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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!

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2 Answers 2

up vote 2 down vote accepted

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.

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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.


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.

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Thanks alot Clive! –  ccc Jan 7 '13 at 14:00
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This doesn't work this way without more effort. The image of $S$ need not be closed and the inverse of $S$ need not be bounded (with respect to the subspace topology of $Z$). Take $S$ to be the inclusion of $\ell^1$ in $\ell^2$, for example. –  Martin Jan 7 '13 at 14:07
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@ccc: Yes, but I think Robert Israel gave you some good hints. There is little I can add without giving the problem away. Keep working on it! –  Martin Jan 7 '13 at 14:16
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@Johan: no. But it's not very hard: the space of sequences with finitely many non-zero terms is dense in both spaces, so the image of $\ell^1$ under the inclusion in $\ell^2$ is dense. Now take any element of $\ell^2 \setminus \ell^1$, say $x = (1,1/2,1/3,\dots)$. Then the sequence of truncates $x_n = (1,1/2,\dots,1/n, 0,0,\dots)$ converges to $x$ in $\ell^2$, while the sequence $(S^{-1}(x_n))_{n \in \mathbb{N}}$ is unbounded in $\ell^1$. –  Martin Jan 8 '13 at 11:04
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@Martin Thanks! did not think about that :) –  Johan Jan 8 '13 at 12:16

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