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Is it true that an open linear map necessarily is reversely bounded, i.e.: $c \lVert x\rVert\leq \lVert T x\rVert$
If so how to proof without the existence of an inverse?

freishahiri
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    No. Take any non-injective, bounded, onto map between Banach spaces. Such as the left shift operator on $\ell_2$. – David Mitra Jan 31 '14 at 23:28
  • Ah, right, thanks! – freishahiri Jan 31 '14 at 23:35
  • A characterization of being bounded below is given in Abramovich and Aliprantis, An Invitation to Operator Theory, Theorem 2.5: A bounded operator between Banach spaces is bounded below iff it is injective and has closed range. That is a nice application of the open mapping theorem answering the question on the open mapping theorem itself. – freishahiri Jun 02 '14 at 17:27

2 Answers2

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If $T$ is an open map, then we have that $T(B_r)=U$ is open and contains $0$ (where $B_r=\{x:\|x\|<r\}$), and thus contains some ball $B_s$, for some $s>0$. If the map is injective, then $Tx\in X\backslash T(B_{\|x\|/2})\subseteq X\backslash B_c$ for some $c$, which you can show can be taken to be independent of $x$ by linearity.

If your Banach space $X$ is finite-dimensional (and over $\mathbb{R}$ or $\mathbb{C}$), then the invariance of dimension theorem tells you that if $T$ is open then it must be an isomorphism.

However if $X$ is infinite dimensional, I think you can find open, non-injective mappings, and such maps are obviously not reversely bounded. I'll let you know if I find a counterexample.

Edit: As @DavidMitra commented, the left-shift operator on $\ell^2$ is such an open, non-injective mapping.

Daniel Robert-Nicoud
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Is it true that an open linear map necessarily is reversely bounded, i.e.: $c \|x\| ≤ \|Tx\|$

No. Let $T$ be the projection of $\mathbb R^2$ onto $\mathbb R$ defined by $T(x,y)=x$. Set $z=x+iy$. It is clear that $c \lvert z\rvert \leq \lvert x\rvert$ does not hold.

It seems that we can define a branch $R$ of $T^{-1}$ such that $\lvert Ry\rvert \leq c \lvert y\rvert$.

epimorphic
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Miodrag Mateljevic
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