Let $\{f_{n}\}$ be continuous functions on $[0,1]$ which converge to a measurable function $f$ in the following sense: $$\lim_{n \to \infty} \int_{0}^{1} |f_{n}(x) – f(x)|^{2} dx \to 0$$
Show $f$ is continuous.
I'm still new at this, but my understanding is that the hypothesis is that the sequence $\{f_{n}\}$ converges to $f$ with respect to the $L^{2}$ norm. This should imply that $\lim_{n \to \infty} f_{n}(x) = f(x)$ (but I think I might need to make use of some theorem to justify the limit and the integral commuting?) From there, I believe you can justify that this convergence is actually uniform (almost everywhere) by Egorov's theorem?
I'm probably a bit off, my I would appreciate if someone could correct my misunderstandings.
Best Answer
The standard counterexample goes along the lines of
$$f_n(x) = \begin{cases}\qquad 0 &, x < \frac{1}{2} - \frac{1}{2^{n+1}} \\ 2^n\left(x-\frac{1}{2}+\frac{1}{2^{n+1}}\right) &, \frac{1}{2}-\frac{1}{2^{n+1}} \leqslant x \leqslant \frac{1}{2}+\frac{1}{2^{n+1}}\\ \qquad 1 &, x > \frac{1}{2} + \frac{1}{2^{n+1}}. \end{cases}$$
The $L^2$-limit of that sequence is
$$f(x) = \begin{cases}0 &, x < \frac{1}{2}\\ 1 &, x \geqslant \frac{1}{2} \end{cases}$$
(or any function almost everywhere equal to that), which has no continuous representative.