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For $-\infty<s<\infty$ 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

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  • $\begingroup$ This is by the fact that the Fourier transform defines an isomorphism from $\mathcal{S}'$ to itself. $\endgroup$
    – Jose27
    Apr 2 at 21:38
  • $\begingroup$ 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 $(x_1,\ldots, x_n)$. $\endgroup$
    – eraldcoil
    Apr 2 at 21:54
  • $\begingroup$ 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). $\endgroup$
    – Jose27
    Apr 10 at 12:38

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