Show $F:[0,\infty] \to [0,\infty]$, $F(x) = \log(1+x)$ is not a contraction mapping.
Attempt:
Assume $F$ is a contraction mapping, then we have that $\forall x,y \in [0,\infty)$, $|F(x) – F(y) | \leq \lambda |x-y| $ for $\lambda \in (0,1)$. In particular let $y = 0$ then we have $|F(x)| = \lambda |x|$ so $\left | \dfrac{\log(1+x)}{x} \right | \leq \lambda$, and taking the limit of both sides as $x \to 0$ we get $1 \leq \lambda$. I don't see anything wrong with this argument but I'm sure something is as there is a hint ot use the mean value theorem.
With the mean value theorem I get that $|F(x)| = \dfrac{|x|}{1+\theta}$ for some $\theta$ between $0$ and $x$, I don't see how I can use this to get a contradiction.
Any help please.
Best Answer
For $F$ to be a contraction mapping, there must be a $C\in (0,1)$ such that $|F(x) - F(y)| \le C|x - y|$ for all $x,y\ge 0$. However, since $\lim_{x\to 0} \log(1 + x)/x = 1$, for every $C \in (0,1)$, there exists $x$ such that $1 - \log(1 + x)/x < 1 - C$, or $\log(1 + x) > Cx$, i.e., $|F(x) - F(0)| > C|x - 0|$. Therefore, $F$ is not a contraction mapping.
If you're set on using the mean value theorem, then suppose $F$ is a contraction with contraction constant $\lambda$, and let $0 < x < (1 - \lambda)/\lambda$. By the mean value theorem, for some $c\in (0,x)$,
$$\frac{1}{1 + c} = \frac{\log(1 + x)}{x} \le \lambda.$$
Thus
$$1 -\lambda \le \lambda(1 + c) - \lambda = \lambda c < \lambda x.$$
This implies
$$x > \frac{1 - \lambda}{\lambda},$$
a contradiction.