Finding the contraction factor of a function

Question

Hi I was working on an exercise where I had to determine the fixed point and the contraction factor of the function : $$F: [1,2]\rightarrow [1,2] : F(x) = \frac{x+2}{x+1}$$ I found the fixed point by asserting that $x=\frac{x+2}{x+1}$ but I struggle finding the contraction factor. My textbook states that the contraction factor is : $\sup_{x,y \in [1,2] \land x\neq y}|{\frac{F(x)-F(y)}{x-y}}|$, but I am not sure how to apply this in this particular case. Does anyone know how I can work out the contraction factor, other methods also welcome as long as they also work for non derivable functions. Thanks in advance.

Answer 1

Observe that $\frac {F(x)-F(y)}{x-y}=\frac {\frac {x+2}{x+1}-\frac {y+2}{y+1}}{x-y}=\frac {-x+y}{(x+1)(y+1)}=\frac {-1}{xy+x+y+1}$. Then $sup_{x,y \in [1,2] \wedge x \neq y}\lvert\frac {F(x)-F(y)}{x-y}\rvert=max_{x,y \in [1,2]}\frac {1}{xy+x+y+1}=\frac {1}{min_{x,y \in [1,2]}(xy+x+y+1)}=\frac {1}{min_{x,y \in [1,2]}(xy+x+y)+1}=\frac {1}{(xy+x+y)\Big|_{x=y=1}+1}=\frac1 4$.