 .
Then
𝑎𝑛 = 𝑠𝑛 − 𝑠𝑛−1 .
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Since
∞
∑
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Introductory Analysis 2017 Lecture Manual
𝑎𝑛 converges,
𝑛=1
lim 𝑠 = 𝑠
𝑛→∞ 𝑛
exists. Since also 𝑛 − 1 → ∞ as 𝑛 → ∞,
lim 𝑠
= 𝑠.
𝑛→∞ 𝑛−1
Hence
lim 𝑎 = lim (𝑠𝑛 − 𝑠𝑛−1 ) = lim 𝑠𝑛 − lim 𝑠𝑛−1 = 𝑠 − 𝑠 = 0.
𝑛→∞ 𝑛
𝑛→∞
𝑛→∞
𝑛→∞
The contrapositive statement to Theorem 5.2 is very useful:
Theorem 5.3 (Test for Divergence). If lim 𝑎𝑛 does not exist or if lim 𝑎𝑛 ≠ 0, then the series
𝑛→∞
𝑛→∞
∞
∑
𝑎𝑛 diverges.
𝑛=1
Note. From lim 𝑎𝑛 = 0 nothing can be concluded about the convergence of the series
𝑛→∞
∞
∑
𝑎𝑛 .
𝑛=1
From Theorem 2.3 we immediately infer
Theorem 5.4 (Sum Laws). Let 𝑐 ∈ ℝ and suppose that
∞
∑
(𝑎𝑛 + 𝑏𝑛 ) and
𝑛=1
1.
2.
3.
∞
∑
∞
∑
𝑎𝑛 and
𝑛=1
∞
∑
𝑏𝑛 both converge. Then also
𝑛=1
(𝑎𝑛 − 𝑏𝑛 ) converge, and
𝑛=1
∞
∑
(𝑐𝑎𝑛 ) = 𝑐
∞
∑
𝑎𝑛 ,
𝑛=1
∞
∑
∞
∑
𝑛=1
∞
∑
𝑛=1
∞
∑
𝑛=1
∞
∑
𝑛=1
𝑛=1
𝑛=1
∞
∑
(𝑎𝑛 + 𝑏𝑛 ) =
(𝑎𝑛 − 𝑏𝑛 ) =
𝑎𝑛 +
𝑎𝑛 −
𝑏𝑛 ,
𝑏𝑛 ,
𝑛=1
Example 5.1.4. Find the sum of the series
∞
∑
𝑛=1
(
)
1
5
+ 𝑛 .
𝑛(𝑛 + 1) 5
Solution. Recall from Calculus I that
𝑚 (
)
∑
1
1
1
1
=
−
=1−
.
𝑛(𝑛 + 1) 𝑛=1 𝑛 𝑛 + 1
𝑚+1
𝑛=1
𝑚
∑
Hence
∞
∑
1
= 1.
𝑛(𝑛 + 1)
𝑛=1
And
∞
∑
1
5𝑛
𝑛=1
is a geometric series with 𝑎 = 51 and 𝑟 = 15 . Hence
∞
∑
1
5𝑛
𝑛=1
Therefore
∞
∑
𝑛=1
(
5
1
+
𝑛(𝑛 + 1) 5𝑛
)
=5
=
1 1
1
= .
51− 1
4
5
∞
∑
∞
∑
1
1
1 21
+
=5+ =
.
𝑛
𝑛(𝑛
+
1)
5
4
4
𝑛=1
𝑛=1
Note. For convergence it does not matter at which index the series starts. But it matters for the sum.
∞
∑
(𝑐𝑎𝑛 ),
𝑛=1
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Introductory Analysis 2017 Lecture Manual
Theorem 5.5. The series
∞
∑
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𝑎𝑛 converges if and only if for each 𝜀 > 0 there is 𝐾 ∈ ℕ such that for all 𝑚 ≥ 𝑘 ≥ 𝐾,
𝑛=1
𝑚
|∑
|
|
|
| 𝑎𝑛 | < 𝜀.
|
|
|𝑛=𝑘 |
Proof. Let 𝑠𝑘 =
𝑘
∑
𝑎𝑛 . Then
𝑛=1
∞
∑
𝑎𝑛 converges ⇔ (𝑠𝑛 )∞
converges
𝑛=1
(by definition)
𝑛=1
⇔ ∀ 𝜀 > 0 ∃𝐾 ∈ ℕ ∀𝑚 ≥ 𝑘 ≥ 𝐾, |𝑠𝑚 − 𝑠𝑘−1 | < 𝜀
𝑚
|∑
|
|
|
⇔ ∀ 𝜀 > 0 ∃𝐾 ∈ ℕ ∀𝑚 ≥ 𝑘 ≥ 𝐾, | 𝑎𝑛 | < 𝜀.
|
|
|𝑛=𝑘 |
(by Theorem 2.12)
Tutorial 5.1.1. 1. Test each of the following series for convergence or divergence:
∞
∞
∞ (
)
( )
∑
∑
∑
1
𝑛2 − 1
1 𝑛
,
(b)
𝑛 sin
,
(c)
.
(a)
1+
2
𝑛
𝑛
𝑛=1
𝑛=1 𝑛 − 50𝑛
𝑛=1
2. Which of the following is valid? Justify your conclusions.
∞
∑
(a) If 𝑎𝑛 → 0 as 𝑛 → ∞, then
𝑎𝑛 is convergent.
(b) If 𝑎𝑛 ̸→ 0 as 𝑛 → ∞, then
∞
∑
(c) If
𝑛=1
∞
∑
𝑎𝑛 is divergent.
𝑛=1
𝑎𝑛 is divergent, then 𝑎𝑛 ̸→ 0 as 𝑛 → ∞.
𝑛=1
5.2
Convergence Tests for Series
Lemma 5.6. Let 𝑎𝑛 ≥ 0 for all 𝑛 ∈ ℕ∗ and let 𝑠𝑘 =
𝑘
∑
𝑛=1
𝑎𝑛 . Then
∞
∑
𝑎𝑛 converges if and only if (𝑠𝑛 ) is bounded.
𝑛=1
Proof. Since 𝑠𝑛+1 = 𝑠𝑛 + 𝑎𝑛+1 ≥ 𝑠𝑛 , it follows that (𝑠𝑛 ) is an increasing sequence. By Theorem 2.10, this sequence
and hence the series converges if and only if the sequence (𝑠𝑛 ) is bounded.
Theorem 5.7 (Comparison Test). Let
∞
∑
𝑎𝑛 and
𝑛=1
for all 𝑛 ∈ ℕ∗ .
∞
∞
∑
∑
(i) If
𝑏𝑛 converges, then also
𝑎𝑛 converges.
𝑛=1
∞
∑
(ii) If
𝑏𝑛 be series with nonnegative terms and assume that 𝑎𝑛 ≤ 𝑏𝑛
𝑛=1
𝑛=1
𝑎𝑛 diverges, then also
𝑛=1
Proof. Let 𝑠𝑘 =
∞
∑
∞
∑
∞
∑
𝑏𝑛 diverges.
𝑛=1
𝑘
∑
𝑛=1
𝑎𝑛 and 𝑡𝑘 =
𝑘
∑
𝑏𝑛 . Then both (𝑠𝑘 ) and (𝑡𝑘 ) are increasing sequences with 𝑠𝑘 ≤ 𝑡𝑘 . Hence, if
𝑛=1
𝑏𝑛 converges, then (𝑡𝑘 ) is bounded, say 𝑡𝑘 ≤ 𝑀 for all 𝑘 ∈ ℕ, and 𝑠𝑘 ≤ 𝑡𝑘 ≤ 𝑀 for all 𝑘 ∈ ℕ. Hence (𝑠𝑘 ) is a
𝑛=1
bounded sequence and thus converges by Lemma 5.6.
(ii) is the contrapositive of (i).
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Example 5.2.1. Test the series
∞
∑
sin2 𝑛 + 10
𝑛 + 2𝑛
𝑛=1
for convergence.
Solution. Putting
sin2 𝑛 + 10
𝑛 + 2𝑛
𝑎𝑛 =
it follows that 0 < 𝑎𝑛 < 11
( )𝑛
1
2
=∶ 𝑏𝑛 . By Theorem 5.1, the series with general term 𝑏𝑛 converges. Hence
∞
∑
sin2 𝑛 + 10
converges by the comparison test.
𝑛 + 2𝑛
𝑛=1
Definition 5.2. 1. A series
converges.
2. A series
∞
∑
𝑎𝑛 is called absolutely convergent if the series of its absolute values
|𝑎𝑛 |
𝑛=1
𝑛=1
∞
∑
∞
∑
𝑎𝑛 is called conditionally convergent if it is convergent but not absolutely convergent.
𝑛=1
Theorem 5.8. Every absolutely convergent series is convergent.
Proof. Let
∞
∑
𝑎𝑛 be absolutely convergent. Let 𝜀 > 0. Then, by Theorem 5.5, there is 𝐾 ∈ ℕ such that for all
𝑛=1
𝑚
∑
𝑚 ≥ 𝑘 ≥ 𝐾,
|𝑎𝑛 | < 𝜀. Because of
𝑛=𝑘
𝑚
𝑚
|∑
| ∑
|
|
|𝑎𝑛 |
| 𝑎𝑛 | ≤
|
|
|𝑛=𝑘 | 𝑛=𝑘
𝑚
∞
|∑
|
∑
|
|
it follows that | 𝑎𝑛 | < 𝜀 for these 𝑘, 𝑚, and therefore
𝑎𝑛 converges by Theorem 5.5.
|
|
|𝑛=𝑘 |
𝑛=1
∞
∞
∑
∑
Alternative proof. By assumption and Theorem 5.4, both
|𝑎𝑛 | and
2|𝑎𝑛 | converge. From
𝑛=1
𝑛=1
0 ≤ 𝑎𝑛 + |𝑎𝑛 | ≤ 2|𝑎𝑛 |
and the comparison test, it follows that also
∞
∑
(𝑎𝑛 + |𝑎𝑛 |) converges. Again by Theorem 5.4 it follows that
𝑛=1
∞
∑
𝑎𝑛 =
𝑛=1
∞
∑
((𝑎𝑛 + |𝑎𝑛 |) − |𝑎𝑛 |)
𝑛=1
converges.
Recall from Calculus I that there are convergent series which are not absolutely convergent.
Definition 5.3. An alternating series is a series of the form
∞
∑
𝑛=1
𝑛
(−1) 𝑏𝑛
or
∞
∑
(−1)𝑛−1 𝑏𝑛
with 𝑏𝑛 ≥ 0.
𝑛=1
∞
∞
∑
∑
𝑛
Theorem 5.9 (Alternating series test). If the alternating series
(−1) 𝑏𝑛 or
(−1)𝑛−1 𝑏𝑛 satisfies
𝑛=1
(i) 𝑏𝑛 ≥ 𝑏𝑛+1 for all 𝑛,
(ii) lim 𝑏𝑛 = 0,
𝑛→∞
then the series converges.
𝑛=1
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Introductory Analysis 2017 Lecture Manual
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Proof. For 𝑘 ∈ ℕ∗ and 𝑚 ∈ ℕ we have
(−1)𝑘
𝑘+2𝑚
∑
(−1)𝑛 𝑏𝑛 = (𝑏𝑘 − 𝑏𝑘+1 ) + (𝑏𝑘+2 − 𝑏𝑘+3 ) + ⋯ + (𝑏𝑘+2𝑚−2 − 𝑏𝑘+2𝑚−1 ) + 𝑏𝑘+2𝑚
𝑛=𝑘
= 𝑏𝑘 − (𝑏𝑘+1 − 𝑏𝑘+2 ) − ⋯ − (𝑏𝑘+2𝑚−1 − 𝑏𝑘+2𝑚 ).
Hence
0 ≤ 𝑏𝑘+2𝑚 ≤ (−1)𝑘
𝑘+2𝑚
∑
(−1)𝑛 𝑏𝑛 ≤ 𝑏𝑘 .
𝑛=𝑘
Similarly,
0 ≤ (−1)
𝑘
𝑘+2𝑚+1
∑
(−1)𝑛 𝑏𝑛 ≤ 𝑏𝑘 − 𝑏𝑘+2𝑚+1 ≤ 𝑏𝑘 .
𝑛=𝑘
Now let 𝜀 > 0. Since 𝑏𝑘 → 0 as 𝑘 → ∞, there is 𝐾 ∈ ℕ such that 𝑏𝐾 < 𝜀. Hence for all 𝑙 ≥ 𝑘 ≥ 𝐾:
|∑
|
| 𝑙
|
| (−1)𝑛 𝑏𝑛 | ≤ 𝑏𝑘 ≤ 𝑏𝐾 < 𝜀.
|
|
|𝑛=𝑘
|
|
|
Hence the alternating series converges.
Note. lim 𝑏𝑛 = 0 is necessary by Theorem 5.3 since lim (−1)𝑛 𝑏𝑛 = 0 ⇔ lim 𝑏𝑛 = 0 ⇔ lim (−1)𝑛−1 𝑏𝑛 = 0.
𝑛→∞
Example 5.2.2. The alternating series
𝑛→∞
∞
∑
(−1)𝑛−1
𝑛=1
𝑛→∞
𝑛→∞
3𝑛
does not converge since
4𝑛 − 1
3𝑛
lim
𝑛→∞ 4𝑛 − 1
3
≠0
4
=
and thus (ii) is not satisfied, which is necessary for convergence. See the note following the statement of the
Alternating Series Test.
Worked Example 5.2.3. Find whether the series
∞
∑
(−1)𝑛−1
𝑛=1
𝑛2
𝑛3 + 1
is convergent.
Solution. Clearly, we have an alternating series with
lim 𝑏𝑛 = lim
𝑛→∞
𝑛2
𝑛→∞ 𝑛3 + 1
= 0.
To show that 𝑏𝑛 ≥ 𝑏𝑛+1 we have at least the following 2 choices:
1. Show the inequality directly by cross-multiplication and simplification.
𝑥2
2. Show that 𝑓 (𝑥) =
has a negative derivative for sufficiently large 𝑥.
𝑥3 + 1
Alternatively, write
1
𝑛2 + 1𝑛
𝑛2
1
1
=
− 𝑛 = −
.
𝑛3 + 1
𝑛3 + 1 𝑛3 + 1 𝑛 𝑛(𝑛3 + 1)
Then
∞
∞
∞
∑
∑
(−1)𝑛−1 ∑ (−1)𝑛
𝑛2
(−1)𝑛−1
+
=
3
𝑛
𝑛3 + 1 𝑛=1
𝑛=1
𝑛=1 𝑛(𝑛 + 1)
is the sum of two alternating series which converge by the alternating series test; hence the series converges.
∞
∑
| 𝑎𝑛+1 |
| = 𝐿 < 1, then the series
Theorem 5.10 (Ratio Test). (i) If lim sup ||
𝑎𝑛 converges absolutely.
|
𝑛→∞ | 𝑎𝑛 |
𝑛=1
∞
∑
| 𝑎𝑛+1 |
| = 𝑙 > 1, then the series
(ii) If lim inf ||
𝑎𝑛 diverges.
𝑛→∞ | 𝑎𝑛 ||
𝑛=1
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Proof. Note that the ratio test assumes 𝑎𝑛 ≠ 0 for all 𝑛 ∈ ℕ.
(i) Let 𝜀 > 0 such that 𝐿 + 𝜀 < 1.
| 𝑎𝑛+1 |
| < 𝐿 + 𝜀 for all 𝑛 ≥ 𝐾. Hence for 𝑚 > 𝐾:
Then there is 𝐾 ∈ ℕ such that ||
|
| 𝑎𝑛 |
|
|
|
|𝑎
|𝑎𝑚 | = |𝑎𝐾 | | 𝐾+1 | ⋅ ⋯ ⋅ | 𝑎𝑚 | < |𝑎𝐾 |(𝐿 + 𝜀)𝑚−𝐾 .
|
|𝑎
| | | || 𝑎 |
| 𝑚−1 |
| 𝐾 |
Since
∞
∑
(*)
|𝑎𝐾 |(𝐿 + 𝜀)𝑚−𝐾 is a convergent geometric series, it follows from (*) and the Comparison Test, Theorem
𝑚=𝐾
∞
∑
5.7, that
𝑎𝑚 converges absolutely. Hence also
𝑚=𝐾
(ii) Let 𝑙′ ∈ (1, 𝑙).
∞
∑
𝑎𝑛 converges absolutely.
𝑛=1
| 𝑎𝑛+1 |
| > 𝑙′ for all 𝑛 ≥ 𝐾. Hence for 𝑚 > 𝐾:
Then there is 𝐾 ∈ ℕ such that ||
|
| 𝑎𝑛 |
|
|
|
|𝑎
|𝑎𝑚 | = |𝑎𝐾 | | 𝐾+1 | ⋅ ⋯ ⋅ | 𝑎𝑚 | > |𝑎𝐾 |,
| | | || 𝑎 |
|𝑎
|
| 𝑚−1 |
| 𝐾 |
so that 𝑎𝑛 ̸→ 0 as 𝑛 → ∞. Hence
∞
∑
𝑎𝑛 diverges by the Test for Divergence, Theorem 5.3.
𝑛=1
∞
∑
√
Theorem 5.11 (Root Test). (i) If lim sup 𝑛 |𝑎𝑛 | = 𝐿 < 1, then the series
𝑎𝑛 converges absolutely.
𝑛→∞
(ii) If lim sup
√
𝑛
𝑛→∞
|𝑎𝑛 | = 𝐿 > 1, then the series
𝑛=1
∞
∑
𝑎𝑛 diverges.
𝑛=1
Proof. (i) Let 𝜀 > 0 such that 𝐿 + 𝜀 < 1.
∞
∑
√
𝑛
𝑛
(𝐿 + 𝜀)𝑛 is
Then there is 𝐾 ∈ ℕ such that |𝑎𝑛 | < 𝐿 + 𝜀 for all 𝑛 ≥ 𝐾. Hence |𝑎𝑛 | < (𝐿 + 𝜀) for 𝑛 ≥ 𝐾: Since
a convergent geometric series, it follows from the comparison test, Theorem 5.7, that
Hence also
∞
∑
∞
∑
𝑛=𝐾
𝑎𝑛 converges absolutely.
𝑛=𝐾
𝑎𝑛 converges absolutely.
𝑛=1
(ii) Let 𝐿′ ∈ (1, 𝐿).
√
Then for each 𝐾 ∈ ℕ there is 𝑚 ≥ 𝐾 such that 𝑚 |𝑎𝑚 | > 𝐿′ . Hence |𝑎𝑚 | > (𝐿′ )𝑚 > 1 for this 𝑚, and we conclude
𝑎𝑛 ̸→ 0 as 𝑛 → ∞.
Indeed, if 𝑎𝑛 → 0 as 𝑛 → ∞, for 𝜀 = 1 there would be 𝐾 ∈ ℕ such that |𝑎𝑛 | < 1 for all 𝑛 > 𝐾.
∞
∑
𝑎𝑛 diverges by the Test for Divergence, Theorem 5.3.
Hence
𝑛=1
{( )𝑛
( )𝑛 }
( )
1
3
1
Tutorial 5.2.1. 1. Test
+5
𝑛 sin
for convergence.
2
4
𝑛
𝑛=1
∞
∑
2. Prove that the sequence (𝑎𝑛 )∞
converges if and only if
𝑛=1
(i) (𝑎2𝑛 )∞
converges,
𝑛=1
(ii) (𝑎2𝑛−1 )∞
converges,
𝑛=1
(iii) (𝑎𝑛 − 𝑎𝑛−1 ) → 0 as 𝑛 → ∞.
3. Use Tutorial 2 to prove the Alternating Series Test.
Hint. Show that (𝑠2𝑛 )∞
and (𝑠2𝑛−1 )∞
are monotonic sequences.
𝑛=1
𝑛=1
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4. Use the alternating series test, ratio test or root test to test for convergence:
)2𝑛
∞ (
∞
∑
∑
(−1)𝑛 𝑛
𝑛!2𝑛
(a)
,
(b)
,
2𝑛 + 1
(2𝑛)!
𝑛=1
𝑛=1
(
∞
∞
(
) )
∑
∑
1 𝑛
𝑛𝑛
𝑛
(−1) 𝑒 − 1 +
(c)
,
, (d)
2
𝑛
𝑛=1
𝑛=1 (𝑛!)
(e)
∞ 𝑛
∑
2
𝑛=1
5.3
𝑛
.
Power Series
Definition 5.4. Let 𝑎 and 𝑐𝑛 , 𝑛 ∈ ℕ, be real numbers. A series
∞
∑
𝑐𝑛 (𝑥 − 𝑎)𝑛
𝑛=0
is called a power series in (𝑥 − 𝑎) or a power series centred at 𝑎 or a power series about 𝑎.
Note that (𝑥 − 𝑎)0 = 1. For 𝑥 = 𝑎, all terms from the second onwards are 0, so the series converges to 𝑐0 for 𝑥 = 𝑎.
Each power series defines a function whose domain is those 𝑥 ∈ ℝ for which the series converges.
Notation. With each power series
∞
∑
𝑐𝑛 (𝑥 − 𝑎)𝑛 we associate a number 𝑅 or ∞, called the radius of convergence
𝑛=0
of the series, which is defined as follows:
√
(i) 𝑅 = 0 if lim sup 𝑛 |𝑐𝑛 | = ∞,
𝑛→∞
(ii) 𝑅 =
1
if 0 < lim sup
√
𝑛→∞
lim sup 𝑛 |𝑐𝑛 |
√
𝑛
|𝑐𝑛 | ∈ ℝ,
𝑛→∞
√
(iii) 𝑅 = ∞ if lim sup 𝑛 |𝑐𝑛 | = 0.
𝑛→∞
Theorem 5.12. There are three alternatives for the domain of a power series
∞
∑
𝑐𝑛 (𝑥 − 𝑎)𝑛 :
𝑛=0
(i) If 𝑅 = 0, then the series converges only for 𝑥 = 𝑎.
(ii) If 𝑅 = ∞, then the series converges absolutely for all 𝑥 ∈ ℝ.
(iii) If 0 < 𝑅 ∈ ℝ, then the series converges absolutely if |𝑥 − 𝑎| < 𝑅 and diverges if |𝑥 − 𝑎| > 𝑅.
Note that (iii) says that the series converges for 𝑥 in the interval (𝑎 − 𝑅, 𝑎 + 𝑅) and diverges outside [𝑎 − 𝑅, 𝑎 + 𝑅].
