Is dirac comb a tempered distribution?

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Prove or disprove the statement $$ III:=\sum_{k=-\infty}^\infty\delta_k\in \mathscr{S}'(\mathbb{R}^n)$$ where $\delta_k\varphi:=\varphi(k),\,\,\forall\varphi\in\mathscr{S}(\mathbb{R}^n)$

I tried to prove linearity and I think it holds, but I am not so sure about the continouity.

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Yes, it is. Notice that $$|\langle III,\varphi \rangle| = \left |\sum_{k\in \mathbb Z} \varphi(k)\right| = \left |\sum_{k\in \mathbb Z} (1 + k^2)\frac{\varphi(k)}{1 + k^2}\right|\leq \sup_{x\in \mathbb R} |(1 + x^2) \varphi(x)| \sum_{k\in \mathbb Z} \frac{1}{1 + k^2} $$ Also $\sum_{k\in \mathbb Z} \frac{1}{1 + k^2}<\infty$ and $\sup_{x\in \mathbb R} |(1 + x^2) \varphi(x)|$ is clearly a seminorm in the Schwartz space so $III$ is continuous.

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I think yes; if $p_n(x)$ converges to $0$ in the Schwartz space, then in particular, for any $\alpha>0$ the sequence of numbers $\sup_x |x^\alpha p_n(x)|$ converges to $0$ for $n\to\infty$.

In particular, for any $\epsilon>0$ there exists $N$ s.t. for $n>N$, $|p_n(x)|$ is smaller then $\epsilon$ on $x\in [-1,1]$ and smaller than $\epsilon x^{-2}$ for $x\in \mathbb{R}\setminus [-1,1]$. But then $\sum_{k\in\mathbb{Z}} p_n(k)$ is smaller than $\epsilon (3 + \frac{\pi^2}{3})$ or something like that, so $III(p_n)$ converges to $0$.