8/11/2019 Homework 7.pdf
1/2
1. A sequence of functions{fn}, withfn: E R (E R) is pointwise
bounded if there is a function(x) : E R such tha|fn(x)| < (x)
for allx E. The sequence is uniformly bounded if there is a numberM
such that|fn(x)| < M for allx En N.
a)Give an example of a sequence which is pointwise bounded but
not uniformly bounded.
Solution: Consider the sequence of functions fn(x) = 1
nx, on (0, 1) where n N. Then, there is a function (x) = 1
x. It follo
that|fn(x)| = | 1
nx| < 1
x for all x (0, 1). Hence, {fn(x)} is pointwise bounded.
b)Prove that every uniformly convergent sequence of bounded
functions is uniformly bounded.
Proof: Let{fn(x)}n=1 be a uniformly convergent sequence of
bounded functions, say |fn(x)| Mn for all x and all n. Sincethe
sequence converges uniformly, it is a uniformly Cauchy sequence.
Hence there exists N such that |fm(x) fn(x)| 0. ChooseN1 so large
that |fm(x) f(x)| <
2 for all m > N1. Then, since f is continuous at x, choose
>
small |f(y) f(x)| < 2
if|y x| < . Finally choose N2 so large that |xn x| < ifn
> N2. Then ifn >max(N1, N2) we h
|fn(xn) f(x)| |fn(xn) f(xn)| + |f(xn) f(x)| <
Hence, limn fn(xn) = f(x).
d Is the converse to c) true?
Solution: The converse is not true in general. For example let
fn(x) be given by fn(x) = sin2(x) forn |x| n + 1 and
fn(x) = 0 for |x| n or |x| n + 1. Thus, fn(x) tends to zero,
since fn(x) = 0 ifn |x|, butfn(x) does not convergeuniformly, since
fn(n +
1
2) = 1. Then for any convergent sequence, say xn x, letN
max(|x|, |x1|, |x2|, . . . , |xn|, . . .). We
have fn(xn) = 0 for all n N, and so fn(xn) f(x).
2. Letfn be a sequence of continuous functions on[a, b].
a)Suppose that for eachx [a, b], fn(x) is a decreasing sequence
of real numbers, andfn(x) 0 pointwise. Then, fn 0uniformly on[a,
b].
Proof: Since [a, b] is closed and bounded, it is a compact set
by the Heine-Borel theorem. Let >0 be given. LetKn be theof all
x [a, b] withfn(x) . Sincefn is continuous, Kn is closed (from the
theorem a mapping f :X Y is continuous oif and only iff1(C) is
closed inXfor every closed setCinY). HenceKnis compact (from the
theorem closed subsets of com
ets are compact). Since fn> fn+1, we have Kn Kn+1. Fix x K.
Since fn(x) 0, we see that x / Kn ifn is sufficiently laThus x
/
Kn. In other words,
Kn is empty. Hence KNis empty for some N (from the theorem
if{K}a collection of com
ubsets of a metric space Xsuch that intersection of every finite
subcollection of{K} is nonempty, then
K is nonemptyollows that 0 fn(x)< for all x Kand for all n N.
This proves the statement.
b)Suppose that for eachx [a, b], fn is an increasing sequence of
real numbers, andfn fpointwise withf continuous. Tfn f uniformly
on[a, b].
Proof: The proof is similar to that of the problem above. Let
> 0 be given. For each n, let gn = f fn and let Kn beet of those
x [a, b] such that gn . Each gn is continuous, and so each Kn is
open (from the theorem a mapping f : Xs continuous if and only
iff1(V) is open in X for every open set V in Y). Sincefn is
increasing, gn is decreasing, so it folhat the sequence of Kn is
increasing. Sincefn converges pointwise to f, it follows that the
collection {Kn} is an open cova, b]. By compactness, we obtain that
there is some positive integer N such that KN = [a, b]. That is,
ifn > N and x [a, b], fn(x) f(x)| . Hence fn funiformly on [a,
b].
