How prove this integral equation has a unique solution Question:
Consider the equation 
$$3u(x)=x+(u(x))^2+\int_{0}^{1}|x-u(y)|^{1/2}dy$$
show that it has a continous solution $u$ satisfying $0\le u(x)\le 1$ for $0\le  x\le 1$
this problem From page 376 ,he says 
Hint: Krasnoselskii’s ﬁxed point theorem
My idea: 
let $f(x_{1}),f(x_{2})$ is solution,then $f=f_{1}(x)-f_{2}(x)$
I
Thank you

It can be written equivalently as
$$
u(x)=\frac{3}{2}-\left(\frac{9}{4}-x-\int_{0}^{1}\vert x-u(y)\rvert^{1/2}dy\right)^{1/2}=:\big({\mathcal F}(u)\big)(x)
$$
Clearly, the nonlinear functional maps the closed unit ball of $C[0,1]$ to itself, and in fact to a compact subset of it (use Arzelà–Ascoli theorem), and due to Schauder Fixed Point Theorem it has a fixed point.
