= Solution
Using <multi-index notation>, the <Schwartz space> is
$$
\mathcal S(\mathbb R^n)=\left\{\varphi\in C^\infty(\mathbb R^n):p_{\alpha,\beta}(\varphi)=\sup_{x\in\mathbb R^n}|x^\alpha\partial^\beta\varphi(x)|<\infty\text{ for all }\alpha,\beta\right\}.
$$
A <sequence> $\varphi_m$ converges to $\varphi$ in this <Fréchet space>[Fréchet space] when $p_{\alpha,\beta}(\varphi_m-\varphi)\to0$ for every $\alpha,\beta$. The space $\mathcal S'(\mathbb R^n)$ of <tempered distributions> is the <continuous dual space>[continuous dual] of $\mathcal S(\mathbb R^n)$, and $u_m\to u$ there means <weak convergence of distributions>, namely $\langle u_m,\varphi\rangle\to\langle u,\varphi\rangle$ for every $\varphi\in\mathcal S$.
Continuity of a <linear functional> immediately implies that $\varphi_m\to0$ entails $\langle u,\varphi_m\rangle\to0$. Conversely, enumerate the Schwartz <seminorms> as $p_1,p_2,\ldots$. If $u$ were not continuous, then for each $m$ one could choose $\varphi_m$ such that
$$
p_j(\varphi_m)\leq\frac1m\quad(1\leq j\leq m),
\qquad
|\langle u,\varphi_m\rangle|\geq1.
$$
Every fixed seminorm tends to zero along this sequence, so $\varphi_m\to0$ in $\mathcal S$, contradicting the assumed sequential property. This is the <sequential continuity criterion for a linear map on a metrizable topological vector space>.
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