7
$\begingroup$

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.

1 Answers 1

5

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?).

  • 0
    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.2012-06-01
  • 0
    @LeonidKovalev that's correct, thanks for pointing that out. Of course one would take balls of increasing integer radius, say. Sloppy me.2012-06-01
  • 0
    @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 $p=2$.)2012-06-02
  • 0
    @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 :-)2012-06-02
  • 0
    @Thomas: Well, in the _first_ step you don't need a subsequence, that was my point.2012-06-02
  • 0
    @HendrikVogt What do you mean by first step? Compactness of the embedding into $L^p$ only guarantees convergence of a subsequence, if I recall correctly (I may not, it's been 12 years by now).2012-06-02
  • 0
    @Thomas: Oh, 12 years is a long time `:-)` Let me recall: a compact operator maps weakly convergent sequences into strongly convergent sequences. And the embedding is a compact map, so $(u_m)$ is strongly convergent in $L^p$.2012-06-02
  • 0
    @HendrikVogt Ok, then I see your point. Thank you!2012-06-02