Let $f(x,y)$ be a bounded function on $[0,1]\times \mathbb R$ (so that $\exists M<0$ such that $|f(x)|\le M$ for all $x\in [0,1]$). Let $S:=\{v\in C[0,1]:|v(x_2)-v(x_1)|\le \frac M2 |x_2-x_1|, \forall x_1, x_2\in [0,1], v(0)=0\}$. It is known that $S$ is a compact convex set in $(C,\|\cdot \|_\infty)$. I want to show that $$T(u):=\int_0^1 g(x,\xi)f(\xi, u(\xi))d\xi$$ is a continuous mapping of $S$ to $S$, where
$g(x,\xi)=\cases{\xi(x-1), & $0\le \xi \le x \le 1$\\x(\xi-1), & $0\le x \le \xi \le 1$}$
Let $u\in S$. We can see that, since $T(u)$ is related to the boundary value problem below, $T(u(0))=0$. We also need to show that $|T(u(x_2))-T(u(x_1))|\le \frac M2 |x_2-x_1|$ and we need to show that $T$ is continuous on $[0,1]$. But how can $T$ be continuous on $[0,1]$ if $g$ is discontinuous at $\xi$?
Moreover, $|T(u(x_1)-T(u(x_2))|=\left| \int_0^1 (g(x_1,\xi)-g(x_2,\xi))f(\xi, u(\xi))d\xi \right|\le M \int_0^1 |g(x,\xi)-g(x_2,\xi)|d\xi$. But what's next?
Also, how does one deduce from all this that $u''=f(x,u), u(0)=u(1)=0$ has a solution ($T(u)$ is the integral equation for this boundary value problem)?
I'm completely stuck. Would appreciate some help.