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I read that for $j\in L^1({\mathbb R}^n)$ with $\|j\|_1=1$ and $f\in L^1_{\rm loc}({\mathbb R}^n)$ the mollifiers $j_\epsilon(x):=\epsilon^{-n}j(x/\epsilon)$ exhibit $j_\epsilon\ast f\in L^1({\mathbb R}^n)$ and $j_\epsilon\ast f\rightarrow f$ in $L^1({\mathbb R}^n)$ as $\epsilon\rightarrow 0$. But convergence in $L^1({\mathbb R}^n)$ does not imply convergence almost everywhere, yet I couldn't find a counterexample for this case, so do we additionally have $j_\epsilon\ast f\rightarrow f$ pointwise almost everywhere or not?

Svetoslav
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fweth
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  • What is $F^1$? Also, I think we need $|j,L^1|=1$. Also, if $j$ is regular, then the convolution also has some regularity. If you want to biuld some counterexamples, take $j$ very irregular. – TZakrevskiy Jan 08 '15 at 14:32
  • Sorry, $F^1=L^1$, edited! – fweth Jan 08 '15 at 14:40
  • Here is some work related to this, the question seems trickier than I anticipated. – fweth Jan 08 '15 at 14:42
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    If that article completely answers your question, I suggest posting that link as an answer and outline the major results form the article; like this people will see what is going on (even if the article is removed from that site or their server is down). – TZakrevskiy Jan 08 '15 at 14:45
  • It's not answering it completely but I'll still post it as an answer because it gives some interesting related theorems. I just have to work through it first... – fweth Jan 08 '15 at 14:50

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