## Every sequence of bounded functions that is uniformly converent is uniformly bounded

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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 Recognitions: Gold Member Homework Help Science Advisor The most important clue here is uniform convergence. Without that property, the theorem isn't true. I'll look over your proof later on, if necessary.
 Recognitions: Gold Member Science Advisor Staff Emeritus Since you were wondering about "sequence of bounded functions" (yes, for every n, there exist number Mn such that |fn(x)|< Mn), as you clear on "uniformly bounded"? That simply means that "there exist an number M such that, for all n, |fn(x)|< M". That is, that you can choose a single number M rather than a different Mn for each n. Notice that this does NOT ask you to prove anything about the limit of the sequence- and, in particular, |f(x)|< infinity does NOT mean the function is bounded! The simple function f(x)= x satisfies the condition that |f(x)|< infinity, but is not a bounded function.

## Every sequence of bounded functions that is uniformly converent is uniformly bounded

 Quote by HallsofIvy Notice that this does NOT ask you to prove anything about the limit of the sequence- and, in particular, |f(x)|< infinity does NOT mean the function is bounded! The simple function f(x)= x satisfies the condition that |f(x)|< infinity, but is not a bounded function.
That is an important point that you brought up. Thanks, for that. I solved the problem by showing that |f(x)| < M(N+1)+1 for e=1 and |fn| < Mn. And, since for n >=N, the function is uniformly bounded, we have |f(x)| < 1 + M(N+1). Henc,e |f(x)| is bounded.

Thanks, again.

 Quote by HallsofIvy Since you were wondering about "sequence of bounded functions" (yes, for every n, there exist number Mn such that |fn(x)|< Mn), as you clear on "uniformly bounded"? That simply means that "there exist an number M such that, for all n, |fn(x)|< M". That is, that you can choose a single number M rather than a different Mn for each n. Notice that this does NOT ask you to prove anything about the limit of the sequence- and, in particular, |f(x)|< infinity does NOT mean the function is bounded! The simple function f(x)= x satisfies the condition that |f(x)|< infinity, but is not a bounded function.

That is an important point that you brought up. Thanks, for that. I solved the problem by showing that |f(x)| < M(N+1)+1 for e=1 and |fn| < Mn. And, since for n >=N, the function is uniformly bounded, we have |f(x)| < 1 + M(N+1). Henc,e |f(x)| is bounded.

Thanks, again.

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