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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.

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2 Answers 2

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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.

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

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  • $\begingroup$ Why does the given estimate imply injectivity? That's what I've been trying to figure out. $\endgroup$
    – Johnny Apple
    Jun 25, 2014 at 16:12
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    $\begingroup$ @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! $\endgroup$
    – Johnny Apple
    Jun 25, 2014 at 16:22

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