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I'm studying linear functionals that are not distributions, and I came across this post: link.

In one of the comments it is claimed that the functional $u:C_c^\infty\to\mathbb{C}$ given by $$ u= \sum_{n\geq 0} \frac{R^n}{n!}\delta^{(n)} $$ is not even well-defined unless the test function is analytic with radius of convergence around 0 greater than $R$. How does one prove this? I was able to prove that the sum does not converge absolutely if we take a non-analytic test function with the help of the Cauchy estimate, but this does not imply that the sum is not well-defined.

MSDG
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  • I do not think this is true, consider a smooth cutoff of the usual $e^{-1/x^2}$ example. A better example would be something like a smooth cutoff of $1/(x-1)$ whose derivatives grow like factorials. This example illustrates their point because it involves a function that "should" have a pole singularity. – Ian Apr 19 '18 at 16:48
  • Yes, I was a bit skeptical about the claim as well. Could you elaborate your example a bit further and put it as an answer to the question? – MSDG Apr 19 '18 at 16:56

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Strictly speaking this statement is not true. Fix a smooth function $\phi$ with compact support equal to $1$ on a neighborhood of zero and define

$$f(x)=\begin{cases} \phi(x) e^{-1/x^2} & x \neq 0 \\ 0 & x=0 \end{cases}.$$

Then $u(f;R)=0$ for all $R$, but $f$ is not analytic on any neighborhood of zero.

This statement is sort of "morally true" however, in the sense that you should only expect $u(f;R)$ to make sense if $f$ is analytic on a disk centered at the origin of radius greater than $R$, even though there can be exceptions. To see this, consider now a smooth function $\phi$ with support compactly contained in $[-1,1]$ and which is equal to $1$ on a neighborhood of zero. Then look at $f(x)=\frac{\phi(x)}{x-1}$. For this, $u(f;R)$ diverges if $R \geq 1$, even though $f$ is clearly a $C^\infty_c$ function. What is the problem? The problem is that $f$ is a non-analytic continuation of the meromorphic function $\frac{1}{z-1}$ which has a pole at $z=1$. But $u$ only sees the behavior of $f$ on an infinitesimal neighborhood of $0$, so $u$ "thinks" that $f$ is actually that meromorphic function. This prohibits convergence of the series for $R \geq 1$.

In any case, the main point is that the topology on $D$ forces all distributions to have some finite "order", they cannot take derivatives above this order. This allows us to handle non-analytic smooth functions within the framework of distribution theory.

Ian
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