For 𝑥 = 𝑎 − 𝑅 and 𝑥 = 𝑎 + 𝑅 anything can happen, see Calculus I. In any case, the domain of the series is an
interval, called the interval of convergence.
√
Proof. (i): We note that if lim sup 𝑛 |𝑐𝑛 | = ∞, then for 𝑥 ≠ 𝑎 also
𝑛→∞
√
√
lim sup 𝑛 |𝑐𝑛 (𝑥 − 𝑎)𝑛 | = lim sup 𝑛 |𝑐𝑛 | |𝑥 − 𝑎| = ∞.
𝑛→∞
𝑛→∞
In view of the root test,
√ this shows that the power series diverges if 𝑥 ≠ 𝑎.
(ii), (iii): If lim sup 𝑛 |𝑐𝑛 | ∈ ℝ, then
𝑛→∞
√
√
lim sup 𝑛 |𝑐𝑛 (𝑥 − 𝑎)𝑛 | = |𝑥 − 𝑎| lim sup 𝑛 |𝑐𝑛 |.
𝑛→∞
𝑛→∞
Hence the power series converges for all 𝑥 ∈ ℝ if lim sup
𝑛→∞
√
𝑛
|𝑐𝑛 | = 0, proving (ii).
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√
If finally 0 < lim sup 𝑛 |𝑐𝑛 | ∈ ℝ, then, by the Root Test, the series converges if
𝑛→∞
√
1
,
|𝑥 − 𝑎| lim sup 𝑛 |𝑐𝑛 | < 1, i. e., |𝑥 − 𝑎| <
√
𝑛→∞
lim sup 𝑛 |𝑐𝑛 |
𝑛→∞
and the series diverges if
1
|𝑥 − 𝑎| >
.
√
lim sup 𝑛 |𝑐𝑛 |
𝑛→∞
√
Tutorial 5.3.1. 1. (a) Let 𝑐 > 1 and put 𝑐𝑛 = 𝑛 𝑐 − 1.
(i) Show that 𝑐𝑛 ≥ 0.
(ii) Show that lim sup 𝑐𝑛 ≤ 0. Hint. Use Bernoulli’s inequality.
𝑛→∞
√
(iii) Conclude that lim 𝑛 𝑐 = 1.
𝑛→∞
√
(b) Use (a) to show that lim 𝑛 𝑐 = 1 for all 𝑐 > 0.
𝑛→∞
2. Consider
∞
∑
𝑎𝑛 with 𝑎𝑛 ≠ 0 for all 𝑛 ∈ ℕ. Show that
𝑛=1
√
√
| 𝑎𝑛+1 |
|𝑎 |
| ≤ lim inf 𝑛 |𝑎𝑛 | ≤ lim sup 𝑛 |𝑎𝑛 | ≤ lim sup | 𝑛+1 | .
lim inf ||
|
|
|
𝑛→∞ | 𝑎𝑛 |
𝑛→∞
𝑛→∞
𝑛→∞ | 𝑎𝑛 |
| 𝑎𝑛+1 |
| exists or is ∞?
What can you say if lim ||
𝑛→∞ | 𝑎𝑛 ||
∞
∑
3. Consider the power series
𝑎𝑛 (𝑥 − 𝑎)𝑛 with 𝑎𝑛 ≠ 0 for all 𝑛 ∈ ℕ. Using tutorial problem 2 above or otherwise,
𝑛=1
| 𝑎 |
|
|
prove that if 𝑅 = lim | 𝑛 | exists or is ∞, then 𝑅 is the radius of convergence of the power series.
𝑛→∞ | 𝑎𝑛+1 |
|
|
√
4. Prove that lim 𝑛 𝑛 = 1.
𝑛→∞
5. Find the radius and interval of convergence for each of the following power series:
∞
∞
∞
∑
∑
∑
(2𝑥)𝑛
𝑥𝑛
(a)
,
(b)
,
(c)
𝑛𝑛 𝑥𝑛 ,
𝑛
𝑛𝑛
𝑛=1
𝑛=0
𝑛=0
(d)
(g)
∞
∑
(𝑥 − 1)𝑛
√ ,
𝑛 𝑛
𝑛=1 3
∞ (
∑
𝑛=0
𝑛
𝑛+1
)
𝑛2
(e)
∞
∑
(−2𝑥)𝑛
𝑛=1
𝑥𝑛 ,
(h)
∞
∑
𝑛3
,
(𝑥 + 3)𝑛
,
𝑛3
𝑛=1
(f)
∞
∑
(−1)𝑛 𝑥𝑛 ,
𝑛=0
(i)
∞
∑
(𝑛𝑥)𝑛
𝑛=0
(2𝑛)!
.
Chapter 6
Riemann Integration
6.1
Suprema and Infima (self study)
Recall that if a nonempty set 𝐴 ⊂ ℝ is bounded above, then the supremum of 𝐴 is the least upper bound of 𝐴,
and if 𝐴 is bounded below, then the infimum of 𝐴 is the greatest lower bound of 𝐴. Hence for all 𝑥 ∈ 𝐴 we
have inf 𝐴 ≤ 𝑥 ≤ sup 𝐴 and for each 𝜀 > 0 there exist 𝑝, 𝑞 ∈ 𝐴 such that 𝑝 < 𝜀 + inf 𝐴 and 𝑞 > −𝜀 + sup 𝐴, see
Theorems 1.9 and 1.12.
Lemma 6.1. Let 𝐴 and 𝐵 be non-empty bounded sets of real numbers, let 𝑐 ∈ ℝ, and define
𝑐𝐴 ∶= {𝑐𝑥 ∣ 𝑥 ∈ 𝐴},
𝑐 + 𝐴 ∶= {𝑐 + 𝑥 ∣ 𝑥 ∈ 𝐴}
and
𝐴 + 𝐵 ∶= {𝑥 + 𝑦 ∣ 𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵}.
Then 𝑐𝐴, 𝑐 + 𝐴 and 𝐴 + 𝐵 are bounded, and
(a) sup(𝑐𝐴) = 𝑐 sup 𝐴, inf (𝑐𝐴) = 𝑐 inf 𝐴, if 𝑐 ≥ 0,
(b) sup(𝑐𝐴) = 𝑐 inf 𝐴, inf (𝑐𝐴) = 𝑐 sup 𝐴, if 𝑐 ≤ 0,
(c) sup(𝑐 + 𝐴) = 𝑐 + sup 𝐴, inf (𝑐 + 𝐴) = 𝑐 + inf 𝐴,
(d) sup(𝐴 + 𝐵) = sup 𝐴 + sup 𝐵, inf (𝐴 + 𝐵) = inf 𝐴 + inf 𝐵.
Proof. (a) Let 𝑀 = sup 𝐴. Then 𝑥 ≤ 𝑀 for all 𝑥 ∈ 𝐴 and therefore 𝑐𝑥 ≤ 𝑐𝑀 for all 𝑐𝑥 ∈ 𝑐𝐴. Hence 𝑐𝐴 is
bounded above with bound 𝑐𝑀. This shows that 𝐿 = sup(𝑐𝐴) satisfies 𝐿 ≤ 𝑐𝑀.
If 𝑐 = 0 then 𝐿 = sup{0} = 0 = 0 ⋅ 𝑀 = 𝑐𝑀, and if 𝑐 > 0, the the above inequality gives
𝑐𝑀 = 𝑐 sup(𝑐 −1 (𝑐𝐴)) ≤ sup(𝑐𝐴) = 𝐿.
Hence 𝐿 ≤ 𝑐𝑀 and 𝑐𝑀 ≤ 𝐿, which proves (a) for supremum.
The proof for the infimum is similar, or we can use Proposition 1.10, with 𝐴 = −(−𝐴).
(b) Since 𝑐𝐴 = −((−𝑐)𝐴) and −𝑐 ≥ 0, this follows from (a) and Proposition 1.10.
(d) This is Tutorial 1.2.1, part 7.
(c) is a special case of (d) if one observes 𝑐 + 𝐴 = {𝑐} + 𝐴 and inf {𝑐} = sup{𝑐} = 𝑐.
Lemma 6.2. If ∅ ≠ 𝐴 ⊂ 𝐵 ⊂ ℝ, then
sup 𝐴 ≤ sup 𝐵,
inf 𝐴 ≥ inf 𝐵.
Proof. Assume 𝐵 is bounded. Let 𝑎 ∈ 𝐴. Then 𝑎 ∈ 𝐵, and hence inf 𝐵 ≤ 𝑎 ≤ sup 𝐵. It follows that 𝐴 is bounded,
that inf 𝐵 is a lower bound of 𝐴 and that sup 𝐴 is an upper bound of 𝐴. Hence inf 𝐴 ≥ inf 𝐵 and sup 𝐴 ≤ sup 𝐵.
The cases that inf 𝐵 = −∞ or sup 𝐵 = ∞ are similar.
Lemma 6.3. Let 𝐴 and 𝐵 be nonempty subsets of ℝ such that 𝑥 ≤ 𝑦 for all 𝑥 ∈ 𝐴 and 𝑦 ∈ 𝐵. Then 𝐴 is bounded
above, 𝐵 is bounded below, and
sup 𝐴 ≤ inf 𝐵.
Proof. Let 𝑏 ∈ 𝐵. Then 𝑎 ≤ 𝑏 for all 𝑎 ∈ 𝐴. So 𝑏 is an upper bound of 𝐴. Hence 𝐴 is bounded above and
sup 𝐴 ≤ 𝑏 since sup 𝐴 is the least upper bound. Then sup 𝐴 ≤ 𝑏 for all 𝑏 ∈ 𝐵, and therefore sup 𝐴 is a lower bound
for 𝐵. Thus sup 𝐴 ≤ inf 𝐵 follows.
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Riemann Sums
The aim of this part of your course is to formalize the Riemann integral and consequently prove the major theorems
concerning its properties.
Definition 6.1. A partition, 𝑃 , of the interval [𝑎, 𝑏] is a finite set of points 𝑎 = 𝑥0 < 𝑥1 < ⋯ < 𝑥𝑁 = 𝑏 which for
brevity will be denoted
𝑃 = {𝑎 = 𝑥0 < 𝑥1 < ⋯ < 𝑥𝑁 = 𝑏}.
The set of all partitions of the interval [𝑎, 𝑏] will be denoted by (𝑎, 𝑏).
Next we consider the upper and lower Riemann sums with respect to a partition.
Definition 6.2. Let 𝑓 be a bounded function on [𝑎, 𝑏], (i.e. 𝑓 ∶ [𝑎, 𝑏] → ℝ and there exists a constant 𝐾 such that
|𝑓 (𝑥)| ≤ 𝐾 for all 𝑥 ∈ [𝑎, 𝑏]) and let 𝑃 = {𝑎 = 𝑥0 < ⋯ < 𝑥𝑁 = 𝑏} be a partition of the closed finite interval
[𝑎, 𝑏].
(a) We define the upper sum of 𝑓 over [𝑎, 𝑏] with respect to the partition 𝑃 by
𝑈 (𝑓 , 𝑃 ) ∶=
𝑁
∑
(𝑥𝑗 − 𝑥𝑗−1 ) sup{𝑓 (𝑥) ∣ 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ]}.
𝑗=1
(b) We define the lower sum of 𝑓 over [𝑎, 𝑏] with respect to the partition 𝑃 by
𝐿(𝑓 , 𝑃 ) ∶=
𝑁
∑
(𝑥𝑗 − 𝑥𝑗−1 ) inf {𝑓 (𝑥) ∣ 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ]}.
𝑗=1
𝑓
𝑦
𝑥0
𝑥2
𝑥3
𝑥𝑛
𝑥
=
=
𝑎
𝑥1
𝑏
Observe that if 𝑓 (𝑥) ≥ 0 on [𝑎, 𝑏], then both 𝐿(𝑓 , 𝑃 ) and 𝑈 (𝑓 , 𝑃 ) are approximations of the area under the curve
𝑦 = 𝑓 (𝑥) for 𝑥 ∈ [𝑎, 𝑏]. Here 𝐿(𝑓 , 𝑃 ) under approximates the area, while 𝑈 (𝑓 , 𝑃 ) over approximates the area.
Worked Example 6.2.1. Find 𝑈 (𝑓 , 𝑃 ) and 𝐿(𝑓 , 𝑃 ) for 𝑓 (𝑥) = 𝑥2 − 1 and 𝑃 = {0 < 1∕4 < 1∕2 < 1}.
Solution.
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1
4
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Upper Sum:
1
2
𝑓
1
𝑥
sup 𝑓 (𝑥) =
𝑥∈[0, 14 ]
sup 𝑓 (𝑥) =
𝑥∈[ 14 , 12 ]
1
−15
−1=
16
16
1
−3
−1=
4
4
sup 𝑓 (𝑥) = 1 − 1 = 0
𝑥∈[ 12 ,1]
⇒ 𝑈 (𝑓 , 𝑃 ) =
(
(
)
(
)
)
1
27
−15 1
−3 1 1
+0 1−
=−
−0 +
−
16 4
4 2 4
2
64
Lower Sum:
−15
−3
1
1
−1=
,
inf 𝑓 (𝑥) = − 1 =
1
16
16
4
4
𝑥∈[ 2 ,1]
(
(
)
(
)
)
1
−3
1
55
−15 1 1
⇒ 𝐿(𝑓 , 𝑃 ) = −1
+
1−
=−
−0 +
−
4
16 2 4
4
2
64
inf 𝑓 (𝑥) = 0 − 1 = −1,
𝑥∈[0, 14 ]
inf
𝑥∈[ 14 , 12 ]
1
Remark. From 1st year:
∫
(𝑥2 − 1) 𝑑𝑥 =
0
𝑓 (𝑥) =
]1
[ 3
2
𝑥
− 𝑥 = − so we observe that
3
3
0
1
55
27
−
= 𝐿(𝑥2 − 1, 𝑃 ) ≤ (𝑥2 − 1) 𝑑𝑥 ≤ 𝑈 (𝑓 , 𝑃 ) = − .
∫
64
64
0
The following lemma gives some useful bounds.
Lemma 6.4. Let 𝑓 be a bounded function on the closed interval [𝑎, 𝑏] and let 𝑃 be a partition of [𝑎, 𝑏]. Then
(𝑏 − 𝑎) inf {𝑓 (𝑥) ∣ 𝑥 ∈ [𝑎, 𝑏]} ≤ 𝐿(𝑓 , 𝑃 ) ≤ 𝑈 (𝑓 , 𝑃 ) ≤ (𝑏 − 𝑎) sup{𝑓 (𝑥) ∣ 𝑥 ∈ [𝑎, 𝑏]}.
Proof. From the definition of supremum and infimum:
inf
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
𝑓 (𝑥) ≤
sup
𝑓 (𝑥).
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
Also, since [𝑥𝑗−1 , 𝑥𝑗 ] ⊂ [𝑎, 𝑏], it follows from Lemma 6.2 that
sup
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
𝑓 (𝑥) ≤ sup 𝑓 (𝑥),
𝑥∈[𝑎,𝑏]
inf 𝑓 (𝑥) ≤
𝑥∈[𝑎,𝑏]
inf
𝑓 (𝑥).
sup
𝑓 (𝑥) ≤ sup 𝑓 (𝑥).
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
Thus
inf 𝑓 (𝑥) ≤
𝑥∈[𝑎,𝑏]
inf
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
𝑓 (𝑥) ≤
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
𝑥∈[𝑎,𝑏]
Multiplying by (𝑥𝑗 − 𝑥𝑗−1 ) (> 0) and summing up over 𝑗 gives
𝑛
∑
(𝑥𝑗 − 𝑥𝑗−1 )inf 𝑓 (𝑥) ≤
𝑥∈[𝑎,𝑏]
(𝑥𝑗 − 𝑥𝑗−1 ) inf 𝑓 (𝑥) ≤
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
(𝑥𝑗 − 𝑥𝑗−1 ) sup 𝑓 (𝑥) ≤
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
(𝑥𝑗 − 𝑥𝑗−1 )sup 𝑓 (𝑥)
𝑥∈[𝑎,𝑏]
=
𝑗=1
=
𝑗=1
𝑛
∑
=
𝑗=1
𝑛
∑
=
𝑗=1
𝑛
∑
(𝑏 − 𝑎) inf 𝑓 (𝑥)
𝐿(𝑓 , 𝑃 )
𝑈 (𝑓 , 𝑃 )
(𝑏 − 𝑎) sup 𝑓 (𝑥)
𝑥∈[𝑎,𝑏]
𝑥∈[𝑎,𝑏]
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Worked Example 6.2.2. Let 𝑓 (𝑥) = 𝑥2 and 𝑃 = {1 < 4∕3 < 5∕3 < 2}. With 𝑎 = 1 and 𝑏 = 2 we have
inf 𝑓 = 1,
[1,2]
sup 𝑓 = 4,
𝑏 − 𝑎 = 1.
[1,2]
Thus we conclude
1 ≤ 𝐿(𝑓 , 𝑃 ) ≤ 𝑈 (𝑓 , 𝑃 ) ≤ 4.
Note to lecturers: This calculation may be skipped. In fact, since 𝑓 is increasing on [1, 2], infimum and supremum
on each subdivision interval are taken at the left and right endpoint, respectively. Therefore
1 16 1 25 1
77
⋅
+ ⋅
+ ⋅4=
3 9
3 9
3
27
1 16 1 25 50
1
+ ⋅
=
,
𝐿(𝑓 , 𝑃 ) = ⋅ 1 + ⋅
3
3 9
3 9
27
𝑈 (𝑓 , 𝑃 ) =
i. e.
≤
≤
4
=
77
27
=
≤
=
50
27
=
1
(𝑏 − 𝑎) inf 𝑓
𝐿(𝑓 , 𝑃 )
𝑈 (𝑓 , 𝑃 )
(𝑏 − 𝑎) sup 𝑓
[1,2]
[1,2]
Tutorial 6.2.1. 1. Let 𝑓 (𝑥) = 𝑥 − 21 and 𝑃 = {0 < 1∕2 < 1 < 3∕2 < 2}. Find 𝑈 (𝑓 , 𝑃 ) and 𝐿(𝑓 , 𝑃 ) and compare
2
them with ∫0 𝑓 (𝑥) 𝑑𝑥.
2. Let 𝑓 (𝑥) = 𝑥 and 𝑃𝑁 = {−1 < −1 + 𝑁1 < −1 + 𝑁2 < ⋯ < −1 + 3𝑁
= 2}, 𝑁 ∈ ℕ∗ .
𝑁
]}
{
[
]}
{
[
𝑗−1
𝑗
𝑗
and
inf
𝑓
(𝑥)
∣
𝑥
∈
−1
+
.
,
−1
+
,
−1
+
(i) Find sup 𝑓 (𝑥) ∣ 𝑥 ∈ −1 + 𝑗−1
𝑁
𝑁
𝑁
𝑁
(ii) Hence compute 𝑈 (𝑓 , 𝑃𝑁 ) and 𝐿(𝑓 , 𝑃𝑁 ).
(iii) Find lim 𝑈 (𝑓 , 𝑃𝑁 ) and lim 𝐿(𝑓 , 𝑃𝑁 ).
𝑁→∞
𝑁→∞
2
(iv) Compare these limits with ∫−1 𝑥 𝑑𝑥.
⎧1
⎪
3. Verify Lemma 6.4 for the function sgn(𝑥) ∶= ⎨ 0
⎪ −1
⎩
𝑃 = {−2 < −1 < 0 < 1 < 2 < 3}.
6.3
if 𝑥 > 0
if 𝑥 = 0
on the interval [−2, 3] with partition
if 𝑥 < 0
Refinements
Definition 6.3. Let 𝑃 = {𝑎 = 𝑥0 < 𝑥1 < ⋯ < 𝑥𝑚 = 𝑏} and 𝑄 = {𝑎 = 𝑦0 < 𝑦1 < ⋯ < 𝑦𝑛 = 𝑏} be partitions of
[𝑎, 𝑏]. Then 𝑄 is said to be a refinement of 𝑃 if 𝑃 ⊂ 𝑄 i.e. {𝑥0 , 𝑥1 , … , 𝑥𝑚 } ⊂ {𝑦0 , 𝑦1 , … , 𝑦𝑛 }.
𝑄 is called a refinement because 𝑄 chops [𝑎, 𝑏] into “smaller” pieces than 𝑃 did. Or you can think of 𝑃 as first
cutting [𝑎, 𝑏] into pieces and then 𝑄 additionally chops some of those pieces into smaller pieces.
Worked Example 6.3.1. 1. 𝑃 = {0 < 1∕2 < 1} and 𝑄 = {0 < 1∕3 < 2∕3 < 1} are both partitions of [0, 1] but
𝑄 is not a refinement of 𝑃 as 12 belongs to the partition 𝑃 but not to 𝑄.
2. 𝑃 = {0 < 1∕2 < 1} and 𝑄 = {0 < 1∕4 < 2∕4 < 3∕4 < 1} are both partitions of [0, 1] and 𝑄 is a refinement of
𝑃.
𝑛
3. Let 𝑃𝑛 = {0 < 21𝑛 < 22𝑛 < ⋯ < 22𝑛 }. Then 𝑃𝑛 is a partition of [0, 1] for each 𝑛 ∈ ℕ and 𝑃𝑛 ⊂ 𝑃𝑛+1 since
𝑗
= 22𝑗
𝑛+1 .
2𝑛
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Lemma 6.5. Let 𝑓 be a bounded function on [𝑎, 𝑏] and let 𝑃 and 𝑅 be partitions of [𝑎, 𝑏] with 𝑅 a refinement of
𝑃 . Then
𝐿(𝑓 , 𝑃 ) ≤ 𝐿(𝑓 , 𝑅) ≤ 𝑈 (𝑓 , 𝑅) ≤ 𝑈 (𝑓 , 𝑃 ).
