Find Smallest contraction coefficient

Question

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 $0<q<1$.

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.

Answer 1

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.

Answer 2

Hint: Because of the mean value inequality, you want to maximize $|f'(x)|$ for $x \in [-1,1]$.

Answer 3

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.