# Functions belong to $L^{\frac{2n}{n+1}}$ whose Fourier transforms are infinite on $S^{n-1}$

I'm looking for functions $f\in L^{\frac{2n}{n+1}}$ such that $\hat{f}=\infty$ on $S^{n-1}$. Is there any explicit expression of such kind of examples?

This seems to be a well-known result, but I can not find it in standard references such as Stein's Harmonic Analysis and Grafakos's classical Fourier Analysis.

• Can you please clarify what you mean by $\widehat{f}=\infty$ on $S^{n-1}$? – Christian Remling Sep 10 '18 at 15:55
• You are perhaps referring to the restriction problem: terrytao.wordpress.com/tag/restriction-conjecture But failure of restriction means that the restriction operator, initially defined on Schwartz functions, say, is unbounded, and not what you're asking for. – Christian Remling Sep 10 '18 at 16:01

As mentioned in the comments, the Fourier transform $\hat f$ of a function in $L^{\frac{2n}{n+1}}({\bf R}^n)$ is a priori only defined as an element of the dual space $L^{\frac{2n}{n-1}}({\bf R}^n)$ (as per the Hausdorff-Young inequality), and so cannot immediately be restricted to the measure zero set $S^{n-1}$ unless one possesses a restriction theorem (which fails at the endpoint $L^{\frac{2n}{n+1}}$, as should be stated in the references you supplied, though I don't have them handy right now).

Nevertheless, as long as one is willing to proceed formally, one can answer your question as follows. If one restricts attention to spherically symmetric functions $f(x) = F(|x|)$, then the Fourier-Bessel transform (or Hankel transform) gives the formula $$\hat f(\xi) = (2\pi)^{n/2} \int_0^\infty r^{\frac{n}{2}} F(r) J_{\frac{n-2}{2}}(r)\ dr \quad (1)$$ for $\xi$ on the unit sphere (the factors of $2\pi$ may show up elsewhere, depending on exactly how one normalises the Fourier transform, but they are not too relevant for this discussion). Technically, this formula is only valid for "nice" functions $f$ (e.g. absolutely integrable), but we are working formally and will pretend that this formula applies for more general functions. The Bessel function $J_{\frac{n-2}{2}}(r)$ decays like $r^{-1/2}$ as $r \to \infty$ (and also oscillates sinusoidally). So if one sets $$F(r) := \frac{(2+r)^{-\frac{n}{2}}}{\log(2+r)} J_{\frac{n-2}{2}}(r)$$ (say) in order to remove the cancellation, then the integral in (1) is now nonnegative and will just barely fail to converge, and so formally one has $\hat f(\xi) = \infty$ on the unit sphere. On the other hand, by working in polar coordinates and again using the decay of Bessel functions we see that $f$ barely manages to lie in $L^{\frac{2n}{n+1}}$ (it decays a tiny bit faster than $(2+r)^{-\frac{n+1}{2}}$).

I am not sure to whom to attribute this simple example to; I am sure it was known to Stein for instance when he formulated his restriction conjecture in

Stein, Elias M., Some problems in harmonic analysis, Harmonic analysis in Euclidean spaces, Part 1, Williamstown/ Massachusetts 1978, Proc. Symp. Pure Math., Vol. 35, 3-20 (1979). ZBL0445.42006.

(but again I do not have the content of this reference available currently). There is also a similar counterexample of Herz in

Herz, Carl S., On the mean inversion of Fourier and Hankel transforms, Proc. Natl. Acad. Sci. USA 40, 996-999 (1954). ZBL0059.09901.

for the closely related problem of Bochner-Riesz summation. Finally, there is a more advanced counterexample to the restriction theorem in the Lorentz space $L^{\frac{2n}{n+1},1}$ (based on the Besicovitch set construction) in

Beckner, W.; Carbery, A.; Semmes, S.; Soria, F., A note on restriction of the Fourier transform to spheres, Bull. Lond. Math. Soc. 21, No. 4, 394-398 (1989). ZBL0681.42006.

• I suppose this example can be made even more convincing by showing that the distributional Fourier transform of $F(|x|)$, or the distributional Hankel transform of $F(r)$, coincides with a continuous function (in the extended sense, that is, with values in $[-\infty, \infty]$). The usual integration by parts trick (a.k.a. Abel–Dirichlet test) should work here: $r^{n/2} (2+r)^{-n/2} / \log(2+r)$ eventually decreases to zero, while the indefinite integral of $J_{n/2-1}(r) J_{n/2-1}(r s)$ is bounded unless $s = 1$. But I did not attempt to work out the details. – Mateusz Kwaśnicki Sep 10 '18 at 20:47