Contractive mappings and fixed points

Question

Hy friends,

In a metric space, we now that $d(f(x),f(y))<d(x,y)$ is not sufficient for the existence of a fixed point for $f$.

However, the results of Rakotch (A Note on Contractive Mappings, 1962) shows that if there is a function $\alpha(d(x,y))$ such that

1. $\alpha(x,y)=\alpha(d(x,y))$
2. $0 \leq \alpha(\rho) <1, \forall \rho>0$
3. $\alpha(\rho)$ is monotonically decreasing funtion of $\rho$,

we garantee the existence of a fixed point of a function $f$ if $$d(f(x),f(y))<\alpha(x,y)d(x,y). \ \ \ \ (I)$$

My question: the function $f(x)=\ln(1+e^x)$ has no fixed point even if it is a weak contraction (note that $f'(x)=\frac{e^x}{1+e^x}<1$).

How to prove directly that there is no $\alpha$ satisfying (I) for this function?

Thanks a lot!

Answer 1

Notice that $$d(f(x),f(x+1))\leq\alpha(1)$$ but since $$\lim_{x\to\infty}\frac{\log(1+e^x)}x=1,$$ $d(f(x),f(x+1))>1-\epsilon$ for any $\epsilon$. This implies $\alpha(1)=1$, a contradiction.