Answer to Question #122940 in Real Analysis for Ruksan

Question

Come up with a definition of uniform convergence for a sequence of functions {fn} on a set A taking values in a normed linear space W over R.
Show that if A = [0, 1], and if {fn} is a sequence of continuous W-valued
functions on [0, 1] which converges uniformly to f : [0, 1] → W, then f is
continuous. (You will have to look at the notes on Lecture 1 that I posted
to learn the definition of continuous functions from a subset C of a normed
space V to a normed space W.)

Answer 1

Let $X$ be a normed linear space (such as linear product space) and let {$f_n$ } $_n$ $\epsilon$ $_N$ be a sequence of elements of X.

We say that {$f_n$ }$_n\epsilon _N$ converges to $f\epsilon X$ and write $f_n\mapsto f$ if $lim//f-f_n//$ $=0$

$n\to \infin$

$\forall _\epsilon \gt 0$ $N\gt 0$ such that $n\gt N\to //f-f_n//\lt\epsilon$

A function $f:A\to R$ where $A\subset R$

And suppose that $C\epsilon A$ . Then $f$ is continuous at $C$ if for every $\epsilon \gt 0$ there exist a $\delta \gt0$ such that

$\mid x-c \mid \lt\delta$ and $x\epsilon A$ implies that $\mid f(x)-f(c)\mid\lt\epsilon$

$f_n : E\to R$ continuous $\forall n$

$fn\to f$ uniformly.

Let $\epsilon \gt 0, \exists N_{\epsilon}$ such that $\forall n\gt N_{\epsilon}$

$\mid f_n(x)-f(x)\mid \lt$ $\epsilon\over 3$ $\forall$ $x\epsilon E$

Fix $x_0 \epsilon E$ which means $\exists \delta\gt 0$

Such that

$\mid x-x_0 \mid$ $\lt \delta$ tends to $\mid f_n(x)-f_n(x_0)$ $\mid$ $\lt$ $\epsilon\over 3$ Is said to be continuous. We therefore show that $\mid f(x)-f(x_0)$

$=\mid f(x)-f_n(x)-f_n(x_0)+f_n(x_0)-f(x_0)\mid$

$\le$ $\mid f(x)-f_n(x)\mid +\mid f_n(x)-f_n(x_0)\mid +\mid f_n(x_0)-(fx_0)\mid$

$\lt$ $\epsilon\over 3$ $+$ $\epsilon \over 3$ $+$ $\epsilon\over 3$ $=$ $\epsilon$