First of all, I am aware there are many questions about this very topic, yet none of the answers is enough down to earth for me.
Let $\mathcal{D}'$ be the dual space of smooth function on $\mathbb{R}$ with compact support and let $T : \mathcal{D}' \rightarrow \mathcal{D}'$ be the application given by the adjoint identity $\langle\phi, T \Lambda \rangle = \langle S \phi, \Lambda\rangle$, where $S$ is a continuous map $\mathcal{D} \rightarrow \mathcal{D}$.
What does it mean that $T$ is continuous exactly?
I believe that sequential continuity should be equivalent to continuity, so if $\Lambda_n$ is a sequence of distribution converging to $0$, so that for any test function $\phi$ $\langle\phi, \Lambda_n\rangle \rightarrow \langle\phi, 0\rangle$ I get $\langle\phi, T \Lambda_n \rangle = \langle S \phi, \Lambda_n\rangle$, but since $\langle\phi, \Lambda_n\rangle$ converges to $0$ for any test function, isn't it true for $ S \phi$ making every linear map $T$ continuous?
Let $V,W$ be a topological vectorspace and let $V^*,W^*$ denote their dual spaces, ie the spaces of continuous linear maps $V\to\Bbb K$ resp. $W\to\Bbb K$. Suppose $T:V\to W$ is continuous and let $T^*: W^*\to V^*$ be its adjoint (continuity of $T$ is necessary for the existence of the adjoint).
If $w'_\alpha$ is some net in $W^*$, converging in weak* topology to $w'$, then for all $v\in V$: $$\langle T^* w'_\alpha, v\rangle = \langle w_\alpha', Tv\rangle \to \langle w',Tv\rangle = \langle T^*w',v\rangle.$$ Since this holds for all nets $v\in V$ you get that $T^*w_\alpha'$ converges in weak* topology to $T^*w'$. Since the convergent net was arbitrary you find $T^*$ is weak* continuous.