= Solution
A bounded sequence $(x_n)$ lies in a scalar multiple of the closed unit ball of the <reflexive Banach space> $X$, which is <weakly compact set>[weakly compact]. Part (a) gives a subsequence $x_{k_n}\rightharpoonup z$ for some $z\in X$.
The <compact operator> $T$ is weak-to-weak continuous, so $Tx_{k_n}\rightharpoonup Tz$. We claim that the convergence is in norm. Otherwise some further subsequence would satisfy $\|Tx_{k_n}-Tz\|\geq\varepsilon$. Compactness supplies a norm-convergent subsubsequence; its norm limit must also be its weak limit $Tz$, a contradiction. Hence
$$
\boxed{Tx_{k_n}\longrightarrow Tz\quad\text{in norm}.}
$$
Back to article page
