# fa.functional analysis – Poincare Inequality for \$H^2\$ function satisfying homogeneous Robin boundary conditions

Let $$Omegasubsetmathbb{R}^3$$ 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 $$uin Xsubset H^1(Omega)$$ 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 $$X={uin H^1(Omega)| int_Omega u,dx=0}$$.

Here’s my question. Suppose, we’d like to say that there exists $$C$$ such that for all $$uin X={uin H^2(Omega)|(partial_n u+u)_{|partialOmega}=0}subset H^2(Omega)$$ 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 $$(partial_n u+u)_{|partialOmega}=0$$ should at least formally rule out constant translates, since $$(partial_n u+u)_{|partialOmega}=0$$ 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 $$u_nin H_0^1$$ 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 $$vin H^1$$. 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 $$H^frac{1}{2}(partialOmega)$$ 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 $$H^frac{1}{2}(partialOmega)$$. In particular, if $$v$$ is the weak $$H^1$$ limit of a sequence $$v_nin H^2$$, then $$v$$ could be in $$H^1$$ but not $$H^2$$ and thus the notion of the normal derivative $$(partial_n v)_{|partialOmega}$$ may not even make sense for $$v$$. And without being able to say $$(partial_n v+v)_{|partialOmega}=0$$, 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.