= Solution
Assume weakly null sequences have norm-null images. Let $(y_n)$ be any <sequence> in $T(K)$, where $K$ is weakly compact, and choose $x_n\in K$ with $y_n=Tx_n$. By the weak subsequence result, some $x_{n_j}\rightharpoonup x\in K$. Then $x_{n_j}-x\rightharpoonup0$, and the assumed property gives
$$
\|y_{n_j}-Tx\|=\|T(x_{n_j}-x)\|\longrightarrow0.
$$
Thus every sequence in $T(K)$ has a norm-convergent subsequence with limit in $T(K)$. In a <metric space>, sequential compactness is equivalent to compactness. Therefore \b[$\boxed{T(K)\text{ is norm compact}}$]. This is one characterization of a <completely continuous operator>.
Back to article page
