# Understanding Robin Laplacian definition through Friedrichs extension on compact manifolds

Let's consider $$\Omega$$ to be either a compact manifold with boundary (as good as needed) or a bounded domain in $$\mathbb{R}^n$$). Also $$H^k(\Omega)$$ is the Sobolev Hilbert space of order $$k$$.

There are some places (for instance, here) where I've seen the Robin Laplacian can be defined in two ways, and I can't see they are equivalent.

First, it can be defined as the operator acting as the Laplacian on the domain $$D = \{u \in H^2(\Omega): \nu \cdot \nabla u + \alpha u = 0 \text{ on } \partial\Omega\},$$ where $$\nu \cdot \nabla u$$ is the normal derivative of $$u$$ and $$\alpha< 0$$ the Robin parameter.

Second, they say that it can be defined as using the Friedrichs Theorem via the quadratic form $$Q(u) = \Vert\nabla u\Vert_{L^2(\Omega)}^2 + \alpha\Vert \gamma(u) \Vert_{L^2(\partial\Omega)}^2,$$ defined on $$H^1(\Omega)$$, where $$\gamma: H^1(\Omega) \to H^{1/2}(\Omega)$$ is the trace map.

As far as I know, one needs to close the form domain under the norm $$\Vert u \Vert_Q = \sqrt{Q(u) + M \Vert u \Vert^2_{L^2}},$$ where $$M$$ is the lower bound of the form $$Q$$ (which is semibounded from below) to get the closure $$\bar{Q}$$ of $$Q$$. Then the associated self-adjoint operator $$T$$ is given by Kato's representation theorem: $$u \in D(\bar{Q})$$ is in $$D(T)$$ if there is $$v \in L^2(\Omega)$$ such that $$Q(u,w) = \langle v, w\rangle$$ for every $$w \in D(\bar{Q})$$, and for that $$u$$ we have $$Tu = v$$.

So my question is the following. Since $$\alpha < 0$$ it could be that $$Q(u) = 0$$ for some $$u$$, and in that case when one add to $$D(Q)$$ limit points respect $$\Vert \cdot \Vert_Q$$ there is no guarantee that $$D(\bar{Q})$$ is still a subset of $$H^1(\Omega)$$ which I think is crucial to prove that $$D(T) = D$$. How can one prove that without $$D(\bar{Q}) \subset H^1$$, or how can one see that actually the inclusion holds?

• Your definition of $Q$ is missing some squares. Commented Feb 28, 2019 at 14:50
• Yep, sorry for that. I'll edit it. Commented Feb 28, 2019 at 19:07

Ok, I finally got a reference where they explain it. The following inequality is needed: $$||\gamma(u)||_{L^2(\partial \Omega)} \leq C ||u||_{L^2(\Omega)} ||u||_{H^1(\Omega)}. \tag{1}\label{eq1}$$ This inequality is proven in the linked reference for $$\Omega$$ with $$C^1$$ boundary (Thm. 7.9) and it is proven for $$\Omega$$ Lipschitz in Thm. 1.5.1.10 of P. Grisvard's Elliptic Problems in Nonsmooth Domains.
Once that inequality is available, the rest follows easily using Young's inequality (with $$\varepsilon$$), $$ab \leq \frac{1}{2}(\varepsilon a^2 + \frac{1}{\varepsilon} b^2)$$, which holds for any $$a, b, \varepsilon > 0$$.
Theorem. If $$\Omega$$ is Lipschitz, then norm associated with $$Q$$ (see the question for the definition) is equivalent to the $$H^1$$ norm: $$||\cdot||_Q \sim ||\cdot||_{H^1(\Omega)}.$$ Proof. It is clear from the definition that $$||u||_Q \leq ||u||_{H^1(\Omega)}$$, so we just need to prove the other inequality. Applying Young's inequality to (the square of) \eqref{eq1} it follows $$|\alpha| ||\gamma(u)||_{L^2(\partial \Omega)}^2 \leq \frac{1}{2} ||u||_{H^1(\Omega)}^2 + c ||u||_{L^2(\Omega)}^2,$$ and for $$\alpha < 0$$ (which was the case on the question), $$\alpha ||\gamma(u)||_{L^2(\partial \Omega)}^2 \geq -\frac{1}{2} ||u||_{H^1(\Omega)}^2 - c ||u||_{L^2(\Omega)}^2 = -\frac{1}{2} ||\nabla u||_{L^2(\Omega)}^2 - \left(\frac{1}{2} + c\right) ||u||_{L^2(\Omega)}^2.$$ Substitution into the definition of $$Q$$ leads to $$Q(u) \geq \frac{1}{2} ||\nabla u||_{L^2(\Omega)}^2 - \left(\frac{1}{2} + c\right) ||u||_{L^2(\Omega)}^2$$ and therefore $$Q(u) + (c + 1) ||u||_{L^2(\Omega)}^2 \geq \frac{1}{2} ||u||_{H^1(\Omega)}^2.$$ This, together with the fact that taking any $$M' > M$$ for the definition of $$||\cdot||_Q$$ leads to an equivalent norm, concludes the proof. $$\square$$
Hence, all the limit points of $$D(Q)$$ with respect to $$||\cdot||_Q$$ need to be in $$H^1(\Omega)$$, which implies $$D(\bar{Q}) \subset H^1(\Omega)$$ whenever $$\Omega$$ is good enough (Lipschitz or $$C^1$$).