# Poincare Inequality for $H^2$ function satisfying homogeneous Robin boundary conditions

Let $$\Omega\subset\mathbb{R}^3$$ be a bounded smooth domain. In general, for a Poincare inequality of the type $$\|u\|_{L^2}\le C \|\nabla u\|_{L^2}$$ to hold for all $$u\in X\subset H^1(\Omega)$$ and $$C$$ independent of $$u$$, then $$X$$ needs to be such that it doesn't contain constant translates. That is, if we consider $$u+M$$ for large $$M>0$$, the left hand side of the inequality increases indefinitely while the right hand side is unchanged, so we need some extra constraint in the definition of $$X$$. So common choices are $$X=H^1_0(\Omega)$$ or $$X=\{u\in H^1(\Omega)| \int_\Omega u\,dx=0\}$$.

Here's my question. Suppose, we'd like to say that there exists $$C$$ such that for all $$u\in X=\{u\in H^2(\Omega)|(\partial_n u+u)_{|\partial\Omega}=0\}\subset H^2(\Omega)$$ we have $$\|u\|_{L^2}\le C\|\nabla u\|_{L^2}.$$ First, is this true? If so, how does one prove such a statement? Essentially the requirement that $$u$$ satisfies the homogeneous Robin condition $$(\partial_n u+u)_{|\partial\Omega}=0$$ should at least formally rule out constant translates, since $$(\partial_n u+u)_{|\partial\Omega}=0$$ is not invariant under translation of $$u$$ by constants.

My guess is that it IS true, however, the usual proof I know of such statements usually relies on some compactness argument. For example, if $$X$$ were simply $$H_0^1(\Omega)$$, then for the sake of contradiction, if we assume that there exists a sequence $$u_n\in H_0^1$$ such that $$\|u_n\|_{L^2}\ge n\|\nabla u_n\|_{L^2}$$ then, defining $$v_n=u_n/\|u_n\|_{L^2}$$, we have $$\frac{1}{n}\ge \|\nabla v_n\|_{L^2}.$$ Thus we have a bounded sequence in $$H^1$$ and a subsequence that converges strongly in $$L^2$$ and weakly in $$H^1$$ to some $$v\in H^1$$. Because $$\|\nabla v_n\|_{L^2}\to 0$$, $$v$$ is constant. And since the trace map is continuous (and weakly continuous) from $$H^1(\Omega)$$ to $$H^\frac{1}{2}(\partial\Omega)$$ we have that $$v$$ is in fact in $$H^1_0(\Omega)$$ and therefore $$v=0$$. Then we have a contradiction because $$\|v_n\|_{L^2}=1$$ for each $$n$$ implies that $$\|v\|_{L^2}=1$$.

Now this argument doesn't work for Robin boundary conditions because now the relevant (Robin) trace operator is continuous and weakly continuous from $$H^2(\Omega)$$ to $$H^\frac{1}{2}(\partial\Omega)$$. In particular, if $$v$$ is the weak $$H^1$$ limit of a sequence $$v_n\in H^2$$, then $$v$$ could be in $$H^1$$ but not $$H^2$$ and thus the notion of the normal derivative $$(\partial_n v)_{|\partial\Omega}$$ may not even make sense for $$v$$. And without being able to say $$(\partial_n v+v)_{|\partial\Omega}=0$$, we can't necessarily say that $$v=0$$ like we did in the previous paragraph. So this is where I'm stuck. Any help would be appreciated.

• You are not dense, I was wrong. The counterexample by gerw is fine. Commented Jun 9, 2021 at 13:30
• Fozz, you might want to check Section 6.16 in "Linear functional analysis" by Alt where a quite general theorem on the validity of Poincare inequalities for subsets $X$ of $W^{1,p}$ is proven. (There it is also required that $X$ is convex and closed, so weakly closed!) Commented Jun 9, 2021 at 14:44

This is not true. Take $$\Omega = (-1,1)$$ and functions $$u_M$$ like (I hope that I got the constants right) $$u_M(x) =\begin{cases} -M^2 (|x|-1)(|x|-1+1/M) + M & \text{for } |x| > 1-1/(2M) \\ 1/4 + M & \text{else.}\end{cases}$$ Then, $$\|u_M\|_{L^2}$$ grows like $$M$$, whereas $$\|\nabla u_M\|_{L^2}$$ grows like $$\sqrt{M}$$. By modifying the definition of $$u_M$$ it should also be possible to get a bounded norm of the derivative.

The idea is to take an constant function and to modify it locally such that it satisfies the boundary condition. This is possible since $$u \mapsto \partial_\nu u$$ cannot be bounded on $$H^1(\Omega)$$.

• Is $u_M$ in $H^2$? Commented Jun 9, 2021 at 13:06
• Yes, if I did not introduce an error, $u_M$ should belong to $C^{1,1} = W^{2,\infty}$.
– gerw
Commented Jun 9, 2021 at 13:07
• Yes, you are right, $Du$ is piecewise linear. The factor to scale everything by is $\sqrt{M/6}$. Commented Jun 9, 2021 at 13:27
• Thanks, looks good. Just a sanity check here: if we were simply going for a statement like $\|u\|_{L^2(\Omega)}\le C(\|\nabla u\|_{L^2(\Omega)}+\|u\|_{L^2(\partial\Omega)})$, then I think we can take $X=H^1(\Omega)$?
– Fozz
Commented Jun 9, 2021 at 15:40
• @Fozz: Yes, this is possible.
– gerw
Commented Jun 9, 2021 at 16:07