I have a doubt concerning the trace of Sobolev functions. Let $C=\Omega\times(0,\infty)$ an infinite cylinder of basis a smooth domain $\Omega$ of $\mathbb{R}^{N}$ and consider the classical Sobolev space $H^{1}(C)$. If for a function $u\in H^{1}(\mathcal{C})$ we denote by $u(\cdot,0)$ the trace of $u$ over $\Omega$, is it true that the range of this trace operator coincides with the Sobolev space $H^{1/2}(\Omega)$, i.e. $$\left\{u(\cdot,0):u\in H^{1}(\mathcal{C})\right\}=H^{1/2}(\Omega)$$ and, if yes, where I can find the proof of such result?

Thank you very much for any help!

  • $\begingroup$ Should be true, since this is a Lipschitz domain. I think you can reduce to the case of bounded domain by using a partition of unity along the unbounded direction. I don't know of a reference. $\endgroup$ – user147263 Sep 17 '14 at 16:13
  • $\begingroup$ Thank you so much, I'll try to figure out by following your arguments! $\endgroup$ – user40033 Sep 24 '14 at 21:17
  • 1
    $\begingroup$ @user40033 Did you succeed with this? I am interested in the case $\Omega$= compact manifold. $\endgroup$ – riem Jan 1 '15 at 17:29
  • $\begingroup$ @NormalHuman Does the general trace theorem and trace inequality hold on an unbounded Lipschitz domain? $\endgroup$ – Ellya Jul 20 '15 at 20:41

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