Let $X$ be a dense subspace of a Banach space $Y$. The restriction map $$r: Y^* \to X^*:\omega \mapsto \omega \vert_X$$ is then an isometric isomorphism that is weak$^*$-continuous.
I am wondering if it is true if the map $r$ is a weak $^*$-homeomorphism? My intuition tells me that this is not true.
Of course, on norm-bounded subsets, the inverse map is weak$^*$-continuous, so the problem (if any) exists because of the unbounded nature of things.
Any insight is appreciated!
Theorem 3.10 in Rudin Functional Analysis "Suppose $X$ is a vector space and $X'$ is a separating vector space of linear functionals on $X$. Then the $X'$-topology $\tau'$ makes $X$ into a locally convex space whose dual space is $X'$."
Now consider the situation you're talking about. $X$ is a dense subspace of $Y$. If $\alpha_1,\alpha_2\in Y^*$ and $\alpha_1\rvert_X=\alpha_2\rvert_X$ then $\alpha_1=\alpha_2$ because $X$ is dense. This means that $X$ separates points on $Y^*$. Of course, $Y$ also separates points on $Y^*$. You're considering two topologies on $Y^*$. $\tau_X$ is the topology from $X$ as linear functionals on $Y^*$. $\tau_Y$ is the topology from $Y$ as linear functionals on $Y^*$. By the theorem quoted above, the dual of $(Y^*,\tau_X)$ is $X$ and the dual of $(Y^*,\tau_Y)$ is $Y$, so the topologies $\tau_X,\tau_Y$ cannot be identical except when $X=Y$.
Here is something more explicit: Assume $X\subsetneq Y$ and let $y\in Y\setminus X$. Then $y\ne 0$. The linear functional $p:Y^*\to\Bbb{C}$ defined by $p(\alpha)=\alpha(y)$ is weak-$Y^*$ continuous on $Y^*$. Assume $r^{-1}$ is continuous from the weak-$X^*$ topology to the weak-$Y^*$ topology. Let $q:X^*\to\Bbb{C}$ be the linear functional $q(\beta)=p(r^{-1}(\beta))$. Since $q$ is a composition of continuous maps, it is continuous in the weak-$X^*$ topology on $X^*$.
Since $q$ is weak-$X^*$ continuous, there exist $x_1,\ldots,x_n\in X$ and $C>0$ such that $$|q(\beta)|\le C\sum_{i=1}^n|\beta(x_i)|\tag{1}$$ for all $\beta\in X^*$ (see here).
Since $\operatorname{span}\{x_1,\ldots,x_n\}\subseteq X$ and $y\not\in X$, there is an $\alpha_0\in Y^*$ such that $\alpha_0(x_i)=0$ for all $i, 1\le i\le n$ and $\alpha_0(y)=\|y\|\ne 0$ (this is by Hahn-Banach.) Now let $\beta_0=r(\alpha_0)=\alpha_0\rvert_X$. Then $\beta_0(x_i)=(r(\alpha_0))(x_i)=\alpha_0\rvert_X(x_i)=\alpha_0(x_i)=0$ for all $i,1\le i\le n$. And $q(\beta_0)=p(r^{-1}(\beta_0))=p(\alpha_0)=\alpha_0(y)\ne 0$. This contradicts $(1)$ (taking $\beta=\beta_0$). Therefore $r^{-1}$ cannot have the continuity presumed.