Show uniform convergence of bounded functions implies uniform boundness.

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.

Solution 1:

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?