How to prove the contraction of $f: [0,1] \to \mathbb{R}, x \mapsto \frac{1}{1+x^2}$?

Question

I am currently studying for my analysis exam by working through older exams.

One question that I am stuck on however, is the following: "Using the Banach fixed-point theorem, prove that for $f(x) = \frac{1}{1+x^2}$ there exists a fixed point $x_*$ in $I = [0,1]$ such that $f(x_*) = x_*$."

To my understanding, 3 criteria need to be met:

1. $I$ needs to be a closed set
2. $f(I) \subset I$
3. $f$ is a contraction, that is: $||f(x)-f(y)|| \leq L \space ||x-y||$ with $x,y \in I$ and $L \in [0,1)$

Omitting the first two, this is what I did so far:

$$\begin{align} |f(x) - f(y)| &= \left| \frac{1}{1+x^2} - \frac{1}{1+y^2} \right| \\ \\ &= \left| \frac{y^2-x^2}{(1+x^2)(1+y^2)} \right| \\ \\ & = \left| \frac{y^2-x^2}{x^2y^2+x^2+y^2+1} \right| \end{align}$$

By $(a+b)(a-b) = a^2-b^2$ we know that $a-b = \frac{a^2-b^2}{a+b}$. Hence, setting $x=b$ and $y=a$, we know that

$$\frac{y^2-x^2}{x^2y^2+x^2+y^2+1} \leq y-x$$

since $x^2y^2 + x^2+y^2+1 > x+y$ and therefore

$$\left| \frac{y^2-x^2}{x^2y^2+x^2+y^2+1} \right| \leq \left| y-x \right|$$

However, I am unsure how to find a constant $L$ that fits the criteria, as I can't pull out any constant factors of the term on the left and as such, I am unsure how to prove the proposition.

Answer 1

For these kinds of questions the mean value theorem is key: Let $f \colon [a, b] \to \mathbb R$ be differentiable. Then the mean value theorem implies $$\lvert f(x) - f(y) \rvert \leq \sup_{\xi \in [a, b]} \lvert f'(\xi)\rvert \cdot \lvert x - y \rvert$$ for all $x,y \in [a, b]$. So just calculate the derivate of $f'$ and calculate its local maxima and minima. Since $f'(x) \to 0$ as $x \to \pm \infty$, you will see that those are actually global extrema. And then you just need to show that their absolute values with respect to $f$ are sufficiently small, i.e., $< 1$.

Answer 2

For $x\ne y$ we have $\dfrac {|f(x)-f(y)|}{|x-y|}=\dfrac {x+y}{(1+x^2)(1+y^2)}\le\dfrac 4 5.$ Proof:

We have $x^2-x+1/2=(x-1/2)^2+1/4\ge 1/4$, and similarly $y^1-y+1/2\ge 1/4.$ Therefore $$(1+x^2)(1+y^2)\ge 1+x^2+y^2=$$ $$=(x^2-x+1/2)+(y^2-y+1/2)+(x+y)\ge$$ $$\ge 1/2+(x+y)\ge$$ $$\ge (x+y)/4+(x+y)=\frac 5 4 (x+y).$$