Let $\phi, W:\mathbb{R} \to \mathbb{R}$. How can one see that $$ \int_{\mathbb{R}^2} \int_{\mathbb{R}^2} W(z-w)\phi(z)\phi(w)dzdw= \int_{\mathbb{R}^2} \hat W(\xi) |\hat\phi(\xi)|^2 \tag{1} $$ holds? (perhaps up to a factor of $2\pi$ or something like that, depending on your convention for the Fourier transform).
I didn't state precisely what my assumptions on $W,\phi$ are; I am fine with assuming whatever suffices here.
I see that $$ W*\phi(z)=\int_{\mathbb{R}^2} W(z-w)\phi(w)dw, $$ so the LHS of $(1)$ equals $$ \int_{\mathbb{R}^2} (W*\phi)(z) \phi(z) dz. $$ I am not sure how to proceed from here. I thought applying the Plancherel identity, but this doesn't seem to give the required result.
Note that $\phi$ is real-valued.
Plancherel says that \begin{align*} \int(W\ast\phi)(z)\overline{\phi(z)}dz=\int\widehat{W\ast\phi}(z)\overline{\widehat{\phi}(z)}dz, \end{align*} then \begin{align*} \int(W\ast\phi)(z)\phi(z)dz=\int\widehat{W\ast\phi}(z)\overline{\widehat{\phi}(z)}dz=\int\widehat{W}(z)\widehat{\phi}(z)\overline{\widehat{\phi}(z)}dz=\int\widehat{W(z)}|\widehat{\phi}(z)|^{2}. \end{align*}