 Solution

Continuity of $u$ at zero supplies finitely many Schwartz seminorms and a constant $C$ bounding $|u[\varphi ]|$. Increasing their two indices to common maxima $N,k$ givesThus $u$ is bounded for the $X_{N,k}$ norm. Since $\mathcal{S}$ is dense in the Banach space $X_{N,k}$, it has a unique continuous extension $\tilde{u}\in X_{N,k}^{\prime }$.