Integral equation and metric spaces Let $C([0,\frac{\pi }{2}])$ be the set off all continuous functions defined on $[0,\frac{\pi }{2}]$ . Prove that this integral equation $$
f(t) = \int\limits_0^{\frac{\pi }{2}} {\arctan } (\frac{{f(s)}}{2} + t)\,ds
$$ has an unique solution on $C([0,\frac{\pi }{2}])$.
Any ideas ? I just started studying metric spaces and an older peer suggested this problem, but I don't even know where to start. I'd be very grateful to anyone who could show me how such a pretty problem can be solved .

Daniel is right, Banach's fixed point theorem is the way to go.  Defining $A$ as he did, we see that $A: C([0,\pi/2]) \rightarrow C([0,\pi/2])$.  In addition, 
$||Af-Ag||_{C([0,\pi/2])} = \sup_{t \in [0,\pi/2]} \left|\int_0^{\pi/2} \arctan\left(\frac{f(s)}{2} +t\right) - \arctan\left(\frac{g(s)}{2} + t\right) ds\right|$
Since arctan is Lipschitz continuous with Lipschitz constant 1 (this follows from the fact that its derivative is uniformly bounded), we have
$||Af-Ag||_{C([0,\pi/2])} \leq \left|\int_0^{\pi/2} \frac{f(s)}{2} - \frac{g(s)}{2} ds\right| \leq \int_0^{\pi/2} \frac{1}{2} ||f - g||_{C([0,\pi/2])} ds = \frac{\pi}{4} ||f-g||_{C([0,\pi/2])}$
This shows that $A$ is a strict contraction, and thus there exists a unique fixed point, i.e. a solution to the given integral equation.
