I am trying to prove that the Fourier Transform of $f=\chi_{[-1,1]}$ is not in $L^1[\mathbb{R}]$.
I computed $\hat{f}$ and got $\hat{f}(w)=\sqrt{\frac{2}{\pi}}\frac{\sin w}{w}$.
My problem with this is that $\hat{f}$ is not even defined at 0. We certainly define it to be 1 at 0 in order to obtain continuity, but this is convenient, not mandatory. And we should not integrate over a set that supersedes the domain of $\hat{f}$, I believe.
And a related question is: where does the Fourier transform map to? In this, case, since $f\in L^1[\mathbb{R}]$ and $f\in L^2[\mathbb{R}]$, then $f\in L^2[\mathbb{R}]$. Can we say anything else?
If I was not clear, please let me know.
Of course $\hat{f}(0) = \int_{-\infty}^\infty f(x)dx = 2$ is well-defined, and $\hat{f}(\omega) = 2 \frac{\sin \omega}{\omega}$.
The Fourier transform is a bounded (continuous) operator $L^1 \to C_0^0$ (where $C_0^0$ are the uniformly continuous functions vanishing at $\infty$ with the $\sup$ norm) and $L^2 \to L^2$. Since $f \in L^1 \cap L^2$ then $\hat{f} \in C_0^0 \cap L^2$.
Conversely, if $\hat{f} \in L^1$ then $\check {\hat{f}}=f \in C_0^0$. But here $f = \chi_{[-1,1]} \not \in C_0^0$ thus $\hat{f} \not \in L^1$.