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 $g \in C^\infty([0,\infty))$. 


Let $(a\_n), (b\_n)$ are real sequences 
s.t.


* $b\_n<0$, $b\_n \rightarrow -\infty$,
* $ \sum\_{n=1}^\infty |a\_n| |b\_n|^m <\infty \ for m=0,1,2....$,
* $ \sum\_{n=1}^\infty a\_n b\_n^m =1 \ for \ m=0,1,2....$


and let $h\in C\_c(\mathbb R)$ be s. t. $h(x)=1$ for $x \in [0,1]$, $h(x)=0$ for $x \geq 2$. 


Operator
$
E(g)(x)=\sum\_{n=1}^\infty a\_n h(b\_n x) g(b\_n y) \ for \ x<0
$
and
$E(g)(x)=g(x) \ for \ x\geq 0$
is an example of Seeley extension operator.


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 $g \in C^\infty ([0,c)$ ?



 Extending operator
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.


