If I have a function $f \in H^s(\mathbb R^n)$, what can I say about $f^2$? In particular, I'm interested in the case when $s < 0$ and would like to conclude that, for example, $f \in H^{2s}(\mathbb R^n)$. Is something like this true?
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No, something like this is not true. In any negative Sobolev space, you can find a function f with a point singularity such that $f^2$ is not integrable. Hence $f^2$ is not even a distribution, and certainly not in $H^{2s}$.
answered Jul 10, 2017 at 18:27
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1$\begingroup$ Thanks! Would the additional hypothesis that $f$ is regular help at all? $\endgroup$ Commented Jul 10, 2017 at 19:49
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1$\begingroup$ Yes. If $f$ is in $H^s$ with $s>n/2$, then $f^2$ is also in $H^s$. But not in $H^{2s}$. $\endgroup$ Commented Jul 10, 2017 at 21:44
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$\begingroup$ To clarify: is there something more we could assume about $f$ such that it would not imply is was in some $H^s$ with $s > 0$, but would let us say that $f^2 \in H^{s'}$ for $s' < 0$? You stated the issue in general with $f \in H^s$, $s < 0$, is that $f$ could have a singularity. But this I assume you mean some akin to a delta function. Is that right? What if we assume $f$ doesn't have any such singularities? E.g. $f$ is a distribution that can be represented by a function. Would some such assumption suffice to ensure $f^2$ is a well-defined distribution and, moreover, in a Sobolev space? $\endgroup$ Commented Jul 11, 2017 at 1:24
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$\begingroup$ Well, yes. For instance if f is in $L^2=H^0$, then $f^2$ is in $L^1$, and by Sobolev embedding $f^2$ is in $H^{-r}$ for every $r>n/2$. You can argue along similar lines if $f$ is in $H^s$ for $0<s\le n/2$. $\endgroup$ Commented Jul 11, 2017 at 13:43
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$\begingroup$ This also provides a partial answer to my follow-up question: Is square of Delta function defined somewhere? $\endgroup$ Commented Jul 11, 2017 at 13:57