## Convolution of a continuous function and a finitely additive measure

Let $f$ be a continuous function on $\mathbb R$ with compact support and $\mu$ a finitely additive measure which is in the dual space of $L^\infty(\mathbb R)$. Is the convolution $f\ast \mu(x)=\int_{\mathbb R} f(x-y)d\mu(y)$ a continuous function in $x$? This is really an update of a question I asked, where I took $f$ to be only $L^\infty$ and I received the answer that in that case $f\ast \mu$ may not be continuous.

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You can use continuity in order to apply dominated convergence theorem (this differs from the case $f\in L^{\infty}$). Can you provide a link to the previous question? – Davide Giraudo Aug 14 at 8:18
@Davide: dominated convergence does not hold in general for finitely additive measures. – Wolfgang Loehr Aug 14 at 10:04
@Wolfgang: Space $C_{c}(\mathbb{R})$ is contained in $C_{0}(\mathbb{R})$ and by Riesz representation theorem we are really dealing with an honest, countably additive measure. It may be more interesting to ask, whether this holds for space of bounded continuous functions whose dual space is not so nice. – Mateusz Wasilewski Aug 14 at 10:24
@Mateusz & Davide: Yes, you are right, thank you. – Wolfgang Loehr Aug 14 at 10:43
Given Wolfgang's answer below, I think this result might be a special case of a general result in abstract harmonic analysis about LUC being a so-called introverted subspace of L^\infty. I will try to check this and find sources if time oermits – Yemon Choi Aug 14 at 17:42

Since $f$ has compact support, it is uniformly continuous. Let $h$ be a uniform modulus of continuity. If $|x'-x| < \varepsilon$, then $|f(x-y)-f(x'-y)| < h(\varepsilon)$ for all $y$, hence $|f*\mu(x)-f*\mu(x')| < h(\varepsilon)\|\mu\|$ (where $\|\cdot\|$ is variational norm) and $f*\mu$ is uniformly continuous.
Edit: As Mateusz pointed out, it becomes more interesting if $f$ does not need to vanish at $\infty$. For uniformly continuous $f$, the above still works and for $\sigma$-additive $\mu$ we can use dominated convergence as suggested by Davide. For arbitrary, bounded continuous $f$ and non-$\sigma$-additive $\mu$, $f*\mu$ need not be continuous:
Let $\mu$ be defined by $\mu(f) = \lim_{n\to\infty, n\in\mathbb{N}} f(n)$ if the limit exists and extend $\mu$ by Hahn Banach to a positive linear functional (a Banach-Mazur limit). Let $f$ be a continuous function which is zero on $[n-\frac1{|n|}, n+\frac1{|n|}]$ for every $n\in\mathbb{Z}$ and 1 for points which are more than $\frac2{|n|}$ away from every $n\in \mathbb{Z}$. Then $f*\mu(0) = 0$ but $f*\mu(x) = 1$ for $x$ close to but unequal zero.