Question

(12) A function f: A R is called a step function if ran(f) is finite. Prove that for every R, there is a sequence fn: [a, b-R of step functions such continuous function f [a, b that fn(x)f(x) for all e la,b and fnfuniformly [a, b). on

Answer 1

Without loss of generality we can take a=0 and b=1. Therefore f is a function that is continuous.

R f: 0, 1

Define
n
as follows. Let
くam) = 1 (n) (n) n)く くI 0 i n
where
7 (n)
.

Set
f (x) inf re[i,i+1)
when
Ei, it1
and
(1f(1)
.

Now since f is continuous and [0,1] is compact, the function is bounded in every interval (Say
f(x) M
) and the infimum is attained at some point in the interval . Therefore
n
is well defined.

By definition $f_n(x) \leq f(x)$ .Since [0,1] is compact, f is actually uniformly continuous. Let
L
be given. Since f is uniformly continuous,  $\exists \delta >0$ such that $|f(x) - f(y)|<\epsilon$ when $|x-y|<\delta$. Now take N large enough such that
1/NS
.

Now $\forall n \geq N$ , for some $x^{'} \in [x_i,x_{i+1}]$ where $x \in [x_i,x_{i+1}]$ . Since $|x^{'} - x| \leq |x_{i+1} - x_i| \leq \frac{1}{n} < \delta$

$|f_n(x) - f(x)| = |f(x^{'} ) - f(x)| <\epsilon$

Hence $f_n \rightarrow f$ uniformly.