I was looking at my definition of Fourier transformation of a function $u \in L^1(\mathbb{R}^N)$, and the following question came up.
Question: Why is it defined for $ u \in L^1(\mathbb{R}^N)$? Why not for $u \in L^2(\mathbb{R}^N)$... $L^p(\mathbb{R}^N)$ with $p \in [1,\infty]$? What is the intuition of letting $u$ be in $L^1(\mathbb{R}^N)$?
Thanks for any possible explanation.
The first and simplest reason might be that the Fourier transform $\widehat{u}$ exists if $u \in L^{1}(\mathbb{R}^{N})$ and - this should be the second reason - has some nice properties which can be easily proven. For instance if $u \in L^{1}(\mathbb{R}^{N})$ then $\widehat{u} \in C_{0}(\mathbb{R}^{N})$ - so it defines a continuous bounded function that vanishes at infinity.
On the other hand, the Fourier transform $\widehat{\cdot} : L^{1}(\mathbb{R}^{N}) \to C_{0}(\mathbb{R}^{N})$ has also 'bad' properties. For example it is not bijective as a mapping defined on $L^{1}(\mathbb{R}^{N})$ to $C_{0}(\mathbb{R})$.
By using the results on $L^{1}(\mathbb{R}^{N})$ it can be shown that for all $u \in L^{1}(\mathbb{R}^{N}) \cap L^{2}(\mathbb{R}^{N})$ we have $\widehat{u} \in L^{2}(\mathbb{R}^{N})$ with $\left\|u\right\|_{L^{2}} = \left\| \widehat{u} \right\|_{L^{2}}$. Therefore it can be extended in a unique way to an unitary operator on $L^{2}(\mathbb{R}^{N})$.
So in this sense the Fourier transform can be defined on $L^{2}(\mathbb{R}^{N})$, but have in mind that this actually is an extension of the 'original' Fourier transform on $L^{1}(\mathbb{R}^{N})$.