= Solution
Assume instead that $T$ sends weakly compact sets to norm-compact sets. If $x_n\rightharpoonup0$, then $K=\{0,x_1,x_2,\ldots\}$ is weakly compact: an open cover has a member containing zero, which contains all but finitely many terms of the convergent sequence. Hence $T(K)$ is norm compact. A <bounded linear operator> is weak-to-weak continuous, because each $g\in F^*$ pulls back to $g\circ T\in E^*$. Thus $Tx_n\rightharpoonup0$. If $\|Tx_n\|$ did not tend to zero, a subsequence with norms bounded below by some $\varepsilon>0$ would have a norm-convergent further subsequence with limit $y$. That further subsequence also converges weakly to both $y$ and zero; uniqueness of limits in the Hausdorff <weak topology> forces $y=0$, contradicting the lower bound. Therefore \b[$\boxed{\|Tx_n\|\to0}$], proving the equivalence and the characterization of <completely continuous operators>.
Now consider the inclusion of the <space of continuous functions on a compact space> $C(\mathbb T)$ into <L2 space> for normalized <Lebesgue measure>. It is a <bounded linear operator>, since $\|f\|_2\le\|f\|_\infty$. For a weakly null sequence $(f_n)$ in $C(\mathbb T)$, point evaluations are <continuous linear functionals>, so $f_n(t)\to0$ for every $t$. The <Uniform boundedness principle>, applied to the canonical embeddings $f_n\in C(\mathbb T)^{**}$, gives a uniform bound $\|f_n\|_\infty\le M$: each dual functional is bounded on the weakly convergent sequence. The <dominated convergence theorem> now yields
$$
\|j_2f_n\|_2^2=\int_{\mathbb T}|f_n|^2\,d\mu\longrightarrow0.
$$
Thus \b[$j_2$ satisfies both equivalent conditions].
It is nevertheless \b[not a <compact operator>]. Over complex scalars the functions $f_n(t)=e^{int}$ lie in the unit ball of $C(\mathbb T)$ and form an <orthonormal sequence> in $L^2(\mu)$, so
$$
\boxed{\|j_2f_n-j_2f_m\|_2=\sqrt2\quad(n\ne m).}
$$
There is no norm-convergent subsequence. Over real scalars use $\cos(nt)$, whose distinct images have distance $1$. Complete continuity controls weakly compact sets, rather than all bounded sets.
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