# Show that $f(\phi)=\sum_{k=0}^\infty \phi^{(k)}(k)$ for $\phi \in D$ has no finite order

Let $\phi \in D:=D(K)=C^\infty_0(K)$ be a test function and let $f \in D^*=\{f:D \to \mathbb{R} : f \text{ bounded and linear} \}$ be a distribution.

A distribution has finite order if:

$$\exists m \in \mathbb{N}_0 ~\forall K^{\text{compact}} \subset \mathbb{R}~ \exists C>0 ~\forall \phi \in D: |f(\phi)| \leq C \sum_{i=0}^m ||\phi^{(i)}||_{\infty}$$

Show: $f(\phi)=\sum_{k=0}^\infty \phi^{(k)}(k)$ has no finite order.

So, I have to show the negation, i.e. $$\forall m \in \mathbb{N}_0 ~\exists K^{\text{compact}} \subset \mathbb{R}~ \forall C>0~\exists \phi \in D: |f(\phi)| > C \sum_{i=0}^m ||\phi^{(i)}||_{\infty}$$

It would be fine if I could find one $\phi \in D$ s.t. this is fulfilled. But I don't know a lot test functions. And computing the n-th derivative seems strenuous.

I know $$\phi(x)=\begin{cases} \exp\left(-\frac{1}{1-x^2}\right) & ,|x|<1 \\ 0 &,\text{else} \end{cases}$$ as the bump function with compact support on $[-1,1]=:K$. Here I have $$f(\phi)=\sum_{k=0}^\infty \phi^{(k)}(k)=\phi(0)=e^{-1}.$$ If I set $m=0$ I have: $$C \sum_{i=0}^0 ||\phi^{(i)}||_{\infty}=C \sup_{x \in K} |\phi(x)|=C \phi(0)=C e^{-1}$$ and obviously $e^{-1} \ngtr Ce^{-1} ~\forall C>0$. So this example does not even work for $m=0$ :(

Yeah, I don't really have a leading approach, sorry for that. Has someone got an idea?

Thanks a lot, Marvin

EDIT: corrected tags EDIT2: corrected quantors as suggested in the hint

• First of all, you should include a quantor for $\phi\in D(K)$ in your definition of infinite order. And it is not very difficult to specify a test function whose $n$-th derivate at a certain point has a specified value ... – Vobo May 1 '16 at 18:14
• @Vobo Thank you for your comment, I added the quantor. Hmm, I can't really find one. Can you help me? – Fritz May 1 '16 at 18:28
• You got the order of the quantifiers wrong. $f$ having finite order means $(\exists m)(\forall K)(\exists C)(\forall \phi)\dotsc$. The quantification over the test function is innermost. – Daniel Fischer May 2 '16 at 11:27
• Yes. And you have to let $\phi$ depend on $m$. Because the restriction of $f$ to $D(L)$ has finite order for every (relatively) compact $L$, so in particular $f\lvert_{D(\operatorname{supp} \phi)}$ has finite order for every $\phi$. – Daniel Fischer May 2 '16 at 11:35
• I should have indexed after all. But I didn't want to write $\psi_k^{(k)}(k)$. Then $f(\psi_k) = \psi_k^{(k)}(k) \neq 0$, and one can select a $\psi_k$ such that all lower-order derivatives are very small. – Daniel Fischer May 2 '16 at 12:39

The behaviour of $f(\phi)$ is easiest to control when only one of the terms in the defining sum can be nonvanishing. To this end, for every $k\in \mathbb{N}$ we define $K_k := [k-1,k+1]$. For $\phi \in D(K_k)$ we then have $f(\phi) = \phi^{(k)}(k)$.

It is straightforward to see that the order of the restriction of $f$ to $D(K_k)$ is at most $k$, and when we have established that the order is no smaller than $k$, we have achieved our goal. Given an arbitrary $m\in \mathbb{N}$, we choose a $k > m$, and looking at functions in $D(K_k) \subset D(\mathbb{R})$ then shows that $f$ is not of order $m$.

So we want to show that the distribution $\phi \mapsto \phi^{(k)}(k)$ has exact order $k$. Since translations obviously don't affect the order, let us for simplicity of notation look at $\phi \mapsto \phi^{(k)}(0)$ instead.

