Extending operator - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:07:55Z http://mathoverflow.net/feeds/question/106895 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106895/extending-operator Extending operator B-B 2012-09-11T08:58:03Z 2012-10-08T10:01:40Z <p>Seeley's extension operator it is a linear continuous mapping $E: C^\infty([0, \infty)) \rightarrow C^\infty(\mathbb R)$ with the property $E(g)|_{[0, \infty))}=g$ for $g \in C^\infty([0,\infty))$. </p> <p>Let $(a_n), (b_n)$ are real sequences s.t.</p> <ul> <li><p>$b_n&lt;0$, $b_n \rightarrow -\infty$,</p></li> <li><p>$\sum_{n=1}^\infty |a_n| |b_n|^m &lt;\infty \ for m=0,1,2....$,</p></li> <li>$\sum_{n=1}^\infty a_n b_n^m =1 \ for \ m=0,1,2....$</li> </ul> <p>and let $h\in C_c(\mathbb R)$ be s. t. $h(x)=1$ for $x \in [0,1]$, $h(x)=0$ for $x \geq 2$. </p> <p>Operator $E(g)(x)=\sum_{n=1}^\infty a_n h(b_n x) g(b_n y) \ for \ x&lt;0$ and $E(g)(x)=g(x) \ for \ x\geq 0$ is an example of Seeley extension operator.</p> <p>Does there exist an operator $E$, with similar properties, from $C^\infty([0,c))$ into $C^\infty(-c,c)$ which is linear continuous s.t. $E(g)|_{[0, c))}=g$ for $g \in C^\infty ([0,c)$ ?</p> http://mathoverflow.net/questions/106895/extending-operator/106912#106912 Answer by Rafe Mazzeo for Extending operator Rafe Mazzeo 2012-09-11T13:51:24Z 2012-09-11T13:51:24Z <p>Why isn't this trivial? For example, take a smooth cutoff function $\chi(t)$ which equals $1$ for $t \leq c/2$ and vanishes for $t \geq 3c/4$. Then let $E_{c,\chi}(g) = E (\chi g) + (1-\chi)g$ where $E$ is Seeley's original operator.</p>