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Suppose $Y$ is a separable Banach space and $(y^*_n)_{n \in > \mathbb{N}} \subset B_{Y^*}$. Let $T : X \rightarrow Y$ be an operator. Since $Y$ is separable, we have the unit dual ball $B_{Y^*}$ is metrisable in its weak$^*$ topology, hence there exists a weak$^*$-convergent subsequence $(y^*_{n_j})_{j \in \mathbb{N}}$ of $(y^*_n)_{n \in \mathbb{N}}$. Then, $(T^*(y^*_{n_j}))_{j \in > \mathbb{N}}$ is also weak$^*$-convergent as $T^*$ is w$^*$-to-w$^*$ convergent.

The statement above is the proof of Proposition $6.2$.

Questions:

$(1)$ Why there exists weak$^*$-convergent subsequence $(y^*_{n_j})_{j \in \mathbb{N}}$?

$(2)$ Why $T^* : Y^* \rightarrow X^*$ is weak$^*$-to-weak$^*$ continuous if $T:X \rightarrow Y$ is continuous?

My thought:

$(1)$ Since $B_{Y^*}$ is metrisable in its weak$^*$ topology, we have $B_{Y^*}$ is finite-dimensional. Hence, $B_{Y^*}$ is compact. In metric space, compact is equivalent with sequential compact, hence the existence of the subsequence.

$(2)$ Suppose $T$ is continuous. Let $(y^*_n)_{n \in \mathbb{N}}$ to be a sequence in $Y^*$ which converges to $y^*$ in $w^*$ topology. Then for any $y \in Y$, we have $$|y_n^*(y) - y^*(y)| < \epsilon.$$

In particular, for any $Tx \in Y$, we have $$|y_n^*(Tx) - y^*(Tx)| < \epsilon.$$

Note that $|(T^*y_n^*)(x) - (T^*y^*)(x)| = |y_n^*(Tx) - y^*(Tx)|<\epsilon.$

Is my proof correct?

Idonknow
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1 Answers1

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Separability of $Y$ implies that the weak star topology on the dual unit ball is metrizable, eg, suppose that $x_n$ is a dense set of $B_Y$. Then $d(x^*, y^*) = \sum_{n=1}^{\infty} \frac{|(x^* - y^*)(x_n)|}{2^n}$ provides a metric. One can check that this is equivalent. The fact that this is non zero for every $x,y$ follows, eg, as $(x^* - y^*)$ has some element $z$ in the unit ball for which $(x^* - y^*)(z) = \|x^* - y^*\|$, so if $x_M$ is sufficiently close to $z$ the two things differ.

The unit ball of $Y^*$ is compact in the weak star topology (Banach-Alaoglu) and metrizable, so every sequence has a convergent subsequence.

$T$ is continuous from $X$ to $Y$ and so $T^*$ is weak star to weak star continuous from $Y^*$ to $X^*$. This is easy to establish.

James Kilbane
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  • Regarding the continuity implies its adjoint is $w^$ to $w^$ continuous, is my proof correct? – Idonknow Nov 06 '15 at 13:13
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    The closed unit ball of $Y^{\ast}$ is metrisable in the weak$^{\ast}$ topology. $Y^{\ast}$ itself is not metrisable in the weak$^{\ast}$ topology generally (if $Y$ is Banach, then if and only if $Y$ is finite-dimensional). – Daniel Fischer Nov 06 '15 at 13:18
  • @DanielFischer: can explain your sentence in bracket? 'if $Y$ is Banach, then if and only if $Y$ is finite dimensional'. Do you mean $Y$ is metrisable if and only if $Y$ is finite dimensional? – Idonknow Nov 06 '15 at 13:25
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    @Idonknow If $Y$ is a Banach space, then $Y^{\ast}$ is metrisable in the weak$^{\ast}$ topology if and only if $Y$ is finite-deminsional. – Daniel Fischer Nov 06 '15 at 13:28
  • @Idonknow: Daniel Fischer is right, James's claim is false and his answer is not correct. – Alex M. Dec 11 '15 at 19:39