2
$\begingroup$

by Schwartz’s theorem, any generalized function from $S'$ has a finite singularity order. In this example, it is infinite and I want to show that the series $\notin S'$.
($g^{(l)}$ means $l$th derivative)

Let $$f=\sum\limits_{n=-\infty}^{\infty}\delta^{(|n|)}(x-n)$$ then for each $\phi \in S',$ $(f,\phi)<\infty$ and f is linear.

I try to give an example when $f$ is not continuous.
(I remind you that $\phi$ are rapidly decreasing functions and $\phi_k \to \phi\:$ iff $\:\forall l \: \phi_k^{(l)} \rightrightarrows \phi^{(l)} $ in $ \mathbb{R}$).

So I try to take $\phi = 0$ (maybe it's easyer), then $\phi_k \to 0$ and$$(f,\phi_k)=\sum\limits_{n=-\infty}^{\infty}\phi_k^{(|n|)}(n)$$ I want to $(f,\phi_k) \nrightarrow 0$.
It looks like $\phi_k$ should look something like $\frac{n^ae^{-bx}}{k}$ but it doesn't work...
Please give a hint how to find the right $\phi_k$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.