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I'm now reading Pazy's book about the semi-group operator. To prove the existence of the solution of KdV equation. He define the Hilbert space $H^s(\mathbb{R})$ $$ \Vert u\Vert_s=\left(\int(1+\xi)^s|\widehat{u}(\xi)|^2d\xi\right)^{\frac{1}{2}}$$ the inner product $$(u,v)_s=\int(1+\xi)^s\widehat{u}(\xi)\overline{\widehat{v}}(\xi)d\xi)$$ where $\widehat{u}$ is the Fourier Transform of $u$. In his book, I have the following question,when $s\geq 3$:

1.If $u\in H^s(R)$,why $Du\in H^{s-1}(\mathbb{R})$? ($Du$ denote $\frac{d}{dx}$)

2.Why $Du\in L^{\infty}(\mathbb{R})$? And $ \Vert Du \Vert_{\infty}\leq C \Vert Du \Vert_{s-1}\leq C \Vert u \Vert_s$?

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The norm should be defined as $\lVert u\rVert_s^2:=\int_{\mathbb R}(1+\xi^2)^s|\widehat u(\xi)|^2d\xi$.

  1. We use the fact that $\widehat{Du}(\xi)=i\xi\widehat{u}(\xi)$. Then \begin{align} \lVert Du\rVert_{s-1}^2&=\int_{\Bbb R}(1+|\xi|^2)^{s-1}|\xi|^2|\widehat u(\xi)|^2d\xi\\ &\leq \int_{\Bbb R}(1+|\xi|^2)^{s}\widehat u(\xi)|^2d\xi\\ &=\lVert u\rVert_s^2. \end{align}
  2. We use the inversion formula for Fourier transform. \begin{align} |Du(x)|&=C\left|\int_{\Bbb R}e^{itx}it\widehat u(t)dt\right|\\ &\leq C\int_{\Bbb R}|\xi||\widehat u(\xi)|d\xi\\ &=C\int_{\Bbb R}\sqrt{|\xi|}(1+\xi^2)^{-(s-1)/2}|\sqrt{|\xi|}\widehat u(\xi)|(1+\xi^2)^{(s-1)/2}d\xi\\ &\leq C\sqrt{\int_{\Bbb R}\xi(1+\xi^2)^{-(s-1)}d\xi}\lVert Du\rVert_{s-1} \end{align} (the last integral is convergent as $2(s-1)-1>1$).
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How can I prove $L^2(R)=H^0(R)$?That is to say is $|\widehat{u}(\xi)|^2=|u(\xi)|^2$? –  Joe May 28 '12 at 8:41
    
It's a result about Fourier transform: we define it on $L^1\cap L^2$, show that it's an isometry and extend it to $L^2$. –  Davide Giraudo May 28 '12 at 8:54

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