Math 501
Introduction to Real Analysis
Instructor: Alex Roitershtein
Iowa State University
Department of Mathematics
Summer 2015
Exam #1
Solutions
This is a take-home examination. The exam includes 8 questions. The total mark is 100
points. Please show all the work, not only the answers.
1. [12 points] Fix any d ∈ N and let k · k denote the usual Euclidean norm in Rd . That is,
v
u d
uX
x2i for a vector
x = (x1 , . . . , xd ) ∈ Rd .
kxk = t
i=1
Suppose that u, v ∈ Rd . Find w ∈ Rd and r > 0 such that
kx − uk = 2kx − vk
if and only if kx − wk = r.
Hint: Prove that w =
1
3
4v − u and r =
2
3
v−u .
First solution: This is a short solution based on a “completion to square” idea.
Start with the equation
kx − uk2 = 4kx − vk2 .
(1)
Observe that it is equivalent to
kxk2 − 2x · u + kuk2 = 4kxk2 − 8x · v + 4kvk2 ,
and hence to
kxk2 − 2x ·
4v − u
kuk2 − 4kvk2
=
3
3
Let w = 31 4v − u . Then the left-hand side is kxk2 − 2x · w. Add to both the sides kwk2 ,
to obtain
kuk2 − 4kvk2
+ kwk2
3
kuk2 − 4kvk2 1
4kvk2 − 8v · u + 4kuk2
=
+ k4v − uk2 =
3
9
9
4
=
kv − uk2 = r2 ,
9
kx − wk2 = kxk2 − 2x · w + kwk2 =
where r =
2
3
v−u .
1
Remark. This solution is using the hint, in particular it enjoys the a-priori knowledge of w.
To guess the value of w and r one can find two pints satisfying equation (1) on the straight
line connecting u and v and then observe that w must be the center of the segment connecting
these two points. The location of the two points is given by Equation (8) below. An easy
way to find them is to set x = v + t(u − v) = u + (1 − t)(v − u), plug these expressions
into, correspondingly, the right- and left-hand sides of equation (1), and solve the ressulting
quadratic equation to find two possible values of t.
Second solution: This solution is longer than the previous one. It exploits the same
orthogonal decomposition x = v − t(v − u) + z⊥ as the solution of a similar problem in the
sample, and in fact is mimicking the latter.
Assume that
kx − uk2 = 4kx − vk2 .
(2)
x = v − t(v − u) + z⊥ ,
(3)
Write
where
(v − u) · z⊥ = 0
and t ∈ R is a suitable constant. Thus −t(v − u) and z⊥ are projection of the vector x − v
into, respectively, the subspace spanned by v − u and its orthogonal complement.
Notice that (3) is equivalent to
x = u + (1 − t)(v − u) + z⊥ .
(4)
Thus, plugging the above expressions for x into (2) (respectively, (3) into the left-hand side
and (4) into the right-hand side) and using the Pythagoras theorem, we obtain
(1 − t)2 kv − uk2 + kz⊥ k2 = 4t2 kv − uk2 + 4kz⊥ k2 .
It follows that (2) is equivalent to
kz⊥ k2 =
(1 + t)(1 − 3t)
kv − uk2 .
3
(5)
For a fixed value of the parameter t, the right-hand side of (5) is a constant independent of
x. In general, in view of (3), for any constant a > 0,
kz⊥ k = a
is an equation of a sphere with center at v − t(v − u) in the
(d − 1)-dimensional subspace orthogonal to the vector v − u.
(6)
Since the right-hand side expression in the equation (5) must be non-negative, we obtain
that (1 + t)(1 − 3t) ≥ 0, and hence
−1 ≤ t ≤ 1/3.
2
(7)
Notice that kx − wk = r is an equation of the d-dimensional sphere with center at w and
radius equal r. In particular, the values of w and r, if exist, are uniquely determined by (2).
It follows from (6) and (7) that if such a sphere exists its center must correspond to the
middle of the segment with the endpoints (corresponding to the endpoints of the interval in
(7) describing the range of t):
x1 = v − (−1)(v − u)
and
1
x2 = v − (v − u).
3
(8)
By the symmetry, the center w of the sphere kx − wk = r should be located in the mid-point
of this segment, namely at v − t0 (v − u) with t0 = −1+1/3
= − 31 being the mid-point of the
2
interval (−1, 1/3). Thus
1
1
w=v− −
· (v − u) = 4v − u .