Proof. Since 𝑅 can be formed by adding a finite number of partition points to 𝑃 , proceeding inductively it suffices
to consider the case where 𝑅 is formed by introducing one extra point to 𝑃 :
Suppose 𝑃 = {𝑥0 < 𝑥1 < ⋯ < 𝑥𝑁 } and 𝑅 = {𝑥0 < 𝑥1 < ⋯ < 𝑥𝑘−1 < 𝑦 < 𝑥𝑘 < ⋯ < 𝑥𝑁 }. We have
𝑈 (𝑓 , 𝑃 ) =
=
𝑛
∑
𝑗=1
𝑛
∑
(𝑥𝑗 − 𝑥𝑗−1 )
(𝑥𝑗 − 𝑥𝑗−1 )
𝑗=1
𝑗≠𝑘
sup
𝑓 (𝑥)
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
sup
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
𝑓 (𝑥) + (𝑥𝑘 − 𝑥𝑘−1 )
sup
𝑓 (𝑥)
𝑥∈[𝑥𝑘−1 ,𝑥𝑘 ]
But 𝑄 = {𝑥𝑘−1 < 𝑦 < 𝑥𝑘 } is a partition of [𝑥𝑘−1 , 𝑥𝑘 ], and Lemma 6.4 gives
(𝑦 − 𝑥𝑘−1 )
sup
𝑥∈[𝑥𝑘−1 ,𝑦]
𝑓 (𝑥) + (𝑥𝑘 − 𝑦) sup 𝑓 (𝑥) ≤ (𝑥𝑘 − 𝑥𝑘−1 )
𝑥∈[𝑦,𝑥𝑘 ]
sup
𝑓 (𝑥).
𝑥∈[𝑥𝑘−1 ,𝑥𝑘 ]
Hence
𝑈 (𝑓 , 𝑃 ) ≥
𝑛
∑
(𝑥𝑗 − 𝑥𝑗−1 )
𝑗=1
𝑗≠𝑘
sup
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
𝑓 (𝑥) + (𝑦 − 𝑥𝑘−1 )
sup
𝑥∈[𝑥𝑘−1 ,𝑦]
𝑓 (𝑥) + (𝑥𝑘 − 𝑦) sup 𝑓 (𝑥)
𝑥∈[𝑦,𝑥𝑘 ]
= 𝑈 (𝑓 , 𝑅).
The inequality 𝐿(𝑓 , 𝑃 ) ≤ 𝐿(𝑓 , 𝑅) follows in a similar way, and 𝐿(𝑓 , 𝑅) ≤ 𝑈 (𝑓 , 𝑅) has been shown in Lemma
6.4.
Definition 6.4. If 𝑃 and 𝑄 are partitions of [𝑎, 𝑏], then 𝑃 ∪ 𝑄 (with elements listed in increasing order and without
repetitions) is called the common refinement of 𝑃 and 𝑄.
Worked Example 6.3.2. Let 𝑃 = {−1 < 0 < 1} and 𝑄 = {−1 < −1∕3 < 1∕3 < 1}. Then the common
refinement of 𝑃 and 𝑄 is 𝑃 ∪ 𝑄 = {−1 < −1∕3 < 0 < 1∕3 < 1}.
Lemma 6.6. If 𝑃 and 𝑄 are partitions of [𝑎, 𝑏] and 𝑓 is a bounded function on [𝑎, 𝑏], then
𝐿(𝑓 , 𝑄) ≤ 𝑈 (𝑓 , 𝑃 ).
Proof. Since
𝑃 ⊂ 𝑃 ∪ 𝑄 and
𝑄 ⊂ 𝑃 ∪ 𝑄,
Lemma 6.5 gives
𝐿(𝑓 , 𝑄) ≤ 𝐿(𝑓 , 𝑃 ∪ 𝑄),
𝑈 (𝑓 , 𝑃 ∪ 𝑄) ≤ 𝑈 (𝑓 , 𝑃 ) and
𝐿(𝑓 , 𝑃 ∪ 𝑄) ≤ 𝑈 (𝑓 , 𝑃 ∪ 𝑄).
Hence
𝐿(𝑓 , 𝑄) ≤ 𝐿(𝑓 , 𝑃 ∪ 𝑄) ≤ 𝑈 (𝑓 , 𝑃 ∪ 𝑄) ≤ 𝑈 (𝑓 , 𝑃 ).
Tutorial 6.3.1. 1. Let 𝑃𝑛 = {0 < 1𝑛 < 2𝑛 < ⋯ < 𝑛𝑛 }. Then 𝑃𝑛 is a partition of [0, 1] for each 𝑛 ∈ ℕ. Show that
𝑃𝑛+1 is not a refinement of 𝑃𝑛 for 𝑛 ≥ 2.
2. Let 𝑓 (𝑥) = 𝑥3 + 𝑥2 , 𝑃 = {−1 < 0 < 1}, 𝑄 = {−1 < −1∕2 < 0 < 1}. Compute 𝐿(𝑓 , 𝑃 ), 𝐿(𝑓 , 𝑄), 𝑈 (𝑓 , 𝑃 ),
𝑈 (𝑓 , 𝑄) and compare their values.
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The Riemann Integral
Definition 6.5. Let 𝑓 be a bounded function on the interval [𝑎, 𝑏].
(a) We define the lower integral of 𝑓 over [𝑎, 𝑏] as
𝑏
𝑓 (𝑡) 𝑑𝑡 ∶= sup{𝐿(𝑓 , 𝑃 ) ∣ 𝑃 a partition of [𝑎, 𝑏]}.
∫
𝑎
(b) We define the upper integral of 𝑓 over [𝑎, 𝑏] as
𝑏
∫
𝑓 (𝑡) 𝑑𝑡 ∶= inf {𝑈 (𝑓 , 𝑃 ) ∣ 𝑃 a partition of [𝑎, 𝑏]}.
𝑎
(c) We say that 𝑓 is Riemann integrable on [𝑎, 𝑏] if
𝑏
𝑏
𝑓 (𝑡) 𝑑𝑡 =
∫
𝑎
𝑓 (𝑡) 𝑑𝑡,
∫
𝑎
𝑏
in which case the common value is denoted by
∫
𝑓 (𝑡) 𝑑𝑡 and is called the Riemann integral of 𝑓 over [𝑎, 𝑏].
𝑎
Note. By the properties of suprema and infima, see Lemmas 6.6 and 6.3,
𝑏
𝑏
𝑓 (𝑡) 𝑑𝑡 ≤
∫
𝑎
𝑓 (𝑡) 𝑑𝑡.
∫
𝑎
Hence 𝑓 is Riemann integrable if and only if
𝑏
∫
𝑎
𝑏
𝑓 (𝑡) 𝑑𝑡 ≤
∫
𝑓 (𝑡) 𝑑𝑡.
𝑎
Theorem 6.7 (Integrability Condition). A bounded function 𝑓 on the interval [𝑎, 𝑏] is Riemann integrable if and
only if for each 𝜀 > 0 there exists a partition 𝑃 of [𝑎, 𝑏] such that
𝑈 (𝑓 , 𝑃 ) − 𝐿(𝑓 , 𝑃 ) < 𝜀.
Proof. ⇒: Let 𝑓 be Riemann integrable on [𝑎, 𝑏] and let 𝜀 > 0. Then there is a partition 𝑃 of [𝑎, 𝑏] such that
𝑏
∫
𝑏
𝑓 (𝑡) 𝑑𝑡 ≤ 𝑈 (𝑓 , 𝑃 ) <
𝑎
∫
𝑓 (𝑡) 𝑑𝑡 +
𝜀
2
𝑎
and there is a partition 𝑄 of [𝑎, 𝑏] such that
𝑏
𝑏
𝜀
𝑓 (𝑡) 𝑑𝑡 − < 𝐿(𝑓 , 𝑄) ≤ 𝑓 (𝑡) 𝑑𝑡.
∫
∫
2
𝑎
𝑎
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Let 𝑅 = 𝑃 ∪ 𝑄. Then, in view of Lemma 6.5,
𝑏
𝑏
𝜀
𝜀
𝑓 (𝑡) 𝑑𝑡 − = 𝑓 (𝑡) 𝑑𝑡 − < 𝐿(𝑓 , 𝑄) ≤ 𝐿(𝑓 , 𝑅)
∫
2 ∫
2
𝑎
𝑎
𝑏
𝑏
𝜀
𝜀
𝑓 (𝑡) 𝑑𝑡 + .
≤ 𝑈 (𝑓 , 𝑅) ≤ 𝑈 (𝑓 , 𝑃 ) < 𝑓 (𝑡) 𝑑𝑡 + =
∫
2 ∫
2
𝑎
𝑎
In particular,
𝑏
𝑈 (𝑓 , 𝑅) <
𝜀
𝑓 (𝑡) 𝑑𝑡 + ,
2
∫
𝑎
𝑏
−𝐿(𝑓 , 𝑅) < −
𝜀
𝑓 (𝑡) 𝑑𝑡 + .
2
∫
𝑎
Summing up gives
𝑈 (𝑓 , 𝑅) − 𝐿(𝑓 , 𝑅) < 𝜀.
⇐: Suppose that there are 𝜀 > 0 and a partition 𝑃𝜀 of [𝑎, 𝑏] such that
𝑈 (𝑓 , 𝑃𝜀 ) − 𝐿(𝑓 , 𝑃𝜀 ) < 𝜀.
Then from
𝑏
𝑓 (𝑡) 𝑑𝑡 ≥ 𝐿(𝑓 , 𝑃𝜀 ),
∫
𝑎
𝑏
∫
𝑓 (𝑡) 𝑑𝑡 ≤ 𝑈 (𝑓 , 𝑃𝜀 )
𝑎
we get
𝑏
∫
𝑏
𝑓 (𝑡) 𝑑𝑡 −
𝑎
𝑓 (𝑡) 𝑑𝑡 ≤ 𝑈 (𝑓 , 𝑃𝜀 ) − 𝐿(𝑓 , 𝑃𝜀 ) < 𝜀.
∫
𝑎
Hence
𝑏
∀𝜀 > 0
0≤
∫
𝑏
𝑓 (𝑡) 𝑑𝑡 −
𝑎
∫
𝑓 (𝑡) 𝑑𝑡 ≤ 𝜀,
𝑎
giving
𝑏
∫
𝑎
𝑏
𝑓 (𝑡) 𝑑𝑡 =
∫
𝑓 (𝑡) 𝑑𝑡.
𝑎
Theorem 6.8. If 𝑓 is an increasing function on the interval [𝑎, 𝑏], then 𝑓 is Riemann integrable on [𝑎, 𝑏].
Proof. By assumption, 𝑓 (𝑎) ≤ 𝑓 (𝑥) ≤ 𝑓 (𝑏) for all 𝑥 ∈ [𝑎, 𝑏]. Hence 𝑓 is bounded.
Let 𝜀 > 0 and choose 𝑁 ∈ ℕ such that
𝑓 (𝑏) − 𝑓 (𝑎)
(𝑏 − 𝑎)
< 𝜀.
𝑁
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Consider the partition
𝑃 =
{
}
2(𝑏 − 𝑎)
(𝑁 − 1)(𝑏 − 𝑎)
𝑁(𝑏 − 𝑎)
𝑏−𝑎
𝑎<𝑎+
<𝑎+
<⋯<𝑎+
<𝑎+
=𝑏 .
𝑁
𝑁
𝑁
𝑁
Observing that over each subinterval of [𝑎, 𝑏] the increasing function 𝑓 takes its minimum at the left endpoint and
its maximum at the right endpoint of the interval, it follows that
{
]}
(
)
[
𝑁
|
∑
𝑗(𝑏 − 𝑎)
𝑗(𝑏 − 𝑎)
(𝑗 − 1)(𝑏 − 𝑎)
𝑏−𝑎
|
𝑈 (𝑓 , 𝑃 ) =
sup 𝑓 (𝑥) | 𝑥 ∈ 𝑎 +
,𝑎 +
=
𝑓 𝑎+
,
|
𝑁
𝑁
𝑁
𝑁
𝑁
|
𝑗=1
𝑗=1
{
]}
(
)
[
𝑁
𝑁
|
∑
∑
𝑗(𝑏 − 𝑎)
(𝑗 − 1)(𝑏 − 𝑎)
(𝑗 − 1)(𝑏 − 𝑎)
𝑏−𝑎
𝑏−𝑎
|
𝐿(𝑓 , 𝑃 ) =
inf 𝑓 (𝑥) | 𝑥 ∈ 𝑎 +
,𝑎 +
=
𝑓 𝑎+
.
|
𝑁
𝑁
𝑁
𝑁
𝑁
|
𝑗=1
𝑗=1
𝑁
∑
𝑏−𝑎
Hence
𝑈 (𝑓 , 𝑃 ) − 𝐿(𝑓 , 𝑝) =
𝑁
∑
𝑏−𝑎
𝑁
𝑗=1
=
(
𝑎+
𝑓
𝑗(𝑏 − 𝑎)
𝑁
)
−
𝑁−1
∑
𝑏−𝑎
𝑓
𝑁
𝑗=0
(
𝑎+
𝑗(𝑏 − 𝑎)
𝑁
)
𝑏−𝑎
(𝑓 (𝑏) − 𝑓 (𝑎)) < 𝜀.
𝑁
Hence 𝑓 is integrable by Theorem 6.7.
1
Worked Example 6.4.1. 1. Let 𝑓 (𝑥) = 𝑥. Then 𝑓 is Riemann integrable on [0, 1] and ∫0 𝑓 (𝑡) 𝑑𝑡 =
this let 𝑃𝑛 = {0 < 1𝑛 < 2𝑛 < ⋯ < 𝑛𝑛 }. Then
𝐿(𝑓 , 𝑃𝑛 ) =
𝑛
∑
1 𝑗−1
𝑗=1
𝑛
𝑛
and
𝑈 (𝑓 , 𝑃𝑛 ) =
𝑛
∑
1 𝑗
𝑗=1
𝑛 𝑛
=
=
1
. To see
2
(
)
1 (𝑛 − 1)𝑛 1
1
=
1−
2
2
𝑛
𝑛2
(
)
1 𝑛(𝑛 + 1) 1
1
=
1+
.
2
2
𝑛
𝑛2
1
, which can be made arbitrarily small but taking 𝑛 sufficiently large. Thus 𝑓 is
𝑛
Riemann integrable. Also the integral is sandwiched between 𝐿(𝑓 , 𝑃𝑛 ) and 𝑈 (𝑓 , 𝑃𝑛 ), both of which converge to
1
1
1
, giving ∫0 𝑓 (𝑡) 𝑑𝑡 = .
2
2
{
0 if 𝑥 ∉ ℚ
. Then 𝑓 is not Riemann integrable on [0, 1], since for each partition 𝑃 of [0, 1],
2. Let 𝑓 (𝑥) =
1 if 𝑥 ∈ ℚ
𝑈 (𝑓 , 𝑃 ) = 1, while 𝐿(𝑓 , 𝑃 ) = 0.
Thus 𝑈 (𝑓 , 𝑃𝑛 ) − 𝐿(𝑓 , 𝑃𝑛 ) =
2
2
and 𝐿(sgn, 𝑃𝑛 ) = − .
𝑛
𝑛
4
4
Thus 𝑈 (sgn, 𝑃𝑛 ) − 𝐿(sgn, 𝑃𝑛 ) = → 0 as 𝑛 → ∞. For each 𝜀 > 0 there exists 𝑛 ∈ ℕ such that < 𝜀. Hence sgn
𝑛
𝑛
1
is Riemann integrable on [−1, 1]. Finally, lim 𝑈 (sgn, 𝑃𝑛 ) = lim 𝐿(sgn, 𝑃𝑛 ) = 0 gives ∫−1 sgn(𝑥) 𝑑𝑥 = 0.
3. Let 𝑓 (𝑥) = sgn(𝑥) on [−1, 1]. Let 𝑃𝑛 ∶= {−1 < − 1𝑛 < 1𝑛 < 1} then 𝑈 (sgn, 𝑃𝑛 ) =
𝑛→∞
𝑛→∞
Tutorial 6.4.1. Let 𝑓 ∶ [𝑎, 𝑏] → ℝ be bounded. For 𝑛 ∈ ℕ∗ let 𝑃𝑛 ∈ (𝑎, 𝑏) such that lim 𝐿(𝑓 , 𝑃𝑛 ) =
𝑛→∞
𝑏
lim 𝑈 (𝑓 , 𝑃𝑛 ). Show that 𝑓 is Riemann integrable on [𝑎, 𝑏] and that ∫𝑎 𝑓 (𝑥) 𝑑𝑥 = lim 𝐿(𝑓 , 𝑃𝑛 ) = lim 𝑈 (𝑓 , 𝑃𝑛 ).
𝑛→∞
𝑛→∞
1
Tutorial 6.4.2. Show that 𝑓 (𝑥) = 𝑥2 is Riemann integrable on [0, 1] and that ∫0 𝑓 (𝑡) 𝑑𝑡 =
𝑛→∞
1
.
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Properties of the Integral
Theorem 6.9 (Homogeneity). If 𝑓 is a bounded Riemann integrable function on the interval [𝑎, 𝑏] and if 𝑐 ∈ ℝ,
then 𝑐𝑓 is Riemann integrable and
𝑏
𝑏
𝑐𝑓 (𝑡) 𝑑𝑡 = 𝑐
∫
𝑎
∫
𝑓 (𝑡) 𝑑𝑡.
𝑎
Proof. Note that whenever 𝑓 is bounded, then also 𝑐𝑓 is bounded.
Case I: 𝑐 ≥ 0.
Let 𝑃 = {𝑎 = 𝑥0 < 𝑥1 < ⋯ < 𝑥𝑁 = 𝑏} be a partition of [𝑎, 𝑏]. Then, by Lemma 6.1,
𝑈 (𝑐𝑓 , 𝑃 ) =
𝑁
∑
(𝑥𝑗 − 𝑥𝑗−1 ) sup{𝑐𝑓 (𝑥) ∣ 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ]}
𝑗=1
=
𝑁
∑
(𝑥𝑗 − 𝑥𝑗−1 )𝑐 sup{𝑓 (𝑥) ∣ 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ]}
𝑗=1
=𝑐
𝑁
∑
(𝑥𝑗 − 𝑥𝑗−1 ) sup{𝑓 (𝑥) ∣ 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ]}
𝑗=1
= 𝑐𝑈 (𝑓 , 𝑃 ).
Thus
𝑏
𝑐𝑓 (𝑡) 𝑑𝑡 = inf {𝑈 (𝑐𝑓 , 𝑃 ) ∣ 𝑃 ∈ (𝑎, 𝑏)}
∫
𝑎
= inf {𝑐𝑈 (𝑓 , 𝑃 ) ∣ 𝑃 ∈ (𝑎, 𝑏)}
= 𝑐 inf {𝑈 (𝑓 , 𝑃 ) ∣ 𝑃 ∈ (𝑎, 𝑏)}
𝑏
=𝑐
𝑓 (𝑡) 𝑑𝑡.
∫
𝑎
Interchanging sup and inf we get 𝐿(𝑐𝑓 , 𝑃 ) = 𝑐𝐿(𝑓 , 𝑃 ) and then
𝑏
∫
𝑏
𝑐𝑓 (𝑡) 𝑑𝑡 = 𝑐
𝑎
𝑏
But
∫
𝑎
∫
𝑓 (𝑡) 𝑑𝑡.
𝑎
𝑏
𝑓 (𝑡) 𝑑𝑡 =
∫
𝑓 (𝑡) 𝑑𝑡 now gives
𝑎
𝑏
∫
𝑏
𝑐𝑓 (𝑡) 𝑑𝑡 = 𝑐
𝑎
∫
𝑎
𝑏
𝑓 (𝑡) 𝑑𝑡 = 𝑐
∫
∫
𝑎
𝑓 (𝑡) 𝑑𝑡 = 𝑐
𝑎
𝑏
𝑐𝑓 (𝑡) 𝑑𝑡 = 𝑐
∫
𝑎
𝑏
Hence 𝑐𝑓 is Riemann integrable and
𝑏
∫
𝑎
𝑓 (𝑡) 𝑑𝑡.
𝑏
𝑓 (𝑡) 𝑑𝑡 =
∫
𝑎
𝑐𝑓 (𝑡) 𝑑𝑡.
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Case II: 𝑐 = −1. Let 𝑃 = {𝑎 = 𝑥0 < 𝑥1 < ⋯ < 𝑥𝑁 = 𝑏} be a partition of [𝑎, 𝑏]. Then, using the results of Section
6.1,
𝑈 (−𝑓 , 𝑃 ) =
𝑁
∑
(𝑥𝑗 − 𝑥𝑗−1 ) sup{−𝑓 (𝑥) ∣ 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ]}
𝑗=1
=−
𝑁
∑
(𝑥𝑗 − 𝑥𝑗−1 ) inf {𝑓 (𝑥) ∣ 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ]}
𝑗=1
= −𝐿(𝑓 , 𝑃 ).
Thus
𝑏
∫
(−𝑓 (𝑡)) 𝑑𝑡 = inf {𝑈 (−𝑓 , 𝑃 ) ∣ 𝑃 ∈ (𝑎, 𝑏)}
𝑎
= inf {−𝐿(𝑓 , 𝑃 ) ∣ 𝑃 ∈ (𝑎, 𝑏)}
= − sup{𝐿(𝑓 , 𝑃 ) ∣ 𝑃 ∈ (𝑎, 𝑏)}
𝑏
= − 𝑓 (𝑡) 𝑑𝑡.
∫
𝑎
Since so far we have only used that 𝑓 and −𝑓 are bounded, we can replace 𝑓 by −𝑓 and obtain
𝑏
∫
𝑏
𝑓 (𝑡) 𝑑𝑡 =
∫
𝑎
𝑏
(−(−𝑓 (𝑡))) 𝑑𝑡 = − (−𝑓 (𝑡)) 𝑑𝑡,
∫
𝑎
𝑎
and therefore
𝑏
∫
𝑏
(−𝑓 (𝑡)) 𝑑𝑡 = − 𝑓 (𝑡) 𝑑𝑡.
∫
𝑎
𝑎
Thus
𝑏
∫
𝑎
𝑏
𝑏
𝑏
𝑏
(−𝑓 (𝑡)) 𝑑𝑡 = − 𝑓 (𝑡) 𝑑𝑡 = − 𝑓 (𝑡) 𝑑𝑡 = − 𝑓 (𝑡) 𝑑𝑡 = (−𝑓 (𝑡)) 𝑑𝑡.
∫
∫
∫
∫
𝑎
𝑎
𝑎
This shows that −𝑓 is Riemann integrable and that
𝑏
∫
𝑏
(−𝑓 (𝑡)) 𝑑𝑡 = −
𝑎
𝑓 (𝑡) 𝑑𝑡.
∫
𝑎
Case III: 𝑐 < 0.
