I'm learning some applications of the Banach Fixed Point Theorem and I have the following question:
Consider the integral equation $\displaystyle x(t)=\int_{0}^{\frac{\pi}{2}}\arctan \left(\frac{x(s)}{2}+t\right)ds \ , \ \ t \in \mathbb{R}$.
(a) Show that this equation has a unique solution in $C([0,\pi/2])$.
(b)Explain why this equation also has a unique solution in $C(\mathbb{R})$?
For (a) I want to use Banach fixed point theorem. So, it is enough to show that the map $T: C([0,\pi/2]) \to C([0,\pi/2])$ with $$Tx(t)=\int_{0}^{\frac{\pi}{2}}\arctan\left(\frac{x(s)}{2}+t\right)ds$$
is a contraction. Here I define $\|Tx-Ty\|=\sup_{\substack{t\in[0,\pi/2]}}|Tx(t)-Ty(t)|$. But here I cannot take the integral. Thanks for your help!
HINT: Use the mean value theorem to show $|\arctan a - \arctan b| \leq |a-b|.$ Use this and the triangle inequality to see $$ \| Tx-Ty \| = \sup_{t\in [0,\pi/2]} \left| \int^{\pi/2}_0 \arctan\left( \frac{x(s)}{2} + t\right) - \arctan\left( \frac{y(s)}{2} + t\right) \right| ds $$
$$ \leq \sup_{t\in [0,\pi/2]} \int^{\pi/2}_0 \left| \frac{x(s)-y(s)}{2} \right| ds = \frac{ \pi}{4}\| x - y \|$$