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We recall that if $f_1\in L^p(\mathbb R)$ and if $f_2\in L^q(\mathbb R)$ where
$1 \lt p \lt \infty$ and $\frac 1p+\frac1q=1$ then the function $f_1\ast f_2(x)=\int_{\mathbb R} f_1(x-y) f_2(y)dy$ is a continuous function in $x$. Now if we take $f$ to be a $L^\infty$ function on $\mathbb R$ and $\mu$ an element of the dual space of $L^\infty(\mathbb R)$ which is a finitely additive measure, is $f\ast \mu(x)=\int_{\mathbb R} f(x-y)d\mu(y)$ a continuous function?

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I'm fairly sure the answer is no, but would have to look up the proof or example (Hewitt and Ross vol. 1 is a natural place to look) – Yemon Choi Aug 10 '12 at 10:14
Yemon, thank you for comment, I removed my stupid answer. – Yulia Kuznetsova Aug 10 '12 at 10:36
If you have a modified question, then either ask it as a new question, or add it as an update to your existing question. Don't ask in the comments to an answer – Yemon Choi Aug 11 '12 at 18:49
Sorry. I would prefer to update the question. But I do not know how to do it. So I will make it another question. – spr Aug 14 '12 at 6:48
up vote 5 down vote accepted

This should be a comment on Yulia Kuznetsova's answer, but I didn't have the points and it's gone now anyway:

Can't you take $\mu$ as corresponding to a functional $f \mapsto f(0)$ if $f$ is an $L^\infty$ function continuous at $0$, extended to all of $L^\infty(\mathbb{R})$ by Hahn-Banach? Then if you take e.g. a step function $g = \mathbb{1}_{(0,\infty)}$ which is $0$ to the left of $0$ and $1$ to the right, $g \star \mu$ should be $0$ a.e. on $(-\infty,0)$ and $1$ a.e. on $(0,\infty)$, and so wouldn't have a continuous representative. What am I missing?

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I think this extension can be 0 on $g$ and on all its translations. – Yulia Kuznetsova Aug 10 '12 at 10:45
It can't be $0$ on translations of $g$ to the left. Any (nonzero) translation of $g$ is continuous at $0$. – Jakob Katz Aug 10 '12 at 10:50
I think it's correct; it is natural to approximate Dirac delta as well as one can. I have also tried to define a measure $\mu$ and a set $A \subset \mathbb{R}$ such that $x \mapsto \mu(x-A)$ is not continuous and that is exactly what we have witnessed here. – Mateusz Wasilewski Aug 10 '12 at 11:25
Jacob's answer is correct, and the answer to the original question is "no". – Alexandre Eremenko Aug 10 '12 at 13:15
Thanks to all who answered and commented. I am wondering now what happens if we restrict to a continuous $L^\infty$-function. Precisely if $f\in L^\infty(\mathbb R)$ is also continuous, is $f\ast \mu$ continuous where $\mu$ is such a finitely additive measure? – spr Aug 11 '12 at 11:35

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