I can neither prove nor disprove the following claim.
Let $u_k \in \mathcal{S}'(\mathbb{R})$ and $u_k \to 0$ in $\mathcal{D}'(\mathbb{R})$. Then, $u_k \to 0$ in $\mathcal{S}'(\mathbb{R})$.
I expect this is false. Define $u_k := \frac{d^k\delta_k}{dx^k} \in \mathcal{S}'(\mathbb{R})$, i.e. $\langle u_k, \varphi \rangle := (-1)^k\frac{d^k\varphi}{dx^k}(k)$ for all $k \in \mathbb{N}$. Then, it is clear that $u_k \to 0$ in $\mathcal{D}'(\mathbb{R})$. I expect $u_k \nrightarrow 0$ in $\mathcal{S}'(\mathbb{R})$, however, I cannot prove it.
I would be grateful if anyone could give me some advice.
$\frac1k e^{t^2} 1_{|t|< k^2}$ converges to $0$ in $D'$ but not in $S'$ (pairing with $e^{-x^2}$).
It is true that $\delta^{(k)}(x-k)$ doesn't converge to $0$ in $S'$, take $\phi \in C^\infty_c$ constant $=1$ on $[-1/2,1/2]$ and $=0$ on $|x|>3/2$, let $\psi(x) = \frac{\phi(x)}{1-x}$ and look at $\varphi(x)=\sum_{k\ge 0}2^{-k} \psi(x-k)$, all the derivatives are rapidly decreasing, but $\varphi^{(k)}(k)=2^{-k}\psi^{(k)}(0) = 2^{-k}k!$