TITLE: Is Fourier transform defined on $L^p(\mathbb{R})$ only for $p \in [1, 2]$?
QUESTION [7 upvotes]: Is  Fourier transform defined on $L^p(\mathbb{R})$ only for $p \in [1, 2]$?
From Lieb and Loss's Analysis, they extend the definition of Fourier transform from $L^1(\mathbb{R})$ to $L^p(\mathbb{R}), p \in (1, \infty)$, using 
$$
\| FT(f) \|_q \leq C_{p,q} \|f\|_p
$$
which they said only holds when $p \in (1,2]$. Does that mean that FT cannot be defined on $L^p(\mathbb{R})$ with $p \in (2, \infty)$, possibly via other means?
Thanks and regards!

REPLY [8 votes]: Fourier transform is a linear operator. When defining an operator, we should think not only of the domain space, but also of the target. It is true that Fourier transform is not a bounded operator from $L^p(\mathbb R)$ to $L^q(\mathbb R)$ ($1/p+1/q=1$) when $p>2$. The issue is that large $p$ allows the function to have a pretty heavy "tail", which on the Fourier side results in bad local behavior. 
But it is possible to define the Fourier transform of any function in $L^p(\mathbb R)$, by considering $f$ as a tempered distribution. Then the transform becomes an automorphism of the topological vector space of tempered distributions. In fact, the Wikipedia article on Fourier transform covers this, and links to an article on distributions.  
Related MSE question: Fourier transform in $L^p$