The Plancherel Theorem for the Fourier transform $\hat{f}(s)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-ist}dt$ on $\mathbb{R}$ states that $$ \int_{-\infty}^{\infty}|\hat{f}(s)|^2ds = \int_{-\infty}^{\infty}|f(t)|^2dt,\;\;\;\; f \in L^2(\mathbb{R})\cap L^1(\mathbb{R}). $$ Without appealing to classical convergence results, can this result be used to show that $$ \lim_{u,v\rightarrow\infty}\left\|\frac{1}{\sqrt{2\pi}}\int_{-u}^{v}\hat{f}(s)e^{isx}ds-f\right\|_{L^2(\mathbb{R})} = 0. \;\;\; ? $$ Does anyone know of a nice way to show this?
Background: This is not a homework problem or something I found in a text. It would seem reasonable to expect this result because of the discrete version where Parseval's equality for the Fourier series implies $L^2$ convergence of the Fourier series; or, if $\{ e_n \}_{n=1}^{\infty}$ is an orthonormal set in an inner product space, then $$\sum_{n=1}^{\infty}|(f,e_n)|^2=\|f\|^2 \iff \lim_{N\rightarrow\infty} \left\|\sum_{n=1}^{N}(f,e_n)e_n -f \right\|=0. $$
Yes: letting $\chi_{-a,b}$ be the characteristic function of the interval $[-a,b]$, and knowing that $\widehat{f}$ exists as an $L^2$ function whether or not $f$ was also in $L^1$, certainly $\chi_{-a,b}\cdot \widehat{f}$ converges to $\widehat{f}$ in $L^2$. Since (by Plancherel) (isometrically-extended) inverse Fourier transform is a continuous map $L^2\to L^2$, the inverse Fourier transforms of $\chi_{-a,b}\cdot \widehat{f}$ converge in $L^2$ to the inverse Fourier transform of $\widehat{f}$, namely, $f$.