Express Dirichlet energy $E_\mu(f) := \int \|\nabla f(x)\|^2 d\mu(x)$ in terms of Fourier information alone Let $\mathbb R^d$ and let $\mu = p(dx)$ be a probability distribution thereupon, with density $p$ (which maybe assumed bounded, etc.). For a continuously differentiable function $f:\mathbb R^d \to \mathbb R$, let $E_\mu(f) := \|\nabla f\|_{L^2(\mu)}^2 := \int_{\mathbb R^d} \|\nabla f(x)\|^2 d\mu(x)$ be its Dirichlet energy w.r.t $\mu$.

Question. How can $E_\mu(f)$ be written in terms of the Fourier transform of $f$, and information on $\mu$?

Important cases

*

*$\mu$ $d$-dimensional standard Gaussian, i.e $\mu = p(dx)$, where $p(x) \propto e^{-\|x\|^2/2}$.

*$\mu$ is the uniform distribution on the unit-sphere in $\mathbb R^d$.

 A: Disclaimer. Below, I give a solution for the case of the $d$-dimensional gaussian distribution. I'm not 100% sure of all my arguments. I've hand checked with a few choices of the function $f$ (e.g linear $f(x) \equiv x^\top w$, etc.). Just posting here to start a discussion.

One can always write $\|\nabla f(x)\|^2 d\mu(x) = H(x)^2dx$, where $H(x): = F(x)G(x)$, with $F(x):=\|\nabla f(x)\|$ and $G(x):=\sqrt{p(x)}$.
By the Plancherel Theorem, we may simplify like so
$$
E_\mu(f) = \int_{\mathbb R^d} H(x)^2 dx = \int_{\mathbb R^d} |\hat{H}(z)|^2dz,
$$
where $\hat{H}$ is the Fourier transform of $H$. Now, the convolution property of the Fourier transform, we have $\hat{H} = \hat{F} \star \hat{G}:z \mapsto \int_{\mathbb R^d} \hat{F}(t)\hat{G}(t-z)dt$.
On the other hand, note that $\hat{F}(z) = \widehat{\|\nabla f\|}(z) = -i\|z\| \hat{f}(z)$. It remains to compute the Fourier transform of $G := \sqrt{p}$, and simplify...
Partial answer for gaussian case
In this case, the density of $\mu$ is $p = \gamma_{d,x}(x)$ defined by $\gamma_{d,\sigma}(x) = (2\pi\sigma^2)^{-d/2} e^{-\|x\|^2/(2\sigma^2)}$. This gives $\hat{G}(z) = (2\pi)^{-d/2}e^{-\sigma^2\|z\|^2/2}=\sigma^d\gamma_{d,1/\sigma}(z)$. Thus, $\hat{H} = T_{K_{d,1/\sigma}}\widehat{\|\nabla f\|} = T_{K_{d,1/\sigma}} \tilde{f}$, where $T_{K_{d,1/\sigma}}$ is the kernel integral operator corresponding to the psd radial kernel $$
K_{d,1/\sigma}(z,z'):= \gamma_{d,1/\sigma}(z-z'),
$$
and the function $\tilde{f}:\mathbb R^d \to \mathbb C$ is defined by
$$
\tilde{f}(z) := \|z\| \hat{f}(z).
$$
Because $T_{K_{d,1/\sigma}}$ is an isometry on $L^2(\gamma_{d,1/\sigma}) := L^2(\mathbb R^d,\gamma_{d,1/\sigma})$ (due to spherical symmetry), we deduce that
$$
E_{\gamma_{d,\sigma}}(f) = \sigma^d \|\tilde{f}\|_{L^2(\gamma_{d,1/\sigma})}^2.
$$
A: I'm not sure how useful the following is to you or not.
dohmatob's answer can certainly work in the case where $\mu$ has a density w.r.t. Lebesgue measure, but the volume form on the unit sphere does not.
I don't know enough distribution theory to tell you whether or not a surface measure can be given a meaningful square root and maybe it can.
In any case, if you know what the Fourier transform of $\mu$ is, then you ought to be able to say something more based on two (possibly dubious) applications of Fubini's theorem.
First, observe that,
$$\begin{align}
\|\nabla f\|^2 & = \sum_j(\partial_jf)^2 \\
& = (2\pi)^{-d}\sum_j\left(\int i\xi_j e^{ix\cdot\xi}\hat f(\xi)d\xi\right)\left(\int i\zeta_j e^{ix\cdot\zeta}\hat f(\zeta)d\zeta\right) \\
& = -(2\pi)^{-d}\iint\xi\cdot\zeta\hat f(\xi)\hat f(\zeta)e^{ix\cdot(\xi + \zeta)}\,d\xi\, d\zeta
\end{align}$$
which we can then use to write
$$\begin{align}
\int\|\nabla f\|^2\mu(dx) & = -(2\pi)^{-d}\iiint \xi\cdot\zeta\,\hat f(\xi)\hat f(\zeta)e^{ix\cdot(\xi + \zeta)}\mu(dx)\,d\xi\,d\zeta \\
 & = -(2\pi)^{-d/2}\iint\xi\cdot\zeta\,\hat f(\xi)\hat f(\zeta)\hat\mu(-\xi - \zeta)\,d\xi\,d\zeta.
\end{align}$$
(I think this is getting all the factors of $2\pi$ correct but maybe worth checking again.)
For the Gaussian case, the Fourier transform of $\mu$ has the same form as $\mu$.
When $\mu$ is the volume form on the unit sphere, $\hat\mu$ has a closed-form expression in terms of Bessel functions; see for example this question.
All this is just an elementary application of the convolution theorem so you might have thought of it already anyway.
