3
$\begingroup$

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.

$\endgroup$

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Where did you use the estimate? $\endgroup$ Jun 25, 2014 at 16:13
  • $\begingroup$ To deduce that $\lVert x_n-u\rVert\leqslant \varepsilon^{-1}\lVert Tx_n-Tu\rVert$. $\endgroup$ Jun 25, 2014 at 16:15
  • $\begingroup$ You are welcome. $\endgroup$ Jun 25, 2014 at 16:21
2
$\begingroup$

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?).

Why does this help you?

$\endgroup$
3
  • $\begingroup$ Why does the given estimate imply injectivity? That's what I've been trying to figure out. $\endgroup$ Jun 25, 2014 at 16:12
  • 1
    $\begingroup$ @JohnnyApple $0 \leq \|x-y\| \leq \epsilon \|T(x-y)\|=\|Tx-Ty\|= 0$ for every $x,y$ with $Tx = Ty$. $\endgroup$
    – Surb
    Jun 25, 2014 at 16:16
  • $\begingroup$ I appreciate the response! $\endgroup$ Jun 25, 2014 at 16:22

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.