7
$\begingroup$

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 ?

$\endgroup$
2
  • 3
    $\begingroup$ 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. $\endgroup$
    – Norbert
    Commented Apr 25, 2013 at 18:45
  • 1
    $\begingroup$ If it were really simple, then why does he go on and prove things like Theorem 5.5(i) ?? $\endgroup$
    – GEdgar
    Commented Apr 25, 2013 at 19:11

2 Answers 2

6
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – user74376
    Commented Apr 26, 2013 at 8:50
-1
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .