Why $\dot{H}^{s}(\mathbb{R}^{d})\subset L^{2}_{loc}(\mathbb{R}^{d}), s\in]0,1[$? from Bahouri-Chemin-Danchin book Fourier analysis and nonlinear pde

Question

Why $\dot{H}^{s}(\mathbb{R}^{d})\subset L^{2}_{loc}(\mathbb{R}^{d}), s\in]0,1[$? a detail from Bahouri-Chemin-Danchin book fourier analysis and nonlinear partial differential equations Page28, Proposition 1.37.

In this page, the authors say "In order to see that $u$ is in $L^{2}_{loc}(\mathbb{R}^{d})$, it is suffices to write $u=\mathscr{F}^{-1}(1_{B(0,1)}\hat{u})+\mathscr{F}^{-1}(1_{\mathbb{R}^{d}-B(0,1)}\hat{u}).$"

I'm so confused about this decomposition of $u$. If we want to prove this Proposition, we need to prove $$\int_{B(0,R)}u^{2}dx<\infty,\forall R>0.$$ According to authors' hint, we need to prove that $$\mathcal{F}^{-1}(1_{B(0,1)}\hat{u})\in L^{2}(B(0,R))$$ and $$\mathcal{F}^{-1}(1_{B(0,1)^{c}}\hat{u})\in L^{2}(B(0,R)),$$ but what should we do next?

Answer 1

By definition, $u\in\dot{H}^s(\Bbb{R}^d)$ means $\hat{u}\in L^1_{\text{loc}}(\Bbb{R}^d)$ and $\int_{\Bbb{R}^d}|\xi|^2|\hat{u}(\xi)|^2\,d\xi<\infty$. Let $B$ denote the unit ball.

• Since $\hat{u}$ is locally integrable, we have $1_{B}\hat{u}\in L^1(\Bbb{R}^d)$ and so its inverse Fourier transform $\mathcal{F}^{-1}(1_B\hat{u})$ is defined by the usual integral formula and it belongs to $C_0(\Bbb{R}^d)$ (continuous function which vanishes at infinity). In fact all we need is that this is a continuous function (which easily follows by dominated convergence) and so it is locally bounded and hence it belongs to $L^p_{\text{loc}}(\Bbb{R}^d)$ for all $1\leq p\leq\infty$, in particular for $p=2$.
• Next, $\int_{\Bbb{R}^d}|1_{B^c}\hat{u}|^2\,d\xi=\int_{B^c}|\hat{u}(\xi)|^2\,d\xi\leq\int_{B^c}|\xi|^2|\hat{u}(\xi)|^2\,d\xi\leq\int_{\Bbb{R}^d}|\xi|^2|\hat{u}(\xi)|^2\,d\xi<\infty$. I’m simply using the fact that if $\xi\in B^c$ then $1\leq |\xi|^2$, and then I’m trivially enlarging the domain of integration. This implies $1_{B^c}\hat{u}$ belongs to $L^2(\Bbb{R}^d)$, and since the Fourier transform is an isometry here, the inverse Fourier transform belongs to $L^2(\Bbb{R}^d)$ as well.

So, with this decomposition of $u$, we have actually shown that $\dot{H}^s(\Bbb{R}^d)\subset C_0(\Bbb{R}^d)+L^2(\Bbb{R}^d)\subset L^2_{\text{loc}}(\Bbb{R}^d)$, i.e every function in $\dot{H}^s$ can be written as a sum of a $C_0$ function and an $L^2$ function, and hence belongs to $L^2_{\text{loc}}(\Bbb{R}^d)$ (and hence by Holder’s inequality, you can also deduce, if you wanted, that it belongs to $L^p_{\text{loc}}$ for all $1\leq p\leq 2$).