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I know that $F_2:L^2 \rightarrow L^2$ is of course unitary, whereas $F_1:L^1 \rightarrow C_0$ is injective but not surjective. This can be seen by looking at the dual map.

Riesz-Thorin gives us that there is also $F_p: L^p \rightarrow L^q$ for $p \in (1,2).$ Here, the dual map trick does not work, so this transform has a chance of being surjective. Since every $f \in L^q$ is also in $S'$ we can also define a promising candidate $F_{S'}^{-1}(f).$ Unfortunatly, this does not really tell me whether this $F_{S'}^{-1}(f) \in L^p$ again.

This raises the question whether $F_p$ is actually surjective or not?

Comment on the discussion below: Thanks to everybody participating in the disccusion. Actually this question came to my mind while I was thinking about this problem from PDEs, which would have an easy solution in this case. I have to admit that the fact that $L^p$ is not isomorphic to $L^q$ is indeed something I know, but I have never actually used it, as I am not primarily active in analysis. Probably I should give my questions more thought in the future.Sorry for any inconvenience my question caused.

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    $\begingroup$ $F_p$ is injective, and $L_p$ is not isomorphic to $L_q$ unless $p=q=2$, so the answer is no. $\endgroup$
    – Yemon Choi
    Commented May 12, 2016 at 19:52
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    $\begingroup$ I think it is better to add it in the Original post. $\endgroup$
    – Amir Sagiv
    Commented May 12, 2016 at 19:52
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    $\begingroup$ The fact cited by @YemonChoi, that $L_p$ and $L_q$ are not isomorphic, is true but not obvious: see mathoverflow.net/questions/79713/lp-mathbbr-vs-lq-mathbbr for more details. But I think this should be posted as an answer! $\endgroup$
    – Nate Eldredge
    Commented May 12, 2016 at 20:04
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    $\begingroup$ More explicitly, there are purely singular measures whose FT is in $L^p$ for a given $p>2$ (and you can even get pointwise decay with a power rate). These can be constructed using Riesz products, for example. $\endgroup$
    – Christian Remling
    Commented May 12, 2016 at 21:02
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    $\begingroup$ @ChristianRemling Well, MO is for people to get answers to things which are easy to those with the right knowledge but not so obvious to people in a different area, right? I mean, all my questions about finite group theory and group characters are the kinds of thing that algebraists might learn in "graduate education" but which I simply didn't $\endgroup$
    – Yemon Choi
    Commented May 12, 2016 at 21:38

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If $1\leq p<2$ then $\mathscr{F}: L^p \to L^{p'}$ is not surjective. I had this as a homework problem a week back.

The reason is the bounded inverse theorem: $\mathscr{F}: L^p \to L^{p'}$ is injective, (by fourier inversion on the dense subspace of schwarz functions). If the map were surjective then there would be an inverse that would be continuous, since $\mathscr{F}$ is an open map under this assumption.

Thus we just need to prove that there is no bounded inverse: For $f \in \mathscr{S}$, there is no constant $c$ such that $ ||f||_{p} \leq c ||\hat f||_{p'}$ for $f \in \mathscr{S}$ with the constant only depending on $p$.

This is easy: The function $f_\lambda=e^{-\pi i \lambda x^2-\pi x^2}$ satisfies $||f_\lambda||_p=c$ independent of $\lambda$, whereas $||\hat f_\lambda||_{p'} \leq c \lambda^{1/p'-1/2}$. But there is no constant such that $c \leq \lambda^{1/p'-1/2} $ for all $\lambda >0$. Therefore the fourier transform is not surjective from $L^p \to L^{p'}$ for $1\leq p<2$

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