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
$$\sum_{n=1}^{\infty}|(f,e_n)|^2=\|f\|^2 \iff \lim_{N\rightarrow\infty} \left\|\sum_{n=1}^{N}(f,e_n)e_n -f \right\|=0.
$$
Best Answer
$f \in L^2$
$\hat{f}= \lim\limits_{u,v\mapsto \infty}\frac{1}{\sqrt{2\pi}}\int_{-u}^v fe^{-isx}dx$
multiply $f$ by a bump function $g_n$ with compact support in $[-u,v]$
such that $ \lim\limits_{n\mapsto \infty}g_nf=f$ on $[-u,v]$
let $h_n=g_nf$$\quad$ $h_n \in L^1 \bigcap L^2$ and $\hat{h}_n \in L^2 $(Plancherel theorem)
Also $\lim\limits_{n\mapsto \infty}\hat{h}_n=\frac{1}{\sqrt{2\pi}}\int_{-u}^v fe^{-isx}dx$
The good part is that the sequence {$h_n$} is uniformly integrable over $[-u,v]$: this follows from property of bump function. These imply $\hat{f} \in L^2$
Fourier Inversion theorem:$$ \lim_{u,v\rightarrow\infty}\left\|\frac{1}{\sqrt{2\pi}}\int_{-u}^{v}\hat{h}_n(s)e^{isx}ds-h_n\right\|_{L^1(\mathbb{R})} = 0. \;\;\; $$
Plancherel theorem:$$ \lim_{u,v\rightarrow\infty}\left\|\frac{1}{\sqrt{2\pi}}\int_{-u}^{v}\hat{h}_n(s)e^{isx}ds-h_n\right\|_{L^2(\mathbb{R})} = 0. \;\;\; $$
for sufficiently large $n,u,v$:$$\int_{-\infty}^{\infty}(\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\hat{h}_n(s)e^{isx}ds-\int_{-u}^{v}\frac{1}{\sqrt{2\pi}}\hat{f}_n(s)e^{isx}ds)^2dx \le \delta$$
$$\int_{-\infty}^{\infty}({h}_n-f)^2dx=\int_{-\infty}^{\infty}(\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\hat{h}_n(s)e^{isx}ds-f)^2dx \le \epsilon$$
writing $(a-b)^2=(a-c+c-b)^2=(a-c)^2+(c-b)^2+2(a-c)(c-b)$ and by Cauchy–Schwarz inequality:
$$\int_{-\infty}^{\infty}(f-\int_{-u}^{v}\frac{1}{\sqrt{2\pi}}\hat{f}(s)e^{isx}ds)^2 dx\le \int_{-\infty}^{\infty}((\int_{-u}^{v}\frac{1}{\sqrt{2\pi}}\hat{f}(s)e^{isx}ds-\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\hat{h}_n(s)e^{isx}ds)^2+(\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\hat{h}_n(s)e^{isx}ds-f)^2)dx +(2\delta \epsilon)^{1/2}\le\delta+\epsilon+2(\delta \epsilon)^{1/2}$$