The classic Closed Range theorem states that for a linear bounded operator $T:X\to Y$ between Banach spaces, and its transpose $T^*:Y^*\to X^*$, the four conditions: $T(X)$ is $s$-closed; $T(X)$ is $w$-closed; $T^*(Y^*)$ is $s$-closed; $T^*(Y^*)$ is $w^*$-closed, are all equivalent.
I was wondering if this can be generalised to an equality for the closure of images: for $T\in L(X,Y)$ it is certainly true that $\overline{T(X)}^s= \overline{T(X)}^ {w} $, but is it true that $$\overline{T^*(Y^*)}^s= \overline{T^*(Y^*)}^ {w^*} $$$$\bf ?$$
Since $\overline{T^*(Y^*)}^s\subset \overline{T^*(Y^*)}^ {w^*} = (\ker T )^{\perp}$, the problem reduces to the opposite inclusion, and the question can be reformulated as:
Assume $f\in(\ker T )^{\perp}$, that is $f=g\circ T$ for a linear form $g$ on $Y$. Does there exist a sequence $(g_n)_{n\in\mathbb N}$ of continuous linear forms on $Y$ such that $g_n\circ T$ converges to $g\circ T$ strongly in $X^*$ $$\bf ?$$
(I also mention that for a while I thought that a possible counterexample could be the second transpose of the inclusion $i:X\to X^{**}$ of a non-reflexive Banach in its bi-dual, the embedding $i^{**}:X^{**}\to X^{****}$, since by the Goldstine theorem $X^{**}$ is $w^*$-dense and not $w^*$-closed in its bi-dual $X^{****}$, whereas it is $s$-closed in it. But the argument is flawed, because $i_X^{**}$ is not the same as $i_{X^{**}}$, although both are right inverses to $i_{X^*}^*:X^{****}\to X^{**}$: in fact $i_{X^{**}}$ can't be the transpose of any operator $X^{***}\to X^*$: if it were, exactly by the mentioned CRThm, we could say that it has $w^*$-closed image because it has $s$-closed image, but being $w^*$-dense, it would be surjective, so $X^{**}$, hence $X$, would be reflexive).
