Bijectiveness, injectiveness and surjectiveness of Fourier transformation defined on $L^p(\mathbb{R}^n), p\in [1, 2]$ For Fourier transformation defined from $L^p(\mathbb{R}^n)$ to $L^q(\mathbb{R}^n)$, $p\in [1, 2]$, and $1/p+1/q=1$,
I heard that when $p=2$, FT is bijective.
Is $p=2$ iff


*

*FT is bijective, 

*FT is injective, or/and

*FT is surjective?


Thanks and regards!
 A: The Fourier transform is always injective. In greatest generality, it's bijective as a map on the space of tempered distributions. Indeed, the transform of distribution $T$ is the distribution $\widehat T$ such that $\widehat{T}\varphi = T\widehat \varphi$ for all test functions $\varphi$ in the Schwartz space. The Fourier transform is  a linear isomorphism of the Schwartz space onto itself. Therefore, $T\mapsto \widehat T$ is a bijection on the space of tempered distributions.
The Fourier transform is not surjective as a map $L^p\mapsto L^q$ for $1\le p<2$. When $p=1$, this follows from the fact that the Fourier transform of an integrable function is continuous. (Alternatively, from the fact that $L^1$ is separable by $L^\infty$ isn't.) For $1<p<2$ one can argue as follows: if $g\in L^q$ is a Fourier transform of $f\in L^p$, then $\widehat g\in L^p$ (because $\mathcal{F}^{-1}$ is just $\mathcal F$ composed with a flip). Thus, if $\mathcal{F}:L^p\to L^q$ were surjective, it would follows that   $\mathcal F$   maps $L^q$ to $L^p$ when $q>2$. The latter is known to be false (see here).
A: Adding to the answer before that the Fourier transform $T : L^2 \rightarrow L^2$ is indeed surjective, the answer is: the Fourier transform is bijective iff $p=2$ (considering it on all $L^p, p \in [1,2]$).
The most important igredient to show this is Plancherel's Theorem (unitarity of $T$) which is first used to extend $T$ continuously from $L^1$ to $L^2$ and later to show invertibility of $T$ on all of $L^2$.
