Discussion
Let X, Y and Z be Banach spaces. Suppose that \(T : X \to Y\) is linear and \(S : Y \to Z\) is linear, bounded and injective. In addition, if \(S \circ T : X \to Z\) is bounded, then, which of the following statements is TRUE?
Answer: C
The conclusion is T is bounded.
Use the closed graph theorem: a linear map between Banach spaces with a closed graph is bounded. Suppose \(x_{n} \to x\) in X and \(Tx_{n} \to y\) in Y. Since S is bounded, \(STx_{n} \to Sy\); since S ∘ T is bounded, \(STx_{n} \to STx\). Limits are unique, so Sy = STx, and because S is injective, y = Tx. The graph of T is closed, so T is bounded.
'T is not bounded' contradicts this. 'Bounded but not continuous' is impossible, since for a linear map between normed spaces bounded and continuous mean the same thing. And nothing forces T to be surjective: the zero map satisfies every hypothesis.
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