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I am studying Functional analysis from Kreyszig book and can somebody please help with this problem

Problem: Let $X$ and $Y$ be Banach Spaces and $T : X \to Y$ be an injective bounded linear operator. Show that $ T^{-1} : \mathscr R(T) \to X$ is bounded iff $\mathscr R(T)$ is closed in $Y$.

Can somebody please give hints on how to proceed through this question.

Ben Grossmann
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    What are your thoughts on the problem? What have you tried? – Ben Grossmann Nov 03 '19 at 07:52
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    @Omnomnomnom for => I tried using bounded linear operator property of inverse of T and I assumed an x in R( T) closure such that there exists a sequence ( $x_n$) ----> x but I am not able to show x belongs to R(T) –  Nov 03 '19 at 07:58
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    @Omnomnomnom <= I assumed R(T) to be closed but I don't know how to prove $ T^-1 $ to be bounded using boundedness of T –  Nov 03 '19 at 08:00
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    Great! Sounds like you're moving in the right direction. When posting questions on the site in the future, please include your ideas in your post along with your question. – Ben Grossmann Nov 03 '19 at 08:05
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    @Omnomnomnom Will keep in mind. Thank you –  Nov 03 '19 at 08:08

2 Answers2

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Hints:

  • If $T^{-1}$ is bounded and $(y_n) \subset \mathscr R(T)$ converges to $y \in Y$, consider the sequence $(T^{-1}y_n) \subset X$.
  • If $\mathscr R(T)$ is closed, then $T$ defines a bijective and bounded linear map between the Banach spaces $X$ and $\mathscr R(T)$. Consider the theorems that are proven in 4.12.
Ben Grossmann
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  • If R(T) is closed how does an injective map becomes bijective ? –  Nov 03 '19 at 08:25
  • and even if I assume it then how inverse of T becomes bounded can you please elaborate a bit? –  Nov 03 '19 at 08:27
  • Assuming Inverse of T to be bounded then R(T) is closed this side is clear to me. –  Nov 03 '19 at 08:37
  • I mean if we change the codomain from $Y$ to $\mathscr R(T)$, then the map is now also surjective – Ben Grossmann Nov 03 '19 at 08:43
  • sorry was very silly to ask for it. Yes now by bounded inverse theoram I got it. Thanks –  Nov 03 '19 at 08:47
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Adding clarity to the hints/comments that were correctly stated.

Note that $R(T) \subset Y$.

For $(\Rightarrow)$ assume that $T^{-1}$ is bounded. Then, by definition, there exists a $M > 0 $ such that : $$\left\|T^{-1}y\right\| \leq M \left\|y\right\|, \; \forall y \in R(T) \subset Y$$

Take $\{y_n\}_n^\infty \subseteq R(T)$ with $y_n \to y \in Y$. Then, there exists an $\varepsilon >0$ such that $\left\|y_n - y \right\| < \varepsilon, \; $ for all $n \geq n_0 \in \mathbb N$. But, this also means :

$$\left\|T^{-1}(y_n - y) \right\| \leq M\left\|y_n - y\right\| < \varepsilon, \forall n \geq n_0 \in \mathbb N$$ Thus, the sequence $\left\{T^{-1}y_n\right\}_n^\infty \subset X$ is convergent as well with $T^{-1}y_n \to T^{-1}y$ which means that $y \in R(T)$, thus $R(T)$ is closed in $Y$.

For $(\Leftarrow)$, assume that $R(T)$ is closed in $Y$. Since $R(T)$ is a subspace of the Banach space $Y$, being closed implies that it is also a Banach space. This is easy to prove, as taking a Cauchy Sequence in $R(T)$ will straight-forwardly lead to the sequence converging to a point in $R(T)$, since it's closed. Considering now that since $R(T)$ is closed in $Y$, the expression of the operator $T$ can be "renamed" as $T : X \to R(T)$. Thus, we have a bounded linear bijection on our hands and by a consequence of the Open Mapping Theorem (Bounded inverse theorem - Rudin 1973, Corollary 2.12) we have that $T^{-1}$ is also continuous (bounded).

Rebellos
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  • Notably, the bounded inverse theorem is also given as theorem 2 of section 4.12 in Kreyszig's book (the section from which these problems are taken), hence the second hint. – Ben Grossmann Nov 03 '19 at 08:54
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    Also, my approach to $\implies$ would be to avoid the $\epsilon$-$n$ proof by simply remarking that a linear operator is bounded iff it is continuous. – Ben Grossmann Nov 03 '19 at 08:56
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    By writing $T^{-1}(y_n-y)$, you're assuming $y$ is in the range of $T$, which is what you're trying to show. No? – David Mitra Nov 03 '19 at 08:58
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    @Rebellos Thanks a lot. –  Nov 03 '19 at 08:59
  • @Omnomnomnom Pretty sure about it, just don't have a view of the book atm so I cited by the Rudin copy that I have (also a wikipedia cite I think). Sure, you can avoid it by the remark you noted. – Rebellos Nov 03 '19 at 09:02