3
3
It follows then from (5) that
1 2
r2 = kx − wk2 = t +
ku − vk2 + kz⊥ k2
3
(1 + t)(1 − 3t)
4
1 2
ku − vk2 +
kv − uk2 = kv − uk2 ,
= t+
3
3
9
and hence r = 23 kv − uk. As we mentioned before, the values of w and r are uniquely
determined by (2), and hence the problem is solved once one admissible pair (w, r) is found.
2. [14 points] Solve Exercise 35 in Chapter 1 of the textbook.
Remark. Solutions of Exercises 2 and 3 are somewhat similar, both are based on a version of
Cantor’s diagonal trick. However, the solution to the former is slightly more technically involved, while the solution to the latter relies on the main idea in a more direct and transparent
way. It therefore might be of a benefit to students to first read the solution to Exercise 3.
Solution:
(a) Functions with the values in the set {0, 1} are commonly called binary functions. If f
is a binary function and A = {s ∈ S : f (s) = 1}, then f is called the indicator of set
A and is denoted by 1A . Let G : P → F be a function mapping the subsets of S into
their indicators:
G(A) = 1A .
Notice that any binary function is an indicator of some (possibly empty) set. Furthermore, 1A = 1B clearly implies A = B. Thus G is a bijection from P onto F. The
inverse function G−1 : F → P defined by G−1 (1A ) = A is the desired bijection from
F onto P. The existence of the bijection shows that F and P are equivalent (have the
same cardinality).
3
(b) First, notice that if S is empty or finite than
card(P) = card(P) = 2card(S) .
The first equality is the result in (a) while the second can be easily verified, for instance,
by induction on the number of elements in S.
In general, there is a trivial injection from S onto P defined by γ(s) = {s}, thus
mapping s ∈ S into the singleton {s} ∈ P. To show that S is not equivalent to F (and
hence is not equivalent to P) one can utilize a version of Cantor’s diagonal trick as
follows.
Assume that there is a bijection β : S → F. It will be convenient to denote β(s) as f (s) .
Then f (s) : S → {0, 1} is a binary function for each s ∈ S, and, since β is a bijection,
each element in F is f (s) for some s ∈ S. To see that this is in fact impossible, define
a binary function h : S → {0, 1} by setting
h(s) = not f (s) (s) = 1 − f (s) (s).
(9)
Since β is a bijection, we must have h = f (t) for some t ∈ S. However, if h = f (t) then
h(t) = f (t) (t) contradicting (9). This shows that h 6= f (t) = β(t) for any t ∈ S, and
hence β : S → F is actually not a bijection.
Remark. Here is an instructive alternative solution to 2(b). It is based on a similar but
different idea from the one hinted in the textbook. In particular, rather than compare S to F
it directly compares S to P.
Assume that there is a bijection α : S → P. Define E ∈ P as follows
E = {s ∈ S : s 6∈ α(s)}.
(10)
Since α is a surjection, E = α(s) for some s ∈ S. However,
• s ∈ E = α(s) directly contradicts (10).
• s 6∈ α(s) implies, by virtue of (10), that s ∈ E = α(s).
Either way we get a contradiction showing that, in fact, a bijection α : S → P doesn’t exist.
3. [12 points] Let E be the set of all real numbers x ∈ [0, 1] whose decimal expansion
contains only the digits 4 and 7. Is E countable?
Solution: One can utilize a version of Cantor’s diagonal trick as follows. Assume that E
is countable and let f : N → E be a bijection from N to E (enumeration of the elements in
E). Using decimal expansions write
f (n) = 0.an,1 an,2 an,3 · · · =
∞
X
k=1
4
10−k an,k ,
n ∈ N,
where an,k ∈ {4, 7}, k ∈ N. Consider the element of E
g = 0.b1 b2 b3 · · · =
∞
X
10−k bk ,
k=1
where bk ∈ {4, 7} are defined as follows:
bk = not ak,k = 11 − ak,k ,
k ∈ N.
(11)
Since f is a bijection, we must have g = f (n) for some n ∈ N. However, if g = f (n) then
bn = an,n contradicting (11). This shows that g 6= f (n) for any n ∈ N, and hence f is
actually not a bijection.
4. [14 points] Fix any real number α > 1. Take x1 >
xn+1 =
√
α − x2n
α + xn
= xn +
,
1 + xn
1 + xn
α and define recursively
n ∈ N.
(a) Prove that x1 > x3 > x5 > . . . .
(b) Prove that x2 < x4 < x6 < . . . .
√
(c) Prove that limn→∞ xn = α.
√
(d) Let εn = |xn − α|. Show that εn < cβ n for some constants c > 0 and β ∈ (0, 1).
