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If $A, \phi \in \mathsf{GL}(V)$ where $V$ is a normed vector space and we are using the operator norm on $\mathsf{GL}(V)$, I'm trying to show that

$$ \frac{ \|\phi^{-1}\|^{2} \| A - \phi \| }{1 - \| \phi^{-1}(A - \phi) \|} < \| \phi^{-1} \|^2 \|A - \phi\| $$


$$ \|A - \phi \| < \frac{1}{ \|\phi^{-1}\|} $$

I have tried various approaches to this but none of them seem to work out. About the only thing I've been able to conclude is that the denominator is nonnegative but this doesn't seem to help with the estimation. I'm sure there's some algebra trick I could employ to see this but, if so, I cannot see it and would appreciate any constructive pointers on how to proceed.

If it helps, the context of this is a proof to show that the function that carries an operator to its inverse is continuous.

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This is obviously false as stated. – scineram Mar 11 '12 at 0:12
If somebody on here tells me one more time that something is "obvious", I'm going to change my username to ItsNotFreakingObvious. In fact, I'm going to do that anyway, no need to wait. If it were "obvious" to me I would not have asked the question in the first place. – ItsNotObvious Mar 11 '12 at 3:11
up vote 1 down vote accepted

Observe that $\|\phi^{-1}(A-\phi)\|\leq \| \phi^{-1}\| \| A-\phi\|<1$. Thus, for $A\ne \phi$, it follows that $0<\| \phi^{-1}(A-\phi)\|<1$. Hence the opposite inequality holds by multiplying each side by the denominator and cancelling common factors.

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I don't understand what you mean by "hence the opposite inequality holds by multiplying each side by the denominator and cancelling common factors". Each side of what? Which denominator? What inequality is opposite of what? – ItsNotObvious Mar 11 '12 at 3:08
The equation you are trying to prove has the term $\| \phi^{-1}\|^2\| A-\phi\|$ on both sides of the inequality. So you can cancel them if they are positive. – azarel Mar 11 '12 at 3:38

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