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:

  1. $Tx_n' \overset{*}{\rightharpoonup} Tx'$ in $Y$ $\Leftrightarrow Tx_n' \rightharpoonup Tx'$ in $Y'$ weakly.
  2. $x_n' \overset{*}{\rightharpoonup} x'$ 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?

  • $\begingroup$ Not all $T$s are a transpose of an $A \in L(Y,X)$. $T$ is a transpose if and only if it is weak$^{\ast}$-continuous. But you're asking about $T$ being weak$^{\ast}$-sequentially continuous, which is potentially a strictly weaker condition. I think $T$ need not be weak$^{\ast}$-sequentially continuous, but I don't immediately see an example. $\endgroup$ – Daniel Fischer Jul 14 '18 at 18:04

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 $c_0$.

  • $\begingroup$ Thanks! How is $\{e_n\} \overset{*}{\rightharpoonup} 0$ defined? Generally, I am not sure how weak or weak* convergence are defined if we only have the "ground" space $X$ and a space isomorphic to $X'$. The same problem occurs with $L^p$ spaces ($1 < p < \infty$): why is the dual product defined to be the integral? $\endgroup$ – FunkTheo Jul 15 '18 at 8:59
  • $\begingroup$ In the above scenario, the dual pairing is given by $\langle x, y\rangle=\sum_k x(k)y(k)$ for $x\in\ell^1$ and $y\in c_0$. So we say a sequence $\{x_n\}$ in $\ell^1$ is weak$^*$ convergent to $x$ if $\sum_kx_n(k)y(k)=\sum_kx(k)y(k)$ for all $y\in c_0$. $\endgroup$ – Aweygan Jul 15 '18 at 14:52
  • 1
    $\begingroup$ For your last question, the dual product is defined that way because that's how the math works out: For conjugate exponents, the map $L^q\to (L^p)^*$ given by $g\mapsto(f\mapsto\int fg)$ is an isometric isomorphism. $\endgroup$ – Aweygan Jul 15 '18 at 14:56

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