Question #122940
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.)
1
Expert's answer
2020-06-23T19:54:57-0400

Let XX be a normed linear space (such as linear product space) and let {fnf_n } n_n ϵ\epsilon N_N be a sequence of elements of X.

We say that {fnf_n }nϵN_n\epsilon _N converges to fϵXf\epsilon X and write fnff_n\mapsto f if lim//ffn//lim//f-f_n// =0=0

nn\to \infin

ϵ>0\forall _\epsilon \gt 0 N>0N\gt 0 such that n>N//ffn//<ϵn\gt N\to //f-f_n//\lt\epsilon

A function f:ARf:A\to R where ARA\subset R

And suppose that CϵAC\epsilon A . Then ff is continuous at CC if for every ϵ>0\epsilon \gt 0 there exist a δ>0\delta \gt0 such that

xc<δ\mid x-c \mid \lt\delta and xϵAx\epsilon A implies that f(x)f(c)<ϵ\mid f(x)-f(c)\mid\lt\epsilon

fn:ERf_n : E\to R continuous n\forall n

fnffn\to f uniformly.

Let ϵ>0,Nϵ\epsilon \gt 0, \exists N_{\epsilon} such that n>Nϵ\forall n\gt N_{\epsilon}

fn(x)f(x)<\mid f_n(x)-f(x)\mid \lt ϵ3\epsilon\over 3 \forall xϵEx\epsilon E

Fix x0ϵEx_0 \epsilon E which means δ>0\exists \delta\gt 0

Such that

xx0\mid x-x_0 \mid <δ\lt \delta tends to fn(x)fn(x0)\mid f_n(x)-f_n(x_0) \mid <\lt ϵ3\epsilon\over 3 Is said to be continuous. We therefore show that f(x)f(x0)\mid f(x)-f(x_0)

=f(x)fn(x)fn(x0)+fn(x0)f(x0)=\mid f(x)-f_n(x)-f_n(x_0)+f_n(x_0)-f(x_0)\mid

\le f(x)fn(x)+fn(x)fn(x0)+fn(x0)(fx0)\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 ϵ3\epsilon\over 3 ++ ϵ3\epsilon \over 3 ++ ϵ3\epsilon\over 3 == ϵ\epsilon


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