Solution:
(a, b) We have:
√
√
√
α + xn √
( α − 1)( α − xn )
xn+1 − α =
− α=
1 + xn
1 + xn
(12)
Hence, since α > 1,
The sequence yn = xn −
√
α alternates the signs
(13)
n
α + α+x
2α + xn (1 + α)
α + xn+1
1+xn
=
=
,
α+xn =
1 + xn+1
2xn + 1 + α
1 + 1+xn
(14)
Furthermore,
xn+2
and hence
xn+2 − xn =
2α + xn (1 + α)
2(α − x2n )
− xn =
.
2xn + 1 + α
2xn + 1 + α
The claims in (a) and (b) follow from (13) and (15) combined together.
5
(15)
(c) It follows from the results in (a) and (b) that the sequences x2n and x2n+1 are monotone.
Furthermore, (13) implies that
√
x2n < α < x2n−1 ,
∀ n ∈ N.
Therefore the sequence are bounded and following limits exist:
t1 = lim x2n
n→∞
and
t2 = lim x2n+1 .
n→∞
Moreover, it follows from (14) that t1 and t2 are non-negative roots of the equation
t=
2α + t(1 + α)
.
2t + 1 + α
2
The last identity
√ yields 2t + t(1 + α) = 2α + t(1 + α), which has a unique non-negative
solution t = α. The proof of part (c) is complete.
√
(d) Let εn = |xn − α|. It follows from (12) that
√
( α − 1)εn
.
εn+1 =
1 + xn
Therefore,
√
εn+1
α−1
lim
=√
,
n→∞ εn
α+1
(16)
which implies that the error εn decays asymptotically with an exponential rate. More
precisely, pick any
√α − 1 β∈ √
,1 .
α+1
It follows from (16) and the definition of the limit that there is N ∈ N such that n > N
implies
εn+1
< β < 1.
εn
Thus, for any m ∈ N,
N +m−1
Y εk+1
εN +m
=
< β m.
εN
ε
k
k=N
Using the substitution n = N + m we can rewrite this as
εN
n > N.
εn < N β n ,
β
Therefore, setting
c = max
1≤k≤N
nε o
k
βk
,
we obtain
εn ≤ cβ n ,
∀ n ∈ N.
6
5. [12 points] Fix any d ∈ N and let k · k denote the usual Euclidean norm in Rd . That is,
v
u d
uX
x2i for a vector
x = (x1 , . . . , xd ) ∈ Rd .
kxk = t
i=1
We say that a sequence (xn )n∈N in Rd converges to x ∈ Rd and write limn→∞ xn = x if for
any ε > 0 there exists Nε ∈ N such that
n ∈ N and m, n > Nε
⇒
kxn − xk < ε.
Call two converging sequences X = (xn )n∈N and Y = (yn )n∈N in Rn equivalent and write
X ∼ Y if
lim kxn − yn k = 0.
n→∞
Let E be the set of all converging sequences in Rn . Show that ∼ is an equivalence relation
in E.
Solution: To show that ∼ is an equivalence relation we must verify the following three
property of the relation:
1. X ∼ X for any X ∈ E (reflexivity)
2. X ∼ Y implies Y ∼ X for any X, Y ∈ E (reflexivity)
3. X ∼ Y and Y ∼ Z imply together X ∼ Z for any X, Y, Z ∈ E (transitivity)
Take any X = (xn )n∈N ∈ E. Clearly, limn→∞ kxn − xn k = 0 and hence X ∼ X. Furthermore,
if Y = (yn )n∈N ∈ E and limn→∞ kxn − yn k = 0 then limn→∞ kyn − xn k = 0, and hence
X ∼ Y implies Y ∼ X. Finally, if Z = (zn )n∈Z ∈ E, the triangle inequality yields
kxn − zn k ≤ kxn − yn k + kyn − zn k,
and hence X ∼ Y and Y ∼ Z imply together X ∼ Z.
6. [12 points]
(a) Let (sn )n∈R be a sequence of reals such that
sn + sn−1
.
2
Show that sn is a Cauchy sequence and hence converges.
sn+1 =
(b) Let (sn )n∈R be a sequence of reals defined recursively by
s2n−1
,
2
Find lim supn→∞ sn and lim inf n→∞ sn .
s1 = 0, ,
s2n =
s2n+1 =
1
+ s2n .
2
Hint: Consider the sequences un = s2n and vn = s2n−1 separately.
7
Solution:
(a) Let In be the closed interval with endpoints sn−1 and sn . If sn−1 = sn the interval is
degenerate and consists of a single point. Let Ln be the length of the interval In , that
is Ln = |sn − sn−1 |. Then
Ln+1 =
sn + sn−1
1
− sn = Ln .
