How can $\ln(x+2)$ have a fixed point in $(-2,-1]$?

Question

I have to determine the fixed points of $\ln(x+2)$. So as a first step I plotted $\ln(x+2)-x$ and found that it does have two fixed points. One between $[0, \infty]$, which is perfectly fine and one in $(-2,-1]$.

The fixed point in the first interval makes sense to me since it fulfills all requirements for the Banach fixed point theorem:

1. $(\mathbb{R},d)$ where d is the Euclidean metric is a metric space
2. The derivative in the metric is $|f'(x)| \leq \frac{1}{2} < 1$, thus it is bounded and Lipschitz-continuous, thus it is a contraction.

But in the interval $(-2,-1]$, its derivative is unbounded (isn't it?) or at least $L > 1$, thus it is not Lipschitz-continous in that interval.

Can someone explain to me how it is then possible that $\ln(x+2)$ does have a fixed point in that interval, if one of the conditions is clearly not fulfilled? Or am I missing something/made a mistake?

Answer 1

The Banach Fixed Point theorem gives a sufficient condition for the existence (and uniqueness) of a fixed point, but that condition is by no means necessary.

Answer 2

Your confusion is the following one:

• If a map satisfies the conditions of the Banach fix point theorem, it has a fix point.
• However the converse is not true. A map may have a fix point without satisfying the Banach fix point hypothesis.

This is the case here. $g(x) = \ln(x+2)-x$ is continuous in the interval $(-2,1]$ and $g(1)>0$ while $\lim\limits_{x \to -2^+} g(x)=-\infty$. Hence $g$ vanishes in that interval, which means that $\ln(x+2)$ has a fix point (even if the hypothesis of Banach fix point theorem are not met).