Is operator from $X^{*}$ to $c_{0}$ compact

Question

$X$ - Banach space. Let $X \ni (x_n)$ be such that $x_n \stackrel{\omega}{\rightarrow}0 $ weakly. Define $T: X^{*} \rightarrow c_0$ by $T=\left(x^{*}(x_n) \right)_{n=1}^{\infty}$. Is operator $T$ compact?

Answer 1

Not necessarily. For example take $X=c_0$ and $(x_n)$ to be the sequence of standard unit vectors in $c_0$. Then $T$ is the non-compact identity injection of $\ell_1$ into $c_0$.

(What is true is the following: The bounded operators from a Banach space $X$ into $c_0$ correspond to the weak* null sequences $(x_n^*)$ in $X^*$ via $Tx=(x_n^*x)$. $T$ is compact if and only if $(x_n^*)$ is norm null.)