Continuity of a convolution (Version 2) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T01:19:42Z http://mathoverflow.net/feeds/question/76111 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76111/continuity-of-a-convolution-version-2 Continuity of a convolution (Version 2) Anand 2011-09-22T08:19:20Z 2011-09-30T10:54:50Z <p>Hello,</p> <p>This problem bothers me for some time. Suppose that</p> <ol> <li>$\mu$ is a non-negative Radon measure (or positive linear functional of the space of continuous functions with compact support);</li> <li>$\psi$ is a continuous function, vanishing at infinity and integrable, i.e., $\psi\in C^0_0(R)\cap L^1(R)$;</li> <li>$\sup_{x \in R}|(\psi*\mu)(x)|&lt;+\infty$.</li> </ol> <p>Then, we would like to prove that the function $x\mapsto (\psi*\mu)(x)$ is continuous.</p> <p>Thank you very much for your help and any hints!</p> <p>Anand</p> <hr> <p>Version 2. If we add an additional property, </p> <ol> <li>$\sup_{x\in R} |(G*\mu)(x)|&lt;+\infty$, where $G(x)=\frac{1}{\sqrt{2\pi}}\exp(-x^2/2)$.</li> </ol> <p>Then, is it possible to prove that the function $x\mapsto (\psi*\mu)(x)$ is continuous?</p> <p>Thanks</p> <p>Anand</p> http://mathoverflow.net/questions/76111/continuity-of-a-convolution-version-2/76112#76112 Answer by Igor Rivin for Continuity of a convolution (Version 2) Igor Rivin 2011-09-22T08:59:07Z 2011-09-22T08:59:07Z <p>This is a theorem of I. Glicksberg.See <a href="http://dl.dropbox.com/u/5188175/glickrev.pdf" rel="nofollow">http://dl.dropbox.com/u/5188175/glickrev.pdf</a> for the math review.</p> http://mathoverflow.net/questions/76111/continuity-of-a-convolution-version-2/76119#76119 Answer by Matthew Daws for Continuity of a convolution (Version 2) Matthew Daws 2011-09-22T12:08:09Z 2011-09-22T12:08:09Z <p>Here's a counter-example.</p> <p>Let $\mu$ be counting measure supported on $\mathbb Z$; so $\int f(x) \ d\mu(x) = \sum_{m\in\mathbb Z} f(m)$ for $f$ continuous with compact support.</p> <p>Choose a very rapidly decreasing sequence of positive reals $\delta_n$. Let $\psi$ be the piecewise linear function which is $1$ at $n+1/n$ for $n\geq 10$ (say), and is $0$ at $n+1/n \pm \delta_n$ (and is zero at all $x&lt;10+1/10-\delta_{10}$). By definition, $\psi$ is continuous (and could be made smooth by the use of a bump function) and $\psi$ is Lebesgue integrable (if $\delta_n$ decreases fast enough).</p> <p>Then set <code>$\alpha(x) = (\psi*\mu)(x) = \int \psi(x-y) \ d\mu(y) = \sum_{m\in\mathbb Z} \psi(x+m).$</code> A priori, this sum might diverge, but only to $+\infty$. Clearly $\alpha$ is periodic in that $\alpha(x+k)=\alpha(x)$ for any $k\in\mathbb Z$.</p> <p>Fix $x\in[-1/2,1/2)$, and consider which $m\in\mathbb Z$ are such that $\psi(x+m)>0$. This occurs iff there is $n\geq 10$ with $1/n-\delta_n &lt; x+m-n &lt; 1/n+\delta_n$. As $m-n\in\mathbb Z$ and $n\geq 10$ and $|x|\leq 1/2$, this can only occur if $m=n$. So $1/n-\delta_n &lt; x &lt; 1/n+\delta_n$. Choosing $(\delta_n)$ suitably, we can arrange that $1/(k+1)+\delta_{k+1} &lt; 1/k-\delta_k$ for all $k$, and then $n$ is unique for any given $x$.</p> <p>We conclude that for every $x$, there is at most one $m\in\mathbb Z$ with $\psi(x+m)>0$. In particular, $\alpha(x)\in[0,1]$ for all $x$.</p> <p>However, clearly $\alpha(1/n)\geq 1$ for all $n\geq 10$, but $\alpha(0) = 0$ as if $\psi(m)>0$ for some $m\in\mathbb Z$, then there is $n\geq 10$ with $-\delta_n &lt; m-n-1/n &lt; \delta_n$ which forces $m=n$, but then $-\delta_n &lt;-1/n&lt;\delta_n$, which we can avoid by a suitable choice of $(\delta_n)$. So $\alpha = \psi*\mu$ is not continuous at $0$.</p>