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?