By Case I, (−𝑐)𝑓 is Riemann integrable, and
𝑏
∫
𝑏
(−𝑐)𝑓 (𝑡) 𝑑𝑡 = (−𝑐)
𝑎
∫
𝑓 (𝑡) 𝑑𝑡.
𝑎
Then, by Case II, −(−𝑐)𝑓 is Riemann integrable, and
𝑏
∫
𝑎
𝑏
(−(−𝑐)𝑓 (𝑡)) 𝑑𝑡 = −
∫
𝑎
(−𝑐)𝑓 (𝑡) 𝑑𝑡.
𝑎
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Thus 𝑐𝑓 is Riemann integrable, and
𝑏
∫
𝑏
𝑐𝑓 (𝑡) 𝑑𝑡 = −
∫
𝑎
𝑏
(−𝑐)𝑓 (𝑡) 𝑑𝑡 = −(−𝑐)
∫
𝑎
𝑏
𝑓 (𝑡) 𝑑𝑡 = 𝑐
𝑎
∫
𝑓 (𝑡) 𝑑𝑡.
𝑎
Lemma 6.10. If 𝑓 and 𝑔 are bounded functions on [𝑎, 𝑏] and 𝑃 is a partition of [𝑎, 𝑏], then
𝑈 (𝑓 + 𝑔, 𝑃 ) ≤ 𝑈 (𝑓 , 𝑃 ) + 𝑈 (𝑔, 𝑃 ),
𝐿(𝑓 + 𝑔, 𝑃 ) ≥ 𝐿(𝑓 , 𝑃 ) + 𝐿(𝑔, 𝑃 ).
Proof. Let 𝑃 = {𝑥0 < 𝑥1 < ⋯ < 𝑥𝑁 } be a partion of [𝑎, 𝑏]. For each 𝑗 ∈ {1, … , 𝑁} and 𝑡 ∈ [𝑥𝑗−1 , 𝑥𝑗 ] we have
𝑓 (𝑡) + 𝑔(𝑡) ≤ sup{𝑓 (𝑥) ∣ 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ]} + sup{𝑔(𝑥) ∣ 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ]},
whence
sup{𝑓 (𝑥) + 𝑔(𝑥) ∣ 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ]} ≤ sup{𝑓 (𝑥) ∣ 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ]} + sup{𝑔(𝑥) ∣ 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ]}.
Multiplying with (𝑥𝑗 − 𝑥𝑗−1 ) and summing up over 𝑗 gives
𝑈 (𝑓 + 𝑔, 𝑃 ) ≤ 𝑈 (𝑓 , 𝑃 ) + 𝑈 (𝑔, 𝑃 ).
𝐿(𝑓 +𝑔, 𝑃 ) ≥ 𝐿(𝑓 , 𝑃 )+𝐿(𝑔, 𝑃 ) can be shown similarly, or we may use the estimate for upper sums and 𝐿(𝑓 , 𝑃 ) =
−𝑈 (−𝑓 , 𝑃 ).
Theorem 6.11 (Additivity). If 𝑓 and 𝑔 are bounded Riemann integrable functions on the interval [𝑎, 𝑏], then 𝑓 + 𝑔
is Riemann integrable and
𝑏
∫
𝑏
(𝑓 (𝑡) + 𝑔(𝑡)) 𝑑𝑡 =
𝑎
𝑏
𝑓 (𝑡) 𝑑𝑡 +
∫
𝑎
𝑔(𝑡) 𝑑𝑡.
∫
𝑎
Proof. Let 𝜀 > 0 and choose 𝑃1 , 𝑃2 ∈ (𝑎, 𝑏) such that
𝑏
∫
𝜀
𝑓 (𝑡) 𝑑𝑡 ≥ 𝑈 (𝑓 , 𝑃1 ) − ,
2
𝑎
𝑏
∫
𝜀
𝑔(𝑡) 𝑑𝑡 ≥ 𝑈 (𝑔, 𝑃2 ) − .
2
𝑎
Let 𝑃 = 𝑃1 ∪ 𝑃2 . By Lemmas 6.5 and 6.10,
𝑈 (𝑓 + 𝑔, 𝑃 ) ≤ 𝑈 (𝑓 , 𝑃 ) + 𝑈 (𝑔, 𝑃 ) ≤ 𝑈 (𝑓 , 𝑃1 ) + 𝑈 (𝑔, 𝑃2 ).
Thus
𝑏
∫
(𝑓 (𝑡) + 𝑔(𝑡)) 𝑑𝑡 ≤ 𝑈 (𝑓 + 𝑔, 𝑃 ) ≤ 𝑈 (𝑓 , 𝑃1 ) + 𝑈 (𝑔, 𝑃2 )
𝑎
𝑏
≤
∫
𝑏
𝑓 (𝑡) 𝑑𝑡 +
𝑎
∫
𝑔(𝑡) 𝑑𝑡 + 𝜀.
𝑎
Since 𝜀 > 0 was arbitrary, it follows that
𝑏
∫
𝑎
𝑏
(𝑓 (𝑡) + 𝑔(𝑡)) 𝑑𝑡 ≤
∫
𝑎
𝑏
𝑓 (𝑡) 𝑑𝑡 +
∫
𝑎
𝑔(𝑡) 𝑑𝑡.
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Similarly,
𝑏
𝑏
(𝑓 (𝑡) + 𝑔(𝑡)) 𝑑𝑡 ≥
∫
𝑓 (𝑡) 𝑑𝑡 +
∫
𝑎
𝑏
𝑔(𝑡) 𝑑𝑡.
∫
𝑎
𝑎
But since 𝑓 and 𝑔 are integrable over [𝑎, 𝑏]
𝑏
𝑏
𝑓 (𝑡) 𝑑𝑡 +
∫
𝑎
𝑏
𝑔(𝑡) 𝑑𝑡 =
∫
𝑎
𝑏
𝑓 (𝑡) 𝑑𝑡 +
∫
𝑔(𝑡) 𝑑𝑡 =
∫
𝑎
𝑏
𝑓 (𝑡) 𝑑𝑡 +
∫
𝑎
𝑏
𝑎
∫
𝑔(𝑡) 𝑑𝑡,
𝑎
and therefore
𝑏
∫
𝑏
(𝑓 (𝑡) + 𝑔(𝑡)) 𝑑𝑡 ≤
𝑎
∫
𝑏
𝑓 (𝑡) 𝑑𝑡 +
∫
𝑎
𝑏
𝑔(𝑡) 𝑑𝑡 ≤
(𝑓 (𝑡) + 𝑔(𝑡)) 𝑑𝑡,
∫
𝑎
𝑎
𝑏
𝑏
which shows that 𝑓 + 𝑔 is integrable, and
𝑏
(𝑓 (𝑡) 𝑑𝑡 + 𝑔(𝑡)) 𝑑𝑡 =
∫
𝑎
∫
𝑓 (𝑡) 𝑑𝑡 +
𝑎
∫
𝑔(𝑡) 𝑑𝑡.
𝑎
Theorem 6.12. If 𝑓 is a bounded Riemann integrable function on the interval [𝑎, 𝑏] and on the interval [𝑏, 𝑐], then
𝑓 is Riemann integrable on [𝑎, 𝑐] and
𝑐
∫
𝑏
𝑓 (𝑥) 𝑑𝑥 =
𝑎
∫
𝑐
𝑓 (𝑥) 𝑑𝑥 +
𝑎
∫
𝑓 (𝑥) 𝑑𝑥.
𝑏
Proof. Let 𝑃 be a partition of [𝑎, 𝑏] and 𝑄 be a partition of [𝑏, 𝑐]. Writing
𝑃 = {𝑎 = 𝑥0 < 𝑥1 < ⋯ < 𝑥𝑚 = 𝑏},
𝑄 = {𝑏 = 𝑥𝑚 < 𝑥𝑚+1 < ⋯ < 𝑥𝑛 = 𝑐},
we have that 𝑅 = {𝑎 = 𝑥0 < 𝑥1 < ⋯ < 𝑥𝑛 = 𝑐} is a partition of [𝑎, 𝑐] and
𝑈 (𝑓 , 𝑅) =
=
𝑛
∑
𝑗=1
𝑚
∑
(𝑥𝑗 − 𝑥𝑗−1 ) sup{𝑓 (𝑥) ∣ 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ]}
(𝑥𝑗 − 𝑥𝑗−1 ) sup{𝑓 (𝑥) ∣ 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ]} +
𝑗=1
𝑛
∑
(𝑥𝑗 − 𝑥𝑗−1 ) sup{𝑓 (𝑥) ∣ 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ]}
𝑗=𝑚+1
= 𝑈 (𝑓 , 𝑃 ) + 𝑈 (𝑓 , 𝑄).
Hence, by Lemmas 6.1 and 6.2,
𝑐
∫
𝑓 (𝑥) 𝑑𝑥 = inf {𝑈 (𝑓 , 𝑆) ∣ 𝑆 ∈ (𝑎, 𝑐)}
𝑎
≤ inf {𝑈 (𝑓 , 𝑅) ∣ 𝑅 ∈ (𝑎, 𝑐), 𝑏 ∈ 𝑅}
= inf {𝑈 (𝑓 , 𝑃 ) + 𝑈 (𝑓 , 𝑄) ∣ 𝑃 ∈ (𝑎, 𝑏), 𝑄 ∈ (𝑏, 𝑐)}
= inf {𝑈 (𝑓 , 𝑃 ) ∣ 𝑃 ∈ (𝑎, 𝑏)} + inf {𝑈 (𝑓 , 𝑄) ∣ 𝑄 ∈ (𝑏, 𝑐)}
𝑏
=
∫
𝑎
𝑐
𝑓 (𝑥) 𝑑𝑥 +
∫
𝑏
𝑓 (𝑥) 𝑑𝑥.
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Similarly
𝑐
∫
𝑓 (𝑥) 𝑑𝑥 = sup{𝐿(𝑓 , 𝑆) ∣ 𝑆 ∈ (𝑎, 𝑐)}
𝑎
≥ sup{𝐿(𝑓 , 𝑃 ) + 𝐿(𝑓 , 𝑄) ∣ 𝑃 ∈ (𝑎, 𝑏), 𝑄 ∈ (𝑏, 𝑐)}
𝑏
=
𝑐
𝑓 (𝑥) 𝑑𝑥 +
∫
𝑎
𝑓 (𝑥) 𝑑𝑥.
∫
𝑏
Since 𝑓 is integrable over [𝑎, 𝑏] and over [𝑏, 𝑐], it follows from above that
𝑐
∫
𝑏
𝑓 (𝑥) 𝑑𝑥 ≤
𝑎
∫
𝑐
𝑓 (𝑥) 𝑑𝑥 +
∫
𝑎
𝑐
𝑓 (𝑥) 𝑑𝑥 ≤
∫
𝑓 (𝑥) 𝑑𝑥.
𝑎
𝑏
Therefore 𝑓 is integrable over [𝑎, 𝑐] and
𝑐
∫
𝑏
𝑓 (𝑥) 𝑑𝑥 =
𝑎
∫
𝑐
𝑓 (𝑥) 𝑑𝑥 +
𝑎
∫
𝑓 (𝑥) 𝑑𝑥.
𝑏
Lemma 6.13. Let 𝑓 be a continuous function on [𝑎, 𝑏] and let 𝜀 > 0. Then there is a partition  = {𝑥0 < 𝑥1 <
⋯ < 𝑥𝑁 } of [𝑎, 𝑏] such that |𝑓 (𝑥) − 𝑓 (𝑥𝑗−1 )| ≤ 𝜀 for all 𝑗 ∈ {1, … , 𝑁} and all 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ].
Proof. Let 𝜀 > 0 and put 𝑥0 = 𝑎. For 𝑗 ∈ ℕ∗ we define numbers 𝑥𝑗 recursively as follows: Let
𝐸𝑗 = {𝑥 ∈ [𝑥𝑗−1 , 𝑏] ∣ |𝑓 (𝑥) − 𝑓 (𝑥𝑗−1 )| ≥ 𝜀}.
Note that 𝐸𝑗 = ∅ if 𝑥𝑗−1 = 𝑏.
If 𝐸𝑗 = ∅, let 𝑥𝑗 = 𝑏.
If 𝐸𝑗 ≠ ∅, let 𝑥𝑗 = inf 𝐸𝑗 .
Claim 1. If 𝑥𝑗−1 < 𝑏, then 𝑥𝑗−1 < 𝑥𝑗 .
If 𝐸𝑗 = ∅, then 𝑥𝑗−1 < 𝑏 = 𝑥𝑗 . Now let 𝐸𝑗 ≠ ∅. Since 𝑓 is continuous at 𝑥𝑗−1 , there is 𝛿 > 0 such that
|𝑓 (𝑥) − 𝑓 (𝑥𝑗−1 )| < 𝜀 whenever 𝑥 ∈ [𝑎, 𝑏] with |𝑥 − 𝑥𝑗−1 | < 𝛿. Hence 𝑥 ∈ 𝐸𝑗 implies 𝑥 ≥ 𝑥𝑗−1 and |𝑥 − 𝑥𝑗−1 | ≥ 𝛿,
and therefore 𝑥 ≥ 𝑥𝑗−1 + 𝛿. Thus 𝑥𝑗−1 + 𝛿 is a lower bound of 𝐸𝑗 and therefore
𝑥𝑗 = inf 𝐸𝑗 ≥ 𝑥𝑗−1 + 𝛿 > 𝑥𝑗−1 .
Claim 2. If 𝑥𝑗−1 < 𝑏, then |𝑓 (𝑥) − 𝑓 (𝑥𝑗−1 )| < 𝜀 for all 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ), |𝑓 (𝑥𝑗 ) − 𝑓 (𝑥𝑗−1 )| ≤ 𝜀, and |𝑓 (𝑥𝑗 ) −
𝑓 (𝑥𝑗−1 )| = 𝜀 if 𝐸𝑗 ≠ ∅.
Indeed, if 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ), then 𝑥 ∈ [𝑥𝑗−1 , 𝑏] and 𝑥 < inf 𝐸𝑗 , and so 𝑥 ∉ 𝐸𝑗 . By definition of 𝐸𝑗 it follows that
|𝑓 (𝑥) − 𝑓 (𝑥𝑗−1 )| < 𝜀, and, since 𝑥𝑗−1 < 𝑥𝑗 by Claim 1, the continuity of 𝑓 and the absolute value function give
|𝑓 (𝑥𝑗 ) − 𝑓 (𝑥𝑗−1 )| = lim− |𝑓 (𝑥) − 𝑓 (𝑥𝑗−1 )| ≤ 𝜀.
𝑥→𝑥𝑗
If 𝐸𝑗 ≠ ∅, then there is a sequence (𝑎𝑘 ) in 𝐸𝑗 such that
lim 𝑎𝑘 = 𝑥𝑗 .
𝑘→∞
Observing |𝑓 (𝑎𝑘 ) − 𝑓 (𝑥𝑗−1 )| ≥ 𝜀, the same continuity argument as above gives
|𝑓 (𝑥𝑗 ) − 𝑓 (𝑥𝑗−1 )| = lim |𝑓 (𝑎𝑘 ) − 𝑓 (𝑥𝑗−1 )| ≥ 𝜀.
𝑘→∞
Claim 3. There is an index 𝑛 such that 𝑥𝑛 = 𝑏.
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Assume that Claim 3 is false. Then, by Claim 1, (𝑥𝑗 ) is an increasing sequence, bounded by 𝑏, which therefore has
a limit, say 𝑥, see Theorem 6.11, and 𝑥 ∈ [𝑎, 𝑏]. Then the continuity of 𝑓 at 𝑥 gives
|𝑓 (𝑥𝑗 ) − 𝑓 (𝑥𝑗−1 )| ≤ |𝑓 (𝑥𝑗 ) − 𝑓 (𝑥)| + |𝑓 (𝑥) − 𝑓 (𝑥𝑗−1 )| → 0 as 𝑗 → ∞.
Hence there is 𝑗 such that |𝑓 (𝑥𝑗 ) − 𝑓 (𝑥𝑗−1 )| < 𝜀, contradicting Claim 2 since 𝑥𝑗 ≠ 𝑏 implies 𝐸𝑗 ≠ ∅.
Putting 𝑁 = min{𝑗 ∈ ℕ ∣ 𝑥𝑗 = 𝑏} completes the proof.
Theorem 6.14. If 𝑓 is a continuous function on [𝑎, 𝑏], then 𝑓 is Riemann integrable on [𝑎, 𝑏].
Proof. From Theorem 3.17, we know that a continuous function on a closed interval [𝑎, 𝑏] is bounded. Let 𝜀 > 0
and let 𝑃 = {𝑥0 < ⋯ < 𝑥𝑁 } be a partition of [𝑎, 𝑏] according to Lemma 6.13. Then, for 𝑗 ∈ {1, … , 𝑁}, with
Lemma 6.1,
sup
𝑓 (𝑥) −
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
inf
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
𝑓 (𝑥) =
sup
[𝑓 (𝑥) − 𝑓 (𝑥𝑗−1 )] −
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
=
sup
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
≤2
[𝑓 (𝑥) − 𝑓 (𝑥𝑗−1 )] +
sup
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
inf
[𝑓 (𝑥) − 𝑓 (𝑥𝑗−1 )]
sup
[𝑓 (𝑥𝑗−1 ) − 𝑓 (𝑥)]
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
|𝑓 (𝑥) − 𝑓 (𝑥𝑗−1 )|
≤ 2𝜀
in view of Lemma 6.13, Claim 2. Hence
𝑁
∑
(𝑥𝑗 − 𝑥𝑗−1 )
𝑈 (𝑓 , 𝑃 ) − 𝐿(𝑓 , 𝑃 ) =
𝑗=1
≤
[
]
sup
𝑓 (𝑥) −
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
inf
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
𝑓 (𝑥)]
𝑁
∑
(𝑥𝑗 − 𝑥𝑗−1 )2𝜀 = 2(𝑏 − 𝑎)𝜀.
𝑗=1
By Theorem 6.7, 𝑓 is Riemann integrable.
Theorem 6.15 (Fundamental Theorem of Calculus). Let 𝑓 be a differentiable function on [𝑎, 𝑏] such that 𝑓 ′ is
continuous on [𝑎, 𝑏]. Then 𝑓 ′ is Riemann integrable on [𝑎, 𝑏] and
𝑏
∫
𝑓 ′ (𝑡) 𝑑𝑡 = 𝑓 (𝑏) − 𝑓 (𝑎).
𝑎
Proof. Since 𝑓 ′ is continuous, 𝑓 ′ is Riemann integrable by the previous theorem. Let 𝑃 = {𝑥
0 < ⋯ < 𝑥𝑁 } be a
partition of [𝑎, 𝑏]. By the First Mean Value Theorem, Theorem 4.5, for each 𝑗 ∈ {1, … , 𝑁} there is 𝑐𝑗 ∈ (𝑥𝑗−1 , 𝑥𝑗 )
such that
𝑓 (𝑥𝑗 ) − 𝑓 (𝑥𝑗−1 ) = (𝑥𝑗 − 𝑥𝑗−1 )𝑓 ′ (𝑐𝑗 ).
Then
𝐿(𝑓 ′ , 𝑃 ) =
𝑁
∑
(𝑥𝑗 − 𝑥𝑗−1 )
𝑗=1
≤
inf
𝑥∈[𝑥𝑗−1 ,𝑥𝑗 ]
𝑓 ′ (𝑥)
𝑁
∑
(𝑥𝑗 − 𝑥𝑗−1 )𝑓 ′ (𝑐𝑗 )
𝑗=1
=
𝑁
∑
𝑓 (𝑥𝑗 ) − 𝑓 (𝑥𝑗−1 ) = 𝑓 (𝑏) − 𝑓 (𝑎).
𝑗=1
Hence
𝑏
∫
𝑎
𝑓 ′ (𝑥) 𝑑𝑥 = sup{𝐿(𝑓 ′ , 𝑃 ) ∣ 𝑃 ∈ (𝑎, 𝑏)} ≤ 𝑓 (𝑏) − 𝑓 (𝑎).
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Similarly,
𝑏
𝑓 ′ (𝑥) 𝑑𝑥 ≥ 𝑓 (𝑏) − 𝑓 (𝑎).
∫
𝑎
Since 𝑓 ′ is Riemann integrable, the upper and lower Riemann integrals are equal, and
𝑏
∫
𝑓 ′ (𝑥) 𝑑𝑥 = 𝑓 (𝑏) − 𝑓 (𝑎)
𝑎
follows.
Tutorial 6.5.1. Let 𝑓 be a bounded Riemann integrable function on [𝑎, 𝑏] and let 𝑃 be a partition of [𝑎, 𝑏]. Then
prove that
𝑏
(𝑏 − 𝑎) inf {𝑓 (𝑥) ∣ 𝑥 ∈ [𝑎, 𝑏]} ≤ 𝐿(𝑓 , 𝑃 ) ≤
∫
𝑓 (𝑡) 𝑑𝑡 ≤ 𝑈 (𝑓 , 𝑃 ) ≤ (𝑏 − 𝑎) sup{𝑓 (𝑥) ∣ 𝑥 ∈ [𝑎, 𝑏]}.
𝑎
*Tutorial 6.5.2. Let 𝑎 < 𝑏, let 𝑓 ∶ [𝑎, 𝑏] → ℝ and let 𝑃 = {𝑎 = 𝑥0 < 𝑥1 < ⋯ < 𝑥𝑛 = 𝑏} ∈ (𝑎, 𝑏). Define
𝑉 (𝑓 , 𝑃 ) ∶=
𝑛
∑
|𝑓 (𝑥𝑗 ) − 𝑓 (𝑥𝑗−1 )|
𝑗=1
and
𝑉𝑎𝑏 𝑓 ∶= sup{𝑉 (𝑓 , 𝑃 ) ∶ 𝑃 ∈ (𝑎, 𝑏)}.
One sets 𝑉𝑎𝑎 𝑓 = 0. The function 𝑓 is said to be of bounded variation on [𝑎, 𝑏] if 𝑉𝑎𝑏 𝑓 < ∞.
1. Let 𝑎 ≤ 𝑥 < 𝑦 ≤ 𝑏. Show that 𝑉𝑎𝑥 𝑓 + |𝑓 (𝑦) − 𝑓 (𝑥)| ≤ 𝑉𝑎𝑦 𝑓 .
2. Assume that 𝑓 is of bounded variation on [𝑎, 𝑏] and for 𝑥 ∈ [𝑎, 𝑏] define
𝑓1 (𝑥) = 𝑉𝑎𝑥 𝑓 ,
𝑓2 (𝑥) = 𝑉𝑎𝑥 𝑓 − 𝑓 (𝑥).
