Functional analysis-Closed graph therem

Question

Let $ X$, $ Y$, $ Z$, be Banach spaces and let $ T:X\to Y $ and $ S:Y\to Z $ be linear transformations.Suppose that $S$ is Bounded and injective and that $ S \circ T $ is bounded.Prove that $T $ is bounded.

So now, $ S \circ T :X \to Z $

take any $ x\in X $

Since $ S \circ T $ is bounded, $ \|S \circ T(x)\| \leqslant\|S \|\cdot\|T \|\cdot\|x \| $ and since $ S $ is bounded $\exists$ $ M>0 $ s.t $ \|S\|\leqslant M $.

How can I use these results to prove that $T$ is bounded? This is given as an application of closed graph theorem.So how can we use it to solve the problem.

Any help is appreciated!

Thanks!

Answer 1

Let $\{x_n\}$ a sequence which converges to $0$ and such that $Tx_n\to y\in Y$. As $S$ is bounded, $STx_n\to Sy$ and since $ST$ is bounded, $STx_n\to 0$. So $Sy=0$, and by injectivity $y=0$, proving that the graph is closed.

Answer 2

Since $S$ is injective it is a bijection onto its image $S(Y) \subseteq Z$ and thus has a well-defined inverse $S^{-1} : S(Y) \to Y$. Then

$$T = \left( X \xrightarrow{S \circ T} (S \circ T)(X) \hookrightarrow S(Y) \xrightarrow{S^{-1}} Y \right) = S^{-1} \circ (S \circ T)$$

By the bounded inverse theorem, which is equivalent to the closed graph theorem, $S^{-1}$ is bounded. And $S \circ T$ is bounded by hypothesis.

I leave it to you to finish the proof.

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If this looks too messy, just pretend (without loss of generality) that $S$ is bijective and that $Z=S(Y)$. This is okay because nowhere in the proof are the elements of $Z \setminus S(Y)$ used.