Does closed image imply the adjoint has weak^* closed image? Question:
If a bounded operator between Banach spaces has closed image, does its adjoint necessarily have weak$^*$-closed image?
The equivalence of these two conditions is claimed in Theorem A.3.48 of Dales' "Banach algebras and automatic continuity", so I would be surprised if it were false, but I cannot find a proof of the above implication anywhere, nor produce one myself.
 A: It's part of the
Closed Range Theorem: Let $E,F$ be Banach spaces and $T\colon E\to F$ a continuous linear operator. Then the following are equivalent:


*

*$\mathcal{R}(T)$ is closed.

*$\mathcal{R}(T^\ast)$ is closed.

*$\mathcal{R}(T^\ast)$ is weak$^\ast$-closed.


Proof: Let $G = E/\ker T$ and $H = \overline{\mathcal{R}(T)}$. Then we can split $T$ as
$$E \xrightarrow{\pi} G \xrightarrow{S} H \xrightarrow{\iota} F$$
where $\iota$ is the inclusion, $\pi$ the canonical projection and $S(x+\ker T) = T(x)$. For $T^\ast$, we therefore have the decomposition $T^\ast = \pi^\ast \circ S^\ast \circ \iota^\ast$.
$\iota^\ast$ is the restriction map $F^\ast \to H^\ast$, which is surjective by the Hahn-Banach theorem(s). $\pi^\ast$ is an isometric embedding of $G^\ast$ into $E^\ast$ whose range is the annihilator $(\ker T)^\perp = \{\lambda\in E^\ast : \lambda\lvert_{\ker T} = 0\}$ of $\ker T$. Annihilators are always weak$^\ast$-closed.
The range of $T$ is closed if and only if $S$ is surjective, by the open mapping theorem, $S$ is then an isomorphism.
So if $\mathcal{R}(T)$ is closed, $S$ is an isomorphism, and therefore $S^\ast$ is also an isomorphism, in particular surjective, and thus $\mathcal{R}(T^\ast) = (\ker T)^\perp$ is weak$^\ast$-closed, showing $1. \implies 3$.
Since the weak$^\ast$ topology is coarser than the norm topology, $3. \implies 2.$ automatically.
Since $\pi^\ast$ is an isometry, $\mathcal{R}(T^\ast) = \pi^\ast(\mathcal{R}(S^\ast))$ is closed if and only if $\mathcal{R}(S^\ast)$ is closed. Since $S^\ast$ is injective ($\ker S^\ast = \mathcal{R}(S)^\perp = \overline{\mathcal{R}(S)}^\perp = H^\perp = \{0\}$), that is the case if and only if there is a $\delta > 0$ such that $\lVert S^\ast\mu\rVert \geqslant \delta\lVert\mu\rVert$ for all $\mu \in H^\ast$ (open mapping theorem). Then $\overline{S(B_G)} \supset \overline{\delta\cdot B_H}$, where $B_X$ denotes the open unit ball in the space $X$: $\overline{S(B_G)}$ is a closed convex and balanced set in $H$. By the Hahn-Banach theorem, for all $y\notin \overline{S(B_G)}$ there is a $\mu\in H^\ast$ such that $\lvert \mu(Sx)\rvert \leqslant 1$ for all $x\in B_G$ and $\lvert\mu(y)\rvert > 1$. Then $\delta\lVert\mu\rVert \leqslant \lVert S^\ast\mu\rVert = \sup \{ \lvert \mu(Sx) \rvert : x \in B_G\} \leqslant 1$, and
$$\delta < \delta\lvert \mu(y)\rvert \leqslant \delta\lVert\mu\rVert\,\lVert y\rVert \leqslant \lVert S^\ast\mu\rVert\lVert y\rVert \leqslant \lVert y\rVert.$$
Hence $\mathcal{R}(S)$ is of the second category in $H$, and by the open mapping theorem, $\mathcal{R}(S) = H$ follows.
