0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The constant $M$ may be different for different $f_n$s. $\endgroup$ – Neal Nov 12 '13 at 3:50
1
$\begingroup$

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 :)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.