For $1<p<2, $ Fourier transform is not onto $L^p(\Bbb T) \to \ell^q(\Bbb Z)$ where $\frac{1}{p}+\frac{1}{q}=1$ For $1&lt;p&lt;2$ , Fourier transform $\mathscr{F}$ is not onto $L^p(\Bbb T) \to \ell^q(\Bbb Z)$ where $\frac{1}{p}+\frac{1}{q}=1$

For $1 \le p \le 2$, Hausdorff-Young inequality implies that $\mathscr{F}:L^p(\Bbb T) \to \ell^q(\Bbb Z)$ where $\frac{1}{p}+\frac{1}{q}=1$
Now for $p=1$ showing that $\mathscr{F}:L^1(\Bbb T) \to \ell^\infty(\Bbb Z)$ is not onto was easy due to Riemann-Lebesgue lemma (I know that it's image is in fact a dense sub-alegbra of $c_0(\Bbb Z)$).
But for $1&lt;p&lt;2$, I'm facing difficulty to show that it's not onto. I know that $\forall 1&lt;p&lt;2, \mathscr{F}(L^p(\Bbb T))\subset \ell^q(\Bbb Z)$ . Hence we have $$\mathscr{F}(L^p(\Bbb T))\subset \ell^q(\Bbb Z) \subsetneq c_0(\Bbb Z)$$ Also considering images of trigonometric polynomials under $\mathscr{F}$ we get that $c_{00}(\Bbb Z) \subsetneq \mathscr{F}(L^p(\Bbb T)$
So basically the scenario is, $$c_{00}(\Bbb Z) \subsetneq \mathscr{F}(L^p(\Bbb T)) \subset \ell^q(\Bbb Z) \subsetneq c_0(\Bbb Z)$$
How to produce an $\ell^q$ sequence which cannot be in the image of $\mathscr{F}$ ? Or if there are other functional analysis proofs which will assert the same?
Take $S=\sum_{n \ge 1} \frac{\cos (2^nx)}{\sqrt n}$. Note that the coefficients of $S$ are in $l^q(\mathbb Z)$ for all $q&gt;2$ (but not in $l^2$!)
Then one can show that the partial series of $S$ and the Caesaro means of said partial series diverge ae on the unit circle so $S$ cannot be a Fourier series. Actually it is easier to show that the partial series/Caesaro means are unbounded ae using the fact that $\cos 2^nx$ and $\cos 2^mx$ are orthogonal for $n \ne m$ (and more generally any $k$ distinct such are orthogonal since for $n_1&lt;n_2&lt;..n_k, \pm 2^{n_1} \pm 2^{n_2}..\pm 2^{n_k} \ne 0$)
Clearly, $\mathscr F$ is one-to-one, and hence, if it were onto, then by virtue of the Open Mapping Theorem, it would be a homeomorphism.