Let $\beta \colon \mathbb{R} \to [0,1]$ be a smooth function with $\beta(x) = 1$ for $x \leqslant \frac{1}{2}$ and $\beta(x) = 0$ for $x \geqslant 1$. Define $\chi(x) := \beta(\lvert x\rvert)$, and $\chi_{\varepsilon}(x) := \chi( \varepsilon^{-1}\cdot x)$ for $\varepsilon > 0$. For $0 < \varepsilon \leqslant 1$, the function

$$\psi_{k,\varepsilon} \colon x \mapsto \chi_{\varepsilon}(x)\cdot \frac{x^k}{k!}$$

then belongs to $D([-1,1])$ and satisfies $\psi_{k,\varepsilon}^{(k)}(0) = 1$. Since also $\psi_{k,\varepsilon}^{(i)}(0) = 0$ for $i < k$, we can hope that

$$\sum_{i = 0}^{k-1} \lVert \psi_{k,\varepsilon}^{(i)}\rVert_{\infty}$$

can be made very small. By Leibniz's rule,

$$\psi_{k,\varepsilon}^{(r)}(x) = \sum_{\rho = 0}^r \binom{r}{\rho} \chi_{\varepsilon}^{(\rho)}(x)\cdot \frac{x^{k-r+\rho}}{(k - r + \rho)!}.$$

Since $\chi_{\varepsilon}^{(\rho)}(x) = 0$ for $\lvert x\rvert \geqslant \varepsilon$, we can estimate

$$\lVert \psi_{k,\varepsilon}^{(r)}\rVert_{\infty} \leqslant \sum_{\rho = 0}^r \binom{r}{\rho} \lVert \chi_{\varepsilon}^{(\rho)}\rVert_{\infty} \cdot \frac{\varepsilon^{k - r + \rho}}{(k - r + \rho)!}.$$

By the chain rule, we have $\lVert \chi_{\varepsilon}^{(\rho)}\rVert_{\infty} = \varepsilon^{-\rho}\cdot \lVert \chi^{(\rho)}\rVert_{\infty}$, and thus

$$\lVert \psi_{k,\varepsilon}^{(r)}\rVert_{\infty} \leqslant \varepsilon^{k-r}\sum_{\rho = 0}^r \binom{r}{\rho} \lVert \chi^{(\rho)}\rVert_{\infty} \frac{1}{(k - r + \rho)!} \leqslant \frac{\varepsilon^{k-r}}{(k - r)!} \sum_{\rho = 0}^r \binom{r}{\rho} \lVert \chi^{(\rho)}\rVert_{\infty}.$$

Finally, we see that

$$\sum_{i = 0}^m \lVert \psi_{k,\varepsilon}^{(i)}\rVert_{\infty} \leqslant M\cdot \varepsilon \tag{\ast}$$

whenever $m \leqslant k-1$, where $M$ can be chosen as

$$\sum_{r = 0}^{k-1}\frac{1}{(k - r)!}\sum_{\rho = 0}^r \binom{r}{\rho}\lVert \chi^{(\rho)}\rVert_{\infty}.$$

From $(\ast)$ it is easy to see that $\phi \mapsto \phi^{(k)}(0)$ doesn't have order $m < k$: Given $C > 0$, we have

$$\psi_{k,\varepsilon}^{(k)}(0) = 1 > C\cdot \sum_{i = 0}^m \lVert \psi_{k,\varepsilon}^{(i)}\rVert_{\infty}$$

for all $0 < \varepsilon < \min \bigl\{ 1, (CM)^{-1}\bigr\}$.

• Wow, thanks a lot. Very detailed, helps me a lot and makes many things clear. Little question: $\psi_{k,\varepsilon}$ belongs to $D([-\varepsilon,\varepsilon])$, doesn't it? Because you scaled $\chi_{\varepsilon}$ to this interval, I think. – Fritz May 2 '16 at 15:40
• Yes, but since I restricted $\varepsilon \leqslant 1$, we have $D([-\varepsilon,\varepsilon]) \subset D([-1,1])$, so we work in one fixed space. – Daniel Fischer May 2 '16 at 15:53
• Ahh, right, I overlooked that. Thank you! – Fritz May 2 '16 at 15:54

You can try induction over m, starting with a j-th power of a bump-function. This yields a sequence of bounded functions that has unbounded first derivatives. From $m\to m+1$ this is a bit more difficult.

Informally speaking, order of a distribution f(.) is the smallest n such that f(fi(.)) does not contain derivatives of fi(.) of order greater than n. A lot of formalism makes it hard to read. By my informal criterion the distribution in question clearly has infinite order.