The following proposition is true. My question is whether we can find a direct, pedagogical proof that can be delivered in class without having to introduce new concepts from functional analysis.
In the following, let $\mathcal S'=\mathcal S'(\mathbb R)$ denote the space of tempered distributions, i.e., the space of continuous linear functionals on the Schwartz class $\mathcal{S}=\mathcal{S}(\mathbb R)$.
Proposition. Let $T\colon \mathcal S'\to \mathcal S'$ be linear and sequentially continuous. Then $T$ is continuous.
Remark. In practice, $T$ will be the Fourier transform. I find this proposition useful for teaching, since it allows for a streamlined proof that the Fourier transform $\mathcal F\colon \mathcal S\to \mathcal S$ extends to a continuous linear map $\mathcal S'\to \mathcal S'$ by taking adjoints.
Proof. This is actually true with $\mathcal S$ replaced by a general separable locally convex space: see, for example, https://math.stackexchange.com/a/2099312/8157. $\Box$
Attempt of a direct proof. A sub-base of neighborhoods of the origin in $\mathcal S'$ is given by the family $V_{f, \epsilon}$, where $\epsilon >0$, $f\in \mathcal S$, and $$ V_{f, \epsilon}:=\{\psi\in\mathcal S'\ :\ \lvert\langle \psi, f\rangle\rvert<\epsilon\}.$$ In particular, $T$ is continuous if and only if, for all $f\in\mathcal S$ and $\epsilon>0$ there are $g_j\in\mathcal S$ and $\delta_j>0$, with $j\in\{1, \ldots N_{f, \epsilon}\}$, such that $$ \psi\in \bigcap_j V_{g_j, \delta_j} \Rightarrow T\psi\in V_{f, \epsilon}.$$ So, assuming this does not hold, i.e. $T$ is not continuous, we need to construct a sequence $\psi_n\in\mathcal S'$ such that $\psi_n\to 0$ but $T\psi_n\not\to 0$, thus proving the contrapositive of the required proposition. The fact that $\mathcal S$ is separable must enter the picture here.
