I am trying to prove that the Fourier transform on $L^1(\mathbb{R}^d)\cap L^2(\mathbb{R}^d)$ preserves the inner product (from $L^2$). I have already proved the result for $f,g\in L^1\cap L^2$ such that $\widehat{f},\widehat{g}$ are also in $L^1\cap L^2$.
In the general case, I know that if $f,g$ are in $L^1\cap L^2$, then the Gaussian convolutions $$f_\sigma:=f*G_\sigma\qquad g_\sigma:=g*G_\sigma$$ are also in $L^1\cap L^2$, and moreover, so are their Fourier transforms (here $G_\sigma$ denotes the density of a Gaussian $\mathcal{N}(0,\sigma^2I_d)$). Using the part that I have already proved, I get $$\langle\widehat{f_\sigma},\widehat{g_\sigma}\rangle=\langle f_\sigma,g_\sigma\rangle.$$ Now, by the dominated convergence theorem, as $\sigma\to 0$ $$\langle\widehat{f_\sigma},\widehat{g_\sigma}\rangle=\int \widehat{f}(x)\widehat{g}(x)e^{-\sigma\|x\|}dx\to\langle\widehat{f},\widehat{g}\rangle.$$ It remains to prove that $\langle f_\sigma,g_\sigma\rangle\to\langle f,g\rangle$. Any hints as to how I can do this?
Note that $\{G_{\sigma}:\sigma>0\}$ is an approximation to identity so $$||f_{\sigma}-f||_2 \to 0$$ $$||g_{\sigma}-g||_2 \to 0$$
Thus $$|<f_{\sigma},g_{\sigma}>-<f,g>| \leq ||f_{\sigma}||_2||g_{\sigma}-g||_2+||g||_2||f_{\sigma}-f||_2$$ by Cauchy-Schwarz.
Also note that $||f_{\sigma}||_2 \to ||f||_2$ thus the above limit is zero.