Sobolev embedding for $H^s$, $s > d/2 + 1$

Question

I am reading a book about Euler equation and at some point we aim to prove Beale-Kato-Majda theorem. We work in $\mathbb R^d$ and we take $s > 0$ such that $s > d/2 + 1$. We consider $u \in L^\infty([0,T], H^s)$ and divergence free. Then the author claims that $$\int_0^t \|\omega(\tau)\|_{L^\infty}d\tau \le C \int_0^t \|u(\tau)\|_{H^s}d\tau $$ by the Sobolev embeddings, for $\omega = \nabla \times u$. However, I really have no idea to which form of the embedding he refers to, I am not really comfortable with this fractional Sobolev space. Could one of you help me with this ?

Answer 1

We have the following inequalities $$||\omega(\tau)||_{L^\infty}\leq C||\omega(\tau)||_{H^{s-1}}\leq C'||u(\tau)||_{H^s}.$$ The first is the sobolev embedding of $H^{s-1}$ into $L^{\infty}$, which holds since $s-1>d/2$ (see e.g. Thm 1.3.5 in Modern Fourier Analysis by Grafakos). The second inequality follows from the definition of $H^s$ for $s>1$, where $||u||_{H^s}=||u||_{H^{\lfloor s\rfloor}}+||\nabla^{\lfloor s\rfloor}u||_{H^{s-\lfloor s\rfloor}}$ with $\lfloor s \rfloor$ the largest integer smaller or equal to $s$. (Or if you define $H^s$ via Fourier transform you can use $|\widehat{\omega}(\tau,\xi)|\leq C|\xi||\widehat{u}(\tau,\xi)|$).