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Let $g$ be the distribution whose Fourier coefficients are given by $$\hat{g}(k) = \begin{cases} 0, & {k=0} \\ |k|^{s-d}, & {k\in \mathbb{Z}^d\setminus\{0\}},\end{cases} \qquad 0\leq s<d,$$ that is the unique solution to $$(-\Delta)^{\frac{d-s}{2}}g = \delta_0 - 1.$$ We will call $g$ the periodic Riesz potential of order $d-s$. It is known (I don't have a ref at the top of my mind) that $g\in L^1(\mathbb{T}^d)$, $g$ is $C^\infty$ away from the origin, and the difference $$g(x)-c_{d,s}|x|^{-s} \in C^\infty(B(0,1/4)), \qquad c_{d,s} := 2^{s-d}\pi^{-d/2}\frac{\Gamma(s/2)}{\Gamma((d-s)/2)}, \ 0<s<d,$$ $$g(x)+c_d\log|x|\in C^\infty(B(0,1/4)), \qquad c_d:=\frac{2}{(4\pi)^{d/2}\Gamma(d/2)}, \ s=0.$$ Here, I am identifying the torus with the cube $[-\frac{1}{2},\frac{1}{2}]^d$ under periodic boundary conditions.

Question. Is there an exact formula for the $L^1$ norm $\|g\|_{L^1(\mathbb{T}^d)}$? I'm really only interested in the case $s=0$, so I would be happy with just an answer for this case.

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  • $\begingroup$ If I am not mistaken, $g$ is simply the periodization of the usual RIesz kernel $c_{d,s}|x|^{-s}$. I do not think, though, that a reasonable expression for $\|g\|_1$ is possible when $d \geqslant 2$. Already the description of the set $\{g < 0\}$ is problematic, I believe. And I the logarithmic case $s = 0$ is does not seem any simpler. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Sep 21, 2022 at 10:36
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    $\begingroup$ @MateuszKwaśnicki Thanks for your comment, Mateusz! I might be wrong, but I don't think one can directly show the periodization is a convergent series in general, since in the case $s<d$, the Riesz potential on $\mathbb{R}^d$ does not decay fast enough. $\endgroup$
    – Matt Rosenzweig
    Commented Sep 21, 2022 at 12:27

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