Homework 2
Math 501
Due September 12, 2014
Exercise 1 (Exercise 14 in Chapter 1)
Given y ∈ R, n ∈ N, and > 0, show that for some δ > 0, if |u − y| < δ then
|un − y n | < . [Hint: Prove the inequality when n = 1, n = 2, and then extend
your argument in general using the identity
un − y n = (u − y)(un−1 + un−2 y + · · · y n−1 .]
Exercise 2 (Exercise 15 in Chapter 1)
Given x >√0 and n ∈ N, prove that there is a unique y > 0 such that y n = x.
That is, y = n x exists and is unique. [Hint: Consider
y = sup{s ∈ R : sn ≤ x}.
Then use the previous exercise to show that y n can be neither less than nor
greater than x.]
Exercise 3 (Exercise 16 in Chapter 1, parts (a),(b), and (c))
Prove that real numbers correspond bijectively to decimal expansions not
terminating in an infinite string of 9’s, as follows. The decimal expansion of
x ∈ R is N.x1 x2 . . . where N is the largest integer less than or equal to x, x1 is
the largest integer less than or equal to 10(x − N ), x2 is the largest integer less
than or equal to 100(x − (N + x1 /10)), and so on.
(a) Show that each xk is a digit between 0 and 9.
(b) Show that for each k there is an ` ≥ k such that x` 6= 9.
(c) Conversely, show that for each such expansion N.x1 x2 . . . not terminating
in an infinite string of 9’s, the set
x1
x2
x1
+
, . . .}
{N, N + , N +
10
10 100
1
is bounded and its least upper bound is a real number x with decimal
expansion N.x1 x2 . . ..
Exercise 4 (Exercise 26 in Chapter 1)
A convex combination of w1 , . . . , wk ∈ Rm is a vector sum
w = s1 w1 + · · · + sk wk
such that s1 + · · · sk = 1 and s1 , . . . , sk ∈ [0, 1].
(a) Prove that if a set E is convex then E contains the convex combination of
any finite number of points in E.
(b) Why is the converse obvious?
Exercise 5 (Exercise 27 in Chapter 1, part (a))
Prove that the ellipsoid
2
y2
z2
3 x
E = (x, y, z) ∈ R : 2 + 2 + 2 ≤ 1
a
b
c
is convex. [Hint: E is the unit ball for a different inner product. What is it?
Does the Cauchy-Schwarz inequality not apply to all inner products?]
Exercise 6
Let (an ) ⊂ R be a sequence. Recall that the supremum of (an ) is simply
the supremum of the set {a1 , a2 , . . .}, which we denote as sup an (or sometimes as supn an ). For any n ≥ 1, let us denote the supremum of the set
{an , an+1 , an+2 , . . .} by supk≥n ak . Likewise, recall that the infimum of (an )
is simply the infimum of the set {a1 , a2 , . . .}, which we denote as inf an (or
sometimes as inf n an ). For any n ≥ 1, we denote the infimum of the set
{an , an+1 , an+2 , . . .} by inf k≥n ak .
The limit supremum, or lim sup, of (an ) as n → ∞ is defined as
lim sup an = lim sup ak .
n→∞
n→∞
k≥n
The limit infimum, or lim inf, of (an ) as n → ∞ is defined as
lim inf an = lim inf ak .
n→∞
n→∞
2
k≥n
One can think of the limit supremum as the limiting upper bound of a sequence,
and correspondingly the limit infimum as the limiting lower bound. It is not
immediately obvious that the limits in these definitions actually exist. We prove
that this is the case in part (b) below.
(a) Define
an =
1 + 1/n
−1 − 1/n
if n is odd,
if n is even.
(i) Show that lim supn→∞ an = 1.
(ii) Show that lim inf n→∞ an = −1.
(b) We show here that the limit supremum is a well defined real number whenever our sequence is bounded. Let (an ) ⊂ R be a bounded sequence.
(i) Suppose A ⊂ B ⊂ R. Show that sup A ≤ sup B.
(ii) For each n ≥ 1, define xn = supk≥n ak . Show that the sequence (xn )
is monotone.
(iii) Show that (xn ) converges.
Remark. If (an ) is not bounded above, it’s not hard to see that xn =
supk≥n ak = ∞ for all n ≥ 1. In this case, we say lim supn→∞ an = ∞. If (an )
is bounded above, but not bounded below, and has infinitely many peaks (in
the language of the previous assignment), then lim supn→∞ an = −∞.
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