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Suppose $\sum_{\xi \in \mathbb{Z}^d}{a_{\xi}}e^{i\langle x, \xi\rangle }$ converges spherically pointwise to $0$ for all $x \in \mathbb{T}^d$, i.e. $\lim_{R \to \infty} \sum_{|\xi| < R}{a_{\xi}}e^{i\langle x, \xi\rangle } = 0$, and that $\sum_{|\xi| = R} |a_{\xi}|^2 \to 0$ as $R \to \infty$ (this second statement actually follows from the first). Denote $$ F(x) = \sum_{\xi \ne 0} \frac{a_{\xi}}{|\xi|^2}e^{i\langle x, \xi\rangle }, $$ the series converging pointwise and in $L^2(\mathbb{T}^d)$.

It is claimed in this paper by Bourgain that "it is easily seen from the property $$ \sum_{|\xi| = R} |a_{\xi}|^2 < c $$ for all $R$, that the set of discontinuities of $F$ has Hausdorff dimension at most $d-2$".

How does one estimate the Hausdorff dimension of the set of discontinuities of a function using this kind of information about its Fourier coefficients?

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    $\begingroup$ I'm already mildly surprised that the first assumption apparently doesn't imply that $a_{\xi}=0$ (?). Is that also "easily seen"? $\endgroup$
    – Christian Remling
    Commented Apr 6 at 14:34
  • $\begingroup$ By the way, there is sometimes a big difference between various ways of forming partial sums in these multi-dimensional problems (cubes versus balls, for example), so perhaps you want to specify how it's done here. $\endgroup$
    – Christian Remling
    Commented Apr 6 at 14:36
  • $\begingroup$ @ChristianRemling thank you for pointing that out. Here the partial sums are formed spherically. $\endgroup$
    – fwd
    Commented Apr 6 at 15:48
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    $\begingroup$ @ChristianRemling the purpose of the paper is to prove that $a_{\xi} = 0$, which is expected, but highly nontrivial to demonstrate. $\endgroup$
    – fwd
    Commented Apr 6 at 15:53

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