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I am reading a book about Euler equation and at some point we aim to prove the existence of global solution in 2D and I struggle with a point in the proof. We have a strong solution of the Euler equation $u \in L^\infty([0,T], H^s)$ with $\nabla \cdot u = 0$ for $s > d/2 + 1 = 2$ (here $d$ is the dimension). By the Sobolev emdedding (as $s - 1 > d/2$), we have $$ \| \omega \|_{L^\infty} \le C\| \omega \|_{H^{s-1}} \le C'\| u\|_{H^{s}} $$ where the last inequality comes from $ |\hat \omega(t, \xi)| \le C|\xi||\hat u(t, \xi)|$. Now we want to prove that $\|\omega(t)\|_{\infty} = \|\omega_0\|_{L^\infty}$ for every $t$. To do this, we prove that, after a few computation that $$\frac{1}{p}\frac{d}{dt}\|\omega\|_{L^p} = 0$$ for every $p \in \mathbb R$. Now the author claims that we obtain the desired result by taking the limit $p \to \infty$. However, I've never heard something like this, could one of you help me with this ?

Falcon
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    see this answer: https://math.stackexchange.com/questions/1450174/if-all-lp-norms-are-bounded-then-the-l-infty-is-bounded – daw May 08 '23 at 15:08
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    The factor $1/p$ is of no significance. If $d/dt|\omega|{L^p}=0$ then $|\omega(t)|{L^p} = |\omega_0|_{L^p}$ for all $t$ and $p$, which implies the desired $L^\infty$ regularity. – daw May 08 '23 at 15:10
  • Ok nice, thanks a lot! – Falcon May 08 '23 at 15:19

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