(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Prove that every sequence of bounded functions that is uniformly convergent is uniformly bounded.

2. Relevant equations

Let {fn} be the sequence of functions and it converges to f. Then for all n >= N, and all x, we have |fn -f| <= e (for all e >0). ---------- (1)

3. The attempt at a solution

This problem is from Rudin, 7.1. I am not clear about the part of "bounded function sequence".

But I suppose this is what I is meant.

|fn(x)| < Mn. , n = 1,2,3....

Also, I am unsure if f(x) (to which the sequene converges is bounded or not). That is is |f(x)| < some real number for all x. I suppose yes. But not sure. Here is my solution anyways.

=> |f1(x)| < M1,

|f2(x)| < M2, ...

|fN-1(x)| < M(N-1).

Also, let e =1 in (1), then n >= N implies that |fn-f| <=1

Hene, for n >=N and for all x , we have |fn| <= |fn-f| + |f| = |f| +1

Now, let M = max {M1,M2,....M(N-1), 1 +|f|}. for all x, where M is a real number.

If I can somehow state that |f|+1 is bounded, then for all n and for all x

|fn(x)| < M. Hence, the sequence is uniformly bounded.

I guess I can safely assume that |f(x)| < infinity for all x. Because

|fn(x)| < Mn. for all n.

Hence, lim (n->infinity) |fn(x)| < infinity for else for some n >N, we shall have an unbounded function in the sequence. But for lim(n->infinity) |fn(x)| = |f(x)|. Hence, |f(x)| < infinity and so is bounded.

Can someone verify? Thanks.

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# Homework Help: Every sequence of bounded functions that is uniformly converent is uniformly bounded

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