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Consider a "weight" function $\rho: \mathbb{R}^{d} \rightarrow \mathbb{R}$ and the following weighted Sobolev norm: \begin{equation} \|h\|_{\dot{H}^{1}(\rho;\mathbb{R}^{d})}:=(\int_{\mathbb{R}^{d}} \|\nabla h(x)\|^{2}\rho(x) d x)^{1/2} \end{equation} where $dx$ stands for the Lesbegue measure on $\mathbb{R}^{d}$. When $\rho(x)=1$ everywhere we recover the "classical" Homogeneous Sobolev norm: \begin{equation} \|h\|_{\dot{H}^{1}(\mathbb{R}^{d})}:=(\int_{\mathbb{R}^{d}} \|\nabla h(x)\|^{2} d x)^{1/2} \end{equation} and in this case we have also a characterization of the dual norm $\dot{H}^{-1}(\mathbb{R}^{d})$ via the Fourier transform, which writes: \begin{equation} \|\nu\|_{\dot{H}^{-1}(\mathbb{R}^{d})}:=\sup_{\|h\|_{\dot{H}^{1}(\mathbb{R}^{d})}\leq 1} |\langle h,\nu\rangle|=(\int \|\omega\|^{-2} |\hat{\nu}(\omega)|^{2} d \omega)^{1/2} \end{equation} for $\nu$ a finite measure and $\hat{\nu}$ denotes its Fourier transform. My question is the following: do we have the same kind of characterization via the Fourier transform for the dual norm of the weighted Sobolev under some conditions on $\rho$ ?

I am looking for examples in simple cases where $\rho$ is the indicator of a compact of $\mathbb{R}^{d}$ or when it is a Gaussian density function.

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  • $\begingroup$ Your third line defines the $\dot{H}^{-1}$ norm (minor typo). For defining your dual norm in the weighted case: do you want the duality pairing to be with respect to $L^2$ or with respect to the weighted $L^2$ norm? $\endgroup$ Commented May 20, 2021 at 12:07
  • $\begingroup$ Oh yes I corrected thanks. Yes indeed I was not precise, I suspect that it should be with a clever weighted $L_2$ norm and I will be very happy if it holds somehow.. $\endgroup$ Commented May 20, 2021 at 14:02
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    $\begingroup$ @TitouanVayer Nice question. I had a similar question, and answered the Gaussian case here mathoverflow.net/a/396157/78539. Main idea: If $\rho(x) = \varphi_d(\|x\|)$, with $\varphi_d(t) := (2\pi)^{-d/2} e^{-t^2/2}$, then $\hat \rho(\zeta) = \varphi_d(\|\zeta\|)$ for all $\zeta \in\mathbb C^d$. Also, not that if $g:\mathbb R^d \to \mathbb R$ is defined by $g(x):=\|\nabla h(x)\|$, then $\widehat g(\zeta)=\widetilde h(\zeta):=\|\zeta\| \hat h(\zeta)$, for all $\zeta \in \mathbb C^d$. We conclude that $$\|\nabla h\|_{L^2(\rho,\mathbb R^d)}^2=\|\widetilde h\|_{L^2(\rho,\mathbb C^d)}^2.$$ $\endgroup$
    – dohmatob
    Commented Nov 10, 2022 at 8:41

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