Weak and strong topologies on tempered distributions

A set $B\subseteq \mathcal{S}({\mathbb{R}}^n)$ is bounded when ${\sup }_{\phi \in B}{p}_{\alpha \beta }(\phi )<\mathrm{\infty }$ for every pair of multi-indices. The weak topology $\sigma ({\mathcal{S}}^{\mathrm{\prime }},\mathcal{S})$ on tempered distributions is generated by

$p_{\phi }(u)=\mathrm{\mid }⟨u,\phi ⟩\mathrm{\mid }(\phi \in \mathcal{S})\mathrm{.}$

The strong topology $\beta ({\mathcal{S}}^{\mathrm{\prime }},\mathcal{S})$ is generated by

$p_B(u)={\sup }_{\phi \in B}\mathrm{\mid }⟨u,\phi ⟩\mathrm{\mid }$

as $B$ ranges over bounded subsets of $\mathcal{S}$. The empty-set supremum is zero. This supremum is finite: continuity of $u$ gives a basic zero-neighborhood $U$ on which $\mathrm{\mid }u\mathrm{\mid }<1$, and the finitely many seminorm bounds defining $U$ imply $B\subseteq tU$ for some finite $t>0$; hence $p_B(u)\le t$. The triangle inequality and homogeneity follow pointwise.

Thus a net $(u_i)$ converges weakly to $u$ exactly when $⟨u_i-u,\phi ⟩\to 0$ for every $\phi \in \mathcal{S}$, and it converges strongly exactly when $p_B(u_i-u)\to 0$ for every bounded $B$. Singletons are bounded, so strong convergence implies weak convergence. Both topologies are Hausdorff, since distinct functionals differ on some test. No identification with either topology on ${\mathcal{D}}^{\mathrm{\prime }}$ is asserted, and no choice axiom is used.