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Let $f\in H^{s_1}(\mathbb{R}^n)$ and $g\in H^{s_2}(\mathbb{R}^n)$, where $s_1, s_2 \in \mathbb{R}$ and can be positive or negative.

It is easy to show that $f *g$ is defined pointwise when $s_1+s_2\ge0$. But what smoothness does this convolution have (what $H^s$ space is it in?)?

Second, what is the most general case possible such that $f*g$ exists? Is there some way of defining such a convolution as a distribution regardless of how small $s_1+s_2$ is?

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  • $\begingroup$ The convolution need not lie in any $H^s$. For that to happen, we need $\widehat{f}\widehat{g}$ to lie in $L^2$ after multiplication by some power, but that won't happen if the product isn't at least locally $L^2$ to start with. $\endgroup$
    – Christian Remling
    Commented Jun 12, 2014 at 22:46
  • $\begingroup$ If $s_1+s_2 \ge 0$, $f*g$ is defined pointwise since it should always be less than or equal to $\|f\|_{H^{s_1}}\|g\|_{H^{s_2}}$, and in that case it is certainly in $L^2_{loc}$. I am trying to determine how smooth it has to be. I don't think that the product of the Fourier transforms has to be an $L^2$ function; when 1 denotes the constant 1 function, $1*f(x)$ exists for all $f$ in $L^2_0(\mathbb{R})$, but the product of the Fourier transforms is not an $L^2$ function in general. The test might be that the product of the Fourier transforms exists (so for example, no convoluting 1 and 1). $\endgroup$
    – Zorgoth
    Commented Jun 12, 2014 at 23:03
  • $\begingroup$ Also, from a practical standpoint, I want to evaluate smoothness and decay rates at infinity separately, so I am interested in the best $H^s_{loc}$ space. $\endgroup$
    – Zorgoth
    Commented Jun 12, 2014 at 23:07
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    $\begingroup$ The trivial answer (for positive indices) is this: using that $\widehat{f*g} = \hat{f} \hat{g}$, you have that $|\widehat{f*g}|\cdot (1 + |\xi|)^{s_1+s_2} \in L^1$ (note that this does not guarantee that $\widehat{f*g} \in L^2_{loc}$). So you have that $f*g \in W^{s_1+s_2,\infty}$. This also guarantees that the convolution is in $H^{s_1+s_2}_{loc}$ using the trivial embedding $L^\infty\subset L^2_{loc}$. $\endgroup$
    – Willie Wong
    Commented Jun 13, 2014 at 8:31
  • $\begingroup$ Thanks for your comment! That is really interesting. It would surprise me a little if that is the best result, since if $f$ is the delta function in $H^{-1}$, $f*g=g$, so it has the smoothness $s_1+s_2+1$. It occurred to me that the Fourier transform might swap smoothness and decay rate according to a defined rule? If that is the case that should imply that $H^{s_1+s_2+1}$ is the general answer to the smoothness question, at least in $\mathbb{R}^1$. $\endgroup$
    – Zorgoth
    Commented Jun 13, 2014 at 15:59

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