# About the Sobolev space $H^{s,p}$ in Wong.

For $$-\infty and $$1\leq p<\infty$$, we define $$H^{s,p}=\left\{u\in \mathcal{S}': \mathcal{F}^{-1}((1+|\xi|^2)^{s/2}\widehat{u}(\xi))\in L^p\right\}$$ The space $$H^{s,p}$$ is a normed vector space if we equip it with the norm $$\left\|u\right\|_{s,p}:=\left\|\mathcal{F}^{-1}((1+|\xi|^2)^{s/2}\widehat{u}(\xi))\right\|_{p}$$ (Definition from the book introduction to pseudo differential operators by Wong)

Question If $$\left\|u\right\|_{s,p}=0$$ then $$u=0$$? why? (I'am confussed with the fact which $$u$$ is a tempered distribution...) Thanks

• This is by the fact that the Fourier transform defines an isomorphism from $\mathcal{S}'$ to itself. Apr 2 at 21:38
• but i can't see the intermediate steps. $\left\|u\right\|_{s,p}=0\Rightarrow \int |\mathcal{F}^{-1}((1+|\xi|^2))^{s/2}\widehat{u})|^p=0\Rightarrow \mathcal{F}^{-1}((1+|\xi|^2))^{s/2}\widehat{u})=0 a.e.\rightarrow (1+|\xi|^2)^{s/2}\widehat{u}=0\Rightarrow \widehat{u}=0\Rightarrow u=0$. Right? My confussion is originated from, if $|\xi|^2\widehat{u}=0$, then $u$ is a polynomial in $(x_1,\ldots, x_n)$. Apr 2 at 21:54
• That's because you don't want to conclude that $(1+|\xi|^2)^{s/2}\hat{u}(\xi)=0$ a.e., but rather that it's zero in the sense of tempered distributions; then since $(1+|\xi|^2)^{-s/2}$ is a smooth function with enough decay, we can multiply by it and obtain another element in $\mathcal{S}'$ (you can't do this with $|\xi|^{-s}$, which is where the polynomials come in; here you can only conclude that the support of $\hat{u}$ is at the origin, and so $\hat{u}$ is a linear combination of derivatives of $\delta$ masses at the origin). Apr 10 at 12:38