9.2 Uniform Convergence 551
(c) In Example 9.1.7 (c), the sequence {fn} of functions
{
l. if lxl < l. ·
fn(x) = l;I if ~ <-1;1'::; 1
converges pointwise on [O, 1] to the limit function f(x) = lxl.
Since 'in EN, llfn - fll = ~' llfn - fll-> 0, and so
fn(x) -> f uniformly on [O, l].
(d) On S = (-oo, oo), define
f(x) = x and
f n ( x) = { x + ~ if x ::;:: 0 or n ::;:: 11; y
~ if x < 0 and 1 < n :::; 10.
(See Figure 9.8.) Notice that fn - f
is not bounded when 1 < n :::; 10.
For n ::;:: 11, fn - f is bounded and
llfn - fll = ~ -> 0. Thus, by Defini- -----r
tion 9.2.3, fn(x) -> f uniformly.
/2
~!10
.--fnn~ll
f '
Note that none of the functions ,, ____
in this example is bounded. Uniform
convergence does not require that the
functions themselves be bounded. D
Figure 9.8
x
Theorem 9.2.7 (Uniform Cauchy Criterion) Suppose {fn} is a sequence
of functions in F(S, JR). Then Un} converges uniformly to some function f E
F(S, JR) if and only if
Ive> 0, :3 no EN 3 m,n ::;:: no* fn -fm is bounded and llfn -fmll < e.1
Proof. Suppose Un} is a sequence of functions in F(S, JR).
Part 1 ( ): Exercise 10.
Part 2 (<=): Suppose that Ve> 0, :3 no EN 3 m,n ::;:: no fn - fm
is bounded and llfn - fmll < e. Then for each x E S, {fn(x)} is a Cauchy
sequence of real numbers, so it converges to some number, call it f(x). In this
way we get a function f E F(S, JR). We shall show that fn-> f uniformly.
Let e > 0. By hypothesis, :3 no E N 3 m, n ::;:: no * fn - fm is bounded
and llfn -fmll < e/2. So,
m, n::;:: no* sup{lfn(x) - fm(x)I: x ES}< e/2.
Fix any n ::;:: no. Then,
'ix ES, lfn(x) - f(x)I = lim lfn(x) - fm(x)I:::; e/2,
m->oo