Carleson-Hunt Theorem on $\Bbb R$. Wikipedia states the following:


If $p\in(1,\infty)$ and $f\in L^p(\Bbb R)$ then $f(x)=\lim_{A\to\infty}\int_{|\xi|<A}\hat f(\xi)e^{i\xi x}\,d\xi$ almost everywhere.


This seems clearly wrong to me, because if $p>2$ then $\hat f$ is just a tempered distribution, so those integrals are undefined. But I realized just now I could be wrong; decided to rollback my edit and see if anyone here knows:
Q. Could it be that $\hat f$ is in fact a locally integrable function for $p<\infty$?
I don't think so, but I don't see an obvious counterexample. (Of course $f=1$ is a counterexample for $p=\infty$.)
 A: This morning I had a counterexample, using this:


Gap. $\sum_1^\infty\frac{\cos(nt)}{n^{1/2}}$ is not the Fourier series of a function in $L^1(\Bbb T)$.


Non-proof: I swear I read yesterday that this was proved by Hardy&Littlewood. I can't find it today. But all I need is a much weaker statement:


Cor. If $p>2$ there exists a sequence $(a_n)\in\ell_p(\Bbb Z)$ such that the series $\sum a_n e^{int}$ does not converge in $L^1(\Bbb T)$.


And that I can prove all by myself. First note that the Hausdorff-Young inequality fails for values of the parameter other than those for which it holds:


Recall that if $q\in[1,2)$ there exists $f\in C(\Bbb T)$ such that $\sum|\hat f(k)|^q=\infty$.


The proof is immediate from the Closed Graph Theorem and the Rudin-Shapiro polynomials (if you can't find the R-S polynomials on Wikipedia lemme know and we'll fix that). Or see the section on the Hausdorff-Young inequality in Complex Made Simple for an explicit gliding-hump construction with no CGT.
Proof of the Corollary: Let $f\sim\sum c_ne^{int}\in C(\Bbb T)$  be as above, with $q=p'$. Since $\sum|c_k|^{p'}=\infty$ there exists a sequence $(a_k)\in\ell_p$ with $a_k\overline{c_k}\ge0$ and $\sum a_k\overline{c_k}=\infty$. Let $s_n(t)=\sum_{k=-n}^n c_ke^{ikt}$.
Now $\langle f, s_n\rangle=\sum_{k=-n}^n a_k\overline{c_k}\to\infty$; since $f$ is bounded this implies that $||s_n||_1\to\infty$.


Stupid Lemma (SL). There exists $\phi\in  C^\infty_c(\Bbb R)$  such that $\hat\phi$ has no zero on $\Bbb R$.


"Stupid" meaning that surely if I simply asserted this nobody would  doubt it. But anyway:
Proof. Say $\psi\in C^\infty(\Bbb R)$, $\psi\ne0$. If you let $\tilde\psi(t)=\overline{\psi(-t)}$ then $\widehat{\psi*\tilde\psi}=|\hat\psi|^2$. Since $\hat\psi$ extends to an entire function it has at most countably many zeroes; hence there exists $a\in\Bbb R$ such that $$|\hat\psi(\xi)|^2+|\hat\psi(\xi-a)|^2>0\quad(\xi\in\Bbb R).$$


Medium-Obvious Lemma (MOL). If $\hat f$ is locally integrable for every $f\in L^p$ then for every $A>0$ there exists $c_A$ such that $\int_{-A}^A|\hat f|\le c_A||f||_p$.


Proof. By the Closed Graph Theorem it's enough to show that if $f_n\to f$ in $L^p(\Bbb R)$ and $\hat f_n\to g$ in $L^1([-A,A])$ then $g=\hat f$. But $f_n\to f$  in $\mathcal S'$ (since $\mathcal S\subset L^{p'}$), hence $\hat f_n\to\hat f$ in $\mathcal S'$. So for every $\phi\in C^\infty_c((-A,A))$ we have $$\int\phi\hat f=\lim\int\phi\hat f_n=\int\phi g.$$
And now we're done:


Theorem. If $p>2$ there exists $f\in L^p(\Bbb R)$ such that the distribution $\hat f$  is not locally integrable.


Proof. Suppose not. It follows from SL that there exists $\phi\in C^\infty_c$ such that $\phi$ is supported in $(-1/2,1/2)$ and $\hat\phi$ has no zero. Let $(a_n)$ be as in the corollary and define $$f(t)=\sum_{ n\in\Bbb Z}a_n{\phi(t-n)}.$$The series defining $f$ converges in $L^p$, so MOL implies that the series $$\hat f(\xi)=\sum_na_n\hat\phi(\xi)e^{in\xi}$$ converges in $L^1([0,2\pi])$. Since $\hat\phi$ is continuous and has no zero this contradicts the Corollary above (because $1/\hat\phi$ is bounded on $[0,2\pi]$).
