Proving that the mapping is a contraction

Question

Prove that mapping $A: C[0,1]\to C[0,1]$, $$ Ax(t) = \frac{1}{4}(t + \int_{0}^{t} \sqrt{1 + x^2(s)} \, {d}s) $$ is contraction.

My solution. Let $x, y \in C[0, 1]$, then \begin{align*} \rho(Ax, Ay) &= \max_{t \in [0,1]}\left|Ax(t)-Ay(t)\right| =\\ &= \frac{1}{4}\max_{t \in [0,1]}\left|\int_{0}^{t}\sqrt{1 + x^2(s)} \, {d}s -\int_{0}^{t}\sqrt{1 + y^2(s)} \, {d}s\right| = \\ &= \frac{1}{4}\max_{t \in [0,1]}\left|\int_{0}^{t}\Big(\sqrt{1 + x^2(s)} -\sqrt{1 + y^2(s)} \, \Big) \, {d}s\right| \le \\ &\le \frac{1}{4}\max_{t \in [0,1]}\int_{0}^{t}\left|\Big(\sqrt{1 + x^2(s)} -\sqrt{1 + y^2(s)} \, \Big)\right|{d}s \end{align*}.

How to show that the last inequality is bounded above by $\frac{1}{4}\rho(x, y)$?

Answer 1

Consider the function $$f(x) = \sqrt{1+x^2}, \,\,\,\,\, x \in \mathbb R.$$ We note that $$\lvert f'(x) \rvert = \frac{\lvert x\rvert}{\sqrt{1+x^2}} \le 1, \,\,\,\,\, x \in \mathbb R.$$ Now for any $x,y \in \mathbb R\,\,\, (x \neq y)$, by the mean value theorem there is $c$ between $x$ and $y$ such that $$\lvert f(x) - f(y) \rvert = \lvert f'(c) \rvert \lvert x -y \rvert \le \lvert x - y \rvert. $$ Thus $$\int_{0}^t \left \lvert \sqrt{1 + x(s)^2} - \sqrt{1+y(s)^2}\right \rvert ds \le \int^t_0 \lvert x(s) - y(s) \rvert ds \le \int^t_0 \rho(x,y)dt \le \rho(x,y).$$ Hence $$\rho(Ax,Ay) \le \frac 1 4 \rho(x,y).$$