# Bounded, surjective linear operator between Banach spaces

How can I show that for a given surjective linear operator $T: X \to Y$ between Banach spaces, if there exists an $\epsilon > 0$ such that $||Tx|| \geq \epsilon||x||$ for all $x \in X$, then $T$ is bounded?

I'm not really sure what to do with the surjectivity here.

Use the closed graph theorem: if $x_n\to x$ and $Tx_n\to y$, hence by surjectivity $y=Tu$ for some $u\in X$. We conclude by the assumption that $\lVert x_n-u\rVert\to 0$ hence $u=x$ and $y=Tx$, proving that the graph is closed.

• Where did you use the estimate? Jun 25, 2014 at 16:13
• To deduce that $\lVert x_n-u\rVert\leqslant \varepsilon^{-1}\lVert Tx_n-Tu\rVert$. Jun 25, 2014 at 16:15
• You are welcome. Jun 25, 2014 at 16:21

The given estimate implies that $T$ is injective. It is thus bijective, so that $T^{-1}$ is a well-defined linear map.

Now the given estimate shows that $T^{-1}$ is in fact bounded (why?).

• @JohnnyApple $0 \leq \|x-y\| \leq \epsilon \|T(x-y)\|=\|Tx-Ty\|= 0$ for every $x,y$ with $Tx = Ty$.