User b-b - MathOverflow

Extending operator

2012-09-11T08:58:03Z

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

Convolution of sequences

2012-09-24T14:16:33Z

Let for given real sequences $(a_n)_{n \in \mathbb Z}, (b_n)_{n \in \mathbb Z}$, $c_n:=\sum_{k\in \mathbb Z} a_k b_{n-k}$ for $n \in \mathbb Z$ be the convolution of sequences $(a_n)$, $(b_n)$.

For classical convolution if one of two functions is in $L^p$, the second in $L^q$, where $1\leq p,q <\infty$ then their convolution $f*g$ is in $L^r$, where $\frac{1}{r}=\frac{1}{p}+\frac{1}{q}-1$. Is it some similar type theorem true for convolution of sequences?

Comment by B-B

2012-09-11T09:57:37Z

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 $-\infty$)?