# fa.functional analysis – Elliptic Estimates with Trace

Say $$Omegasubsetmathbb{R}^3$$ is a bounded domain with smooth boundary. Let $$Bin C^infty(barOmega)$$, and consider the set $$X={u in H^1(Omega)|u_{|partialOmega}=B_{|partialOmega} text{ in the sense of traces}}$$. Due to the Sobolev inequality, we know that $$H^1(Omega)$$ is continuously embedded in $$L^p(Omega)$$ for $$pin(1,6)$$. Say we wanted to prove, for fixed $$pin(1,6)$$, that there exists $$C$$ such that for all $$uin X$$, we have
$$|u|_{L^p}le C(|nabla u|_{L^2}+|B|_{L^infty}).$$
A pretty standard proof probably goes something like this: assume not, then there exists a sequence of functions $$u_nin X$$ such that
$$frac{1}{n}|u_n|_{L^p}ge|nabla u_n|_{L^2}+|B|_{L^infty}.$$
Defining
$$v_n=u_n/|u_n|_{L^p}$$
we have that $$|v_n|_{L^p}=1$$ for each $$n$$ and
$$frac{1}{n}ge |nabla v_n|_{L^2}+|B|_{L^infty}/|u_n|_{L^p}.$$
From this, we see, in particular, that $$v_n$$ is bounded in $$H^1$$. Therefore if $$pin(1,6)$$, then due to Rellich’s theorem, we have that there exists $$vin H^1$$ such that (after passing to a subsequence) $$v_nto v$$ weakly in $$H^1$$ and strongly in $$L^p$$. In addition, since $$|nabla v_n|_{L^2}to 0$$, we conclude that $$v$$ is constant.

On the other hand, by definition, $${v_n}_{|partialOmega}=(B/|u_n|_{L^p})_{|partialOmega}$$. But we see from the above inequality that $$|B|_{L^infty}/|u_n|_{L^p}$$ must also go to zero as $$nto infty$$. Therefore, by the continuity of the trace operator from $$H^1(Omega)$$ to $$H^frac{1}{2}(partialOmega)$$ and the weak convergence of $$v_n$$ to $$v$$ in $$H^1$$, we have $$v_{|partialOmega}=0$$. Therefore, since we already know $$v$$ must be constant, we conclude that $$vequiv 0$$. However, since $$v_nto v$$ strongly in $$L^p$$, we have $$|v|_{L^p}=1$$, a contradiction.

OK, assuming that’s all more or less fine, notice that the contradiction comes for the fact that we were able to extract a subsequence of $$v_n$$ that converges strongly in $$L^p$$, and this followed from Rellich.

My question is, what about the case of $$p=6$$? The embedding $$H^1subset L^6$$ is continuous, but not compact, so we’d only be able to extract a subsequence $$v_n$$ that converges weakly in $$L^6$$. Then, in the last line of the proof above, we’d only be able to conclude $$|v|_{L^6}le 1$$, so no contradiction. I’m inclined to believe that the statement still holds for $$p=6$$, but how does one prove it?