Show that T does not have a finite order

Question

Part A: Show that $\langle T, \psi\rangle=\sum_{n=1}^\infty \psi^{(n)}(n)$ defines a distribution.

Please check: $$|\langle T, \psi\rangle|=|\sum_{n=1}^\infty \psi^{(n)}(n)| \leq \sum_{n=1}^\infty |\psi^{(n)}(n)| \leq \sum_{n=1}^\infty \|\psi^{(n)}(n)\|_\infty =\sum_{n=1}^\infty \|\partial^n \psi\|_\infty$$

Part B: Show that T does not have a finite order

How do I proceed?

Answer 1

Both proofs are done by definition

Definiton of distribution: for each compact $K$ there exists $N_K\ge 0$ and $c_K\ge0$ such that for any test function $\phi $ with support in $K$ we have $|\langle T,\phi\rangle|\le C_K \sum_{j=0}^{N_K}\|\phi^{(j)}\|$.

A: take a compact set $K$ and a test function $\phi$ with support in $K$. Clearly, there exist two integer numbers $m<n$ such that $K\subset [m,n]$.

Now you have two possibilities:

1) $n<0$, then $\langle T,\phi\rangle=0$ by definition of $T$. Fits the definition of distribution with constant $C_K=0$.

2) $n>0$, then $|\langle T,\phi\rangle|\le \sum_{j=0}^{n}\|\phi^{(j)}\|$ (check by definition of $K$). Fits the definitoin with $C_K=1$ and $N_K=n$.

Therefore, $T$ is a distribution.

B: The order of a distribution is the smallest $N$ such that $N\ge N_K$ for all compact $K$.

In you case take $K=[n-\delta,n+\delta]$ for $n\in \mathbb N$, then $|\langle T,\phi\rangle| = |\phi^{n}(n)|\le \|\phi^{(n)}\|$. You can not majorate $|\phi^{n}(n)|$ (irrespectively of $\phi$) by using only the sup-norms of the inferior derivatives (this result should be in your textbook; if it's not, then use some scaling argument to show it), therefore, for this $K$ we have $N_K=n$.

This implies that $T$ doesn't have an order.

Answer 2

Hint for Part B: Let $n \geq 1$. Then it is easily seen that $T = (-1)^n \delta^{(n)}_n$ on a neighbourhood of $n$ and so the order of $T$ is at least the same as that of $\delta^{(n)}_n$, which is equal to the order of $\delta^{(n)}$ since the order is invariant by translation. Now show that this order is exactly $n$. Since $n$ can be chosen arbitrarily it follows that $T$ is of infinite order.

To show that the order of $\delta^{(n)}$ is $n$, argue by contradiction. It should be clear that it is $\leq n$, so suppose that it is $<n$. Using the "estimate-equivalence" for distributions, there would exist a $C$ such that for all $\psi \in C_c^{\infty}(\mathbb{R})$ with $\text{supp } \psi \subset [-1,1]$, $$ |\psi^{(n)}(0)| \leq C \sum_{k=0}^{n-1} \sup |\psi^{(k)}|. $$ The idea is then to find a family of such $\psi$, say $(\psi_t)_{t \geq 1}$, for which the LHS of the last inequality will diverge faster than the RHS as $t \to \infty$. Of course we may as well search nice expressions. So, fix $\phi \in C_c^{\infty}(\mathbb{R})$ with $\text{supp } \phi \subset [-1,1]$ and $\phi^{(n)}(0) = 1$. For $t \geq 1$, let $\psi_t(x) := \phi(tx)$. Finally, obtain a contradiction.