How can I prove that $g$ is a contraction function

Question

Let we have the following function $$h(x)=\frac{1}{1+x^2}$$ where $x \in [0,1] $

How can I prove that $h$ is a contraction function

Answer 1

By the MVT, one has $$ |h(x)-h(y)|=\frac{2c}{(1+c^2)^2}|x-y|, c\text{ is between $x$ and $y$}. $$ Let $$ f(x)=\frac{2x}{(1+x^2)^2}, x\in[0,1]. $$ Then $$ f'(x)=\frac{2(1-3x^2)}{(1+x^2)^3}, f''(x)=\frac{24x(x^2-1)}{(1+x^2)^4}. $$ Letting $f'(x)=0$ gives $x=\frac{1}{\sqrt3}$. Since $f''(\frac{1}{\sqrt3})<0$, $f(x)$ reaches the max $f(\frac{1}{\sqrt3})=\frac{3\sqrt3}8$. So $$ |h(x)-h(y)|\le\frac{3\sqrt3}{8}|x-y|, x,y\in[0,1]. $$ Since $\frac{3\sqrt3}{8}<1$, the map is a contraction.

Answer 2

You need to prove that $|h(x) - h(y)| < |x - y|$ for all $x, y, \in [0,1]$. Let's compute directly: \begin{eqnarray} |h(x) - h(y)| &=& \left| \frac{1}{1 + x^2} - \frac{1}{1+y^2}\right| \\ &=&\left| \frac{1 + y^2 - (1 + x^2)}{(1+x^2)(1+y^2)}\right| \\ &=& \left| \frac{y^2 - x^2}{(1+x^2)(1+y^2)}\right| \\ &=&\left| \frac{(y + x)(y - x)}{(1+x^2)(1+y^2)}\right| \\ &=&|x - y|\cdot \left| \frac{y + x}{(1+x^2)(1+y^2)}\right| \\ \end{eqnarray}

So we are done if we can prove that $ \left| \dfrac{y + x}{(1+x^2)(1+y^2)}\right| < 1$ for all $x, y \in [0, 1]$. This claim is equivalent to the claim that the function $f(x, y) = 1 - x - y + x^2 + y^2 + x^2y^2$ is greater than $0$ on the rectangle $R = [0,1] \times [0,1]$. This can be seen in a variety of ways, the simplest of which might be to observe that $f$ has no local extrema on the interior of the rectangle, and $f$ is always greater than $0$ on the boundary of the rectangle.