Show that 𝑓1 and 𝑓2 are increasing on [𝑎, 𝑏] and that 𝑓 = 𝑓1 − 𝑓2 .
3. Show that 𝑓 is of bounded variation on [𝑎, 𝑏] if and only if 𝑓 is the difference of two increasing functions on
[𝑎, 𝑏].
4. Show that if 𝑓 is of bounded variation on [𝑎, 𝑏], then 𝑓 is Riemann integrable over [𝑎, 𝑏].
5. Give an example of a continuous function on [𝑎, 𝑏] which is not of bounded variation.
Chapter 7
Introduction to Metric Spaces
7.1
Metric Spaces
Definition 7.1. Let 𝑋 be a set and 𝑑 ∶ 𝑋 × 𝑋 → ℝ. 𝑑 is called a metric on 𝑋 if for all 𝑥, 𝑦, 𝑧 ∈ 𝑋 the following
properties are satisfied:
(i) 𝑑(𝑥, 𝑦) = 0 ⇔ 𝑥 = 𝑦,
(ii) 𝑑(𝑥, 𝑦) = 𝑑(𝑦, 𝑥)
(symmetry),
(iii) 𝑑(𝑥, 𝑧) ≤ 𝑑(𝑥, 𝑦) + 𝑑(𝑦, 𝑧)
(triangle inequality).
If 𝑑 is a metric on 𝑋, then (𝑋, 𝑑) or shortly 𝑋 is called a metric space.
Remark 7.1.1. If (𝑋, 𝑑) is a metric space, then 𝑑(𝑥, 𝑦) ≥ 0 for all 𝑥, 𝑦 ∈ 𝑋.
Proof. From
0 = 𝑑(𝑥, 𝑥) ≤ 𝑑(𝑥, 𝑦) + 𝑑(𝑦, 𝑥) = 𝑑(𝑥, 𝑦) + 𝑑(𝑥, 𝑦) = 2𝑑(𝑥, 𝑦)
(i)
(ii)
(iii)
it follows that 𝑑(𝑥, 𝑦) ≥ 0.
In the following, 𝑛 ∈ ℕ∗ = ℕ ⧵ {0} unless stated otherwise.
Worked Example 7.1.1. 1. For 𝑥, 𝑦 ∈ ℝ let 𝑑(𝑥, 𝑦) = |𝑥 − 𝑦|. Then (ℝ, 𝑑) is a metric space.
2. For 𝑥, 𝑦 ∈ ℂ let 𝑑(𝑥, 𝑦) = |𝑥 − 𝑦|. Then (ℂ, 𝑑) is a metric space.
( 𝑛
)1
2
∑
3. For 𝑥 = (𝑥𝑗 )𝑛𝑗=1 , 𝑦 = (𝑦𝑗 )𝑛𝑗=1 ∈ ℂ𝑛 let 𝑑(𝑥, 𝑦) =
|𝑥𝑗 − 𝑦𝑗 |2 . Then (ℂ𝑛 , 𝑑) is a metric space.
𝑗=1
Solution. Parts 1 and 2 are clear if one observes the triangle inequality for real and complex numbers.
For the triangle inequality in 3, we use the notation
‖𝑥‖ = 𝑑(𝑥, 0).
Then ‖𝑥 + 𝑦‖ ≤ ‖𝑥‖ + ‖𝑦‖ can be shown as was done for ℝ𝑛 in Multivariable Calculus. But rather than repeating
that proof we use it to prove the result for ℂ𝑛 . To this end, put 𝑥
̂ = (|𝑥𝑗 |)𝑛𝑗=1 ∈ ℝ𝑛 . Obviously, ‖̂
𝑥‖ = ‖𝑥‖ and
‖𝑥 + 𝑦‖ ≤ ‖̂
𝑥 + 𝑦̂‖ by part 2. Then, from Multivarible Calculus results applicable to 𝑥
̂ and 𝑦̂, it follows that
‖𝑥 + 𝑦‖ ≤ ‖̂
𝑥 + 𝑦̂‖ ≤ ‖̂
𝑥‖ + ‖̂
𝑦‖ = ‖𝑥‖ + ‖𝑦‖.
Then, for 𝑥, 𝑦, 𝑧 ∈ ℂ𝑛 ,
𝑑(𝑥, 𝑧) = ‖𝑥 − 𝑧‖ = ‖(𝑥 − 𝑦) + (𝑦 − 𝑧)‖ ≤ ‖𝑥 − 𝑦‖ + ‖𝑦 − 𝑧‖
= 𝑑(𝑥, 𝑦) + 𝑑(𝑦, 𝑧).
In the following, let 𝔽 be either ℝ or ℂ.
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Worked Example 7.1.2. Let 𝑌 be a nonempty set and let 𝐵(𝑌 ) = 𝐵(𝑌 , 𝔽 𝑛 ) be the set of all bounded functions
from 𝑌 to 𝔽 𝑛 . Here a function 𝑓 from 𝑌 to 𝔽 𝑛 is called bounded if there is a real number 𝑀 such that ‖𝑓 (𝑥)‖ =
( 𝑛
)1
2
∑
|𝑓𝑗 (𝑥)|2
≤ 𝑀 for all 𝑥 ∈ 𝑌 , where 𝑓𝑗 (𝑥) is the 𝑗-th component of 𝑓 (𝑥). For 𝑓 , 𝑔 ∈ 𝐵(𝑌 ) let
𝑗=1
𝑑(𝑓 , 𝑔) = sup ‖𝑓 (𝑥) − 𝑔(𝑥)‖.
𝑥∈𝑌
Then (𝐵(𝑌 ), 𝑑) is a metric space.
Solution. We have for 𝑓 , 𝑔 ∈ 𝐵(𝑌 ) that there are 𝑀, 𝑁 ∈ ℝ such that ‖𝑓 (𝑥)‖ ≤ 𝑀 and ‖𝑔(𝑥)‖ ≤ 𝑁 for all
𝑥 ∈ 𝑌 . Denoting the metric on ℂ𝑛 from Example 7.1.1 by 𝑑𝑛 , it follows for 𝑥 ∈ 𝑌 that
‖𝑓 (𝑥) − 𝑔(𝑥)‖ = 𝑑𝑛 (𝑓 (𝑥), 𝑔(𝑥)) ≤ 𝑑𝑛 (𝑓 (𝑥), 0) + 𝑑𝑛 (𝑔(𝑥), 0) = ‖𝑓 (𝑥)‖ + ‖𝑔(𝑥)‖ ≤ 𝑀 + 𝑁,
so that 𝑑(𝑓 , 𝑔) ∈ ℝ.
Clearly, 𝑑(𝑓 , 𝑓 ) = 0 for all 𝑓 ∈ 𝐵(𝑌 ). If, on the other hand, 𝑓 ≠ 𝑔, then there is 𝑥0 ∈ 𝑌 such that 𝑓 (𝑥0 ) ≠ 𝑔(𝑥0 ).
Hence
𝑑(𝑓 , 𝑔) = sup ‖𝑓 (𝑥) − 𝑔(𝑥)‖ ≥ ‖𝑓 (𝑥0 ) − 𝑔(𝑥0 )‖ > 0.
𝑥∈𝑌
From ‖𝑓 (𝑥) − 𝑔(𝑥)‖ = ‖𝑔(𝑥) − 𝑓 (𝑥)‖ it easily follows that 𝑑(𝑓 , 𝑔) = 𝑑(𝑔, 𝑓 ).
Finally, for 𝑓 , 𝑔, ℎ ∈ 𝐵(𝑌 ),
𝑑(𝑓 , ℎ) = sup ‖𝑓 (𝑥) − ℎ(𝑥)‖
𝑥∈𝑌
≤ sup (‖𝑓 (𝑥) − 𝑔(𝑥)‖ + ‖𝑔(𝑥) − ℎ(𝑥)‖)
𝑥∈𝑌
≤ sup (‖𝑓 (𝑥) − 𝑔(𝑥)‖ + ‖𝑔(𝑦) − ℎ(𝑦)‖)
𝑥,𝑦∈𝑌
= sup ‖𝑓 (𝑥) − 𝑔(𝑥)‖ + sup ‖𝑔(𝑦) − ℎ(𝑦)‖
𝑥∈𝑌
𝑦∈𝑌
= 𝑑(𝑓 , 𝑔) + 𝑑(𝑔, ℎ).
Proposition 7.1. Let 𝑋 = (𝑋, 𝑑) be a metric space and let 𝑌 ⊂ 𝑋 be nonempty. For 𝑥, 𝑦 ∈ 𝑌 let 𝑑𝑌 (𝑥, 𝑦) =
𝑑(𝑥, 𝑦). Then 𝑌 = (𝑌 , 𝑑𝑌 ) = (𝑌 , 𝑑) is a metric space.
Note. When it is clear what the metric is, it is convenient to write briefly 𝑋 for a metric space, and 𝑌 ⊂ 𝑋 is called
a (metric) subspace of 𝑋.
Worked Example 7.1.3. 1. ℝ𝑛 is a subspace of ℂ𝑛 .
2. If 𝑌 is a nonempty set and 𝐴 ⊂ 𝔽 𝑛 , then 𝐵(𝑌 , 𝐴) is a subspace of 𝐵(𝑌 , 𝔽 𝑛 ).
Definition 7.2. A sequence (𝑥𝑘 ) in a metric space (𝑋, 𝑑) is called
(i) convergent if there is 𝑥 ∈ 𝑋 such that
lim 𝑑(𝑥𝑘 , 𝑥) = 0.
𝑘→∞
We then write 𝑥 = lim 𝑥𝑘 and say that (𝑥𝑘 ) converges to 𝑥.
𝑘→∞
(ii) a Cauchy sequence if for all 𝜀 > 0 there is 𝑁 ∈ ℕ such that for all 𝑘, 𝑚 ≥ 𝑁: 𝑑(𝑥𝑘 , 𝑥𝑚 ) < 𝜀.
Theorem 7.2. Let (𝑋, 𝑑) be a metric space and let (𝑥𝑘 ) be a convergent sequence in 𝑋. Then
1. lim 𝑥𝑘 is unique.
𝑘→∞
2. (𝑥𝑘 ) is a Cauchy sequence.
Proof. 1. Let 𝑥, 𝑦 ∈ 𝑋 such that lim 𝑑(𝑥𝑘 , 𝑥) = 0 and lim 𝑑(𝑥𝑘 , 𝑦) = 0. Then
𝑘→∞
𝑘→∞
𝑑(𝑥, 𝑦) ≤ 𝑑(𝑥, 𝑥𝑘 ) + 𝑑(𝑥𝑘 , 𝑦)
= 𝑑(𝑥𝑘 , 𝑥) + 𝑑(𝑥𝑘 , 𝑦)
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gives
0 ≤ 𝑑(𝑥, 𝑦) ≤ lim 𝑑(𝑥𝑘 , 𝑥) + lim 𝑑(𝑥𝑘 , 𝑦) = 0.
𝑘→∞
𝑘→∞
Hence 𝑑(𝑥, 𝑦) = 0, giving 𝑥 = 𝑦.
𝜀
2. Let 𝜀 > 0. Since there is 𝑥 ∈ 𝑋 such that lim 𝑑(𝑥𝑘 , 𝑥) = 0, there is 𝑁 ∈ ℕ such that 𝑑(𝑥𝑘 , 𝑥) < for 𝑘 ≥ 𝑁.
𝑘→∞
2
Then, for 𝑘, 𝑚 ≥ 𝑁,
𝑑(𝑥𝑘 , 𝑥𝑚 ) ≤ 𝑑(𝑥𝑘 , 𝑥) + 𝑑(𝑥𝑚 , 𝑥) < 𝜀.
Tutorial 7.1.1. 1. Let 𝑋 be a set and let 𝑓 ∶ 𝑋 → ℝ be injective.
a) Show that 𝑑(𝑥, 𝑦) = |𝑓 (𝑥) − 𝑓 (𝑦)| defines a metric on 𝑋.
b) Show that a sequence (𝑥𝑘 ) in (𝑋, 𝑑) converges to some 𝑥 ∈ 𝑋 if and only if the sequence (𝑓 (𝑥𝑘 )) converges to
some 𝑦 ∈ range 𝑓 , and that then 𝑦 = 𝑓 (𝑥).
2. Let ℝ = ℝ ∪ {−∞, ∞}, define 𝑓 ∶ ℝ → ℝ by
⎧ arctan 𝑥 if 𝑥 ∈ ℝ,
⎪
⎪ 𝜋
if 𝑥 = −∞ ,
𝑓 (𝑥) = ⎨ −
2
⎪𝜋
if 𝑥 = ∞,
⎪
⎩2
and let 𝑑 be as in part 1. Show that a sequence (𝑥𝑘 ) in ℝ converges in (ℝ, 𝑑) if and only if either (𝑥𝑘 ) converges in
ℝ in the usual sense or 𝑥𝑘 → −∞ or 𝑥𝑘 → ∞.
3. Let 𝑋 be a nonempty set. For 𝑥, 𝑦 ∈ 𝑋 define
{
𝑑(𝑥, 𝑦) =
0
if 𝑥 = 𝑦,
1
if 𝑥 ≠ 𝑦.
a) Show that (𝑋, 𝑑) is a metric space. (𝑋, 𝑑) is called a discrete metric space.
b) Show that a sequence (𝑥𝑘 ) converges in (𝑋, 𝑑) if and only if there is 𝑁 ∈ ℕ such that 𝑥𝑘 = 𝑥𝑁 for all 𝑘 > 𝑁.
4. Let 𝑥, 𝑦 ∈ ℂ𝑛 . Prove directly that |(𝑥, 𝑦)| ≤ ‖𝑥‖‖𝑦‖ and ‖𝑥 + 𝑦‖ ≤ ‖𝑥‖ + ‖𝑦‖.
7.2
Complete Metric Spaces
Definition 7.3. A metric space (𝑋, 𝑑) is called complete if every Cauchy sequence in (𝑋, 𝑑) converges.
Theorem 7.3. ℂ𝑛 is complete.
Proof. Let (𝑧𝑘 ) be a Cauchy sequence in ℂ𝑛 . Write 𝑧𝑘 = (𝑥𝑘,𝑗 + 𝑖𝑦𝑘,𝑗 )𝑛𝑗=1 with 𝑥𝑘,𝑗 , 𝑦𝑘,𝑗 ∈ ℝ. Then
|𝑥𝑘,𝑗 − 𝑥𝑙,𝑗 | ≤ 𝑑(𝑧𝑘 , 𝑧𝑙 ) → 0 as 𝑘, 𝑙 → ∞.
Thus (𝑥𝑘,𝑗 )𝑛𝑗=1 is a Cauchy sequence in ℝ for each 𝑗 = 1, … , 𝑛. By Theorem 2.12, the sequence converges, so that
there is 𝑎𝑗 ∈ ℝ such that
lim 𝑥𝑘,𝑗 = 𝑎𝑗 .
𝑘→∞
Similarly, there is 𝑏𝑗 ∈ ℝ such that
lim 𝑦𝑘,𝑗 = 𝑏𝑗 .
𝑘→∞
Put
𝑧 = (𝑎𝑗 + 𝑖𝑏𝑗 )𝑛𝑗=1 .
Let 𝜀 > 0. Then there is 𝐾 ∈ ℝ such that for all 𝑘 ≥ 𝐾 and all 𝑗 = 1, … , 𝑛,
𝜀
|𝑥𝑘,𝑗 − 𝑎𝑗 | < √ ,
2𝑛
𝜀
|𝑦𝑘,𝑗 − 𝑏𝑗 | < √ .
2𝑛
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Hence, for 𝑘 ≥ 𝐾,
( 𝑛
)1
2
∑(
)
𝑑(𝑧𝑘 , 𝑧) =
|𝑥𝑘,𝑗 − 𝑎𝑗 |2 + |𝑦𝑘,𝑗 − 𝑏𝑗 |2
𝑗=1
<
( 𝑛 (
∑ 𝜀2
𝑗=1
𝜀2
+
2𝑛 2𝑛
)) 12
= 𝜀.
Theorem 7.4. Let 𝑌 be a nonempty set. Then 𝐵(𝑌 ) = 𝐵(𝑌 , 𝔽 𝑛 ) is complete.
Proof. Let (𝑓𝑘 ) be a Cauchy sequence in 𝐵(𝑌 ). For each 𝑥 ∈ 𝑌 ,
‖𝑓𝑘 (𝑥) − 𝑓𝑚 (𝑥)‖ ≤ 𝑑(𝑓𝑘 , 𝑓𝑚 ) → 0 as 𝑘, 𝑚 → 0.
Hence (𝑓𝑘 (𝑥))∞
is a Cauchy sequence in 𝔽 𝑛 . Since 𝔽 𝑛 is complete, there is 𝑓 (𝑥) ∈ 𝔽 𝑛 such that 𝑓𝑘 (𝑥) → 𝑓 (𝑥) as
𝑘=1
𝑛 → ∞.
First we will show that 𝑓 is bounded. Indeed, for each 𝜀 > 0 there is 𝑛𝜀 ∈ ℕ such that for all 𝑘, 𝑚 ≥ 𝑛𝜀 ,
𝑑(𝑓𝑘 , 𝑓𝑚 ) < 𝜀. Also, for all 𝑥 ∈ 𝑌 , there is 𝑚 ≥ 𝑛1 such that
‖𝑓𝑚 (𝑥) − 𝑓 (𝑥)‖ < 1.
Hence
‖𝑓 (𝑥)‖ ≤ ‖𝑓 (𝑥) − 𝑓𝑚 (𝑥)‖ + ‖𝑓𝑚 (𝑥) − 𝑓𝑛1 (𝑥)‖ + ‖𝑓𝑛1 (𝑥)‖
< ‖𝑓𝑛1 (𝑥)‖ + 2.
Hence
sup ‖𝑓 (𝑥)‖ ≤ sup ‖𝑓𝑛1 (𝑥)‖ + 2 < ∞.
𝑥∈𝑌
𝑥∈𝑌
Therefore 𝑓 is bounded.
Finally, for 𝑥 ∈ 𝑌 choose 𝑚 ≥ 𝑛 𝜀 such that
3
‖𝑓𝑚 (𝑥) − 𝑓 (𝑥)‖ <
𝜀
.
4
Then, for 𝑘 ≥ 𝑛 𝜀 ,
4
‖𝑓𝑘 (𝑥) − 𝑓 (𝑥)‖ ≤ ‖𝑓𝑘 (𝑥) − 𝑓𝑚 (𝑥)‖ + ‖𝑓𝑚 (𝑥) − 𝑓 (𝑥)‖
𝜀
< ‖𝑓𝑘 (𝑥) − 𝑓𝑚 (𝑥)‖ +
4
𝜀
≤ 𝑑(𝑓𝑘 , 𝑓𝑚 ) +
4
1
< 𝜀.
2
Hence, for 𝑘 ≥ 𝑛 𝜀 ,
4
𝑑(𝑓𝑘 , 𝑓 ) = sup ‖𝑓𝑘 (𝑥) − 𝑓 (𝑥)‖
𝑥∈𝑌
1
𝜀
2
< 𝜀.
≤
This proves that (𝑓𝑘 ) converges in 𝐵(𝑌 ). Since (𝑓𝑘 ) was an arbitrary Cauchy sequence in 𝐵(𝑌 ), it follows that
𝐵(𝑌 ) is complete.
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Definition 7.4. 1. A subset 𝐴 of a metric space (𝑋, 𝑑) is called open if for each 𝑥 ∈ 𝐴 there is 𝜀 > 0 such that
𝐵(𝑥, 𝜀) ⊂ 𝐴, where 𝐵(𝑥, 𝜀) = {𝑦 ∈ 𝑋 ∶ 𝑑(𝑦, 𝑥) < 𝜀} is the 𝜀-ball with centre 𝑥.
2. A subset 𝐴 of a metric space (𝑋, 𝑑) is called closed if 𝑋 ⧵ 𝐴 is open.
Theorem 7.5. A subset 𝐴 of a metric space (𝑋, 𝑑) is closed if and only if for each sequence (𝑥𝑘 ) in 𝐴 which
converges to some 𝑥 ∈ 𝑋 it follows that 𝑥 ∈ 𝐴.
Proof. Let 𝐴 be closed and let (𝑥𝑘 ) be a sequence in 𝐴 which converges to some 𝑥 ∈ 𝑋. We must show 𝑥 ∈ 𝐴.
Assume that 𝑥 ∉ 𝐴. Since 𝑥 ∈ 𝑋 ⧵ 𝐴 and 𝑋 ⧵ 𝐴 is open, there is 𝜀 > 0 such that 𝐵(𝑥, 𝜀) ⊂ 𝑋 ⧵ 𝐴. Since 𝑥𝑘 ∈ 𝐴
for all 𝑘 ∈ ℕ, it follows that 𝑑(𝑥𝑘 , 𝑥) ≥ 𝜀, which contradicts 𝑥𝑘 → 𝑥.
Conversely, if 𝐴 ⊂ 𝑋 is not closed, then there is 𝑥 ∈ 𝑋 ⧵ 𝐴 such that 𝐵(𝑥, 𝜀) ⊄ 𝑋 ⧵ 𝐴 for all 𝜀 > 0. Therefore, for
positive integers 𝑘, there is 𝑥𝑘 ∈ 𝐴 ∩ 𝐵(𝑥, 𝑘1 ). Thus, for each 𝜀 > 0 and 𝑘 ≥ 1𝜀 we have 𝑑(𝑥𝑘 , 𝑥) < 𝑘1 ≤ 𝜀. Thus
we have found a sequence (𝑥𝑘 ) in 𝐴 which converges to some 𝑥 ∈ 𝑋 which does not belong to 𝐴.
Theorem 7.6. Let (𝑋, 𝑑) be a complete metric space and let 𝐴 ⊂ 𝑋. Then 𝐴 is complete if and only if 𝐴 is closed.
Proof. Assume that 𝐴 is complete and let (𝑥𝑘 ) be a sequence in 𝐴 which converges to some 𝑥 ∈ 𝑋. We must show
that 𝑥 ∈ 𝐴. But since (𝑥𝑘 ) is a convergent sequence in 𝑋, it is a Cauchy sequence in 𝑋 by Theorem 7.2, 2, and
therefore also in 𝐴. Since 𝐴 is complete, (𝑥𝑘 ) converges to some 𝑎 ∈ 𝐴, and therefore also in 𝑋. Since limits are
unique by Theorem 7.2, 1, it follows that 𝑥 = 𝑎 ∈ 𝐴. Hence, by Theorem 7.5, 𝐴 is closed.
