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Suppose you have a function $u$ that lies in the (fractional) Sobolev space $H^s$, with $0<s<1$. Take a smooth cut-off $\varphi$ such that $\varphi(x)=0$ if $|x|>R+1$. What can be said about the norm $\|(1-\varphi)u\|_{H^s}^2$? Does it have a precise decay, as $R \to +\infty$?

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Is it $||(1-\phi)u||$ or $||\phi u||$ ? – Uday Apr 7 '12 at 10:27
Of course not, take any summable sequence $c_k>0$, any nonzero bump function $\chi(x)$ with support in $|x|<1/2$ (just for elegance), and define $u(x)=\sum c_k\chi(x-k)$ – Piero D'Ancona Apr 7 '12 at 14:02

It would not have a decay rate, since the support of fourier transform of $u$ may lies in the infinity, the cut off would not work.

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