Proof. ($\Rightarrow$): Let $\cf \subset F^{*}$ be bounded, then $\cf$ is equicontinuous and hence relatively compact in the weak* topology by the Banach-Alaoglu Theorem. Since $F$ is complete, $\ol{T(B_E(0, 1))}$ is compact. By the Arzelà-Ascoli Theorem, $\cf$ is relatively compact with respect to the topology of uniform convergence on $\ol{T(B_E(0, 1))}$. Thus $T^{*}(\cf) = \bracsn{\phi \circ T|\phi \in \cf}$ is relatively compact with respect to the topology of uniform convergence on $B_{E}(0, 1)$. In other words, $T^{*}(\cf)$ is relatively compact in $E^{*}$, and $T^{*}$ is a compact operator.

($\Leftarrow$): By Proposition 11.13.9, $E^{*}$ is complete. The preceding case then implies that $T^{**}\in L(E^{**}; F^{**})$ is compact. As such, its restriction to $E$, being identified with $T$, is also compact.$\square$