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

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Isn't it easy to reduce to Seeley's operator either by modifying the proof (slightly) or by a simple transformation $C^\infty([0,c)) \to C([0,\infty))$, $g\mapsto g\circ \psi$ with $\psi(t) =\tan(t/c)$ (or so)? – Jochen Wengenroth Sep 11 '12 at 9:39
Thanks. In the first approach, by modyfing the proof, maybe there is a problem with finding seguences $a_n$, $b_n$ satisfying above conditions (with $b_n \rightarrow -c$ instead $-\infty$)? – B-B Sep 11 '12 at 9:57
If I remember right, Seeley gives an explicit example of sequences (a_n) and (b_n). Have you tried modifying them to get sequences that suffice for the general case? – Loop Space Sep 11 '12 at 11:35
(But Jochen's right: the easiest way is via the isomorphisms. Still, if you want to have an extension operator built in the same way as Seeley's then I suggest trying to modify Seeley's sequences.) – Loop Space Sep 11 '12 at 11:36
up vote 2 down vote accepted

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.

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That's right. B-B should accept this answer. – Jochen Wengenroth Sep 12 '12 at 6:19

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