2
2
Iterating, we obtain Ln+1 = 2−n L1 , and hence
lim Ln = 0.
(17)
n→∞
Observe now that sn+1 is the mid-point of the interval In , and hence In+1 ⊂ In . In
particular, sk ∈ In for any n ∈ N and k > n. By virtue of (17), this implies that sn is
a Cauchy sequence and hence converges.
(b) Let un = s2n and vn = s2n−1 . In this notation, the recursions in the statement of the
problem can be rewritten as
un =
vn
2
and
vn+1 =
1
+ un .
2
It follows that
vn+1
1 1
= + un
2
4 2
(18)
1
1 1
+ un = + vn .
2
2 2
(19)
un+1 =
and
vn+1 =
To understand the structure of the sequences un and vn pick any constants α, β ∈ R and
set un = α + xn , vn = β + yn . In the new notation, (18) and (19) become, respectively,
α + xn+1 =
1
1 1
+ α + xn
4 2
2
β + yn+1 =
1 1
1
+ β + yn .
2 2
2
and
Thus, if we could choose parameters α and β in such a way that
α=
1 1
+ α
4 2
(20)
β=
1 1
+ β,
2 2
(21)
and
8
we would have xn+1 = 12 xn and yn+1 = 12 yn . The latter identities clearly imply
lim xn = lim yn = 0.
n→∞
Fortunately, (20) and (21) both have solutions, namely α =
account (22), this yields
lim un = α =
n→∞
1
2
(22)
n→∞
and
1
2
and β = 1. Taking into
lim vn = β = 1.
n→∞
(23)
It is not hard to verify, using the definition of the limit of a sequence and (23), that
any converging subsequence of sn cannot include infinitely many elements from the
sequence un and, at the same time, infinitely many elements from the sequence vn .
This implies that any converging subsequence of sn converges to either 1 or 12 . Using
the sequential definition of the lim sup and lim inf, this immediately yields
lim sup sn = 1
and
n→∞
1
lim inf sn = .
n→∞
2
7. [12 points] Let (sn )n∈R be a sequence of reals and define
n
1X
tn =
si .
n i=1
(a) Prove that if limn→∞ sn = s then limn→∞ tn = s.
(b) Give an example to show that tn can converge even though sn doesn’t.
Solution:
(a) Assume that limn→∞ sn = s. For ε > 0 and let Nε ∈ N be an integer such that n > Nε
implies |sn − s| < ε. Then for n > Nε , we have
n
|tn − s| =
n
1X
1X
(si − s) ≤
|si − s|
n i=1
n i=1
Nε
n
Nε
1 X
1X
1X
|si − s| +
ε≤
|si − s| + ε.
≤
n i=1
n i=N +1
n i=1
ε
Taking n → ∞ while ε and Nε remain fixed in the both sides of the resulting inequality
N
|tn − s| ≤
ε
1X
|si − s| + ε,
n i=1
we obtain
lim sup |tn − s| ≤ ε.
n→∞
9
Since ε > 0 is an arbitrary positive real, lim supn→∞ |tn − s| = 0 and hence
lim |tn − s| = 0.
n→∞
The latter result is equivalent to the desired assertion limn→∞ tn = s.
(b) Consider sn = (−1)n . Clearly, the sequence sn diverges. However, tn = (−1)n+1 /n and
hence limn→∞ tn = 0.
8. [12 points]
(a) Use induction to show that if (x + 1/x) is integer then (xn + 1/xn ) is also integer for
any n ∈ N.
(b) Show that
sup(A ∪ B) = max{sup A, sup B}
for any sets A, B ⊂ R.
Solution:
(a) Let Sn = (xn + 1/xn ), n ∈ N. Verify that Sn · S1 = Sn+1 + Sn−1 , and hence
Sn+1 = Sn · S1 − Sn−1 .
The claim follows from this identity by a version of the induction argument.
(b) The inequalities sup A ≤ sup(A∪B) and sup B ≤ sup(A∪B) follow from the inclusions
A ⊂ (A ∪ B) and B ⊂ (A ∪ B). These two inequalities together imply that
max{sup A, sup B} ≤ sup(A ∪ B).
(24)
On the other hand, sup A bounds from above any element of A while sup B bounds
from above any element of B. It follows that max{sup A, sup B} bounds from above
any element in A ∪ B. Hence
sup(A ∪ B) = max{sup A, sup B}.
The desired result follows from (24) and (25) combined together.
10
(25)