Non-analyticity of convolution - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T10:21:12Z http://mathoverflow.net/feeds/question/106026 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106026/non-analyticity-of-convolution Non-analyticity of convolution Alex A 2012-08-31T11:55:33Z 2012-09-02T14:53:10Z <p>I have posted a similar question in the past but let me make a final try in a simpler framework. </p> <p>Let $g \in C_0 ^\infty (\mathbb{R})$ be smooth and compactly supported. Define <code>$$f(x) = \int \big ((x - y)^2 - 1 \big )^{1/2}(x-y) g (y) \,dy$$</code> where integration is performed over the set where $|y - x|>1$ and $y\in \operatorname {supp}g$. </p> <p>If $g$ fails to be real analytic at some point $x_0$ can we deduce that also $f$ fails to be real analytic at some point depending on $x_0$, like perhaps $x_0 \pm 1$? </p> http://mathoverflow.net/questions/106026/non-analyticity-of-convolution/106032#106032 Answer by Alexandre Eremenko for Non-analyticity of convolution Alexandre Eremenko 2012-08-31T12:54:11Z 2012-09-02T14:53:10Z <p>Your function $g$ in $C_0^\infty$, if not identicaly equal to zero, certainly fails to be real analytic at some point. Because a real analytic function cannot be in $C_0^\infty$. Your convolution is irrelevant for this conclusion.</p> <p>Edit: your comment indicates that you are really asking about the relation of the singular sets of $f$ and $g$. The singular set $S(f)$ (singular support) is the set where the function is not analytic. The general fact here is that $S(f)$ is contained in $(S(g)+1)\cup (S(g)-1)$, the union of shifts of $S(g)$ by one unit left and right. Perhaps this answers your question. This can be found in the second volume of Hormander's Analysis of linear differential operators, sect 16.3, together with a discussion when the equality happens.</p>