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Suppose $\delta\in (0,1)$ and $r<1+\delta.$ Suppose moreover we are given a sequence of functions $u_m\in H^{1/2,2}(\partial B_r(0))$, where $B_r(0)$ denotes the euclidean $n-$dimensional ball. Assume that $u_m\to 0$ strongly in $H^{1/2,2}(\partial B_r(0))$. Then I would like to extend these functions to functions in $H^{1,2}(\mathbb R^n\setminus B_r(0))$ by means of a function $$\eta:H^{1/2,2}(\partial B_r(0))\to H^{1,2}(\mathbb R^n\setminus B_r(0)),$$ defined as follows: setting $v=\eta(u)$ we would have

$v=0$ in $\mathbb R^n\setminus B_2(0),$

$v=u$ on $\partial B_r(0)$,

$-\Delta v=0$ in the annulus $B_2(0)\setminus B_r(0)$.

Is this possible ? Moreover, if such a map existed, would it have some continuity properties? I cannot see if my question is just a consequence of the usual extension theorem because I'm requiring harmonicity in an annulus. Even references are welcomed since I was not able to find any of them.

Best Regards.


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Your questions is confusingly stated. Did you mean $r<1-\delta$? If a function is in $H^{1,2}_0(R^n\backslash B_r)$, then its trace on $\partial B_r$ is zero. – Michael Renardy Dec 11 '12 at 19:05
you are right.. Edited.. -Guido- – user17697 Dec 11 '12 at 19:25
up vote 1 down vote accepted

Yes, this extension works. Your extended function is clearly in $H^1$ on the annulus as well as in $R^n\backslash B_2$. Since the traces on $\partial B_2$ agree, it is then in $H^1$ on $R^n\backslash B_r$.

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I've got a last question... Is it essential that the traces of the $u_m$ on $\partial B_r(0)$ tend to $0$ strongly in $H^{1/2,2}(\partial B_r(0))$ or is this assumption droppable? Finally where I can find some good references on harmonic extensions on annuli? BTW thank you very much -Guido- – user17697 Dec 11 '12 at 21:34
Well, there is not likely to be anything in the literature specific to this, but all you need is basic elliptic regularity theory. I am not sure what you mean by the question whether it is essential. Essential for what? Your question makes no reference to the limit $m \to\infty$, just to extending each $u_m$. If the $u_m$ converge to zero, then so do the corresponding $v_m$. – Michael Renardy Dec 11 '12 at 23:27

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