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If $f \in W^{s,2} (\Bbb R^n) $, then by the Plancherel's theorem, I know that its Fourier transform $ \mathscr F f(\xi) \in L^2 (\Bbb R^n) $. ($ \scr F$ means the Fourier transform). Now I want to show that if $ f \in W^{s,2} (\Bbb R^n)$, for $ \xi = (\xi_1 , \cdots ,\xi_n ), \; s \geqslant 0$, $$ \int_{\Bbb R^n} \xi_j^{2s} | \mathscr F f( \xi) |^2 d \xi < \infty \;\;(\forall j = 1,\cdots,n). $$ How can I prove this, or do I need some more assumptions?

If this holds then I think I can conclude that $(1+ | \xi |^2)^s (\mathscr F f)^2 \in L^1$.

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You need more assumptions. The functions that satisfy this condition are exactly those that belong to the Sobolev spaces $H^s(\mathbb{R}^n)$.

If $s = k$ is an integer, and if $f \in C^k(\mathbb{R}^n)$ such that all derivatives of order $|\alpha| \leq k$ satisfy $D^{\alpha} f \in L^2(\mathbb{R}^n)$ then $f$ satisfies the above condition and belongs to $H^k(\mathbb{R}^n)$ but this space is much larger. It can be characterized as the space of functions that have weak derivatives of all orders $\leq k$ that belong to $L^2(\mathbb{R}^n)$.

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Thank you very much. If $f \in W^{s,2}$ then would you give me some hints to prove the claim? –  Ann Oct 16 '12 at 17:39
    
What is your definition of $W^{s,2}(\mathbb{R}^n)$? –  levap Oct 16 '12 at 17:50
    
It means just general Sobolev space, $W^{s,2} = H^{s}$. –  Ann Oct 16 '12 at 17:51
    
I mean, the usual definition of the Sobolev spaces for real $s$, is that it consists of functions $f \in L^2(\mathbb{R}^n)$ such that for any $\xi = (\xi_1, ..., \xi_n)$ the integral you wrote in the question converges and in that case, there isn't much to prove. If you are talking about integer $s$, then there is the formulation of weak derivatives and then you can prove that this definition is equivalent to the one using the Fourier transform. –  levap Oct 16 '12 at 17:55

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