Homework 12 Solutions
Math 501
Due December 5, 2014
Exercise 1
Let D ⊂ R be a set. Let fn , f : D → R be functions such that fn ⇒ f .
Suppose that (xn ) ⊂ D is a sequence such that xn → x for some x ∈ D.
Let > 0. Observe that by the continuity of fn , it follows that f is continuous. Therefore, since xn → x, it follows that there exists N1 such that
|f (xn ) − f (x)| < /2
for all n ≥ N1 . Moreover, by uniform convergence, there exists N2 such that
|fn (z) − f (z)| < /2
for all n ≥ N2 and all z ∈ D. Let N = max{N1 , N2 } and observe that if n ≥ N
|fn (xn ) − f (x)| ≤ |fn (xn ) − f (xn )| + |f (xn ) − f (x)| < .
It follows that fn (xn ) → f (x).
Exercise 2
Let fn : [a, b] → R be continuous. Let g : [a, b] → R be a Lipschitz function.
Suppose that fn ⇒ f for some f : [a, b] → R.
(a) Note that as a subset of Cb [a, b], the set S = {fn : n ≥ 1} ∪ {f } is bounded.
Indeed, note that there exists N1 such that n ≥ N1 implies that
|fn (x) − f (x)| < 1
for all x ∈ [a, b]. It follows that n ≥ N1 implies |fn (x)| ≤ 1 + |f (x)| ≤
1 + kf k∞ for all x ∈ [a, b], which implies kfn k∞ ≤ 1 + kf k∞ for all n ≥ N1 .
It follows that if we let
M = max{kf1 k∞ , . . . , kfN1 −1 k∞ , 1 + kf k∞ }
1
then M is a bound for S.
Let > 0. Observe that there exists N such that n ≥ N implies
|fn (x) − f (x)| < /2M
for all x ∈ [a, b]. Let x ∈ [a, b] and observe that
|fn (x)2 − f (x)2 | = |fn (x) + f (x)||fn (x) − f (x)| < .
Since x was arbitrary, the result follows.
(b) Let K > 0 be the Lipschitz constant of g. That is,
|g(x) − g(y)| ≤ K|x − y|
for all x, y ∈ [a, b]. Now let > 0 and note that there exists N such that
n ≥ N implies
|fn (x) − f (x)| < /K
for all x ∈ [a, b]. Let x ∈ [a, b] and observe that
|g(fn (x)) − g(f (x))| ≤ K|fn (x) − f (x)| < .
Since x was arbitrary, the result follows.
Exercise 3
Let fn , f : R → R be given by
fn (x) = ex +
1
,
n
f (x) = ex .
(a) Let > 0, let N = d−1 e, and let x ∈ R. Observe that for n ≥ N
|fn (x) − f (x)| =
1
≤ .
n
Since x was arbitrary, it follows that fn ⇒ f .
(b) In order to show that fn2 does not converge uniformly to f , we must show
that there exists > 0 such that for all N ≥ 1 there exist x ∈ R and n ≥ N
such that
|fn (x)2 − f (x)2 | ≥ .
Let = 1 and let N ≥ 1. Observe that if x = log N and n = N , then
|fn (x)2 − f (x)2 | =
2ex
1
2ex
+ 2 ≥
= 2 > .
n
n
n
It follows that fn2 does not converge uniformly to f 2 .
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(c) The uniform boundedness of the sequence of functions was a key tool used
in Exercise 2. This property arose from the fact that the domain of the
(continuous) functions was compact. Here, the domain was not compact.
Exercise 4
Consider the series
f (x) =
∞
X
1
1 + n2 x
n=1
(a) Let D = {0} ∪ {−n−2 : n ≥ 1}. It follows easily from the comparison test
that the series converges for all x ∈ R \ D.
(b) We first claim that the series converges uniformly on any interval of the
form I = [a, b) or [a, b] for a > 0. Indeed, note that for any n ≥ 1 it’s easy
to see that
1
1
=
.
sup
2
1 + n2 a
x∈I 1 + n x
It follows from the Weierstrass M-test that the series converges uniformly on
this interval. Likewise, it is easy to check that the series converges uniformly
on any interval [a, b) or (a, b) for which b < −1.
Moreover, any interval which contains a point of D or has a point of D as
a limit point cannot be an interval on which the series converges uniformly.
It is obvious that if an interval contains a point of D then the series cannot
converge uniformly there. Moreover, if a point of D is a limit point of an
interval, then it follows from Exercise 1 that the series cannot converge
uniformly there.
(c) The function f is continuous on any interval for which the series converges
uniformly.
3