showing the linear functional $\Lambda(\phi)= \sum_{k=1}^\infty k!D^k\phi(k)$ is a distribution

Question

This is an exercise in my functional analysis book. However, I was trying to solve but cannot see how to proceed here to get the solution answers.

Q: Verify the following linear functional is distribution.

• $\Lambda(\phi)= \sum_{k=1}^\infty k!D^k\phi(k)$

The solution says that this is a distribution of infinite order.

• $\Lambda(\phi)= \sum_{k=1}^\infty 2^{-k}D^k\phi(\frac{1}{k})$

The solution says that this is not a distribution on $\Omega$ but distribution on $(0,\infty)$.

I wondered to to have an elaboration this to get a better understanding of the the following definition and general practice in these kind of linear functionals.

Def. So I have to show that for any compact set $A\subset \Omega$ , there is $N\ge 0$ and $C$ , such that the functional is bounded. i.e $|\Lambda(\phi)|\le C \|\phi\|_{C^N}$ for all $\phi \in C^\infty , supp(\phi)\subset A$.

Try

Does the following has any meaning to proceed ;

\begin{align} |\Lambda(\phi)| & \le \sum_{k=1}^\infty |k!D^k\phi(k)|\le \sum_{k=1}^\infty \sup_{k\in \Omega }|k! D^k\phi(k)|\\ & \le \|\phi\|_{C^\infty_c} \sum_{k \in supp(\phi)} |k!| \le \|\phi\|_{C^\infty_c} C \end{align}

Answer 1

Your attempt is flawed: What does $\sup_{k\in \Omega} |k! D^k\phi(k)|$ mean? Are you using the $\Gamma$ function?

The general idea is correct though. Hint: Fix a compact set $A$ and choose $N$ large enough so that $A\subset [-N,N]$ and $C=N!$.

As for why the second one is a distribution on $(0,\infty)$, fix $A$ and choose $N$ so that $0<1/N< \inf(A)$.