Conversely, assume that 𝐴 is closed. Let (𝑥𝑘 ) be a Cauchy sequence in 𝐴. Then (𝑥𝑘 ) is a Cauchy sequence in 𝑋,
and since 𝑋 is complete, there is 𝑥 ∈ 𝑋 such that 𝑥𝑘 → 𝑥. Since 𝐴 is closed, it follows that 𝑥 ∈ 𝐴. Hence each
Cauchy sequence in 𝐴 converges in 𝐴. It follows that 𝐴 is complete.
Definition 7.5. Let 𝐴 ⊂ ℝ and 𝑓 ∶ 𝐴 → 𝔽 𝑛 . Then 𝑓 is uniformly continuous if for each 𝜀 > 0 there is 𝛿 > 0
such that for all 𝑥, 𝑦 ∈ 𝐴 (|𝑥 − 𝑦| < 𝛿 ⇒ ‖𝑓 (𝑥) − 𝑓 (𝑦)‖ < 𝜀).
Remark 7.2.1. It is straightforward to show that a function 𝑓 ∶ [𝑎, 𝑏] → ℂ is (uniformly) continuous if and only if
its real and imaginary part are (uniformly) continuous and that a function 𝑓 ∶ [𝑎, 𝑏] → 𝔽 𝑛 is (uniformly) continuous
if and only if all its component are (uniformly) continuous.
Theorem 7.7. Let 𝑎, 𝑏 ∈ ℝ, 𝑎 < 𝑏, and let 𝑓 ∶ [𝑎, 𝑏] → 𝔽 𝑛 be continuous. Then 𝑓 is uniformly continuous.
Proof. Because of Remark 7.2.1 we only have to consider 𝑓 ∶ [𝑎, 𝑏] → ℝ. Let 𝜀 > 0 and choose 𝑎 = 𝑥0 < 𝑥1 <
⋯ < 𝑥𝑁 = 𝑏 according to Lemma 6.13 with respect to 4𝜀 . Put
𝛿 = min{|𝑥𝑗 − 𝑥𝑗−1 | ∶ 𝑗 = 1, … , 𝑁}.
Now let 𝑥, 𝑦 ∈ [𝑎, 𝑏] such that |𝑥 − 𝑦| < 𝛿. We want to show |𝑓 (𝑥) − 𝑓 (𝑦)| < 𝜀. This is trivial if 𝑥 = 𝑦. Thus let
𝑥 ≠ 𝑦, and we may assume 𝑥 < 𝑦.
There is 𝑗 ∈ {1, … , 𝑁} such that 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ]. Hence
𝑥𝑗−1 ≤ 𝑥 < 𝑦 < 𝑥 + 𝛿 ≤ 𝑥𝑗 + 𝛿.
Case I: 𝑦 ≤ 𝑥𝑗 . Then 𝑦 ∈ [𝑥𝑗−1 , 𝑥𝑗 ], and Lemma 6.13 gives
|𝑓 (𝑥) − 𝑓 (𝑦)| ≤ |𝑓 (𝑥) − 𝑓 (𝑥𝑗−1 )| + |𝑓 (𝑥𝑗−1 ) − 𝑓 (𝑦)| ≤
𝜀 𝜀
+ < 𝜀.
4 4
Case II: 𝑦 > 𝑥𝑗 . Then 𝑗 < 𝑁, and 𝑥𝑗 < 𝑦 < 𝑥𝑗 + 𝛿 ≤ 𝑥𝑗+1 . Again by Lemma 6.13,
|𝑓 (𝑥) − 𝑓 (𝑦)| ≤ |𝑓 (𝑥) − 𝑓 (𝑥𝑗−1 )| + |𝑓 (𝑥𝑗−1 ) − 𝑓 (𝑥𝑗 )| + |𝑓 (𝑥𝑗 ) − 𝑓 (𝑦)| ≤
𝜀 𝜀 𝜀
+ + < 𝜀.
4 4 4
For an interval 𝐼 we denote by 𝐶(𝐼, 𝔽 𝑛 ) the set of continuous functions from 𝐼 to 𝔽 𝑛 .
Theorem 7.8. Let 𝑎, 𝑏 ∈ ℝ and 𝑛 ∈ ℕ⧵{0}. Then 𝐶([𝑎, 𝑏], 𝔽 𝑛 ) is a closed subspaces of 𝐵([𝑎, 𝑏], 𝔽 𝑛 ). In particular,
𝐶([𝑎, 𝑏], 𝔽 𝑛 ) is complete.
Proof. By Remark 7.2.1, it suffices to consider 𝐶([𝑎, 𝑏], ℝ). By Theorem 3.17, each continuous function on [𝑎, 𝑏]
is bounded, so that 𝐶([𝑎, 𝑏], ℝ) ⊂ 𝐵([𝑎, 𝑏], ℝ).
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To show that 𝐶([𝑎, 𝑏], ℝ) is closed in 𝐵([𝑎, 𝑏], ℝ) let (𝑓𝑘 ) be a sequence in 𝐶([𝑎, 𝑏], ℝ) which converges to some
𝑓 in 𝐵([𝑎, 𝑏], ℝ). We must show that 𝑓 is continuous.
𝜀
Let 𝜀 > 0. Then there is 𝑘 ∈ ℕ such that 𝑑(𝑓𝑘 , 𝑓 ) < . Since 𝑓𝑘 is uniformly continuous, there is 𝛿 > 0 such that
3
𝜀
|𝑓𝑘 (𝑥) − 𝑓𝑘 (𝑦)| < for 𝑥, 𝑦 ∈ [𝑎, 𝑏], |𝑥 − 𝑦| < 𝛿.
3
Then, for 𝑥, 𝑦 ∈ [𝑎, 𝑏] and |𝑥 − 𝑦| < 𝛿,
|𝑓 (𝑥) − 𝑓 (𝑦)| ≤ |𝑓 (𝑥) − 𝑓𝑘 (𝑥)| + |𝑓𝑘 (𝑥) − 𝑓𝑘 (𝑦)| + |𝑓𝑘 (𝑦) − 𝑓 (𝑦)|
≤ 𝑑(𝑓 , 𝑓𝑘 ) + |𝑓𝑘 (𝑥) − 𝑓𝑘 (𝑦)| + 𝑑(𝑓𝑘 , 𝑓 )
𝜀 𝜀 𝜀
< + +
3 3 3
= 𝜀.
Thus 𝑓 is uniformly continuous.
By Theorem 7.6, 𝐶([𝑎, 𝑏], 𝔽 𝑛 ) is complete.
Tutorial 7.2.1. 1. Let (𝑋, 𝑑) be the metric space from Tutorial 7.1.1, 1. Show that (𝑋, 𝑑) is complete if and only
if range 𝑓 is closed in ℝ.
2. Show that the metric space (ℝ, 𝑑) from Tutorial 7.1.1, 2, is complete.
3. Show that ℝ is not complete with respect to the metric given in Tutorial 7.1.1, 2.
7.3
Banach’s Fixed Point Theorem
In this section, 𝑋 = (𝑋, 𝑑𝑋 ), 𝑌 = (𝑌 , 𝑑𝑌 ) and 𝑍 = (𝑍, 𝑑𝑍 ) are metric spaces.
Definition 7.6. Let 𝑇 ∶ 𝑋 → 𝑌 . 𝑇 is called bounded if there is a number 𝑀 ≥ 0 such that
𝑑𝑌 (𝑇 𝑥, 𝑇 𝑦) ≤ 𝑀𝑑𝑋 (𝑥, 𝑦) for all 𝑥, 𝑦 ∈ 𝑋.
If 𝑇 is bounded, then
‖𝑇 ‖ = inf {𝑀 ≥ 0 ∶ ∀ 𝑥, 𝑦 ∈ 𝑋, 𝑑𝑌 (𝑇 𝑥, 𝑇 𝑦) ≤ 𝑀𝑑𝑋 (𝑥, 𝑦)}
is called the norm of 𝑇 .
If 𝑇 is not bounded, we may write for convenience ‖𝑇 ‖ = ∞.
Since it is clear that 𝑑𝑌 (𝑇 𝑥, 𝑇 𝑦) ≤ (‖𝑇 ‖ + 𝜀)𝑑𝑋 (𝑥, 𝑦) for all 𝜀 > 0, it follows that
∀ 𝑥, 𝑦 ∈ 𝑋, 𝑑𝑌 (𝑇 𝑥, 𝑇 𝑦) ≤ ‖𝑇 ‖𝑑𝑋 (𝑥, 𝑦).
Proposition 7.9. Let 𝑇 ∶ 𝑋 → 𝑌 . Then
𝑑𝑌 (𝑇 𝑥, 𝑇 𝑦)
.
𝑥,𝑦∈𝑋 𝑑𝑋 (𝑥, 𝑦)
‖𝑇 ‖ = sup
𝑥≠𝑦
Proof. Let
{
𝐴=
}
𝑑𝑌 (𝑇 𝑥, 𝑇 𝑦)
∶ 𝑥, 𝑦 ∈ 𝑋, 𝑥 ≠ 𝑦 , 𝛼 = sup 𝐴.
𝑑𝑋 (𝑥, 𝑦)
If 𝑇 is bounded, then ‖𝑇 ‖ is an upper bound of 𝐴 and thus 𝛼 ≤ ‖𝑇 ‖.
𝑑 (𝑇 𝑥, 𝑇 𝑦)
On the other hand, 𝛼 ≥ 𝑌
for all 𝑥, 𝑦 ∈ 𝑋 with 𝑥 ≠ 𝑦, and thus 𝑑𝑌 (𝑇 𝑥, 𝑇 𝑦) ≤ 𝛼𝑑𝑋 (𝑥, 𝑦) for all 𝑥, 𝑦 ∈ 𝑋.
𝑑𝑋 (𝑥, 𝑦)
Hence 𝛼 ≥ ‖𝑇 ‖.
The case that 𝑇 is not bounded is left as an exercise.
Notation. 1. If 𝑇 ∶ 𝑋 → 𝑌 and 𝑆 ∶ 𝑌 → 𝑍, then we write 𝑆𝑇 = 𝑆◦𝑇 .
2. If 𝑇 ∶ 𝑋 → 𝑋, we define
𝑇 0 by 𝑇 0 𝑥 = 𝑥, 𝑥 ∈ 𝑋,
𝑇 𝑘+1 = 𝑇 𝑇 𝑘 inductively for 𝑘 ∈ ℕ.
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Proposition 7.10. If 𝑇 ∶ 𝑋 → 𝑌 and 𝑆 ∶ 𝑌 → 𝑍 are bounded, then 𝑆𝑇 is bounded, and ‖𝑆𝑇 ‖ ≤ ‖𝑆‖ ‖𝑇 ‖.
Proof. For all 𝑥, 𝑦 ∈ 𝑋 we have
𝑑𝑍 ((𝑆𝑇 )𝑥, (𝑆𝑇 )𝑦) = 𝑑𝑍 (𝑆(𝑇 𝑥), 𝑆(𝑇 𝑦)) ≤ ‖𝑆‖𝑑𝑌 (𝑇 𝑥, 𝑇 𝑦) ≤ ‖𝑆‖ ‖𝑇 ‖𝑑𝑋 (𝑥, 𝑦).
Hence, by definition, 𝑆𝑇 is bounded and ‖𝑆𝑇 ‖ ≤ ‖𝑆‖ ‖𝑇 ‖.
Definition 7.7. Let 𝑇 ∶ 𝑋 → 𝑋 be bounded.
1. 𝑇 is called a contraction if ‖𝑇 ‖ < 1.
2. 𝑇 is called a generalized contraction if there is 𝑚 ∈ ℕ such that ‖𝑇 𝑚 ‖ < 1.
Remark 7.3.1. Let 𝑇 ∶ 𝑋 → 𝑋 be bounded. Then 𝑇 is a generalized contraction if and only if
∞
∑
‖𝑇 𝑗 ‖ < ∞.
𝑗=0
Proof. If the sum converges, then ‖𝑇 𝑗 ‖ → 0 as 𝑗 → ∞, and therefore there is 𝑚 ∈ ℕ such that ‖𝑇 𝑚 ‖ < 1.
Conversely, if ‖𝑇 𝑚 ‖ < 1, then we group the sum into blocks of 𝑚 terms, writing every natural number as a multiple
of 𝑚 plus its remainder, use
‖𝑇 𝑘𝑚+𝑙 ‖ = ‖𝑇 𝑘𝑚 𝑇 𝑙 ‖ ≤ ‖(𝑇 𝑚 )𝑘 ‖ ‖𝑇 𝑙 ‖ ≤ ‖𝑇 𝑚 ‖𝑘 ‖𝑇 𝑙 ‖
and get
∞
∑
‖𝑇 ‖ =
𝑗
𝑗=0
∞ 𝑚−1
∑
∑
‖𝑇 𝑘𝑚+𝑙 ‖
𝑘=0 𝑙=0
≤
∞ 𝑚−1
∑
∑
‖𝑇 𝑚 ‖𝑘 ‖𝑇 𝑙 ‖
𝑘=0 𝑙=0
=
(∞
∑
𝑘=0
=
‖𝑇 𝑚 ‖𝑘
) (𝑚−1
∑
)
‖𝑇 𝑙 ‖
𝑙=0
𝑚−1
∑
1
‖𝑇 𝑙 ‖ < ∞.
1 − ‖𝑇 𝑚 ‖ 𝑙=0
Theorem 7.11 (Banach’s Fixed Point Theorem). Let (𝑋, 𝑑) be a complete metric space and let 𝑇 ∶ 𝑋 → 𝑋 be a
generalized contraction. Then there is a unique 𝑎 ∈ 𝑋 such that 𝑇 𝑎 = 𝑎.
Proof. Since 𝑇 is a generalized contraction, there is 𝑚 ∈ ℕ such that ‖𝑇 𝑚 ‖ < 1.
For uniqueness, let 𝑎, 𝑏 ∈ 𝑋 such that 𝑇 𝑎 = 𝑎 and 𝑇 𝑏 = 𝑏. By induction
𝑇 𝑘 𝑎 = 𝑇 𝑇 𝑘−1 𝑎 = 𝑇 𝑎 = 𝑎 and 𝑇 𝑘 𝑏 = 𝑏,
so that
𝑑(𝑎, 𝑏) = 𝑑(𝑇 𝑘 𝑎, 𝑇 𝑘 𝑏) ≤ ‖𝑇 𝑘 ‖𝑑(𝑎, 𝑏),
and 𝑑(𝑎, 𝑏) ≠ 0 would give the contradiction 1 ≤ ‖𝑇 𝑚 ‖. Hence 𝑑(𝑎, 𝑏) = 0, i. e., 𝑎 = 𝑏.
For existence let 𝑥0 ∈ 𝑋 and consider the sequence (𝑇 𝑘 𝑥0 ). Then, for 𝑗, 𝑘 ∈ ℕ, 𝑘 > 𝑗,
𝑑(𝑇 𝑘 𝑥0 , 𝑇 𝑗 𝑥0 ) ≤ 𝑑(𝑇 𝑘 𝑥0 , 𝑇 𝑘−1 𝑥0 ) + 𝑑(𝑇 𝑘−1 𝑥0 , 𝑇 𝑘−2 𝑥0 ) + ⋯ + 𝑑(𝑇 𝑗+1 𝑥0 , 𝑇 𝑗 𝑥0 )
≤ ‖𝑇 𝑘−1 ‖ 𝑑(𝑇 𝑥0 , 𝑥0 ) + ‖𝑇 𝑘−2 ‖ 𝑑(𝑇 𝑥0 , 𝑥0 ) + ⋯ + ‖𝑇 𝑗 ‖ 𝑑(𝑇 𝑥0 , 𝑥0 )
=
𝑘−1
∑
𝑙=𝑗
‖𝑇 𝑙 ‖ 𝑑(𝑇 𝑥0 , 𝑥0 )
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Since
∞
∑
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‖𝑇 𝑙 ‖ < ∞,
𝑙=0
it follows that (𝑇 𝑘 𝑥0 ) is a Cauchy sequence. Since 𝑋 is complete, this Cauchy sequence has a limit, say 𝑎. Finally,
to show that 𝑇 𝑎 = 𝑎, let 𝜀 > 0. Then there is 𝐾 such that for 𝑗 ≥ 𝐾,
𝑑(𝑇 𝑗 𝑥0 , 𝑎) <
𝜀
.
1 + ‖𝑇 ‖
For 𝑘 ≥ 𝐾 + 1 we get
𝑑(𝑇 𝑎, 𝑎) ≤ 𝑑(𝑇 𝑎, 𝑇 𝑘 𝑥0 ) + 𝑑(𝑇 𝑘 𝑥0 , 𝑎)
≤ ‖𝑇 ‖𝑑(𝑎, 𝑇 𝑘−1 𝑥0 ) + 𝑑(𝑇 𝑘 𝑥0 , 𝑎)
𝜀
𝜀
+
< ‖𝑇 ‖
1 + ‖𝑇 ‖ 1 + ‖𝑇 ‖
= 𝜀.
Hence 𝑑(𝑇 𝑎, 𝑎) = 0, i. e., 𝑇 𝑎 = 𝑎.
Tutorial 7.3.1. Let (𝑋, 𝑑) be a complete metric space and let 𝑇 be a generalized contraction on 𝑋. Let 𝑚 ∈ ℕ∗
such that ‖𝑇 𝑚 ‖ < 1 and let 𝑥 be the fixed point of 𝑇 . Prove the following a priori estimate: For any 𝑥0 ∈ 𝑋,
𝑑(𝑇 𝑥0 , 𝑥0 ) ∑ 𝑗
‖𝑇 ‖.
1 − ‖𝑇 𝑚 ‖ 𝑗=0
𝑚−1
𝑑(𝑥, 𝑥0 ) ≤
Hint. Use that 𝑇 𝑚 𝑥 = 𝑥 and use a "telescoping" sum of distances involving the points 𝑇 𝑚 𝑥, 𝑇 𝑚 𝑥0 , 𝑇 𝑚−1 𝑥0 , … ,
𝑇 𝑥0 .
Tutorial 7.3.2. Newton Raphson method of finding zeros
Let 𝑓 ∶ [𝑎, 𝑏] → ℝ be twice continuously differentiable and assume there are 𝑐, 𝑑 ∈ [𝑎, 𝑏], 𝑐 < 𝑑, such that 𝑓 (𝑐)
and 𝑓 (𝑑) have opposite signs. By the Intermediate Value Theorem, 𝑓 has a zero 𝑦 in (𝑐, 𝑑). Also assume that
𝑓 ′ (𝑥) ≠ 0 for all 𝑥 ∈ [𝑐, 𝑑]. Since 𝑓 ′ is continuous, 𝑓 ′ is either positive or negative by the Internmediate Value
Theorem, so that 𝑓 is strictly monotonic by the Mean Value Theorem. Hence the zero is unique. To find this zero
one can use the Newton Raphson method:
1. Let
𝑓 (𝑥)
𝑇 𝑥 = 𝑥 − ′ , 𝑥 ∈ [𝑐, 𝑑].
𝑓 (𝑥)
Show that for all 𝑥1 , 𝑥2 ∈ [𝑐, 𝑑] with 𝑥1 ≠ 𝑥2 there is 𝜉 between 𝑥1 and 𝑥2 such that
𝑇 𝑥1 − 𝑇 𝑥2 =
𝑓 (𝜉)𝑓 ′′ (𝜉)
(𝑥1 − 𝑥2 ) .
(𝑓 ′ (𝜉))2
2. Show that ‖𝑇 ‖ < 1 if 𝑐 and 𝑑 are sufficiently close to the zero 𝑦 of 𝑓 .
3. Show that one can choose 𝑐 and 𝑑 in such a way that 𝑇 ([𝑐, 𝑑]) ⊂ [𝑐, 𝑑].
Hint. You may show that 𝑇 ([𝑐, 𝑑]) ⊂ [𝑐, 𝑑] if 𝑑 − 𝑦 = 𝑦 − 𝑐, i. e., 𝑦 is the midpoint of the interval [𝑐, 𝑑]. However,
this is not very practical since you do not know 𝑦 (otherwise, the whole procedure would be pointless). The easiest
way out of this is to choose suitable values (by trial and error), 𝑐1 < 𝑑1 in [𝑐, 𝑑] such that 𝑓 (𝑐1 ) and 𝑓 (𝑐2 ) have
opposite sign, 𝑇 𝑐1 , 𝑇 𝑐2 ∈ [𝑐1 , 𝑐2 ], and 𝑓 𝑓 ′′ has no zeros on [𝑐1 , 𝑑1 ] ⧵ {𝑦}.
∗ 4. Assume that statement 2 holds and define
𝑐1 =
𝑐 + ‖𝑇 ‖𝑑
,
1 + ‖𝑇 ‖
𝑑1 =
𝑑 + ‖𝑇 ‖𝑐
.
1 + ‖𝑇 ‖
Show that 𝑐 < 𝑐1 < 𝑦 < 𝑑1 < 𝑑 if 𝑓 (𝑐1 ) and 𝑓 (𝑑1 ) have opposite signs. Then show that 𝑇 ([𝑐, 𝑑]) ⊂ [𝑐, 𝑑].
Hence show that for any 𝑥 ∈ [𝑐, 𝑑], 𝑇 𝑘 𝑥 → 𝑦 as 𝑘 → ∞.
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√
√
√
Tutorial 7.3.3. Let 𝑐 > 0. Show that 𝑐 + 𝑐 + 𝑐 + … converges by using the following proof.
[
)
[
)
√
(a) Let 𝑇𝑐 𝑥 = 𝑐 + 𝑥, 𝑥 ∈ [0, ∞). Show that 𝑇𝑐 𝑥 > 𝑥 for 𝑥 ∈ [0, 1] and that 𝑇𝑐 ∶ 14 , ∞ → 14 , ∞ is a
contraction.
Show that if 𝑥𝑘 ≤ 1 for all 𝑘 ∈ ℕ, then (𝑥𝑘 ) is (strictly) increasing. (Actually, this never happens.)
