Here, $\delta$ denotes the Dirac delta distribution, i.e., $\delta:\mathscr{D}(\mathbb{R})\to\mathbb{R}$ is the continuous linear functional defined as $\langle\delta,\phi\rangle=\phi(0)$ for $\phi\in\mathscr{D}(\mathbb{R})$.
We also denote the $k^{\text{th}}$ derivative of $\delta$ by $\delta^{(k)}$, and $\langle \delta^{(k)},\phi \rangle:=(-1)^{k}\cdot\langle \delta,\phi^{(k)} \rangle$.
We know that a finite linear combination of distributions is again a distribution.
Therefore, $\sum_{k=0}^{n}\frac{(-1)^k}{k!}\delta^{(k)}$ is also a distribution for every $n\geq1$.
First question :
- Is $\sum_{k=0}^{\infty}\frac{(-1)^k}{k!}\delta^{(k)}$ a distribution ? In other words, does the sequence $\left\lbrace\sum_{k=0}^{n}\frac{(-1)^k}{k!}\delta^{(k)}\right\rbrace_{n\geq1}$ converge in the topology of $\mathscr{D}'(\mathbb{R})$ ?
The structure theorem for distributions (roughly) says that every distribution is (at least locally) $\partial^{\alpha}f$ for some continuous function $f$ and some multi-index $\alpha$.
Second question :
- If the above is indeed a distribution then how can we find such continuous function (for locally) for $\sum_{k=0}^{\infty}\frac{(-1)^k}{k!}\delta^{(k)}$ ?
By Borel lemma, any sequence of real numbers is derivatives in $0$ of some smooth function (see this question for references). Multiplying by bump function, we get function from $\mathscr D(\mathbb R)$. So, there is function $f$ in $\mathscr D(\mathbb R)$ s.t. $f^{(n)}(0) = n!$, and on such function your series diverges, thus it does not define a distribution.