# Fixed point existence for a function satisfying a certain condition

Let $$T:\mathbb{R}\to\mathbb{R}$$ satisfying $$|Tx - Ty| \leq \frac{1+\max({|x|,|y|})}{2+\max({|x|,|y|})}|x-y|$$

for every $$x,y\in\mathbb{R}$$. Show that it has a unique fixed point.

The thing is that this function is a weak contraction. I tried to repeat the proof of Banach's Theorem, but I did not get anything from that. Any ideas?

If $$p,q$$ are fixed points, then it must be $$p=q$$ since $$|p-q|\le r|p-q|$$ where $$r=\frac{1+\max\{|p|,|q|\}}{2+\max\{|p|,|q|\}}<1$$. So it remains to show the existence. If $$T(0)=0$$, we are done. Otherwise, we may assume $$T(0)>0$$. If $$T(0)<0$$, consider $$-T(-x)$$ instead. Define $$f(x)=T(x)-x$$ for $$x\ge 0$$. By the given condition, we have $$T(n+1)-T(n)\le \frac{n+2}{n+3},$$hence $$f(n)= \sum_{k=0}^{n-1}\left(T(k+1)-T(k)\right)+T(0)-n \le T(0)- \sum_{k=0}^{n-1}\frac1{k+3} \xrightarrow{n\to\infty} -\infty.$$ Since $$f(0)>0$$ and $$f$$ is continuous, intermediate value theorem implies there exists $$x_0\in (0,\infty)$$ such that $$f(x_0)=T(x_0)-x_0 =0$$. So $$x_0$$ is a fixed point of $$T$$ as wanted.