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I'm reading a paper whose first section discussed $H^1$ maps defined on star-shaped sets, but I got stuck in verifying a claim for quite a while. I'm now thinking the claim is wrong, but it's hard to either prove or give counterexamples.

The following iare the assumptions, immediate consequences and the claim I don't know how to prove.


Let $u_i$ be a bounded sequence in $H^1(B_3)$ where $B_3$ is the ball of radius 3 in $\mathbb{R}^3$, hence by Rellich-Kodrachov Theorem, up to a subsequence $u_i$ converge weakly to some $H^1$ function $u$, and converge strongly to $u$ in $L^2(B_3)$. Let $h_i$ be a sequence of Lipschitz functions defined on $S^2$, the unit sphere in $\mathbb{R}^3$, with uniform Lipschitz constant $C$. We further assume $|h_i| \le 2$ and thus by Arzela-Ascoli Theorem $h_i$ converge to a continuous function $h$ uniformly on $S^2$.

So far so good. However, based on these settings the authors immediately claim the following:

$$u_i(h_i(\theta),\theta) \rightarrow u(h(\theta),\theta) \quad\mbox{strongly in $L^2(S^2)$ under polar coordinates.}$$


The authors didn't give a proof. Maybe they think it's trivial fact, but I really spent a lot of time on understanding this claim. My first try is by observing that $\{u_i(h_i(\theta),\theta)\}$ is a bounded sequence in $H^1(S^2)$, so up to a subsequence it converges to some function $g(\theta)$, but I've no idea why $g(\theta)=u(h(\theta),\theta)$. Then I tried to split the terms: $u(h(\theta),\theta)$ is close to both $u_i(h(\theta),\theta)$ and $u(h_i(\theta),\theta)$, but I've no idea how to estimate $||u_i(h_i(\theta),\theta)-u_i(h(\theta),\theta)||_{L^2(S^2)}$ or $||u_i(h_i(\theta),\theta)-u(h_i(\theta),\theta)||_{L^2(S^2)}$.

Can anyone here give me any hint? I'll really appreciate it.

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  • $\begingroup$ If we take $h_i(theta)$ constant, isn't the result roughly that $u_i \rightarrow u$ in $L^2(S^2)$. I not sure i would believe the result. Consider the same result on $ S_R:= \partial B_R$. I would believe the statement for a.e. $R$ but not sure i believe it for all $R$ ??? $\endgroup$
    – Math604
    Nov 21, 2016 at 0:10
  • $\begingroup$ Assume we can always re-define $u$. Actually all the authors want is the existence of such a $u$ such that the above convergent properties hold. For $h_i$ constant, the claim follows from the fact that trace operator is compact. $\endgroup$
    – student
    Nov 21, 2016 at 1:41

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