Convolution of a continuous function and a finitely additive measure - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T11:09:33Z http://mathoverflow.net/feeds/question/104670 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104670/convolution-of-a-continuous-function-and-a-finitely-additive-measure Convolution of a continuous function and a finitely additive measure spr 2012-08-14T06:55:52Z 2012-08-14T11:01:45Z <p>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. </p> http://mathoverflow.net/questions/104670/convolution-of-a-continuous-function-and-a-finitely-additive-measure/104682#104682 Answer by Wolfgang Loehr for Convolution of a continuous function and a finitely additive measure Wolfgang Loehr 2012-08-14T10:13:33Z 2012-08-14T11:01:45Z <p>Since <code>$f$</code> has compact support, it is uniformly continuous. Let <code>$h$</code> be a uniform modulus of continuity. If <code>$|x'-x| &lt; \varepsilon$</code>, then <code>$|f(x-y)-f(x'-y)| &lt; h(\varepsilon)$</code> for all <code>$y$</code>, hence <code>$|f*\mu(x)-f*\mu(x')| &lt; h(\varepsilon)\|\mu\|$</code> (where <code>$\|\cdot\|$</code> is variational norm) and $f*\mu$ is uniformly continuous.</p> <hr> <p>Edit: As Mateusz pointed out, it becomes more interesting if <code>$f$</code> does not need to vanish at <code>$\infty$</code>. For <em>uniformly continuous</em> <code>$f$</code>, the above still works and for <code>$\sigma$</code>-additive <code>$\mu$</code> we can use dominated convergence as suggested by Davide. For arbitrary, bounded continuous <code>$f$</code> and non-<code>$\sigma$</code>-additive <code>$\mu$</code>, <code>$f*\mu$</code> need not be continuous:</p> <p>Let <code>$\mu$</code> be defined by <code>$\mu(f) = \lim_{n\to\infty, n\in\mathbb{N}} f(n)$</code> if the limit exists and extend $\mu$ by Hahn Banach to a positive linear functional (a Banach-Mazur limit). Let <code>$f$</code> be a continuous function which is zero on <code>$[n-\frac1{|n|}, n+\frac1{|n|}]$</code> for every <code>$n\in\mathbb{Z}$</code> and 1 for points which are more than <code>$\frac2{|n|}$</code> away from every <code>$n\in \mathbb{Z}$</code>. Then <code>$f*\mu(0) = 0$</code> but <code>$f*\mu(x) = 1$</code> for <code>$x$</code> close to but unequal zero.</p>