I'm trying to prove Plancherel's theorem for functions $f\in L^1\cap L^2(\mathbb{R})$. I've included below my attempt and I would really appreciate it if someone could check this for me please, and give me any feedback they might have.
Note: I am working with a slightly different definition of the Fourier transform to usual, namely: $\widehat{f}(u)=\int_{-\infty}^{\infty}e^{iux}f(x)\,dx$. My main worries are that I have not implemented the convergence property of the convolution properly and I'm not sure if I've justified the use of the MCT in the correct way.
My proof goes as follows:
Firstly, I have previously shown that:
$$(G_\lambda\ast f)(x)\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-iux}e^{-\frac{(u/\lambda)^2}{2}}\widehat{f}(u)\,du. \tag{*}$$
Then, writing $|f(x)|=f(x)\overline{f(x)}$, we have:
$$||{f}||_2^2=\int_{-\infty}^{\infty}f(x)\overline{f(x)}\,dx=\lim_{\lambda\to\infty}\int_{-\infty}^{\infty}f(x)\overline{(G_\lambda \ast f)(x)}\,dx.$$
The last inequality holds since $||G_\lambda \ast f-f||_2 \to 0$ in $L^2(\mathbb{R})$ (something I've already proved). Implementing (*) and then applying Fubini's theorem (which I have justified separately), we can write:
$$\begin{align*} ||{f}||_2&=\lim_{\lambda \to \infty}\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x)e^{iux}\overline{\widehat{f}(u)}e^{-\frac{(u/\lambda)^2}{2}}\,du\,dx\\ &=\lim_{\lambda \to \infty}\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x)e^{iux}\,dx\overline{\widehat{f}(u)}e^{-\frac{(u/\lambda)^2}{2}}\,du\\ &=\lim_{\lambda \to \infty}\frac{1}{2\pi}\int_{-\infty}^{\infty}\widehat{f}(u)\overline{\widehat{f}(u)}e^{-\frac{(u/\lambda)^2}{2}}\,du\\ &=\lim_{\lambda \to \infty}\frac{1}{2\pi}\int_{-\infty}^{\infty}|\widehat{f}(u)|^2e^{-\frac{(u/\lambda)^2}{2}}\,du. \end{align*}$$
Define the sequence $\{x_\lambda\}_{\lambda =1}^\infty$ by $x_\lambda=|\widehat{f}(u)|^2e^{\frac{-(u/\lambda)^2}{2}}$. Clearly, we have $\lim_{\lambda\to\infty}x_n=|\widehat{f}(u)|$. We also know $\{x_n\}$ is non-negative and monotone increasing, since $|\widehat{f}(u)|$ is fixed and $e^{\frac{-(u/\lambda)^2}{2}}$ increases in value as $\lambda$ increases. Therefore, by the MCT, we have:
$$||{f}||_2=\frac{1}{2\pi}\int_{-\infty}^{\infty}|\widehat{f}(u)|^2\lim_{\lambda \to\infty}e^{-\frac{(u/\lambda)^2}{2}}\,du=\frac{1}{2\pi}\int_{-\infty}^{\infty}|\widehat{f}(u)|^2\,du=\frac{1}{2\pi}||{\widehat{f}}||_2^2.$$
Any help is really appreciated, thank you in advance!