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

Question

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.

Answer 1

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 $H^1(\Omega)$.