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

1. Dec 9, 2007

### rumjum

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.

Last edited: Dec 9, 2007
2. Dec 9, 2007

### arildno

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.

3. Dec 9, 2007

### HallsofIvy

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.

4. Dec 9, 2007

### rumjum

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.

5. Dec 9, 2007

### rumjum

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.