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## Homework Statement

{f_n} is a sequence of continuous functions on E=[a,b] that converges uniformly on E. for each x in E, set g(x)=sup{f_n(x)}. Prove that g is continuous on E

## Homework Equations

## The Attempt at a Solution

I've got an idea to prove it:

Let e>0 be given, there is a N such that

|f_n(x)-f(x)|<e

for all n>N and all x in E.

Let K_n={x|f_n(x)>=e+f(x)}, n=1,2,3....,N.

each K_n is closed since f_n is continuous.

Let [tex]A=\cup^{N}_{1} K_{i}[/tex] then A is closed.

g(x) is continuous on A (since there are only N functions to deal with. And I find it tedious to prove?)

Let B=E-A.

Then f(x)<=g(x)<f(x)+e, for any x in B.

let p be an interior point in B.

if g(p)=f_k(p), for some k>N.

since f_k(p) is continuous, there is a neighborhood of p such that if q is in it and also in B.

|f_k(p)-f_k(q)|<e

then,

|g(p)-g(q)|<=|f_k(p)-f_k(q)|+|f_k(q)-g(q)|<=e+|f_k(q)-f(q)|+|f(q)-g(q)|<=e+e+e=3e.

hence g(x) is continuous at p.

if g(p)=f(p).

|g(p)-g(q)|<=|f(p)-f(q)|+|f(q)-g(q)|=2e, which implies

g(x) is continuous at p.

It is analogous to say that g(x) is continuous if p is the boundary point of B.

It seems okay. But I'm not satisfied with it, it looks so lenthy and not well organized and I believe there is some other proof much better than this one. Any hint?