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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 $R \to +\infty$ ?

flag	asked Apr 7 2012 at 10:06 siminore 1

	Is it $\|\|(1-\phi)u\|\|$ or $\|\|\phi u\|\|$ ? – Uday Apr 7 2012 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 2012 at 14:02

Wang Ming · Answer 1 · 2012-05-04 03:22:31Z UTC

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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.

answered May 4 2012 at 3:22

Wang Ming
165●6

Decay of Sobolev fractional norms outside a ball

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