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This question has been posted on Mathematics Stack Exchange but got no response, and so I put it here.

When I read the paper "On the attractor for a semilinear wave equation with critical exponent and nonlinear boundary dissipation" by Igor Chueshov, Matthias Eller and Irena Lasiecka, I encounter a difficulty:

Suppose that $\Omega$ is a simple connected domain with smooth boundary, the authors introduced a Robin Laplacian operator $$\Delta_{R}\colon L^{2}(\Omega)\to L^{2}(\Omega).$$This is an unbounded operator with the domain $$D(\Delta_{R})=\bigl\{u\in H^{2}(\Omega)\colon \partial_{\nu}u+u=0\ on\ \partial\Omega\bigr\}.$$ Moreover, the Robin Laplacian can be extended to a continuous operator $\Delta_{R}\colon H^{1}(\Omega)\to H^{1}(\Omega)'$ by $$(-\Delta_{R}u,v)_{L^{2}(\Omega)}=(\nabla u,\nabla v)_{L^{2}(\Omega)}+\langle u,v\rangle_{L^{2}(\partial\Omega)}.$$ Then the authors say "this extension is the duality map $H^{1}(\Omega)$ into $(H^{1}(\Omega))'$" when we equip $H^{1}(\Omega)$ the norm $$\|u\|^* =\sqrt{(\nabla u,\nabla u)_{L^{2}(\Omega)}+\langle u,u\rangle_{L^{2}(\partial\Omega)}}.$$ So, firstly, I want to know if this norm is equivalent to the usual Sobolev norm $$\|u\|=\sqrt{(\nabla u,\nabla u)_{L^{2}(\Omega)}+(u,u)_{L^{2}(\Omega)}}$$ when $u$ satisfies the Robin boundary condition? Next, the authors said $$D((-\Delta_{R})^{\frac{1}{2}})\sim H^{1}(\Omega),$$ why? The authors offered a reference but it is French, but I don't understand French. The reference is "Grisvard, P.: Characterisation de Quelques Espaces d'interpolation. Archives Rational Mechanics and Analysis (1967) #26, pp.40-63"

Any comments and hints are welcome, thank you very much!

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  • $\begingroup$ Whether the norms are equivalent doesn't depend on Robin bdy conditions. It is clear that they are, you just need to know that the embedding $H^1(\Omega) \rightharpoonup L^2(\partial\Omega)$ is compact. One direction is immediate : bounding bdy $L^2$ with $H^1$. In the other direction, supppose bdy $L^2$ +grad goes to zero while $H^1=1$. Take a weakly convergent subsequence. then the $L^2$ bdy norm converges strongly to zero, so the limit is constant (zero gradient in a connected domain) and with zero trace, so it is null, and $L^2$ norm $=1$, a contradiction. The usual proof.. $\endgroup$
    – username
    Commented Sep 26, 2022 at 7:28
  • $\begingroup$ @username you only need that $H^1(\Omega) \hookrightarrow L^2(\Omega)$ is compact, or am I mistaken? //// For the second question of OP, this is the Kato square root property which is always true for operators induced by symmetric forms as in the present case, see Kato's "Perturbation Theory", Chapter 6, §2.6. $\endgroup$
    – Hannes
    Commented Sep 26, 2022 at 13:18
  • $\begingroup$ @Glorfindel Thank you for your correction!!! $\endgroup$
    – monotone operator
    Commented Sep 26, 2022 at 14:42
  • $\begingroup$ @Hannes Let us call $N(u) = \sqrt{ \| \nabla u \|^2_{L^2(\Omega)} + \| u \|^2_{L^2(\partial\Omega)} }$ and $\| u \| = \sqrt{ \nabla u \|^2_{L^2(\Omega)} + \| u \|^2_{L^2(Omega) }}$. To show $N(u)\leq C \| u \|$ yo{u need at least a continuous embedding of $H^1(\Omega)$ in $L^2(\partial \Omega)$. It happens to be compact. In the other direction, yes, you use Rellich-Kondrachov (for $\| u \|_{L^2(\Omega)}=1$). $\endgroup$
    – username
    Commented Sep 26, 2022 at 17:12
  • $\begingroup$ @username That's what I meant, agreed. $\endgroup$
    – Hannes
    Commented Sep 27, 2022 at 12:43

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