A <bounded linear operator> is compact when the closure of the image of its <unit ball> is compact. Equivalently every bounded sequence has a subsequence whose images converge; on a Banach target it also suffices that the image of its <unit ball> be <totally bounded>.

For a <finite-rank operator> that image is bounded in a finite-dimensional subspace, whose closed bounded sets are compact. Thus \b[every <finite-rank operator> is compact].

Solution

Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2007 / A / 4 / i / Solution

 Solution (source code)