Suppose $E$ is a dense subspace of $\mathcal{S}(\mathbb{R}^n)$. Is it true that the topology on $\mathcal{S}’(\mathbb{R}^n)$ induced by the linear functionals $u\mapsto u(\phi)$, $\phi\in \mathcal{S}(\mathbb{R}^n)$ coincides with the usual weak-$\ast$ topology on $\mathcal{S}’(\mathbb{R}^n)$? What about $\mathcal{D}(\mathbb{R}^n)$ and $\mathcal{D}’(\mathbb{R}^n)$?
The specific problem that I have in mind is this. Let $(u_j)_j$ be a sequence in $\mathcal{S}’(\mathbb{R}^n)$. Suppose we can show that $u_j(\phi)\rightarrow 0 $ as $j\rightarrow\infty$ for all $\phi$ that can be factorized as $$\phi(x_1, x_2, \cdots, x_n) = \phi_1(x_1)\phi_2(x_2)\cdots\phi_n(x_n)$$ where $\phi_1, \phi_2, \cdots, \phi_n\in\mathcal{S}(\mathbb{R})$. Does it then follow that $u_j\rightarrow 0 $ in $\mathcal{S}’(\mathbb{R}^n)$?
The answer is no. By (IV.1.2) in [1], there is a one-to-one correspondence between "weak-topologies" and "dual spaces".
In other words, if $\phi$ lies in $\mathcal{S}(\mathbb{R}^n)\setminus E$, then the evaluation at $\phi$, namely the functional $$ u\in \mathcal{S}'(\mathbb{R}^n)\mapsto u(\varphi )\in \mathbb R $$ is not continuous with respect to the topology described in the question, and hence that topology is not the same as the weak-$*$ topology.
[1] Schaefer, Helmut H.; Wolff, M. P., Topological vector spaces., Graduate Texts in Mathematics. 3. New York, NY: Springer. vi, 346 p. (1999). ZBL0983.46002.