# Math Help - sup of equicontinuous is continuous

1. ## sup of equicontinuous is continuous

Let U={f(x)} be a subset of C_0 (the set of continuous functions f:[a,b]->R) that is equicontinuous and bounded. Prove that sup(f(x): feU) is a continuous function of x.

Note: I am not sure if the question means U is itself bounded wrt the sup norm so that sup(|f(x)-g(x)| : xeR) < M for some M or if the question is just poorly worded and really means that each f(x) is point-wise bounded so that sup(f(x): feU) always exists. Then again, f(x) is continuous and [a,b] compact so boundedness is already a given property of the functions...

I am one step short of a proof:

• Define g(x)=sup(f(x): feU)
• Fix x and choose any epsilon.
• By the definition of supremum, there is some f_1 in U s.t. |f_1(x) - g(x)| < epsilon.
• By equicontinuity, there exists a delta such that |x-y|<delta => |f_1(x) - f_1(y)| < epsilon.
• By the definition of supremum, there is there is some f_2 in U s.t. |f_2(y)-g(y)| < epsilon (note that at this step, y determines which f_2 is selected from U).
• Now, to show continuity of g at x, I need to show that f_1(x) is close to f_2(y).

Problems:

• I know that f_1(x) is close to f_1(y) and IF I could show that f_1(x) is close to f_2(x) [or that f_1(y) is close to f_2(y)] then I would be done by triangle inequality.

• I know that f_1(y) shouldn't be far from f_2(y) because h = |f_1 - f_2| = 0 somewhere on (x-delta, x+delta) and h is continuous so it is bounded on (x-delta, x+delta) by some constant C but to make this C as small as epsilon I need to make my delta-neighborhood smaller which will change the f_2 that is selected from U and therefore change the point where h=0 and change the constant C.

I would very much appreciate any help correcting this proof. I knew it was a bad idea to fix y in order to select an f_2 and I also haven't used boundedness

2. ## Re: sup of equicontinuous is continuous

Since $U$ is equicontinuous and bounded in $C_0$, the Arzela-Ascoli theorem implies that $U$ is relatively compact.

So, if $g=supU$, there exists a sequence $(g_n)\subset U$ with $g_n\rightarrow g$ pointwise.
But then, $(g_n)$ must possess a subsequence (denoted still by $(g_n)$) convergent in $C_0$.

This means $g_n\rightarrow g$ in $C_0$, which proves that $g$ is the uniform limit of continuous functions.