$X, Y, Z$ are Banach Space, $f: X \to Y$, $g: Y \to Z$ are bounded linear operators. Show: if $f$ or $g$ is a compact operator , then $g \circ f: X \to Z$ is a compact operator.
Suppose first that $g$ is compact, and let $x_i$ be a bounded sequence in $X$; then $f(x_i)$ is a bounded sequence in $Y$, since $f$ is a bounded operator; then the compactness of $g$ implies that $g \circ f(x_i) = g (f(x_i))$ has a covergent subsequence; hence $g \circ f$ is compact.
Likewise if $f$ is compact, the sequence $f(x_i)$ has a convergent subsequence; thus so does $g \circ f(x_i) = g(f(x_i))$ by the continuity of $g$, which is equivalent to its boundedness. Thus $f \circ g$ is compact in this case as well.