# Prove that $f$ is a bounded function on $S.$

The question is this.

Let$(f_n)$ be a sequence of bounded functions on a set $S$, and suppose that $f_n \rightarrow f$ uniformly on $S$. Prove that $f$ is a bounded function on $S$.

My work is below.

Proof.

Since $(f_n)$ is bounded sequence of functions, we know that $|f_n(x)| < M,\forall x \in S$ and for some real number $M.$ Also, we know that $\forall \epsilon > 0, \exists N \ s.t\ n > N \Rightarrow |f_n(x) - f(x)| < \epsilon, \forall x\in S$ and $\forall n > N.$ Then, $$\begin{eqnarray} -\epsilon&<& f_n(x) - f(x)&<& \epsilon \\ -\epsilon - f_n(x) &<& f(x) &<&\epsilon - f_n(x) \end{eqnarray}$$ Since we know that $-f_n(x) \leq M$ and $-M \leq -f_n(x),$ we have $$-\epsilon - M \leq -\epsilon - f_n(x) < f(x) <\epsilon - f_n(x)\leq \epsilon + M.$$ Setting $\epsilon = 1$, we have $|f(x)| \leq M+1, \forall x \in S. \square$

Is this valid??? My text book says some other thing.

• The constant $M$ may be different for different $f_n$s. – Neal Nov 12 '13 at 3:50

## 1 Answer

Your proof is almost right, unfortunately almost right means wrong.

The problem with your proof is that when you say that $|f_n(x)| < M,\forall x \in S$, the $M$ depends on $n$, different $n$'s lead to different $M$'s.

But you are on the right track, all you have to do is to change the order of the steps. You know that $\epsilon =1$ is what will work, so start by picking $\epsilon =1$, pick then a good $n$ and only last pick $M$ for that chosen $n$.

Corrected version

Pick $\epsilon =1$. Then $\exists N$ s.t for all $n > N$ we have

$$\left|f_n(x) - f(x) \right| < 1, \forall x\in S \,.$$

Pick some fixed $n >N$.

Since $f_n$ is bounded there exists some $M$ such that $$|f_n(x)| < M,\forall x \in S \,.$$

Then, by the triangle inequality we have

$$|f(x) \leq |f(x)-f_n(x)|+|f_n(x) < 1+M \forall x \in S \,.$$

\square

As you see, you had all the right ideas, you just did the steps in the wrong order :)