It's well known that the Fourier transform $\mathcal{F}: L^2(\mathbb{T}) \rightarrow l^2(\mathbb{Z})$ is an isomorphism and $\mathcal{F}: L^1(\mathbb{T}) \rightarrow l^{\infty}(\mathbb{Z})$ is linear and continous (but not surjective even if the image is replaced by $C_0(\mathbb{Z})$).
I would expect by some interpolation argument that:
1) $\mathcal{F}: L^p(\mathbb{T}) \rightarrow l^{p'}(\mathbb{Z})$ is continous for all $1 < p <2$? ($p'$ is the conjugate exponent)
if the first is true:
2) Is it also surjective for all $1< p <2$? (the case $p=2$ is true, but the case $p=1$ is false)
Yes, (1) is true: Hausdorff-Young Inequality .
$\newcommand\FT{\mathcal F}$ But (2) is false. Suppose $1<p<2$ and $\FT:L^\to\ell^{p'}$ is surjective. Since $\FT$ is certainly injective, the Open Mapping Theorem shows that its inverse is bounded: $$||f||_p\le c||\FT f||_{p'}.$$
But let $$f_N(t)=\sum_{k=1}^Ne^{2\pi ikt}.$$Since $f_N$ is "lacunary" it's well known that there exists $c$ with $$||f_N||_p\ge c||f_N||_2=c\sqrt N.$$(I haevn't found a nice reference for this. It "must" be in Kahane "Some Random Series of Functions"; I haven't verified this since the book's miles away...)
Combining two inequalities leads to $$\sqrt N\le c N^{1/p'},$$a contradiction since $1/p'<1/2$.