TITLE: Bijectiveness, injectiveness and surjectiveness of Fourier transformation defined on $L^p(\mathbb{R}^n), p\in [1, 2]$
QUESTION [5 upvotes]: 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!

REPLY [2 votes]: 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 $L^2$.