# Find Smallest contraction coefficient

I have been given the following function $$f:[-1,1]\to \mathbb{R}$$: $$f(x)=\ln(x+2)-x$$ And I have been asked whether it is a contraction or not, and if it is, I have to find the smallest contraction coefficient, such that $$0.

Attempt

Since $$0<|f'(x)|<1$$ for $$(-1,1)$$ we must have that the function is a contraction, and i would intuitively say that $$\left|-\frac{2}{3}\right|=\frac{2}{3}$$ is the smallest contraction coefficient.

Doubts

I have no concrete theorem or example to support my claim, and therefore I am skeptical.

Since $$f$$ is differentiable $$\forall x \in [-1,1], |f'(x)|= |\cfrac{1}{x+2} - 1| = |\cfrac{-x-1}{x+2}| = \cfrac{1+x}{2+x}$$ and $$\forall x \in [-1,1], f''(x) = \cfrac{1}{(x+2)^2} \ge 0$$ so $$f'$$ is growing. And we have $$| f'(1)| = \cfrac{2}{3}$$ So $$\forall x \in [-1,1], |f'(x)| \le \cfrac{2}{3}$$ Therefore $$\cfrac{2}{3}$$ is indeed the smallest contraction coefficient.
Hint: Because of the mean value inequality, you want to maximize $$|f'(x)|$$ for $$x \in [-1,1]$$.
I think that Rasmus' realized that $$L=\frac 23$$ is the maximum value for $$f'$$ and therefore will work as a contraction constant. This was probably not the question. The question, I think, was about the possibility of getting a contraction constant $$\tilde L < L$$. The answer is no, it is not possible. If you consider a smaller constant you immediately get an interval of the form $$[\xi, 1]$$ where the contraction inequality does not hold.