I'm trying to prove the following claim:
$f_n \in C_c$, $C_c$ being the set of continuous functions with compact support, then $\mathrm{lim}_{n \rightarrow \infty} || f_n – f||_{\infty} = 0$ implies $f_n(x) \rightarrow f(x)$ uniformly.
So, according to my understanding,
$$ \mathrm{lim}_{n \rightarrow \infty} || f_n – f||_{\infty} = 0 $$
$$ \Leftrightarrow $$
$$ \forall \varepsilon > 0 \exists N: n > N \Rightarrow |f_n(x) – f(x)| \leq || f_n – f||_{\infty} < \varepsilon$$ $\mu$-almost everywhere on $X$.
Now my problem is that I don't see how uniform convergence follows from pointwise convergence $\mu$ almost everywhere. Can someone give me a hint? I guess I have to use the fact that they have compact support but I don't see how to put this together.
Thanks for your help!
Best Answer
You were not very specific about your hypotheses - I assume you're working on $\mathbb{R}^{n}$ with Lebesgue measure.
Suppose there exists a point $x$ such that $|f_{n}(x) - f(x)| > \varepsilon$. Then there exists an open set $U$ containing $x$ such that for all $y \in U$ you have $|f_{n}(y) - f(y)| > \varepsilon$ by continuity. But this contradicts the a.e. statement you gave.
In case you don't know yet that $f$ is continuous (or rather: has a continuous representative in $L^\infty$), a similar argument shows that $(f_{n})$ is a uniform Cauchy sequence (with the sup-norm, not only the essential sup-norm), hence $f$ will be continuous (in fact, uniformly continuous).
Note that I haven't used compact support at all, just continuity.
If you're working in a more general setting (like a locally compact space), you'd have to require that the measure gives positive mass to each non-empty open set.
Finally, note that $f$ need not have compact support. It will only have the property that it will be arbitrarily small outside some compact set ("vanish at infinity" is the technical term). For instance, $\frac{1}{1+|x|}$ can easily be uniformly approximated by functions with compact support.