why $g(0) < ( 1- k) y$ is contradicts?

by jasmine   Last Updated October 09, 2019 15:20 PM - source

This is orginal post :Yes/No : Is $f$ has fixed point?

But I have some doubts in my mind which i didn't understand

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My doubts is why $g(0) < ( 1- k) y$ is contradicts ?

My attempt : Here $y \le 0$ and $k <1$ , i think this inequality $g(0) < ( 1-k) y$ is true , then why its contradicts

Tags : real-analysis


Answers 1


$g(0) = g(y) - g'(\xi)y$ is from the mean value theorem; here $y < \xi < 0$. By assumption $g'(\xi) \le k - 1$, if we multiply by the positive number $-y$, we obtain $-g'(\xi)y \le (1 - k)y$. So assuming $g(y) < 0$ gets us to $g(0) < (1 - k)y$.

Now the observation is that $|y|$ can be as big as we want. But this inequality says that $\frac{g(0)}{1 - k} < y$—which in terms of absolute value is $|y| < \frac{|g(0)|}{1 - k}$. So if $y = \frac{g(0)}{1 - y}$ or larger (in the negative direction), we have a contradiction.

Trevor Gunn
Trevor Gunn
October 09, 2019 14:57 PM

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