# Confusing definition of homogeneous Sobolev norm of order -1

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 $$h\in C^{1}(\Omega)$$: $$$$\|h\|_{\dot{H}^{1}}:=\left(\int_{\Omega} \|\nabla h(x)\|_2^{2} dx \right)^{1/2}$$$$ where $$dx$$ writes for the Lebesgue measure. They define for a signed measure $$\nu$$ on $$\Omega$$ the norm $$\|\nu\|_{\dot{H}^{-1}(\lambda)}$$ by standard duality arguments: $$$$\|\nu\|_{\dot{H}^{-1}}=\sup_{\|h\|_{\dot{H}^{1}}\leq 1} |\int_{\Omega} h d \nu|= \sup_{\|h\|_{\dot{H}^{1}}\leq 1} |\langle h,\nu\rangle|$$$$ where I noted $$\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: $$$$\|u\|_{\dot{H}_*^{-1}}:= \left(\int_{\Omega} \|\omega\|^{-2}|\hat{u}(\omega)|^{2} d \omega\right)^{1/2}$$$$

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: $$$$\forall \phi \in S(\Omega), \ \langle \nu,\phi\rangle=\int_{\Omega} \phi(x)d \nu(x)=\int_{\Omega} \phi(x)(f(x)-g(x))dx$$$$

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: $$$$\|\nu\|_{\dot{H}^{-1}}= \left(\int_{\Omega} \|\omega\|^{-2}|\hat{f}(\omega)-\hat{g}(\omega)|^{2} d \omega\right)^{1/2}=\|\nu\|_{\dot{H}_*^{-1}}$$$$

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

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.