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Let $B_t(0)$ denote the $n$ dimensional ball of radius $t$ centered at the origin. Does there exist a $\phi\in C(\mathbb{R}^n)$ function with the properties:

$ \phi (x) = \begin{cases} 1&x\in B_r(0) \\\ 0&x\not\in B_{r+3}(0) \end{cases} $

and for any real-valued function $f\in \mathcal{H}^\tau(\mathbb{R^n})$ ($\tau\in\mathbb{R}$, $\tau>d/2$) we have

$ \left\|\phi(\cdot) f(\cdot)\right\|_{\mathcal{H}^\tau(A_1)}\leq C\left\|f\right\|_{\mathcal{H}^\tau(A_2)} $

where $C$ is a constant independent of $f$, $A_1 = B_{r+2}(0)\setminus B_{r+1}(0)$ and $A_2 = B_{r+3}(0)\setminus B_{r}(0)$.

This has a similar feel to extension theory results if we think of $\phi$ as an extension operator which preserves $f$ on $B_r(0)$.

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what is $\mathcal{H}^\tau$? – Anton Petrunin Jun 6 '11 at 15:25
This is the Sobolev space $H^\tau=W^{\tau,2}$. – alext87 Jun 8 '11 at 13:24
I think I am missing something obvious, but why can you not pick any $\phi \in C^2$ and set $C = C(\vert\vert \phi\vert\vert_{C^2(B_{r+2}(0))})$, i.e. pull out the derivatives of $\phi$ in $L^{\infty}$? – Yakov Shlapentokh-Rothman Jun 8 '11 at 15:53
In the above comment, the 2 should be replaced by $s$ where $s > \tau$. – Yakov Shlapentokh-Rothman Jun 8 '11 at 15:55
Sure you can do this for integer $\tau$ just by using Leibniz rule. In fact the details are quite simple. I was more interested in the fractional sobolev space case where the details are escaping me. – alext87 Jun 8 '11 at 16:27
up vote 4 down vote accepted

Short answer: yes.

Let $\psi_\epsilon(x):=\frac{1}{\epsilon^n}\exp{\epsilon^2/(\epsilon^2-|x|^2)}$ for $|x|<\epsilon$, and $\psi_{\epsilon}(x)=0$ for $|x|\geq \epsilon$. Set $\epsilon=2$, and define $\phi$ is the convolution of $C\phi_{\epsilon}$ with the characteristic function of $B_{r+3/2}(0)$, that is,

$\phi(x):= C\psi_\epsilon(x)* \chi_{B_{r+3/2}(0)}$. Here $C$ is a normalizing constant (this may not be needed, but I haven't checked).

This yields a smooth cut-off function which is 1 in the ball $B_{r+1}(0)$, and zero outside $B_{r+2}(x)$.

To see this does the trick, one can use a localization theorem, for example, Theorem 3.20 in 'Strongly Elliptic Systems and Boundary Integral Equations' by W. McLean. This theorem states:

'Suppose that $\phi \in C^r_{comp}(\mathbb{R}^n)$ for some integer $r\geq 1$, and let $|s|\leq r$. If $u\in H^s(\Omega)$ then $\phi u \in H^s(\Omega)$, and $||\phi u||_{H^s(\Omega)} \leq C_r||\phi||_{W^{r,\infty}(\mathbb{R}^n)}Q_u$ where $Q_u=||u||_{H^s(\Omega). }$ (Apologies, I encountered trouble while trying to typeset the LaTeX here).

The same result holds with $H^s(\Omega)$ replaced with $\tilde{H}^s(\Omega)$.'

The proof proceeds using $\Omega = A_2$, and then

either by (a) considering the situation for $s=r$, using duality to see it holds for $s=-r$, and the intermediate $s$ by interpolation. This is suggested by Yakov above.

or (b) by examining $\hat{\phi u}$ and using Peetre's inequality.

Since the constructed $\phi \in C^\infty$ and has compact support, it will satisfy the inequality you seek. In my comment I asked whether you wanted a $\phi$ of minimal regularity (relative to $\tau$); my construction works but may be overkill.

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Is approach (b) supposed to work for $\Omega \neq \mathbb{R}^n$? It is not clear to me how the Fourier transform will be useful unless $\Omega = \mathbb{R}^n$, since that is the only case in which the Sobolev spaces are defined directly in terms of the Fourier transform. – Yakov Shlapentokh-Rothman Jun 15 '11 at 13:50
Yakov, the answer is 'yes'. One defines $H^s(\Omega)$ for any open set as the distributions which are the restrictions to $\Omega$ of some $U\in H^s(\mathbb{R}^n)$,with the norm being the induced norm. This provides the continuous inclusion of $H^s(\Omega) \subset W^{s,2}(\Omega)$ for $s\geq 0$. [To obtain set equality, one needs $\Omega$ to permit an extension operator. Set equality holds for any non-empty open $\Omega$, and $s$ the negative integers.] One then uses Peetre's inequality on the object $U$. MacLean's book has good discussion. – Nilima Nigam Jun 15 '11 at 16:21
Thanks, that makes sense. – Yakov Shlapentokh-Rothman Jun 15 '11 at 17:52

I believe that the fractional Sobolev spaces can be defined as a complex interpolation space between the integer Sobolev spaces (see this for example). As noted in the comments, your question is easily seen to hold on the integer valued spaces. Then we can interpolate to get the result for the fractional spaces.

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This approach works. Thanks. The answer was just a little vague in comparison to Nilima Nigma's response. – alext87 Jun 15 '11 at 7:34

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