We have $f:[0, \infty)$ -> $[0, \infty)$, $f(x)= \frac{x+\frac{1}{2}}{x+1}$
I want to show that $f$ has a fixed-point by using the Banach fixed-point theorem. I already showed that $f$ is contracted but I dont know how to show that every Cauchy-Sequence converges in the given space. I would appreciate any hint.
Actually it is enough to follow the proof of the theorem. Being a closed subset of $\bf R$, the set $X=[0, \infty)$ is complete with the absolute value metric. Notice that $f'(x)=\frac{1}{2(x+1)^2}\leq \frac{1}{2}$ for all $x\geq 0$. Thus $f$ is contraction from $X$ into $X$ by the MVT: $$|f(x)-f(y)|\leq \frac{1}{2}|x-y|\quad (1)$$ for all $x$, $y\in X$. Therefore, if $(x_n)$ is a Cauchy sequence converging to $x\in X$, then by (1), $f(x_n)$ is a Cauchy sequence converging to $f(x)$ (since $f$ is continous). If you set the iteration formula $x_{n+1}=f(x_n)$ for $n=0, 1,\cdots$ and then take the limit as $n\to\infty$, you get $x=f(x)$ (here note that we have used the fact that $x_{n+1}\rightarrow x$.). For the uniquness you should revisit (1) along with method of contradiction. In fact as commented above solving $f(x)=x$ gives the only $x=1/\sqrt 2$.