# 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

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 :

$$g(0) = g(y) - g'(\xi)y$$ 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.