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?

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

1 Answer 1


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. 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 $C^1$).


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