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I am looking for a reference for the proof of the following theorem:

$\textbf{Theorem}.$ Let $X$ and $Y$ be Banach spaces and $T:X\to Y$ be a bounded linear operator. Then $T$ is compact if and only if its adjoin $T^*:Y^*\to X^*$ is $w^*$-norm continuous on $w^*$-compact subsets of $Y^*$.

I obtained the statement from the book Sequences and Series in Banach Spaces by Diestel, it is the exercise 4(d) on Chapter II. The exercise 4(b) is the Schauder's Theorem for adjoint operators ($T$ is compact if and only if $T^*$ is), so I presume the proof I am looking for would be a corollary of Schauder's Theorem.

Thank you all.

Evangelopoulos Foivos
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Jose M Barrientos
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1 Answers1

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Indeed, you can use Schauder's theorem:

Suppose that $T^*$ is weak-star to norm continuous. This implies that $T^*$ is compact: If $(y_i^*)$ is a net in $B_{Y^*}$, the closed unit ball of $Y^*$, then since the latter is weak-star compact (due to Alaoglou) there exists $y^* \in B_{Y^*}$ and a subnet $(y_a^*) \xrightarrow{w^*} y^*$. By the continuity hypothesis $T^* (y_a^*) \to T^*y^*$. This shows that $T^*(B_{Y^*})$ is relatively compact.

For the converse, you only need to assume that $K \subset Y^*$ is norm bounded (but you might as well assume that it is weak-star compact since that is the case for its weak-star closure $\overline{K}^{w^*}$.) So, assume that $T^*$ is compact. Then $T^*(K)$ is compact in $X^*$. You can use the following two facts

On a compact subset $A \subset X^*$ the norm and weak-star topologies coincide.

(Indeed, if $F$ is norm closed in $A$ then it is norm compact and thus weak-star compact. So $F$ is weak-star closed in $A$.)

$T^*$ is weak-star to weak-star continuous

to obtain that $T^*\colon (K,w^*) \to (X^*, \|\cdot\|)$ is continuous.

In fact, suppose that $y^* \in K$ and $B \subset X^*$ is a norm-open neighborhood of $T^*y^*$. Then by the first fact, $B \cap T^*(K)$ is a weak-star open neighborhood of $T^*y^*$ so that by the second fact there exists $U \subset Y^*$ weak-star open neighborhood of $y^*$ such that $T^*(U) \subset B \cap T^*(K)$.

Evangelopoulos Foivos
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