Convolution of a continuous function and a finitely additive measure

2012-08-14T06:55:52Z

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.

Answer by Wolfgang Loehr for Convolution of a continuous function and a finitely additive measure

2012-08-14T10:13:33Z

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.

Convolution of a continuous function and a finitely additive measure - MathOverflow

Convolution of a continuous function and a finitely additive measure

Answer by Wolfgang Loehr for Convolution of a continuous function and a finitely additive measure