I understand that $L^2$ convergence does not imply pointwise convergence and vice versa. But I think that $L^2$ convergence must imply pointwise convergence a. e. $x$? So, since Schwartz functions are dense in $L^2$, then, for every $f\in L^2$, should it be the sequence $\{S_n\}$ of Schwartz functions that converges to $f$ a.e.?

Please, correct me or approve my understanding. Thank you. Marina


No, $L^2$ convergence doesn't imply pointwise convergence a.e .

This function is a counterexemple :

Let $n = 2^i + j$, with $0\leq j<2^i$ and define $f_n = \chi_{[\frac{j}{2^i},\frac{j+1}{2^i}]}$

This converge to $0$ in $L^2$ (as $\| f_n \| = 2^{-i}$), but converge nowhere to 0.

But you have the existence of a subsequence that converge pointwise almost everywhere


$L^2$ convergence does not imply pointwise a.e. convergence. One "well-known" example is the sequence $$ f_1=\chi_{[0,1]},\;f_2=\chi_{[0,\frac{1}{2}]},\; f_3=\chi_{[\frac{1}{2},1]},\; f_4=\chi_{[0,\frac{1}{4}]},\dots $$ in $L^2([0,1])$. This sequence converges to zero in $L^2$, but $\limsup_{n\to\infty}f_n(x)=1$ for all $x\in[0,1]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.