I previously asked this question here
Haussdorff-Young inequality optimal Lebesgue exponents range
and got a comment that referred me to the answer here Fourier transform in $L^p$ but I did not find the answer to my specific question:
What (where can one find) argument/counterexamples that shows the discontinuity of the Fourier transform from $L^{p}$ to $L^{p^{\prime}}$ when $2<p\leq \infty$ ?
I found a counterexample for Fourier series in Complex Made Simple.
Somewhere you can find the construction of the Rudin-Shapiro polynomials. These are trigoometric polynomials $P_n$ such that $$P_n(t)=\sum_{k=1}^{2^n-1}\pm e^{ikt}$$and $$|P_n(t)|\le 2^{(n+1)/2}.$$
It follows that $$||P_n||_p\le 2^{(n+1)/2}$$and $$\left(\sum_k|\hat P_n(k)|^{p'}\right)^{1/p'}=2^{n/p'}.$$
If $p>2$ this shows that $\left(\sum_k|\hat P_n(k)|^{p'}\right)^{1/p'}$ is not bounded by $c||P_n||_p$.
(You could convert this into a counterexample for the Fourier transform by, say, fixing $\phi\in\mathcal S(\Bbb R)$ such that $\hat\phi$ is supported in $(0,1)$ and letting $f_n(t)=P_n(t)\phi(t)$; note that $||f_n||_{L^p(\Bbb R)}\le||P_n||_\infty\,||\phi||_p$ and $\hat f_n(\xi)=\sum_{k=0}^{2^n-1}\pm\hat\phi(\xi-k)$.)