(b) Let 𝑥0 = 𝑐, 𝑥𝑘+1 = 𝑇𝑐 𝑥𝑘 (𝑘 ∈ ℕ).
(c) Show that (𝑥𝑘 ) converges as 𝑘 → ∞.
(d) Find lim 𝑥𝑘 .
𝑘→∞
7.4
Existence of solutions of differential equations
Let 𝐼 ⊂ ℝ be an interval, let 𝑈 ⊂ 𝔽 𝑛 and let 𝑓 ∶ 𝐼 × 𝑈 → 𝔽 𝑛 be continuous. Then
𝑦′ = 𝑓 (𝑥, 𝑦)
(7.4.1)
is called a first-order system of differential equations, and any continuously differentiable function 𝑦 ∶ 𝐼 → 𝑈
satisfying (7.4.1) is called a solution of (7.4.1).
Let 𝑎 ∈ 𝐼 and 𝑏 ∈ 𝔽 𝑛 . Then a solution of (7.4.1) satisfying
𝑦(𝑎) = 𝑏
(7.4.2)
is called a solution of the initial value problem (IVP) (7.4.1), (7.4.2).
From the Fundamental Theorem of Calculus, Theorem 6.15, we obtain
Lemma 7.12. 𝑦 is a a solution of the IVP (7.4.1), (7.4.2) if and only if 𝑦 ∈ 𝐶(𝐼, 𝔽 𝑛 ), 𝑦(𝐼) ⊂ 𝑈 , and
𝑥
𝑦(𝑥) = 𝑏 +
∫
𝑓 (𝑡, 𝑦(𝑡)) 𝑑𝑡
(𝑥 ∈ 𝐼).
(7.4.3)
𝑎
Proof. We only have to note that we can apply the Fundamental Theorem of Calculus componentwise and to real
and imaginary parts separately. Also, if 𝑦 ∈ 𝐶(𝐼, 𝔽 𝑛 ) satisfies (7.4.3), then 𝑓 is differentiable.
We are now going to show that under certain conditions, (7.4.3) and thus the IVP (7.4.1), (7.4.2) has a unique
solution. We write 𝐶 1 (𝐼, 𝔽 𝑛 ) for the set of all continuously differentiable functions from 𝐼 to 𝔽 𝑛 .
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Theorem 7.13. We make the following assumptions:
(i) Let 𝐼 be a closed bounded interval, 𝑎 ∈ 𝐼, 𝐴 = max |𝑥 − 𝑎|, 𝑏 ∈ 𝔽 𝑛 , 0 < 𝐵 ≤ ∞,
𝑥∈𝐼
𝑈 = {𝑦 ∈ 𝔽 𝑛 ∶ ‖𝑦 − 𝑏‖ ≤ 𝐵}, 𝑅 = 𝐼 × 𝑈 .
(ii) Let 𝑓 ∶ 𝑅 → 𝔽 𝑛 be continuous and let 𝑀 = sup ‖𝑓 (𝑥, 𝑦)‖.
(𝑥,𝑦)∈𝑅
(iii) ∃ 𝐿 > 0 ∀ (𝑥, 𝑦1 ), (𝑥, 𝑦2 ) ∈ 𝑅 ‖𝑓 (𝑥, 𝑦1 ) − 𝑓 (𝑥, 𝑦2 )‖ ≤ 𝐿‖𝑦1 − 𝑦2 ‖.
(This is called a Lipschitz condition.)
(iv) 𝐴𝑀 ≤ 𝐵.
Then there is exactly one 𝑦 ∈ 𝐶 1 (𝐼, 𝔽 𝑛 ) such that
𝑦(𝑎) = 𝑏,
(𝑥, 𝑦(𝑥)) ∈ 𝑅 (𝑥 ∈ 𝐼),
𝑦′ (𝑥) = 𝑓 (𝑥, 𝑦(𝑥))
(𝑥 ∈ 𝐼).
Proof. Since 𝑈 is a closed subset of 𝔽 𝑛 , 𝐶(𝐼, 𝑈 ) is a complete metric space. For 𝑦 ∈ 𝐶(𝐼, 𝑈 ) define
𝑥
(𝑇 𝑦)(𝑥) = 𝑏 +
𝑓 (𝑡, 𝑦(𝑡)) 𝑑𝑡
∫
(𝑥 ∈ 𝐼).
(7.4.4)
𝑎
By Lemma 7.12, the theorem is proved if we show that 𝑇 maps 𝐶(𝐼, 𝑈 ) into 𝐶(𝐼, 𝑈 ) and that 𝑇 has exactly one
fixed point.
Thus let 𝑦 ∈ 𝐶(𝐼, 𝑈 ) and 𝑥 ∈ 𝐼. Clearly, 𝑇 𝑦 ∈ 𝐶(𝐼, 𝔽 𝑛 ) and
‖
‖ 𝑥
‖
‖
‖
‖
‖(𝑇 𝑦)(𝑥) − 𝑏‖ = ‖ 𝑓 (𝑡, 𝑦(𝑡)) 𝑑𝑡‖
‖
‖∫
‖
‖𝑎
‖
‖
| 𝑥
|
|
|
|
|
≤ | ‖𝑓 (𝑡, 𝑦(𝑡))‖ 𝑑𝑡|
|∫
|
|𝑎
|
|
|
| 𝑥
|
|
|
|
|
≤ | 𝑀 𝑑𝑡|
|∫
|
|𝑎
|
|
|
= |𝑥 − 𝑎|𝑀
≤ 𝐴𝑀
≤ 𝐵.
Thus we have shown that 𝑇 𝑦 ∈ 𝐶(𝐼, 𝑈 ).
Next we are going to show that 𝑇 is a generalized contraction by proving the following estimate: For all 𝑛 ∈ ℕ,
𝑦1 , 𝑦2 ∈ 𝐶(𝐼, 𝑈 ) and 𝑥 ∈ 𝐼,
‖ 𝑘
‖ 𝐿𝑘 |𝑥 − 𝑎|𝑘
𝑑(𝑦1 , 𝑦2 ).
‖(𝑇 𝑦1 )(𝑥) − (𝑇 𝑘 𝑦2 )(𝑥)‖ ≤
‖
‖
𝑘!
The induction base 𝑘 = 0 is trivial since
‖𝑦1 (𝑥) − 𝑦2 (𝑥)‖ ≤ sup ‖𝑦1 (𝑡) − 𝑦2 (𝑡)‖ = 𝑑(𝑦1 , 𝑦2 ).
𝑡∈𝐼
(7.4.5)
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For the induction step assume that (7.4.5) is true for 𝑘. Then
‖ 𝑥
‖
‖
] ‖
‖ 𝑘+1
‖ ‖ [
‖
𝑘+1
𝑘
𝑘
𝑦1 )(𝑥) − (𝑇
𝑦2 )(𝑥)‖ = ‖
𝑓 (𝑡, (𝑇 𝑦1 )(𝑡)) − 𝑓 (𝑡, (𝑇 𝑦2 )(𝑡)) 𝑑𝑡‖
‖(𝑇
‖
‖ ‖∫
‖
‖𝑎
‖
‖
‖
𝑥
|
|
|
|
| ‖
‖ |
𝑘
𝑘
≤ | ‖𝑓 (𝑡, (𝑇 𝑦1 )(𝑡)) − 𝑓 (𝑡, (𝑇 𝑦2 )(𝑡))‖ 𝑑𝑡|
|∫ ‖
‖ |
|𝑎
|
|
|
| 𝑥
|
|
|
|
‖
‖ |
≤ | 𝐿 ‖(𝑇 𝑘 𝑦1 )(𝑡) − (𝑇 𝑘 𝑦2 )(𝑡)‖ 𝑑𝑡| (by assumption (iii))
|∫
‖
‖ |
|𝑎
|
|
|
|
| 𝑥
|
𝐿𝑘 |𝑥 − 𝑎|𝑘 ||
|
≤| 𝐿
𝑑𝑡| 𝑑(𝑦1 , 𝑦2 ) (by induction hypothesis)
|
|∫
𝑘!
|𝑎
|
|
|
𝑘+1
𝑘+1
𝐿 |𝑥 − 𝑎|
𝑑(𝑦1 , 𝑦2 ).
=
(𝑘 + 1)!
Taking the maximum over all 𝑥 ∈ 𝐼 in (7.4.5) gives
𝑑(𝑇 𝑘 𝑦1 , 𝑇 𝑘 𝑦2 ) ≤
and therefore
‖𝑇 𝑘 ‖ ≤
which leads to
∞
∑
‖𝑇 𝑘 ‖ ≤
𝐿𝑘 𝐴𝑘
,
𝑘!
∞
∑
(𝐿𝐴)𝑘
𝑘=0
𝑘=0
𝐿𝑘 𝐴𝑘
𝑑(𝑦1 , 𝑦2 ),
𝑘!
𝑘!
= 𝑒𝐿𝐴 < ∞.
Hence 𝑇 is a generalized contraction on 𝐶(𝐼, 𝑈 ). An application of Banach’s Fixed Point Theorem completes the
proof.
Worked Example 7.4.1. Consider the initial value problem 𝑦′ = 1 + 𝑦2 , 𝑦(0) = 0.
a) Find the largest 𝑐 > 0 such that the assumptions of Theorem 7.12 are satisfied for 𝐼 = [0, 𝑐].
b) Find a maximal interval on which a solution exists.
Solution. a) We have 𝑓 (𝑥, 𝑦) = 1 + 𝑦2 . Hence
𝑓 (𝑥, 𝑦1 ) − 𝑓 (𝑥, 𝑦2 ) = 𝑦21 − 𝑦22 = (𝑦1 + 𝑦2 )(𝑦1 − 𝑦2 ),
so that
|𝑓 (𝑥, 𝑦1 ) − 𝑓 (𝑥, 𝑦2 )| = |𝑦1 + 𝑦2 | |𝑦1 − 𝑦2 |.
Hence the Lipschitz condition is satisfied if and only if
sup
|𝑦1 |,|𝑦2 |≤𝐵
|𝑦1 + 𝑦2 | < ∞,
which is satisfied if and only if 𝐵 ∈ ℝ, i. e., 𝐵 < ∞.
We have 𝐴 = 𝑐 and
𝑀 = sup |1 + 𝑦2 | = 1 + 𝐵 2 .
|𝑦|≤𝐵
We must satisfy 𝐴𝑀 ≤ 𝐵, i. e.,
𝑐(1 + 𝐵 2 ) ≤ 𝐵
But
or 𝑐 ≤
𝐵
.
1 + 𝐵2
(1 + 𝐵 2 ) − (1 − 𝐵)2
(1 − 𝐵)2
2𝐵
=
=1−
1 + 𝐵2
1 + 𝐵2
1 + 𝐵2
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.
2
)
(
𝜋 𝜋
.
b) The solution of the IVP is 𝑦(𝑥) = tan 𝑥. So the maximal interval where a solution exists is − ,
2 2
Let 𝑀𝑛 (𝔽 ) be the set of 𝑛 × 𝑛 matrices with values in 𝔽 . Then the norm ‖𝑀‖ of 𝑀 ∈ 𝑀𝑛 (𝔽 ) is the norm of
𝑀 ∶ 𝔽 𝑛 → 𝔽 𝑛 . It is easy to see that
has a maximum of 1 for 𝐵 = 1, so that the optimal 𝑐 is 𝑐 =
𝑛
max |𝑚𝑖𝑗 | ≤ ‖𝑀‖ ≤
𝑖,𝑗=1
𝑛
∑
|𝑚𝑖𝑗 |,
where 𝑀 = (𝑚𝑖𝑗 )𝑛𝑖,𝑗=1 .
𝑖,𝑗=1
Corollary 7.14. Let 𝐼 be an interval, 𝑎 ∈ 𝐼, 𝑏 ∈ 𝔽 𝑛 , and consider the linear system of differential equations
𝑦′ = 𝐹 (𝑥)𝑦 + 𝑔(𝑥),
where 𝐹 ∶ 𝐼 → 𝑀𝑛 (𝔽 ) and 𝑔 ∶ 𝐼 → 𝔽 𝑛 are continuous.
Then the initial value problem
𝑦′ = 𝐹 (𝑥)𝑦 + 𝑔(𝑥),
has a unique solution 𝑦 ∈ 𝐶 1 (𝐼, 𝔽 𝑛 ).
Proof. First assume that 𝐼 = [𝑐, 𝑑]. Then, for 𝑦1 , 𝑦2 ∈ 𝔽 𝑛 , 𝑥 ∈ 𝐼,
‖(𝐹 (𝑥)𝑦1 + 𝑔(𝑥)) − (𝐹 (𝑥)𝑦2 − 𝑔(𝑥))‖ ≤ ‖𝐹 (𝑥)‖ ‖𝑦1 − 𝑦2 ‖
(
)
≤ max ‖𝐹 (𝑥)‖ ‖𝑦1 − 𝑦2 ‖.
𝑥∈𝐼
Hence the Lipschitz condition is satisfied with 𝐵 = ∞. Now apply Theorem 7.13.
If 𝐼 is arbitrary, we can choose a decreasing sequence (𝑐𝑘 ) and an increasing sequence (𝑑𝑘 ) such that 𝑐1 ≤ 𝑎 ≤ 𝑑1
and
∞
⋃
𝐼=
[𝑐𝑘 , 𝑑𝑘 ].
𝑘=1
[If, e. g., the right endpoint 𝑑 of 𝐼 belongs to 𝐼, choose 𝑑𝑘 = 𝑑.]
By the first part of the proof, the IVP has a unique solution 𝑦𝑘 on [𝑐𝑘 , 𝑑𝑘 ]. Therefore, 𝑦𝑘+𝑚 (𝑥) = 𝑦𝑘 (𝑥) for all 𝑚 > 0
and 𝑥 ∈ [𝑐𝑘 , 𝑑𝑘 ], so that 𝑦(𝑥) = 𝑦𝑘 (𝑥) for 𝑥 ∈ [𝑐𝑘 , 𝑑𝑘 ] is well-defined and solves the IVP. Because this solution is
unique on [𝑐𝑘 , 𝑑𝑘 ] for each 𝑘, 𝑦 is unique.
For 𝜂 (𝑛) = 𝑓 (𝑥, 𝜂, 𝜂 ′ , … , 𝜂 (𝑛−1) ) define
⎛ 𝜂 ⎞
⎜ 𝜂′ ⎟
𝑦 = ⎜ . ⎟,
⎜ .. ⎟
⎜
⎟
⎝ 𝜂 (𝑛−1) ⎠
⎛ 𝑦2 ⎞
⎜ .. ⎟
𝐹 (𝑥, 𝑦) = ⎜ . ⎟.
⎜ 𝑦
⎟
𝑛
⎜
⎟
⎝ 𝑓 (𝑥, 𝑦) ⎠
Then
𝜂 (𝑛) = 𝑓 (𝑥, 𝜂, 𝜂 ′ , … , 𝜂 (𝑛−1) ) ⇔ 𝑦′ = 𝐹 (𝑥, 𝑦).
Let 𝑏0 , … , 𝑏𝑛−1 ∈ 𝔽 . Then
where
𝜂 (𝑗) (𝑎) = 𝑏𝑗 (𝑗 = 0, … , 𝑛 − 1) ⇔ 𝑦(𝑎) = 𝑏,
⎛ 𝑏0 ⎞
⎜
⎟
𝑏 = ⎜ ... ⎟.
⎜
⎟
⎝ 𝑏𝑛−1 ⎠
Clearly,
‖𝐹 (𝑥, 𝑦) − 𝐹 (𝑥, 𝑧)‖ ≤ (𝑛 − 1)‖𝑦 − 𝑧‖ + |𝑓 (𝑥, 𝑦) − 𝑓 (𝑥, 𝑧)|
and
|𝑓 (𝑥, 𝑦) − 𝑓 (𝑥, 𝑧)| ≤ ‖𝐹 (𝑥, 𝑦) − 𝐹 (𝑥, 𝑧)‖,
so that 𝐹 satisfies a Lipschitz condition if and only if 𝑓 satisfies a Lipschitz condition.
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Corollary 7.15. Let 𝐼 be a bounded closed interval, 𝑎 ∈ 𝐼, 𝑏0 , … , 𝑏𝑛−1 ∈ 𝔽 , 0 < 𝐵 ≤ ∞.
Let 𝑈 = {𝑦 ∈ 𝔽 𝑛 ∶ ‖𝑦 − 𝑏‖ ≤ 𝐵}, 𝑅 = 𝐼 × 𝑈 . Let 𝑓 ∶ 𝑅 → 𝔽 be continuous and assume there is 𝐿 > 0 such that
|𝑓 (𝑥, 𝑦) − 𝑓 (𝑥, 𝑧)| ≤ 𝐿‖𝑦 − 𝑧‖ for all 𝑥 ∈ 𝐼 and 𝑦, 𝑧 ∈ 𝑈 . Then there is 𝐴 > 0 such that the initial value problem
𝜂 (𝑛) = 𝑓 (𝑥, 𝜂, 𝜂 ′ , … , 𝜂 (𝑛−1) ),
𝜂 (𝑗) (𝑎) = 𝑏𝑗 (𝑗 = 0, … , 𝑛 − 1)
has a unique solution on 𝐼 ∩ [𝑎 − 𝐴, 𝑎 + 𝐴].
Note. According to Theorem 7.13 one uses 𝐴𝑀 ≤ 𝐵 where 𝑀 is the supremum of 𝐹 on 𝑅.
Finally, a combination of Corollaries 7.14 and 7.15 gives
Corollary 7.16. Let 𝐼 be an interval, 𝑎 ∈ 𝐼, 𝑏0 , … , 𝑏𝑛−1 ∈ 𝔽 , and consider the linear differential equation
𝜂 (𝑛) = 𝑓𝑛−1 (𝑥)𝜂 (𝑛−1) + 𝑓𝑛−2 (𝑥)𝑦(𝑛−2) + … 𝑓1 (𝑥)𝑦′ + 𝑓0 (𝑥)𝑦 + 𝑔(𝑥),
with initial conditions
𝑦(𝑗) (𝑎) = 𝑏𝑗 ,
(𝑗 = 0, … , 𝑛 − 1),
where 𝑓𝑗 ∶ 𝐼 → 𝔽 (𝑗 = 0, … , 𝑛 − 1) and 𝑔 ∶ 𝐼 → 𝔽 are continuous.
Then the initial value problem has a unique solution 𝜂 ∈ 𝐶 𝑛 (𝐼, 𝔽 ).
Tutorial 7.4.1. 1. Let 𝑛 ∈ ℕ∗ and 𝑀 ∈ 𝑀𝑛 (𝔽 ). Prove that
𝑛
max |𝑚𝑖𝑗 | ≤ ‖𝑀‖ ≤
𝑖,𝑗=1
𝑛
∑
|𝑚𝑖𝑗 |,
where 𝑀 = (𝑚𝑖𝑗 )𝑛𝑖,𝑗=1 .
𝑖,𝑗=1
2. Let 𝑀1 , 𝑀2 ∈ 𝑀𝑛 (𝔽 ).
a) Prove that ‖𝑀1 + 𝑀2 ‖ ≤ ‖𝑀1 ‖ + ‖𝑀2 ‖.
b) Let 𝑑(𝑀1 , 𝑀2 ) = ‖𝑀1 − 𝑀2 ‖. Prove that 𝑑 is a metric on 𝑀𝑛 (𝔽 ).
3. Prove that 𝑀𝑛 (𝔽 ) is complete.
Power Series and the Exponential Function in 𝑀𝑛 (𝔽 )
7.5
In the following let 𝐵 be either 𝔽 𝑛 or 𝑀𝑛 (𝔽 ). We know that 𝐵 is complete, see Theorem 7.3 and Tutorial 7.4.1, 3.
Definition 7.8. Let (𝑎𝑗 ) be a sequence in 𝐵. Then
∞
∑
𝑎𝑗 is called a series in 𝐵.
𝑗=0
The series is said to converge in 𝐵 if the sequence of its partial sums
𝑘
∑
𝑎𝑗 converges in 𝐵 as 𝑘 → ∞.
𝑗=0
The series is said to converge absolutely if
∞
∑
‖𝑎𝑗 ‖ < ∞.
𝑗=0
Since 𝐵 is complete, it follows as in Theorems 5.5 and 5.8:
Theorem 7.17. Let
∞
∑
𝑎𝑗 be a series in 𝐵.
𝑗=0
1. The series
∞
∑
𝑎𝑗 converges in 𝐵 if and only if for all 𝜀 > 0 there is 𝐾 ∈ ℝ such that for all 𝑘 ≥ 𝑚 ≥ 𝐾,
𝑗=0
‖∑
‖
‖ 𝑘
‖
‖
𝑎𝑗 ‖
‖
‖ < ∞.
‖𝑗=𝑚 ‖
‖
‖
2. If
∞
∑
𝑗=0
𝑎𝑗 converges absolutely, then it converges in 𝐵.
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Theorem 7.18. Let
∞
∑
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√
𝑎𝑗 be a series in 𝐵, assume that 𝜌 = lim sup 𝑗 ‖𝑎𝑗 ‖ < ∞ and put 𝑅 = 𝜌1 . Let 𝑎, 𝑥 ∈ ℝ.
𝑗→∞
𝑗=0
Then
1. The series
∞
∑
𝑎𝑗 (𝑥 − 𝑎)𝑗
𝑗=0
converges absolutely if |𝑥 − 𝑎| < 𝑅, and it diverges if |𝑥 − 𝑎| > 𝑅 in case 𝑅 < ∞.
∞
∑
𝑎𝑗 (𝑥 − 𝑎)𝑗 is differentiable for 𝑥 ∈ 𝐼, and
2. Let 𝐼 = ℝ if 𝑅 = ∞ and 𝐼 = (𝑎 − 𝑅, 𝑎 + 𝑅) if 𝑅 ∈ ℝ. The series
𝑗=0
the derivative is represented by the absolutely convergent series
∞
∑
𝑗𝑎𝑗 (𝑥 − 𝑎)𝑗−1
𝑗=1
Proof. The proof of part 1 is similar to that of Theorem ??.
For the proof of part 2 define for 𝑥 ∈ 𝐼:
𝑓 (𝑥) =
𝑔(𝑥) =
∞
∑
𝑗=0
∞
∑
𝑎𝑗 (𝑥 − 𝑎)𝑗 ,
𝑗𝑎𝑗 (𝑥 − 𝑎)𝑗−1 .
