I am wondering whether the following is true:
Let $X, Y$ be normed spaces and $T \in L(X', Y')$.
If $x_n' \overset{*}{\rightharpoonup} x'$ in $X'$, then $Tx_n' \overset{*}{\rightharpoonup} Tx'$ in $Y'$.
Proof in reflexive spaces
This holds at least if $X$ and $Y$ are reflexive. Since then weak and weak* convergence are the same on $X'$ and $Y'$, respectively:
- $Tx_n' \overset{*}{\rightharpoonup} Tx'$ in $Y$ $\Leftrightarrow Tx_n' \rightharpoonup Tx'$ in $Y'$ weakly.
- $x_n' \overset{*}{\rightharpoonup} x'$ in $X$ $\Leftrightarrow x_n' \rightharpoonup x'$ in $X'$ weakly
Thus to prove the claim we prove that $Tx_n' \rightharpoonup Tx'$ in $Y'$ weakly: $$ \left<y'', Tx_n'\right> - \left<y'', Tx'\right> = \left<y'', Tx_n' - Tx'\right> = \left<y'' \circ T, x_n' - x'\right>_{X'' \times X'} \overset{2.}{\to} 0 $$
(We essentially applied reflexivity and preservation of weak convergence under continuous linear maps).
Attempt for general spaces
We try to prove the claim directly:
Let $y \in Y$. We want to show $\left<Tx' - Tx_n', y\right> \to 0$. If $T$ was a of a specific form, namely for a fixed $A \in L(Y, X)$: $T(x')(y) = \left<x', Ay\right>$, then we would be able to easily prove the claim:
$$\left<Tx' - Tx_n', y\right> = \left<T(x' - x_n'), y\right> = \left<x' - x_n', \underbrace{Ay}_{\in X}\right> \to 0$$
Question: Are all $T$s of that form?
This is not true in general. Recall that $(c_0)^*=\ell^1$, where $c_0$ is the space of all sequences convergent to $0$. Define $T:\ell^1\to\ell^1$ by
$$(Tx)(n)=\left\{\begin{array}{cll} \sum_kx(k)&:&n=1\\0&:&\text{otherwise.}\end{array}\right. $$ Then $T$ is linear and bounded, with $\|T\|=1$. Let $\{e_n\}$ be the standard (Schauder) basis of $\ell^1$, that is, $e_n(n)=1$ and $e_n(k)=0$ otherwise. Then $\{e_n\}$ is weak$^*$ convergent to $0$ (as the dual of $c_0$), but $Te_n=e_1$ for all $n$, so $\{Te_n\}$ does not converge weak$^*$ to $T0=0$.
Thus, we cannot have $T$ as the adjoint of an operator on $c_0$.