Fourier transform surjective on $L^p(\mathbb{R}^n)$ for $p \in (1,2)$? 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.
 A: 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 $1\leq p<2$
