Let $\Omega \subset \mathbb{R}^{d}$ and $\|.\|$ is the standard euclidean $2$-norm. I came across a definition of $\dot{H}^{-1}(\Omega)$ which is a bit confusing. In [1] authors define the following semi-norm for $h\in C^{1}(\Omega)$: \begin{equation} \|h\|_{\dot{H}^{1}}:=\left(\int_{\Omega} \|\nabla h(x)\|_2^{2} dx \right)^{1/2} \end{equation} where $dx$ writes for the Lebesgue measure. They define for a signed measure $\nu$ on $\Omega$ the norm $\|\nu\|_{\dot{H}^{-1}(\lambda)}$ by standard duality arguments: \begin{equation} \|\nu\|_{\dot{H}^{-1}}=\sup_{\|h\|_{\dot{H}^{1}}\leq 1} |\int_{\Omega} h d \nu|= \sup_{\|h\|_{\dot{H}^{1}}\leq 1} |\langle h,\nu\rangle| \end{equation} where I noted $\langle h,\nu\rangle=\int_{\Omega} h d \nu$ which defines a inner product. Authors argue that this Homogeneous Sobolev norm is finite for measure having zero total mass.

I was a bit confused: is this definition a particular case of the more "standard" one using tempered distributions and Fourier transform (see Definition 1.31 in [2] e.g.) ? More precisely consider a tempered distribution $u$ over $\Omega$ and: \begin{equation} \|u\|_{\dot{H}_*^{-1}}:= \left(\int_{\Omega} \|\omega\|^{-2}|\hat{u}(\omega)|^{2} d \omega\right)^{1/2} \end{equation}

where $\hat{u}$ writes for the Fourier transform. My idea is that if $\nu$ is a signed measure zero total mass and with density wrt the Lesbegue measure, then it can be written as $\nu= (f-g) dx$ where $f,g$ are positive functions with $\int f dx=\int g dx$. It can be seen as a the following tempered distribution: \begin{equation} \forall \phi \in S(\Omega), \ \langle \nu,\phi\rangle=\int_{\Omega} \phi(x)d \nu(x)=\int_{\Omega} \phi(x)(f(x)-g(x))dx \end{equation}

where $S(\Omega)$ is the Schwartz class. Moreover we would have something like $\hat{\nu}(\omega)= (\hat{f}(\omega)-\hat{g}(\omega))$.

In a dirty way we would have also $\|\nabla h\|_2^{2}=\|\omega\|^{2} |\hat{h}(\omega)|^{2}$ so $\|h\|_{\dot{H}^{1}}=\| \|\omega\| \hat{h} \|_{L_2}$. Moreover $|\langle h,f-g\rangle|=|\langle \hat{h},\hat{f}-\hat{g}\rangle|$ by Plancherel. And also $|\langle \hat{h},\hat{f}-\hat{g}\rangle|=|\langle \|\omega\|^{1} \hat{h},\|\omega\|^{-1}(\hat{f}-\hat{g})\rangle|\leq \| \|\omega\|^{2} \hat{h} \|_{L_2} \ \| \|\omega\|^{-2}|\hat{f}-\hat{g}|^{2}\|_{L_2}$. This would give: \begin{equation} \|\nu\|_{\dot{H}^{-1}}= \left(\int_{\Omega} \|\omega\|^{-2}|\hat{f}(\omega)-\hat{g}(\omega)|^{2} d \omega\right)^{1/2}=\|\nu\|_{\dot{H}_*^{-1}} \end{equation}

Does this reasoning make sense ? Does it requires additional assumptions on $h,f,g$ ? For example it seems that they must be in $L_2(\Omega)$ right ?

[1] Comparison between W2 distance and H˙ −1 norm, and localisation of Wasserstein distance. Rémi Peyre. 2018.

[2] Fourier Analysis and Nonlinear Partial Differential Equations, Hajer BahouriJean-Yves CheminRaphaël Danchin. 2011


1 Answer 1


I think you misunderstood what the author said: "Note that this norm is finite only for measures having zero total mass."

It means that if the measure has a non zero total mass, then the norm is infinite. Indeed if you take $h$ to be big constant function, you can make the sup as big as you want because de Sobolev norm of $h$ is 0.


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