# Do $L^2$ convergence and continuity imply pointwise convergence?

It is said here that $L^2$ convergence and continuity imply pointwise convergence (just before paragraph $5.2$) but I can't find how to prove it. Does anyone see how ?

• It seems wrong to me. Consider triangles of the hieght $1$, with base of the form $[k 2^{-n}, (k+1)2^{-n}]$ (varying over whole segment $[0,1]$). They all continuous and converge in $L_2$ to continuous function $0$ but convergence is not pointwise. Commented Apr 25, 2013 at 18:45
• If it were really simple, then why does he go on and prove things like Theorem 5.5(i) ?? Commented Apr 25, 2013 at 19:11

It's wrong : $f_n(x) = \log(n)e^{-nx}$ over $[0,1]$ is a sequence of continuous functions which converges in $L_2$ toward the zero function. However, $f_n(0) \rightarrow +\infty$
Convergence in $L^2$ does not imply convergence everywhere but it does imply point wise convergence almost everywhere. See http://en.m.wikipedia.org/wiki/Convergence_of_Fourier_series