Does this define a distribution on $\mathbb{R}$? $T=\sum_{k=0}^{\infty}\delta_{k}$ I wonder wether the following series defines a distribution or not:
$$T=\sum_{k=0}^{ \infty}\delta_{k}.$$
Here $\delta_{k} $ is a Dirac delta concentrated in $k$.
Is the following reasoning correct?
$$\langle T,\varphi\rangle=\sum_{k=0}^{\infty}\varphi(k).$$
We have $\varphi\in D(\mathbb{R})$ so there is an interval $[-a,a]$ and a closed set $K$, $K\subset[-a,a]$ such that $\varphi(x)≠0$ if $x\in K$.
Now,
$$\langle T,\varphi\rangle =\sum_{k=0}^{\infty}\varphi(k)= \sum_{k=0}^{k=[a]}\varphi(k)$$
and this sum is convergent.
$T$ is linear and continuous so defines a distribution.But what about $\text{supp}\,T$?
 A: Your reasoning is incomplete. The fact that the sum is convergent is not enough to show the continuity. You have to bound its absolute value with respect to $\phi$ and its derivatives. Here we have the following upper bound for all test functions in the compact set $K$:
$$|\langle T,\varphi \rangle | \le  N_K \| \varphi\|_\infty$$
where $N_K$ is the number of natural integers in $K$.
Actually there is a widely used distribution in sampling theory called the Dirac comb. It is defined as in your example but for $k \in \mathbb{Z}t$ for $t$ some real number, the period of the comb (or the sampling step in the context of sampling theory). In your case $t=1$ and we restrain ourselves to $\mathbb{N}$.
By definition, the support of a distribution $T$ is the complementary of the largest open subset $U$ such that $\langle T,\varphi \rangle =0$ when the support of $\varphi$ is in $U$.
Here $\langle T,\varphi \rangle \ne 0$ iff $\phi(n) \ne 0$ for some $n \in \mathbb{N}$. Since $\mathbb{N}$ is closed in $\mathbb{R}$, its complementary set is open. Hence the support of $T$ is $\mathbb{N}$.
