# closed range implies boundedness of operator

Let $X$ and $Y$ be a Banach spaces and let $T: X\rightarrow Y$ be linear, injective and bounded operator. Denote $R(T):= \left\{ Tx \ : \ x\in X\right\}$. Show that $T^{-1}: R(T)\rightarrow X$ is bounded iff $R(T)$ is closed.

I proved the implication from the right side to the left one. (If range is closed,then it is a Banach space, and using inverse mapping theorem we get the statement.) How to prove this in the second direction?

• This may be useful. – Mark Sep 12 '17 at 16:18
• Honestly I don't see how it works :) – Yelon Sep 12 '17 at 18:16

Suppose $T^{-1}$ is bounded. Then for some $M>0$ we have $\|T^{-1}y\|\leq M\|y\|$ for all $y\in R(T)$. This implies that $T$ is bounded below, which implies $R(T)$ is closed. In case you're not familiar with this result, I'll include a proof:
Suppose $(y_n)$ is a Cauchy sequence in $R(T)$, and write $x_n=T^{-1}y_n$. Given $\varepsilon>0$, there is some $N\in\mathbb N$ such that $\|y_n-y_m\|<\varepsilon/M$ for $n,m\geq N$. Then $$\|x_n-x_m\|\leq M\|y_n-y_m\|<\varepsilon$$ for $n,m\geq N$, so $(x_n)$ is Cauchy, and thus convergent to some $x\in X$. Put $y=Tx$. Now there is some $N\in\mathbb N$ such that $\|x_n-x\|<\varepsilon/\|T\|$ for all $n\geq N$, and thus $$\|y_n-y\|\leq\|T\|\|x_n-x\|<\varepsilon$$ for $n\geq N$, and thus $y_n\to y$. Thus $R(T)$ is complete, and therefore closed.
• @SRJ Off the top of my head, I believe it can be used in the construction of pseudoinverses of Fredholm operators (i.e., if $T$ is Fredholm, this can be used to construct $S:Y\to X$ with $ST-I_X$ and $TS-I_Y$ compact). – Aweygan Jul 3 at 16:38