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

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?

• A fixed point doesn't mean an attractor. It just means $x_0$ such that $f(x_0)=x_0$. – Clement C. Jan 2 at 15:40

## 2 Answers

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.

• Maybe worth addressing the fact that the OP seems to assume fixed points have to be attractors. While a fixed point literally just means "point mapped to itself." (I.e., your answer points that out in a rather implicit way.) – Clement C. Jan 2 at 15:45
• Thank you for the clarification. I really thought that every fixpoint needs to fulfill the conditions – MLK Jan 2 at 16:01

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).

• Thank you very much for the clarification :) – MLK Jan 2 at 16:03