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

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.

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).