I encounter a problem when reading Struwe's book Variational Methods (4th ed). On page 38, it is assumed that $\|u_m\|$ is a minimizing sequence for a functional $E$, i.e. $E(u_m)\rightharpoonup I$ in $L^p(\mathbb{R}^n)$,

and then it assume in addition that

$u_m\rightharpoonup u$ weakly in $H^{1,2}(\mathbb{R}^n)$ and pointwise almost everywhere.

My question is

why the pointwise convergence assumption is reasonable? Since $\mathbb R^n$ is not compact, the embedding theorem is not obviously valid.

Thanks in advance.


For sufficiently small $p$ (more precisely: $p<2n/(n-2)$ for $n\ge 3 $ or $p$ arbitrary otherwise) the space $H^{1,2}(\Omega) $ is compactly embedded in $L^p$ for $\Omega \subset\subset \mathbb{R}^n$ with sufficiently regular boundary (take balls of increasing radius tending to infinity). This implies strong $L^p$ convergence of a subsequence, hence pointwise a.e, on each such $\Omega$, hence a.e.

(If $u_k$ converges pointwise a.e on each open set with compact closure it obviously converges pointwise almost everywhere. You may need to countably often further subsubsequence to make this work, but who cares?).

  • $\begingroup$ Minor point: the boundary of $\Omega$ should not be too weird for embedding to be valid. Lipschitz boundary is enough, and of course we can exhaust $\mathbb R^n$ by balls. $\endgroup$ – user31373 Jun 1 '12 at 17:38
  • $\begingroup$ @LeonidKovalev that's correct, thanks for pointing that out. Of course one would take balls of increasing integer radius, say. Sloppy me. $\endgroup$ – user20266 Jun 1 '12 at 18:16
  • $\begingroup$ @Thomas, minor nitpick: If I'm not mistaken, you get $u_m\to u$ strongly in $L^p$, and then a subsequence converges a.e. (Actually you could just take $p=2$.) $\endgroup$ – Hendrik Vogt Jun 2 '12 at 15:52
  • $\begingroup$ @HendrikVogt you have to take subsequences all the time, that's correct, also in the step mentioned by you. And yes, quite obviously $2 < 2n/(n-2) $ for $n\ge3$. Again, sloppy me :-) $\endgroup$ – user20266 Jun 2 '12 at 16:36
  • $\begingroup$ @Thomas: Well, in the first step you don't need a subsequence, that was my point. $\endgroup$ – Hendrik Vogt Jun 2 '12 at 17:06

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