2

Let $X, Y$ be two Banach spaces and $T:X\to Y$ be a compact operator. $X^*$ is the dual of $X$. It is well known that if $T$ is compact then it maps a weakly convergence sequence to norm convergence. Now, I am struggling with the following questions:

  1. Is $T^*:Y^*\to X^*$ (adjoint of $T$) compact again?

  2. Does $T^*$ map a weak* convergence sequence to norm convergence?

Please help me. I am struggling in these topics. Any help is appreciated. Thanks in advance.

Tuh
  • 826

1 Answers1

4

First of all, the characterization: $T$ is compact iff $T$ is completely continuous (i.e. $x_n \xrightarrow{w}x$ implies $Tx_n \to Tx$) is not true in general spaces. In fact, $$T \text{ compact } \implies T \text{ completely continuous}$$ and the converse is true if $X$ is reflexive.

To answer your questions

  1. Yes. If $T$ is compact then so is $T^*$. This is the Schauder theorem.

  2. Yes. This is essentially the same proof of the implication $T$ compact $\implies $ $T$ completely continuous. To see this, note that $T^*$ is weak-star to weak-star continuous. Let $y_n^* \xrightarrow{w^*} y^*$ in $Y^*$. Then $T^* y_n^* \xrightarrow{w^*} T^* y^*$. Arguing by contradiction suppose that $T^*y_n^* \not \to T^*y^*$ so that $$ \|T^* y_n^* - T^*y\| \ge \epsilon_0 \tag{$\star$}$$ holds for infinitely many $n$, say for $n \in L \subset \mathbb N$. Since $(y^*_n)_{n \in L}$ is bounded, by compactness of $T^*$ we can find $u^* \in X^*$ such that $T^* y_n \to u^*$ strongly in $X^*$, at least for a subsequence. But then $u^* = T^* y^*$ by uniqueness of the weak-star limit, which contradicts $(\star)$.

Evangelopoulos Foivos
  • 4,429