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In Murphy's book it is stated that

''Observe that every invertible linear map is bounded below, as is every isometric linear map.''

It's clear to me that isometries are bounded below but I'm doubtful about invertible linear maps. I can't seem to prove it.

Should the claim really be ''every continuous invertible linear map between Banach spaces is bounded below''? Because this claim I can prove.

Edit

What would be an example of an invertible linear map between Banach spaces that is not bounded below?

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    $\begingroup$ You are right, of course. $\endgroup$ – Jochen May 30 '14 at 10:20
  • $\begingroup$ @Jochen Thank you for your comment. $\endgroup$ – Student May 30 '14 at 12:24
  • $\begingroup$ Regarding your edit: I guess, by "bounded below" you mean $\Vert Tx \Vert \geq C \Vert x \Vert$ for some $C>0$ (otherwise, I only know the term in connection with Hilbert spaces). Then assume that $B$ is infinite dimensional and take a basis $(b_i)_{i \in I}$ of $B$ (using Zorns Lemma). Take a countably infinite family $(i_n)_n$ in $I$ and define $T b_{i_n} = \frac{1}{n} b_{i_n}$ and $Tb_i = b_i$ for $i \in I, i \neq i_n$ for all $n$ and extend $B$ linearly (uniquely) onto $B$. This give an invertible operator that is not bounded below in the above sense. $\endgroup$ – PhoemueX May 30 '14 at 15:04
  • $\begingroup$ @PhoemueX That is not what an "invertible operator" means in functional analysis. $\endgroup$ – user147263 May 30 '14 at 17:56
  • $\begingroup$ Yes, but the OP states that he can prove the statement if $T$ is continuous and invertible (continuity of $T^{-1}$ follows by the open mapping theorem) and continues to ask for a counterexample. For the counterexample, $T$ must then (necessarily) not be bounded. $\endgroup$ – PhoemueX May 30 '14 at 18:29
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The meaning of invertible depends on context. In the context of bounded linear operators, one says that $T$ is invertible if both $T$ and $T^{-1}$ are bounded linear operators. (In particular, $T$ is injective and onto, so that $T^{-1}$ is defined.)

Since $T^{-1}$ is bounded, it follows that $T$ is bounded below.

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  • $\begingroup$ Thank you, I was not aware of such defintion/convention. $\endgroup$ – Student Jun 1 '14 at 9:21

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