I'm trying to solve this question with the following given information; I have $\Omega \in \mathbb{R}$ open set, $C^\infty_o(\Omega)$ is a vector space with usual addition and multiplication, and the space $\mathbb{D}(\Omega)$ equipped with the following convergence is the space of test functions.

define the Convergence $\phi_j \rightarrow \phi$ in $\mathbb{D}(\Omega)$ if:

  1. There exist $K \subset \Omega$ compact such that support$(\phi_j)\subset K,\forall j$.
  2. $\forall \alpha$ multi-index; $\partial^\alpha \phi_j \rightarrow \partial^\alpha \phi$ uniformly on $\Omega$.

$\mathbb{D}^\prime(\Omega)$ is the set of distributions on $\mathbb{D}(\Omega)$

The question is; if $\Omega = (0,2)$ & $u:\mathbb{D}(\Omega) \rightarrow \mathbb{R} , u(\phi) := \sum_{j=1}^\infty \phi^{(j)}(\frac{1}{j})$ where $(j)$ is the $j^{th}$ derivative of $\phi$.

  • (a)Show that $u \in \mathbb{D}^\prime(0,2)$.
  • (b)Does u have finite order?

Solution: (a) apparently $u$ is linear and let K$\subset$ (0,2) compact then there exist $N\in \mathbb{Z}^+$, $i.e \frac{1}{N} \in K$, but $\frac{1}{j} \notin K$ for $j>N$ If support$(\phi) \subset K$ then $|u(\phi)| = |\sum_{j=1}^N \phi^{(j)}(\frac{1}{j})| \leq \sum_{j=1}^N max|\phi^{(j)}|$ then I can conclude it is sequentially continuous then it is a distribution and $u \in \mathbb{D}^\prime(0,2)$

My problem with (b) part, I can see from answer in (a) that I can't assume the finite order because of dependence of $N$ on the set $K$ but also I don't know how to prove it is not finite?

Can anyone help?Thank you.


1 Answer 1


Suppose it was finite order, say order $N$. Let $\phi$ be supported on $U\subset (0,2)$ where $$U\cap \{1,\frac12,\frac13,\dots\} = \{\frac1{N+1}\}$$ Then $u(\phi) = \phi^{(N+1)}(\frac1{N+1})$...


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