Extension theory with bump function - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T23:48:09Z http://mathoverflow.net/feeds/question/67029 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67029/extension-theory-with-bump-function Extension theory with bump function alext87 2011-06-06T11:36:32Z 2011-06-15T04:04:53Z <p>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:</p> <p>$\phi (x) = \begin{cases} 1&amp;x\in B_r(0) \\ 0&amp;x\not\in B_{r+3}(0) \end{cases}$</p> <p>and for any real-valued function $f\in \mathcal{H}^\tau(\mathbb{R^n})$ ($\tau\in\mathbb{R}$, $\tau>d/2$) we have </p> <p>$\left\|\phi(\cdot) f(\cdot)\right\|_{\mathcal{H}^\tau(A_1)}\leq C\left\|f\right\|_{\mathcal{H}^\tau(A_2)}$</p> <p>where $C$ is a constant independent of $f$, $A_1 = B_{r+2}(0)\setminus B_{r+1}(0)$ and $A_2 = B_{r+3}(0)\setminus B_{r}(0)$. </p> <p>This has a similar feel to extension theory results if we think of $\phi$ as an extension operator which preserves $f$ on $B_r(0)$.</p> http://mathoverflow.net/questions/67029/extension-theory-with-bump-function/67369#67369 Answer by Yakov Shlapentokh-Rothman for Extension theory with bump function Yakov Shlapentokh-Rothman 2011-06-09T19:18:47Z 2011-06-09T19:18:47Z <p>I believe that the fractional Sobolev spaces can be defined as a complex interpolation space between the integer Sobolev spaces (see this <a href="http://www.scribd.com/doc/45316527/Sobolev-Spaces-2ed-Robert-a-Adams-John-J-F-Fournier" rel="nofollow">http://www.scribd.com/doc/45316527/Sobolev-Spaces-2ed-Robert-a-Adams-John-J-F-Fournier</a> 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.</p> http://mathoverflow.net/questions/67029/extension-theory-with-bump-function/67825#67825 Answer by Nilima Nigam for Extension theory with bump function Nilima Nigam 2011-06-15T04:04:53Z 2011-06-15T04:04:53Z <p><strong>Short answer: yes.</strong></p> <p>Let $\psi_\epsilon(x):=\frac{1}{\epsilon^n}\exp{\epsilon^2/(\epsilon^2-|x|^2)}$ for $|x|&lt;\epsilon$, 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, </p> <p><strong>$\phi(x):= C\psi_\epsilon(x)* \chi_{B_{r+3/2}(0)}$.</strong> Here $C$ is a normalizing constant (this may not be needed, but I haven't checked). </p> <p>This yields a smooth cut-off function which is 1 in the ball $B_{r+1}(0)$, and zero outside $B_{r+2}(x)$. </p> <p>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:</p> <p>'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 $Q_u=||u||_{H^s(\Omega). }$ <em>(Apologies, I encountered trouble while trying to typeset the LaTeX here).</em></p> <p>The same result holds with $H^s(\Omega)$ replaced with $\tilde{H}^s(\Omega)$.'</p> <p>The proof proceeds using $\Omega = A_2$, and then </p> <p>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.</p> <p>or (b) by examining $\hat{\phi u}$ and using Peetre's inequality.</p> <p>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.</p>