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Suppose that $f_{n} : E\rightarrow \mathbb{R}$ is a sequence of bounded functions that converge uniformly. Prove that there exists $M > 0$ such that for all $n \in \mathbb{N}$ and $x\in E$,

$$\left | f_{n}(x) \right | \leq M$$

I think this has to do with the supremum, but not quite sure.

RFZ
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user72211
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2 Answers2

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First, you might want to show that if each $f_n$ is bounded, the so is $f$. Can you do that?

Second, you might want to use $$|f_n(x)|-|f(x)|\le|f_n(x)-f(x)|$$ What happens if you take $\epsilon =1$ in the definition of uniform convergence?

Also, you can use that for each $\epsilon$ there is an $N$ for which $m,n\geq N$ imply $$|f_n(x)-f_m(x)|<\epsilon$$ for each $x$ in $E$. Assume for each $n$ we have over $E$ that $$|f_n(x)|<M_n$$ In such as case, take $N$ such that $n,m\geq N$ implies for every $x\in E$ that $$|f_n(x)-f_m(x)|<1$$

If we fix say $m=N$, then for every $n\geq N$, and every $x\in E$ $$|f_n(x)-f_N(x)|<1$$

But then whenever $n\geq N$, we have over $E$ that $$|f_n(x)|<1+|f_N(x)|<1+M_N$$

You are left with finitely many $M_1,\dots,M_{N-1}$ left. Can you obtain an uniform bound?

robjohn
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Pedro
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  • If you want, you can actually avoid introducing $f$ by using Cauchy. – Julien Apr 12 '13 at 04:01
  • @julien True. Whatever floats your boat. I have to go now. It would be nice if you added that too. – Pedro Apr 12 '13 at 04:01
  • I haven't been on a boat in a while, I miss that. Seriously, I just commented because you said "you need". I know you knew. – Julien Apr 12 '13 at 04:03
  • @julien I see. I'll change that to "you might want to" – Pedro Apr 12 '13 at 04:20
  • @julien See what I added. – Pedro Apr 12 '13 at 22:34
  • Looks good to me, +1. Just a remark: I think it is slightly more natural to take large inequalities when manipulating upper bounds. – Julien Apr 13 '13 at 21:57
  • @julien What do you mean? – Pedro Apr 13 '13 at 22:08
  • I mean, that if I am told, say, $f_n$ is uniformly bounded, then I take $M$ such that $|f_n(x)|\leq M$ for all $x$ and all $n$. Not $<M$. Of course, it is equivalent here. But in many situations, you will have to take some limit or something afterwards and lose the $<$ for a $\leq $. Also, the definition of an upper bound is with $\leq $. Saying $<$ instead of $\leq$ in such situations is just adding some extra information which is not needed. But I agree this is very nitpicking. – Julien Apr 13 '13 at 22:18
  • @julien Oh, OK. ${}{}{}{}$ – Pedro Apr 13 '13 at 22:19
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Have you tried drawing pictures? Write down the definition of uniform convergence and try to think of an example (e.g. a converging sequence of constant functions)

Also, consider the following simpler problem: Show that a convergent sequence $(x_n) \subset \mathbb R$ is bounded.

Paul Gustafson
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