Show that $u=\sum_{k=0}^\infty k^{-2} \delta_{1/k}$ is a distribution, but $v=\sum_{k=0}^\infty \frac{1}{k} \delta_{1/k}$ is NOT a distribution. Then find the support of u.
The definition of $$|T(\psi)| \leq C_N \sum_{|\alpha| \leq N} ||\partial^\alpha \psi||_\infty$$.
So what I have so far is
$$|<T_u,\psi>=|\int_{-\infty}^\infty \sum_{k=0}^\infty \frac{1}{k^2}\delta_{1/k}\psi(k)dk|$$
Is this the right starting point?
I'm not entirely sure what your work is supposed to convey. One thing to note is that you should never associate delta distributions to integration in this way. It is not a function. Do not treat it as such.
Note that if $f\in C_c^{\infty}(\mathbb{R})$, then $\langle u,f\rangle = \sum_{k=1}^{\infty} \frac{1}{k^2}f\left(\frac{1}{k}\right)$. You know that $f$ is bounded so this supplies the proof.
To see that $v$ is not a distribution, take $f$ to be $1$ on $[-2,2]$. What could you say about $\langle v,f\rangle$ in this case?
