Is it true that $\int \hat{f}g \, dx=\int f\hat{g} \, dx$ for $f,g\in L^2$? It is a very well know fact that the formula holds in $L^1$, I guess that the formula still  holds for $f,g\in L^2$.
 A: Yes it is true. It follows from the fact that the Fourier transform is a unitary transformation in $L_2$.
To prove all this you may approximate $f\in L_2$ by functions in $L_1\cap L_2$, for example $f_k=f\mathbb{1}_{|x|_2\leq k}$, $\|f_k-f\|_2\xrightarrow{k\rightarrow\infty}0$, and $f_k\in L_1\cap L_2$.
Alternatively, you can use the Schwartz space $\mathcal{S}$ and basic fact from the theory of distributions to obtain this result.

Here is a sketch of a proof:
Lemma 1: $L_1\cap L_2$ is dense in $L_2$.
Here is a short proof:
Suppose $f\in\mathcal{L}_2$, and let $C(0;k)=[-k,k]^d$. It is easy to check that $f_k=f\mathbb{1}_{C_k}, h_k=f\mathbb{1}_{B(0;k)}\in \mathcal{L}_1\cap\mathcal{L}_2$  for all $k>0$, and that $\lim_{k\rightarrow\infty}\|f-f_k\|_2=\lim_{k\rightarrow\infty}\|f -h_k\|_2\rightarrow0$.
Lemma 2:
If $f\in\mathcal{L}_1\cap\mathcal{L}_2$, then  $\widehat{f}\in\mathcal{L}_2$ and  $\|\widehat{f}\|_2=\|f\|_2$
Here is a short proof:
If $g(x)=\overline{f(-x)}$ then, $h=f*g\in\mathcal{L}_1\cap \mathcal{C}_0$ since $h$ is the convolution of two $L_2$ function.  As $\widehat{g}=\overline{\widehat{f}}$,
we have that $\widehat{h}=\widehat{f}\widehat{g}=|\widehat{f}|^2\geq0$. By
a well known result in Fourier analysis ($\hat{h}\geq0$ and $h$ continuous at $0$ implies $\hat{h}$ integrable),  $\widehat{h}$ is integrable and
$h(0)=\int_{\mathbb{R}^n}\widehat{h}(y)\,dy$. Thus
$$\begin{align}
\int_{\mathbb{R}^n}|\widehat{f}(y)|^2dy  &= 
\int_{\mathbb{R}^n}\widehat{h}(y)\,dy=h(0)=
\int_{\mathbb{R}^n}f(y)g(0-y)\,dy\\
&=\int_{\mathbb{R}^n}f(y)\overline{f(y)}\,dy=
\int_{\mathbb{R}^n}|f(y)|^2\,dy
\end{align}
$$

Observation:
Lemma 2  shows that the Fourier transform $\mathcal{F}$
maps  $\mathcal{L}_1\cap \mathcal{L}_2$ into  $\mathcal{L}_2$ isometrically. Then  $\mathcal{F}$ admits a unique extension
to all of  $\mathcal{L}_2$ which is also an isometry.
We will keep the notation  $\widehat{f}={\mathcal F}f$ for $f\in\mathcal{L}_2$.

Suppose $f\in L_2$ and $\widehat{f_k},\,\widehat{h_k}$ as in the proof of Lemma 1. Then, by Lemma 2 and the observation following it
$$ \begin{align}
    \lim_{k\rightarrow\infty}\|\widehat{f}-\widehat{f_k}\|_2 =\lim_{k\rightarrow\infty}\|f-f_k\|_2=\lim_{k\rightarrow\infty}\|f-h_k\|_2=
    \lim_{k\rightarrow\infty}\|\widehat{f}-\widehat{h_k}\|_2 =0
    \end{align}
$$
That is,
$$
  \widehat{f}(y)=\lim_{k\rightarrow\infty} \int_{|x|_\infty\leq k}f(x)e^{-2\pi i x\cdot y}\,dx= \lim_{k\rightarrow\infty} \int_{|x|_2\leq k}f(x)e^{-2\pi i x\cdot y}\,dx
$$
in $\mathcal{L}_2$.
To obtain the result in your problem, let  $f,g\in L_2$, and define $f_k=f\mathbb{1}_{B(0;k)}$, $g_k=g\mathbb{1}_{B(0;k)}$. Then, as $f_k,g_k\in \mathcal{L}_1\cap\mathcal{L}_2$,
$$\langle \hat{f_k},g_k\rangle =\langle f_k,\hat{g_k}\rangle$$
Finally, as
$$\begin{align}
\big|\langle \hat{f},g\rangle-\langle \hat{f_k},g_k\rangle\big|&\leq \|\hat{f}-\hat{f_k}\|_2\|g\|_2+\|\hat{f_k}\|_2\|g-g_k\|_2\\
&=\|f-f_k\|_2\|g\|_2+\|f_k\|_2\|g-g_k\|_2\xrightarrow{k\rightarrow\infty}0
\end{align}$$
and
$$\begin{align}
\big|\langle f,\hat{g}\rangle-\langle f_k,\hat{g_k}\rangle\big|&\leq \|f-f_k\|_2\|\hat{g}\|_2+\|f_k\|_2\|\hat{g}-\hat{g_k}\|_2\\
&=\|f-f_k\|_2\|g\|_2+\|f_k\|_2\|g-g_k\|_2\xrightarrow{k\rightarrow\infty}0
\end{align}$$
we have that
$$\begin{align}
\langle \hat{f},g\rangle = \lim_k\langle \hat{f_k},g_k\rangle =\lim_k\langle f_k,\hat{g_k}\rangle =\langle f,\hat{g}\rangle
\end{align}
$$
A: Let $f_r(x) = \chi_{[-r,r]}f(x)$. Then $f_r \in L^1 \cap L^2$ if $f\in L^2$ and $r > 0$, which also gives $\hat{f_r}\in L^2$. Furthermore, $\lim_{r\rightarrow\infty} \|f_r-f\|_{L^2}=0$. The same things are true of $g$ as well. Therefore, if $0 < r,s <\infty$, the $L^1$ identity
$$
           \int \widehat{f_r}g_s\,dx = \int f_r \widehat{g_s} dx
$$
results in
$$
                 \int \widehat{f}g_sdx = \int f\widehat{g_s}dx.
$$
Because $g_s\rightarrow g$ and $\hat{g_s}\rightarrow\hat{g}$ both converge in $L^2$ as $s\rightarrow\infty$, and because $f,\hat{f}$ are in $L^2$, then the above converges as $s\rightarrow\infty$ to the desired result
$$
                 \int\hat{f}g dx = \int f\hat{g}dx,\;\;\; f,g\in L^2.
$$
