Banach Space，pde，compact operator

$$X, Y, Z$$ 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.