𝑗=1
It is straightforward to show that
√
√
√
√
√
lim sup 𝑗 ‖(𝑗 + 1)𝑎𝑗+1 ‖ = lim sup 𝑗 ‖𝑗𝑎𝑗 ‖ = lim 𝑗 ‖𝑗‖ lim sup 𝑗 ‖𝑎𝑗 ‖ = lim sup 𝑗 ‖𝑎𝑗 ‖ = 𝜌,
𝑗→∞
𝑗→∞
𝑗→∞
𝑗→∞
𝑗→∞
so that 𝑔 is an absolutely convergent series.
Now let 𝑥 ∈ 𝐼 and choose ℎ0 > 0 such that [𝑥 − ℎ0 , 𝑥 + ℎ0 ] ⊂ 𝐼. Then 𝑟 = max{|𝑥 − ℎ0 − 𝑎|, |𝑥 + ℎ0 − 𝑎|} < 𝑅.
Let ℎ ∈ ℝ with 0 < |ℎ| < ℎ0 . Since 𝑓 (𝑥), 𝑓 (𝑥 + ℎ) and 𝑔(𝑥) are represented by (absolutely) convergent series,
we have
𝑓 (𝑥 + ℎ) − 𝑓 (𝑥) − 𝑔(𝑥)ℎ = lim
𝑘→∞
= lim
𝑘→∞
𝑘
∑
[
]
𝑎𝑗 (𝑥 + ℎ − 𝑎)𝑗 − (𝑥 − 𝑎)𝑗 − 𝑗(𝑥 − 𝑎)𝑗−1 ℎ
𝑗=1
𝑘
∑
[
]
𝑎𝑗 (𝑥 + ℎ − 𝑎)𝑗 − (𝑥 − 𝑎)𝑗 − 𝑗(𝑥 − 𝑎)𝑗−1 ℎ .
𝑗=2
By the Mean Value Theorem, for each 𝑗 ≥ 2 there is ℎ𝑗 ∈ ℝ with 0 < |ℎ𝑗 | < |ℎ| such that
(𝑥 + ℎ − 𝑎)𝑗 − (𝑥 − 𝑎)𝑗 = 𝑗(𝑥 + ℎ𝑗 − 𝑎)𝑗−1 ℎ.
Another application of the MVT gives ̂
ℎ𝑗 ∈ ℝ with 0 < |̂
ℎ𝑗 | < |ℎ𝑗 | such that
[
]
(𝑥 + ℎ − 𝑎)𝑗 − (𝑥 − 𝑎)𝑗 − 𝑗(𝑥 − 𝑎)𝑗−1 ℎ = 𝑗ℎ (𝑥 + ℎ𝑗 − 𝑎)𝑗−1 − (𝑥 − 𝑎)𝑗−1
= 𝑗(𝑗 − 1)ℎ(𝑥 + ̂
ℎ𝑗 − 𝑎)𝑗−2 ℎ𝑗 ,
so that
‖
‖∑
𝑘
‖ 𝑘
[
]‖ ∑
‖ 𝑎 (𝑥 + ℎ − 𝑎)𝑗 − (𝑥 − 𝑎)𝑗 − 𝑗(𝑥 − 𝑎)𝑗−1 ℎ ‖ ≤
𝑗(𝑗 − 1)|ℎ| |ℎ𝑗 | |𝑥 + ̂
ℎ𝑗 − 𝑎|𝑗−2 ‖𝑎𝑗 ‖
𝑗
‖
‖
‖ 𝑗=2
‖ 𝑗=2
‖
‖
𝑘
∑
≤ ℎ2
𝑗(𝑗 − 1)𝑟𝑗−2 ‖𝑎𝑗 ‖.
𝑗=2
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With a reasoning as above, we have
√
lim sup 𝑗 𝑗(𝑗 − 1)𝑟𝑗−2 ‖𝑎𝑗 ‖ = 𝑟𝜌 < 1.
𝑗→∞
By the Root Test,
𝑐=
∞
∑
𝑗(𝑗 − 1)𝑟𝑗−2 ‖𝑎𝑗 ‖ < ∞,
𝑗=2
and therefore
‖𝑓 (𝑥 + ℎ) − 𝑓 (𝑥) − 𝑔(𝑥)ℎ‖ ≤ ℎ2 𝑐,
which proves that 𝑓 is differentiable at 𝑥 with
𝑓 ′ (𝑥) = 𝑔(𝑥).
Theorem 7.19. Let 𝐴 ∈ 𝑀𝑛 (𝔽 ). Then:
1. The matrix exponential
exp(𝐴) =
∞
∑
1
𝑗!
𝑗=0
𝐴𝑗
converges absolutely in 𝑀𝑛 (𝔽 ).
2. exp(𝐴) is invertible and (exp(𝐴))−1 = exp(−𝐴).
3. If 𝐴, 𝐶 ∈ 𝑀𝑛 (𝔽 ) commute, then
exp(𝐴 + 𝐶) = exp(𝐴) exp(𝐶).
4. 𝑥 → exp(𝐴𝑥) is differentiable on ℝ with
𝑑
exp(𝐴𝑥) = 𝐴 exp(𝐴𝑥) = exp(𝐴𝑥)𝐴.
𝑑𝑥
Proof. 1. We calculate
√
√
√
‖1 ‖
𝑗 ‖ ≤ lim sup 𝑗 1 ‖𝐴‖𝑗 = ‖𝐴‖ lim sup 𝑗 1
lim sup 𝑗 ‖
𝐴
‖ 𝑗! ‖
𝑗!
𝑗!
𝑗→∞
𝑗→∞
𝑗→∞
‖
‖
𝑗!
1
= ‖𝐴‖ lim
= ‖𝐴‖ lim
= 0,
𝑗→∞ (𝑗 + 1)!
𝑗→∞ 𝑗 + 1
and exp(𝐴𝑥) converges for all 𝑥 ∈ ℝ by Theorem 7.18, 1.
4. The differentiability follows from Theorem 7.18, 2, with
∞
∑
𝑗 𝑗 𝑗−1 ∑ 1 𝑗+1 𝑗
𝑑
exp(𝐴𝑥) =
𝐴𝑥
=
𝐴 𝑥
𝑑𝑥
𝑗!
𝑗!
𝑗=1
𝑗=0
=𝐴
Similarly,
𝑑
exp(𝐴𝑥) =
𝑑𝑥
∞
∑
1
𝑗!
𝑗=0
(∞
∑ 1
𝑗!
𝑗=0
𝐴𝑗 𝑥𝑗 = 𝐴 exp(𝐴).
)
𝐴𝑗 𝑥𝑗
𝐴 = exp(𝐴)𝐴.
2 and 3. A direct proof is quite cumbersome. Instead, we consider the matrix function
𝐹 (𝑥) = exp((𝐴 + 𝐶)𝑥) exp(−𝐶𝑥) exp(−𝐴𝑥)
and observe that exp(𝐴 + 𝐶), exp(𝐴) and exp(𝐶) commute with 𝐴 and 𝐶 since 𝐴 + 𝐶, 𝐴 and 𝐶 commute with 𝐴
and 𝐶. The product rule for derivatives and commutativity lead to
𝐹 ′ (𝑥) = exp((𝐴 + 𝐶)𝑥)(𝐴 + 𝐶) exp(−𝐶𝑥) exp(−𝐴𝑥)
− exp((𝐴 + 𝐶)𝑥)𝐶 exp(−𝐶𝑥) exp(−𝐴𝑥) − exp((𝐴 + 𝐶)𝑥) exp(−𝐶𝑥)𝐴 exp(−𝐴𝑥) = 0.
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Since 𝐹 (0) = (exp(0))3 = 𝐼𝑛3 = 𝐼𝑛 , where 𝐼𝑛 is the 𝑛 × 𝑛 identity matrix, and since
of Corollary 7.16 gives 𝐹 (𝑥) = 𝐼𝑛 for all 𝑥 ∈ ℝ.
Putting 𝐶 = 0 and 𝑥 = 1, we get
MATH2022
𝑑
𝐼 = 0, the uniqueness part
𝑑𝑥 𝑛
exp(𝐴) exp(−𝐴) = 𝐼𝑛 = exp(−𝐴) exp(𝐴),
which proves 2. Finally, with 2,
exp(𝐴 + 𝐶) = 𝐹 (1) exp(𝐴) exp(𝐶) = exp(𝐴) exp(𝐶).
How can one find exp(𝐴𝑥)? Recall from Linear Algebra that 𝜒(𝜆) = det(𝐴 − 𝜆𝐼𝑛 ) is the characteristic function
of 𝐴. Note that 𝜒 is a polynomial of degree 𝑛 with leading coefficient (−𝜆)𝑛 . In Complex Analysis you will learn
about the Fundamental Theorem of Algebra which says for this case that there are 𝑛 complex numbers 𝜆1 , … , 𝜆𝑛
such that
𝑛
∏
𝜒(𝜆) =
(𝜆𝑗 − 𝜆).
𝑗=1
Jordan’s canonical form, see Linear Algebra, says that the 𝜆𝑗 can be sorted into 𝑘 groups of equal number 𝜇1 , … , 𝜇𝑘 ,
i. e.,
𝑘
∏
𝜒(𝜆) =
(𝜇𝑗 − 𝜆)𝑝𝑗 ,
𝑗=1
such there is an invertible 𝑛 × 𝑛 matrix 𝑇 so that
𝐴=𝑇
( 𝑘
⨁
)
𝐷𝑗
𝑇 −1 ,
𝑗=1
where
𝐷𝑗 = 𝜇𝑗 𝐼𝑝𝑗 + 𝐽𝑝𝑗
and 𝐽𝜈 is the 𝜈 × 𝜈 matrix which has entries 1 just above the diagonal and 0 elsewhere. If we write, for 𝑙 ∈ ℕ,
𝐽𝜈𝑙 = (𝑚𝑟𝑠 )𝜈𝑟,𝑠=1 , then 𝑚𝑟𝑠 = 1 if 𝑠 − 𝑟 = 𝑙 and 𝑚𝑟𝑠 = 0 otherwise. In particular, 𝐽𝜈𝑙 = 0 if 𝑙 ≥ 𝜈.
Then
)𝑚
)
( 𝑘
( 𝑘
⨁
⨁
𝑚
−1
𝑚
𝐴 =𝑇
𝐷𝑗
𝑇 =𝑇
𝐷𝑗 𝑇 −1 ,
𝑗=1
which gives
exp(𝐴𝑥) = 𝑇
𝑗=1
( 𝑘
⨁
)
exp(𝐷𝑗 𝑥) 𝑇 −1 .
𝑗=1
It remains to find exp(𝐷𝑗 𝑥). For convenience, we drop the index and observe
𝑗 ( )
∑
𝑗 𝑗−𝑙 𝑙
(𝜇𝐼𝜈 + 𝐽𝜈 ) =
𝜇 𝐽𝜈 .
𝑙
𝑙=0
𝑗
Using
( )
𝑗!
1
1 𝑗
=
=
,
𝑗! 𝑙
𝑗!𝑙!(𝑗 − 𝑙)! 𝑙!(𝑗 − 𝑙)!
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it follows that
exp(𝐷𝑥) =
∞
∑
1
𝑗!
𝑗=0
(𝜇𝐼𝜈 + 𝐽𝜈 )𝑗 𝑥𝑗
∞
𝑗 ( )
∑
1 ∑ 𝑗 𝑗−𝑙 𝑙 𝑗
=
𝜇 𝐽𝜈 𝑥
𝑗! 𝑙=0 𝑙
𝑗=0
( )
𝑗 𝑗−𝑙 𝑙 𝑗
=
𝜇 𝐽𝜈 𝑥
𝑗! 𝑙
𝑙=0 𝑗=𝑙
∞
𝜈−1 ∑
∑
1
=
=
=
7.6
𝜈−1
∞
∑
1∑
1
𝜇𝑗−𝑙 𝐽𝜈𝑙 𝑥𝑗
𝑙!
(𝑗
−
𝑙)!
𝑗=𝑙
𝑙=0
𝜈−1
∞
∑
1∑ 1
𝑙! 𝑘=0 𝑘!
𝑙=0
𝜈−1 𝑙
∑
𝑥
𝑙!
𝑙=0
(𝜇𝑥)𝑘 𝐽𝜈𝑙 𝑥𝑙
exp(𝜇𝑥)𝐽𝜈𝑙 .
Solutions of Linear Differential Equations
In this section let 𝐼 be an interval, 𝑛 ∈ ℕ∗ , and 𝐹 ∈ 𝑀𝑛 (𝐶(𝐼)).
Definition 7.9. A matrix function 𝑌 ∈ 𝑀𝑛 (𝐶 1 (𝐼)) is called a fundamental matrix of 𝑦′ = 𝐹 𝑦 if 𝑦′ (𝑥) = 𝐹 (𝑥)𝑦(𝑥)
and if 𝑌 (𝑥) is invertible for all 𝑥 ∈ 𝐼.
Proposition 7.20. 1. 𝑦′ = 𝐹 𝑦 has a fundamental matrix.
2. Let 𝑌 be a fundamental matrix of 𝑦′ = 𝐹 𝑦. Then 𝑍 ∈ 𝑀𝑛 (𝐶 1 (𝐼)) is a fundamental matrix of 𝑦′ = 𝐹 𝑦 if and
only if there is 𝐶 ∈ 𝑀𝑛 (𝔽 ) such that 𝑍 = 𝑌 𝐶.
2
Proof. 1. Let 𝑐 ∈ 𝐼. Identifying 𝑀𝑛 (𝔽 ) with 𝔽 𝑛 , it follows from Corollary 7.14 that there is (a unique) 𝑌 ∈
𝑀𝑛 (𝐶 1 (𝐼)) such that 𝑌 ′ = 𝐹 𝑌 and 𝑌 (𝑐) = 𝐼𝑛 , and there is 𝑍 ∈ 𝑀𝑛 (𝐶 1 (𝐼)) such that 𝑍 ′ = −𝐹 T 𝑍 and 𝑍(𝑐) = 𝐼𝑛 .
Then
( T )′
( )′
𝑍 𝑌 = − 𝑍 T 𝑌 + 𝑍 T𝑌 ′
= −𝑍 T 𝐹 T 𝑌 + 𝑍 T 𝐹 T 𝑌 = 0
shows that 𝑍 T 𝑌 is constant. So we have
𝑍(𝑥)T 𝑌 (𝑥) = 𝑍(𝑐)T 𝑌 (𝑐) = 𝐼𝑛 ,
which implies that 𝑌 (𝑥) is invertible for all 𝑥 ∈ 𝐼.
2. If 𝑍 = 𝑌 𝐶, then 𝑍(𝑥) = 𝑌 (𝑥)𝐶 is the product of invertible matrices and hence invertible for all 𝑥 ∈ 𝐼, and
𝑍 ′ = (𝑌 𝐶)′ = 𝑌 ′ 𝐶 = 𝐹 𝑌 𝐶 = 𝐹 𝑍
shows that 𝑍 is a fundamental matrix of 𝑦′ = 𝐹 𝑦.
Conversely, if 𝑍 is a fundamental matrix of 𝑦′ = 𝐹 𝑦, then 𝑍(𝑥) is invertible for all 𝑥 ∈ 𝐼, and
(𝑌 −1 𝑍)′ = −𝑌 −1 𝑌 ′ 𝑌 −1 𝑍 + 𝑌 −1 𝑍 ′
= −𝑌 −1 𝐹 𝑌 𝑌 −1 𝑍 + 𝑌 −1 𝐹 𝑍 = 0,
which implies that 𝑌 −1 𝑍 is constant. So there is 𝐶 ∈ 𝑀𝑛 (𝔽 ) such that 𝑌 (𝑥)−1 𝑍(𝑥) = 𝐶 for all 𝑥 ∈ 𝐼. In
particular, 𝐶 is invertible since 𝑌 (𝑥)−1 and 𝑍(𝑥) are invertible. Finally, multiplying by 𝑌 (𝑥) from the right, we
get 𝑍(𝑥) = 𝑌 (𝑥)𝐶.
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Theorem 7.21. Assume that 𝐹 (𝑥) = 𝐹 (𝑦) for all 𝑥, 𝑦 ∈ 𝐼. Let 𝑐 ∈ 𝐼 and define
⎛ 𝑥
⎞
𝑌 (𝑥) = exp ⎜ 𝐹 (𝑡) 𝑑𝑡⎟ .
⎜∫
⎟
⎝𝑐
⎠
Then 𝑌 is a fundamental matrix of 𝑦′ = 𝐹 𝑦.
Proof. 𝑌 (𝑐) = exp(0) = 𝐼𝑛 is invertible. Since 𝐹 (𝑥) commutes with 𝐹 (𝑡) for all 𝑥, 𝑡 ∈ 𝐼, it follows that
𝑗
𝑗−1
⎡⎛ 𝑥
⎞⎤
⎛ 𝑥
⎞
𝑑 ⎢⎜
⎟
⎥
𝐹 (𝑡) 𝑑𝑡
= 𝑗𝐹 (𝑥) ⎜ 𝐹 (𝑡) 𝑑𝑡⎟
⎟⎥
⎜∫
⎟
𝑑𝑥 ⎢⎜∫
⎣⎝ 𝑐
⎠⎦
⎝𝑐
⎠
,
so that
𝑗
∞
⎛ 𝑥
⎞
𝑑 ∑ 1 ⎜
𝐹 (𝑡) 𝑑𝑡⎟
𝑌 (𝑥) =
⎟
𝑑𝑥 𝑗=0 𝑗! ⎜∫
⎝𝑐
⎠
′
𝑗
∞
⎡⎛ 𝑥
⎞⎤
∑
1 𝑑 ⎢⎜
=
𝐹 (𝑡) 𝑑𝑡⎟ ⎥
⎟⎥
𝑗! 𝑑𝑥 ⎢⎜∫
𝑗=0
⎣⎝ 𝑐
⎠⎦
⎞
⎛ 𝑥
=
𝐹 (𝑥) ⎜ 𝐹 (𝑡) 𝑑𝑡⎟
⎟
⎜∫
𝑗!
𝑗=1
⎠
⎝𝑐
∞
∑
𝑗
𝑗−1
⎞
⎛ 𝑥
= 𝐹 (𝑥) exp ⎜ 𝐹 (𝑡) 𝑑𝑡⎟
⎟
⎜∫
⎠
⎝𝑐
= 𝐹 (𝑥)𝑌 (𝑥).
Theorem 7.22. Let 𝑌 be a fundamental matrix of 𝑦′ = 𝐹 𝑦, 𝑓 ∈ 𝐶(𝐼, 𝔽 𝑛 ), 𝑐 ∈ 𝐼 and 𝑑 ∈ 𝔽 𝑛 . Then the unique
solution of the initial value problem 𝑦′ = 𝐹 𝑦 + 𝑓 , 𝑦(𝑐) = 𝑑 is
𝑥
𝑦(𝑥) = 𝑌 (𝑥)𝑌
−1
(𝑐)𝑑 + 𝑌 (𝑥)
∫
𝑌 −1 (𝑡)𝑓 (𝑡) 𝑑𝑡.
𝑐
Proof. We could simply substitute 𝑦 into the differential equation. However, we give a constructive proof, the
so-called method of variation of the constants. It is easy to see that any solution of the homogeneous equation is
𝑦(𝑥) = 𝑌 (𝑥)𝑧 with 𝑧 ∈ 𝔽 𝑛 . (we are not going to prove this separately because it is a special case of this theorem
and therefore follows anyway). Now replace the constant vector 𝑧 by a vector function on 𝐼, i. e., 𝑦(𝑥) = 𝑌 (𝑥)𝑧(𝑥),
where 𝑧 is the solution of the initial value problem. Then
𝑦′ (𝑥) = 𝑌 ′ (𝑥)𝑧(𝑥) + 𝑌 (𝑥)𝑧′ (𝑥)
= 𝐹 (𝑥)𝑌 (𝑥)𝑧(𝑥) + 𝑌 (𝑥)𝑧′ (𝑥),
so that
𝑧′ (𝑥) = 𝑌 −1 (𝑥)𝑌 (𝑥)𝑧′ (𝑥)
= 𝑌 −1 (𝑥)[𝑦′ (𝑥) − 𝐹 (𝑥)𝑦(𝑥)]
= 𝑌 −1 (𝑥)𝑓 (𝑥).
Since
𝑧(𝑐) = 𝑌 −1 (𝑐)𝑦(𝑐) = 𝑌 −1 (𝑐)𝑑,
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Introductory Analysis 2017 Lecture Manual
it follows that
87
𝑥
𝑧(𝑥) =
∫
𝑌 −1 (𝑡)𝑓 (𝑡) 𝑑𝑡 + 𝑌 −1 (𝑐)𝑑.
𝑐
Multiplication by 𝑌 (𝑥) completes the proof.
Note. 1. If 𝑛 > 1 and 𝐹 is not constant, then, in general, there is no closed form formula for the fundamental
matrix.
2. With the procedure outlined before Corollary 7.16, the results of this section also apply to linear 𝑛-th order
differential equations. In particular, the elements of the first row of a fundamental system are called a fundamental
system, say 𝑦1 , … , 𝑦𝑛 of the 𝑛-th order differential equation, and any solution of the homogeneous 𝑛-th order
equation is a linear combination of the fundamental system.
Conversely, if 𝑦1 , … , 𝑦𝑛 is a fundamental system of the 𝑛-th order differential equation, then
(
)𝑛
𝑌 = 𝑦(𝑖−1)
𝑗
𝑖,𝑗=1
is a fundamental matrix of the corresponding first order system.
3. In principle, if we consider an 𝑛-th order differential equation with constant coefficients
𝑦(𝑛) =
𝑛−1
∑
𝑎𝑗 𝑦(𝑗) ,
𝑗=0
one can use Jordan’s canonical form to find a fundamental system. However, this is very cumbersome. One can
proceed as follows: Consider the characteristic polynomial
𝜇𝑛 −
𝑛−1
∑
𝑎𝑗 𝜇 𝑗
𝑗=0
and find its distinct (complex) zeros, say 𝜇1 , … , 𝜇𝑘 , with multiplicities 𝑝1 , … , 𝑝𝑘 . Then the 𝑛 functions
𝑥𝑙 𝑒𝜇𝑗 𝑥 ,
𝑗 = 1 … , 𝑘, 𝑙 = 0, … , 𝑝𝑗 − 1
form a fundamental system of the 𝑛-th order linear differential equations.