Why is the dual of the Sobolev space $H_0^1(\Omega)$ denoted $H^{-1}(\Omega)$ ?

For a positive integer $k$, $H^k(\Omega)=W^{k,2}(\Omega)$. What is the motivation behind the $-1$ exponent?

  • 3
    $\begingroup$ I think the answer comes from Sobolev of real exponent, defined with Fourier transform. $\endgroup$ – Davide Giraudo Apr 20 '12 at 19:38

Definition Let $s\in\mathbb R$. $H^s(\mathbb R^n)$ is the vector space which consists of elements $u\in\mathcal S'(\mathbb R^n)$ such that $\widehat u$ is measurable and $(1+|\xi|^2)^{\frac s2}\widehat u \in L^2(\mathbb R^n)$. We endow $H^s(\mathbb R^n)$ with the inner product \begin{equation} \langle u,v\rangle_{H^s(\mathbb R^n)}:=\int_{\mathbb R^n}(1+|\xi|^2)^s\widehat u(\xi) \overline{\widehat v (\xi)}d\xi \end{equation} and we denote
\begin{equation} \lVert u\rVert_{H^s(\mathbb R^n)} :=\left(\int_{\mathbb R^n}(1+|\xi|^2)^s|\widehat u(\xi)|^2d\xi \right)^{\frac 12} \end{equation} the corresponding norm.

Let $s\in \mathbb R$ and $v\in H^{-s}(\mathbb R^n)$. If $u\in H^s(\mathbb R^n)$, the map $\xi\mapsto \widehat u(\xi)\widehat v(-\xi)$ is in $L^1(\mathbb R^n)$.

Indeed, for all $\xi\in \mathbb R^n$, $\widehat u(\xi)\widehat v(-\xi)=\widehat u(\xi) (1+|\xi|^2)^{\frac s2}\widehat v(-\xi)(1+|\xi|^2)^{-\frac s2}$, and the RHS is the product of two map of $L^2(\mathbb R^n)$. By Cauchy-Schwarz inequality, we have
$$\left|\int_{\mathbb R^n}\widehat u(\xi)\widehat v(-\xi)d\xi\right|\leq \lVert u\rVert_{H^s(\mathbb R^n)} \lVert v\rVert_{H^{-s}(\mathbb R^n)}.$$ Therefore, if $v\in H^{-s}(\mathbb R^n)$, the map $L_v$ defined by \begin{equation} u\in H^s(\mathbb R^n)\mapsto L_v(u)=(2\pi)^{-n}\int_{\mathbb R^n}\widehat u(\xi)\widehat v(-\xi)d\xi =\int_{\mathbb R^n}\widehat u(\xi)\overline{\mathcal F}v(\xi)d\xi \end{equation} is a continuous linear functional on $H^s(\mathbb R^n)$ (hence an element of $(H^s(\mathbb R^n))'$ and $\lVert L_v\rVert _{(H^s(\mathbb R^n))'}\leq (2\pi)^{-n}\lVert v\rVert_{H^{-s}(\mathbb R^n)}$. We can define $L\colon H^{-s}(\mathbb R^n)\to (H^s(\mathbb R^n))'$ by $L(v)=L_v$.

Theorem The map $L$ defined above is linear, bijective, and bicontinuous. It allow us to indentify the dual space of $H^s(\mathbb R^n)$ by $H^{-s}(\mathbb R^n)$.

Sketch of proof:

  • The fact that $L$ is linear comes from linearity of the inverse Fourier transform and the integral. Continuity follows from $\lVert L_v\rVert _{(H^s(\mathbb R^n))'}\leq (2\pi)^{-n}\lVert v\rVert_{H^{-s}(\mathbb R^n)}$.
  • $L$ is surjective. Indeed, let $T\in (H^s(\mathbb R^n))'$. We can show that there exists a constant $C$ such that for all $\varphi\in \mathcal S(\mathbb R^n)$ $$|\langle T,\varphi\rangle|\leq C\lVert (1+|\xi|^2)^{s/2}\mathcal F^{-1}\varphi\rVert_{L^2(\mathbb R^n)}.$$ We can see that $(1+|\xi|^2)^{-s/2}\widehat T$ is a linear functional over $L^2(\mathbb R^n)$ and we apply Riesz representation theorem.
  • Injectivity needs only the property $\mathcal F^{-1}\mathcal F(\varphi)=\varphi$ for all $\varphi\in\mathcal S(\mathbb R^n)$.
  • The inverse map is continuous using Banach isomorphism theorem.

When $\Omega$ is a arbitrary open subset of $\mathbb R^n$ we have to use charts (and $\Omega$ needs to be regular enough, since the Sobolev of integer order, defined in a classical way may be not equal to the corresponding one, even if $\Omega$ is bounded), but at least it shows the idea.

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  • $\begingroup$ Dear Sir, OP asked about why dual space of $H_0^1(\Omega)$ is $H^{-1}(\Omega)$. I could not see where do you have used fact that $u\in H_0^1(\Omega)$. Please throw some light on that . Thanks a lot $\endgroup$ – idon'tknow Sep 12 at 6:44

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