1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you so much for your answer :-) $\endgroup$ – Rasmus Jensen May 4 at 13:37
  • $\begingroup$ You're welcome ;-) $\endgroup$ – Monadologie May 4 at 15:54
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you for your answer :-) Can you show the details about the mean value theorem and the interval which does not hold for any smaller interval? I have for long wanted to understand it :-) $\endgroup$ – Rasmus Jensen May 4 at 13:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.