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Let $r_n \to \infty$ as $n \to \infty$. We have that $$\lVert v \rVert_{L^{r_n}(\Omega)} \leq C\lVert v \rVert_{L^{r_0}(\Omega)} < \infty$$ for all $n$, where $C$ is independent of $v$ and $n$.

Can I not just take $n \to \infty$ in this inequality to conclude that $v \in L^\infty(\Omega)$? The reason I ask is, that the author of this paper (see page 300, where my $v$ is his $u^+$) proves after proving the above quoted statement that $v \in L^\infty(\Omega)$ by a proof by contradiction.

StopUsingFacebook
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  • prove by contradiction that you can :-) – reuns Feb 13 '16 at 18:23
  • http://math.stackexchange.com/questions/133773/is-p-mapsto-f-p-continuous or http://math.stackexchange.com/questions/551309/continuity-of-norms-with-respect-to-p – Svetoslav Feb 13 '16 at 18:36

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You totally can, since $C$ is independent of $n$. Sometimes not doing so is illustratively useful, but you can.

Stella Biderman
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  • Thanks.. any idea why the paper doesn't just use this around the inequality (4.19) (I don't see why it's useful to do it differently in this csae ) – StopUsingFacebook Feb 13 '16 at 18:58
  • Ok, maybe the paper is just proving this fact (that the norm converges), and it has nothing to do with the particular problem in consideration. – StopUsingFacebook Feb 13 '16 at 19:03