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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 ?

user74376
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    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. – Norbert Apr 25 '13 at 18:45
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    If it were really simple, then why does he go on and prove things like Theorem 5.5(i) ?? – GEdgar Apr 25 '13 at 19:11

2 Answers2

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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, $f_n(0) \rightarrow +\infty$

Norbert
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xghtfghhgyhhn
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  • This counterexample is just what I needed. For the one given by Norbert in the comments, we have convergence of subsequences, which is not the case here. Thanks a lot. – user74376 Apr 26 '13 at 08:50
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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

Baby Dragon
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