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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?

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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.

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