# A problem in open mapping theoram from Kreyszig Section 4.12 Problem 6

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.

• What are your thoughts on the problem? What have you tried? – Ben Grossmann Nov 3 '19 at 7:52
• @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) – Tim Nov 3 '19 at 7:58
• @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 – Tim Nov 3 '19 at 8:00
• 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 3 '19 at 8:05
• @Omnomnomnom Will keep in mind. Thank you – Tim Nov 3 '19 at 8:08

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.
• If R(T) is closed how does an injective map becomes bijective ? – Tim Nov 3 '19 at 8:25
• and even if I assume it then how inverse of T becomes bounded can you please elaborate a bit? – Tim Nov 3 '19 at 8:27
• Assuming Inverse of T to be bounded then R(T) is closed this side is clear to me. – Tim Nov 3 '19 at 8:37
• I mean if we change the codomain from $Y$ to $\mathscr R(T)$, then the map is now also surjective – Ben Grossmann Nov 3 '19 at 8:43
• sorry was very silly to ask for it. Yes now by bounded inverse theoram I got it. Thanks – Tim Nov 3 '19 at 8:47

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).

• 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 3 '19 at 8:54
• 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 3 '19 at 8:56
• 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 3 '19 at 8:58
• @Rebellos Thanks a lot. – Tim Nov 3 '19 at 8:59
• @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 3 '19 at 9:02