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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 $\mathbb R^d$.

One can always write $\|\nabla f(x)\|^2 d\mu(x) = H(x)^2dx$, where $H(x): = F(X)G(X)$, with $F(x):=\|\nabla f(x)\|^2$ 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 $\hat{F}(z) = \widehat{\|\nabla f\|^2}(z) = -\|z\|^2 \hat{f}(z)$

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 $\mathbb R^d$.

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 $\mathbb R^d$.

One can always write $\|\nabla f(x)\|^2 d\mu(x) = H(x)^2dx$, where $H(x): = F(X)G(X)$, with $F(x):=\|\nabla f(x)\|^2$ and $G(x):=\sqrt{p(x)}$.

By the Plancherel Theorem, we may simplify like so $$ E_\mu(f) = \int_{\mathbb R^d} H(x)^2 dx = \int_{\mathbb R^d} |\hat{H}(z)^2|dz, $$ 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 $\hat{F}(z) = \widehat{\|\nabla f\|^2}(z) = -\|z\|^2 \hat{f}(z)$

Case 1: gaussian

In this case, one we have $G(x) = \gamma_{d,\sigma}(x) = (2\pi\sigma^2)^{-d/2} e^{-\|x\|^2/(2\sigma)^2}$, and so $\hat{G}(z) = (2\pi)^{-d/2}e^{-\sigma^2\|z\|^2/2}=\sigma^d\gamma_{d,\sigma^{-1}}(z)$. Thus, $\hat{H} = T_{K_{d,1/\sigma}}\|\nabla f\|^2 = T_{K_{d,1/\sigma}} \tilde{f}$, 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 $$ \tilde{f}(z) := \|z\|^2 \hat{f}(z). $$ 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

$$ E_{\gamma_{d,\sigma}}(f) = \sigma^d \|\tilde{f}\|_{L^2(\gamma_{d,1/\sigma})}. $$

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 $\mathbb R^d$.

One can always write $\|\nabla f(x)\|^2 d\mu(x) = H(x)^2dx$, where $H(x): = F(X)G(X)$, with $F(x):=\|\nabla f(x)\|^2$ 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 $\hat{F}(z) = \widehat{\|\nabla f\|^2}(z) = -\|z\|^2 \hat{f}(z)$

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 $\mathbb R^d$.

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 $\mathbb R^d$.

One can always write $\|\nabla f(x)\|^2 d\mu(x) = H(x)^2dx$, where $H(x): = F(X)G(X)$, with $F(x):=\|\nabla f(x)\|^2$ and $G(x):=\sqrt{p(x)}$.

By the Plancherel Theorem, we may simplify like so $$ E_\mu(f) = \int_{\mathbb R^d} H(x)^2 dx = \int_{\mathbb R^d} |\hat{H}(z)^2|dz, $$ 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 $\hat{F}(z) = \widehat{\|\nabla f\|^2}(z) = -\|z\|^2 \hat{f}(z)$

Case 1: gaussian

In this case, one we have $G(x) = \gamma_{d,\sigma}(x) = (2\pi\sigma^2)^{-d/2} e^{-\|x\|^2/(2\sigma)^2}$, and so $\hat{G}(z) = (2\pi)^{-d/2}e^{-\sigma^2\|z\|^2/2}=\sigma^d\gamma_{d,\sigma^{-1}}(z)$. Thus, $\hat{H} = T_{K_{d,1/\sigma}}\|\nabla f\|^2 = T_{K_{d,1/\sigma}} \tilde{f}$, 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 $$ \tilde{f}(z) := \|z\|^2 \hat{f}(z). $$ 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

$$ E_{\gamma_{d,\sigma}}(f) = \sigma^d \|\tilde{f}\|_{L^2(\gamma_{d,1/\sigma})}. $$