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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?

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  • $\begingroup$ I changed $< T,\psi >$ to $\langle T,\psi\rangle$. That is standard. Please do it that way. $\endgroup$ Commented Jul 5, 2014 at 2:28
  • $\begingroup$ Your proof does not work. There is no reason for the last series you wrote to converge! Indeed, it is easy to construct examples of test functions for which each of the terms in your series is greater than 1. Your qurstion does not have anything to do with Fourier analysis, by the way... $\endgroup$ Commented Jul 5, 2014 at 2:37
  • $\begingroup$ Younneed to use that test functions have compact support at some point, and bound in terms of the seminorms that define continuity... $\endgroup$ Commented Jul 5, 2014 at 2:38

2 Answers 2

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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.

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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.

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