Let $f \in L^2(\mathbb{R})$ (square integrable functions), I'm trying to prove that his Fourier transform also does: $\mathscr{F}[f] \in L^2(\mathbb{R})$.
I have tried to bound it
\begin{align} \int_{-\infty}^{+\infty}|\ \hat{f}(\omega)\ |^2 d\omega &= \int_{-\infty}^{+\infty} \left| \int_{-\infty}^{+\infty} f(t)\ e^{-i\omega t} \right|^2 d\omega \\ &\leq \int_{-\infty}^{+\infty} \left( \int_{-\infty}^{+\infty}|\ f(t)\ e^{-i\omega t}\ | \right)^2 d\omega \\ &= \int_{-\infty}^{+\infty} \left( \int_{-\infty}^{+\infty}|\ f(t)\ | \right)^2 d\omega \\ \end{align}
but the last expression is not necessarily finite. Any help will be helpful!
Let $f$ be absolutely and absolutely square integrable on $\mathbb{R}$. Then, $$ \overline{\hat{f}(s)}=\overline{\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-ist}dt}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\overline{f(t)}e^{ist}dt = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\overline{f(-t')}e^{-ist'}dt' $$ Let $g(t)=\overline{f(-t)}$. Then $\overline{\hat{f}(s)}=\hat{g}(s)$. By the convolution theorem, $f\star g$ is absolutely integrable $$ |\hat{f}(s)|^2 = \hat{f}(s)\hat{g}(s)=\frac{1}{\sqrt{2\pi}}\widehat{(f\star g)}(s) $$ Then you can evaluate the $L^2$ norm of the Fourier transform \begin{align} \int_{\infty}^{\infty}|\hat{f}(s)|^2ds & =\left.\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\widehat{f\star g}(s)e^{isx}ds\right|_{x=0} \\ & = (f\star g)(0) \\ & = \int_{-\infty}^{\infty}f(t)g(0-t)dt \\ & = \int_{-\infty}^{\infty}|f(t)|^2dt. \end{align} This is an outline of the shortest argument I know. Everything goes through if $f\star g$ is differentiable at $0$ because then you have the classical Fourier inversion result at $0$, and that's what is being used in the last equation string. At least it gives you an idea of why the result is true for the Fourier transform.