Suppose $f_n: [0,1] \mapsto \mathbb{R} $ are continuous functions. I'm interested in knowing under what conditions does $f_n \stackrel{L^2[0,1]}{\to}f$ imply $\sup_{x \in [0,1]} |f_n(x)-f(x)| \to 0$. Clearly one example that implies this is if each $f_n$ is uniformly bounded, and has uniformly bounded first and second derivatives (the result follows from Arzela-Ascoli in this case, Does Lp-convergence and uniform boundedness in $C^2$, imply $C^{1}$ convergence?). Is this basically necessary and sufficient? Or are there weaker conditions under which convergence in mean-square implies convergence uniformly?
When does convergence in $L^2$ imply convergence in $C[0,1]$
functional-analysisprobability theoryreal-analysis
Best Answer
The theorem of Ascoli-Arzelá is necessary and sufficient. By this I mean that, if the sequence of continuous functions $f_n$ converges uniformly to some continuous function, then $f_n$ is uniformly bounded and equicontinuous. This is not hard to prove.
However, the a priori knowledge that $f_n\to f$ in $L^2(0,1)$ does give an important piece of information: if $f_n$ is uniformly bounded and equicontinuous, then $f_n\to f$ uniformly. Indeed, by the theorem of Ascoli-Arzelá, $f_n$ and each of its subsequences have a subsequence that converges uniformly to some continuous function. Such function must be $f$, because uniform convergence implies $L^2$ convergence on bounded intervals. And so we can conclude that $f_n\to f$ uniformly, as claimed.
The point is that Ascoli-Arzelá gives uniform convergence of a subsequence to some continuous function, which is completely unknown a priori. The $L^2$ convergence allows us to conclude that the whole sequence $f_n$, not just a subsequence, converges to $f$, not just to some function.