Why Fourier transform is tempered distribution?

Question

This question refers to the question Fourier transform in $L^p$ In the answer it is clearly written that "If $s > 2$, this question quickly becomes technical and requires the theory of distributions. It is not hard to prove that any $f \in L^p$, $1 \le p \le \infty$ induces a tempered distribution $T_f$ and that the Fourier transform of $T_f$ is itself a tempered distribution." However, I could not prove that both f and its Fourier transform is a tempered distribution. Any help please?

Answer 1

I think you have two questions:

1. If $f\in L^p$, how does $f$ induce a tempered distribution $T_f$?

2. Why is the Fourier transform of a tempered distribution tempered?

Here are the answers:

1. Tempered distributions are the continuous dual of the Schwarz class $\mathcal{S}$. This means that a tempered distribution is a continuous linear functional $T\colon \mathcal S\to \mathbb C$. Here $\mathcal S$ is the family of all functions which decay faster than any polynomial at $\infty$ (think of exponential decay), with a topology that requires functions to be close even after multiplying by polynomials and taking many derivatives. Any function $f\in L^p$ induces the tempered distribution $T_f(h)=\int fh$; the latter integral is finite for all test functions $h\in\mathcal S$ by using the Holder inequality combined with $f\in L^p$ and the rapid decay of $h$ at $\infty$.

2. First, we need to say how the Fourier transform of a tempered distribution is defined. If $T\colon \mathcal S\to \mathbb C$ is a tempered distribution, then define the Fourier transform $\hat T$ by setting $\hat T(h)=T(\hat h)$. Here $\hat h$ is the Fourier transform of $h$, which is well-defined because $h\in \mathcal S$ decays rapidly at $\infty$. By linearity and continuity of the Fourier transform, this gives a continuous linear functional on $\mathcal S$ and hence $\hat T$ is a tempered distribution.