We know that if $f:\mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function such that $|f^{'}(x)|\leq A$ where $0 \leq A <1$. Then $f$ has atleast one fixed point. But the above result is not true if $f^{'}(x)<1$ for example if $f(x)=x+(1+e^{x})^{-1}$ then $f^{'}(x)<1$ but $f $ has no fixed point. Now i am searching an example of a function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $f^{'}(x)\leq A$ where $0 \leq A <1$ but $f$ has no fixed point.Please help me to find such a counterexample if it is possible. I tried it but not found. Thanks for precise time in advance.
[Math] Function with no fixed point.
real-analysis
Best Answer
Yes function will have unique limit point. Uniqueness is obvious as condition on derivative. If $f(0)=0$ then proved. If $f(0)>0$ then consider the function $g(x)=f(x)-x$ . Clearly $g(0)>0.$ we can show that there is a positive real number $a$ such that $g(a)<0$. For if possible let $g(x)>0 $ for all x $\in$ [0,$\infty)$ . Then $g$ is strictly decreasing positive function on $[0,\infty)$ and so $\lim_{x\to\infty} g^{'}(x)=0$ which is not possible as $g^{'}(x)=f^{'}(x)-1 \leq A-1 <0$.Hence $g(a) \leq 0 $for some positive real $a$ and so the result. Similarly we can prove the result if $f(0)<0$.