4
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ What are your thoughts on the problem? What have you tried? $\endgroup$ – Ben Grossmann Nov 3 '19 at 7:52
  • 1
    $\begingroup$ @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) $\endgroup$ – Tim Nov 3 '19 at 7:58
  • 1
    $\begingroup$ @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 $\endgroup$ – Tim Nov 3 '19 at 8:00
  • 1
    $\begingroup$ 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. $\endgroup$ – Ben Grossmann Nov 3 '19 at 8:05
  • 1
    $\begingroup$ @Omnomnomnom Will keep in mind. Thank you $\endgroup$ – Tim Nov 3 '19 at 8:08
4
$\begingroup$

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.
$\endgroup$
5
  • $\begingroup$ If R(T) is closed how does an injective map becomes bijective ? $\endgroup$ – Tim Nov 3 '19 at 8:25
  • $\begingroup$ and even if I assume it then how inverse of T becomes bounded can you please elaborate a bit? $\endgroup$ – Tim Nov 3 '19 at 8:27
  • $\begingroup$ Assuming Inverse of T to be bounded then R(T) is closed this side is clear to me. $\endgroup$ – Tim Nov 3 '19 at 8:37
  • $\begingroup$ I mean if we change the codomain from $Y$ to $\mathscr R(T)$, then the map is now also surjective $\endgroup$ – Ben Grossmann Nov 3 '19 at 8:43
  • $\begingroup$ sorry was very silly to ask for it. Yes now by bounded inverse theoram I got it. Thanks $\endgroup$ – Tim Nov 3 '19 at 8:47
4
$\begingroup$

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).

$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$ – Ben Grossmann Nov 3 '19 at 8:54
  • $\begingroup$ 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. $\endgroup$ – Ben Grossmann Nov 3 '19 at 8:56
  • 1
    $\begingroup$ 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? $\endgroup$ – David Mitra Nov 3 '19 at 8:58
  • 1
    $\begingroup$ @Rebellos Thanks a lot. $\endgroup$ – Tim Nov 3 '19 at 8:59
  • $\begingroup$ @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. $\endgroup$ – Rebellos Nov 3 '19 at 9:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.