Question about applying a combination of Arzela Ascoli and Schauder (also called Krasnoselskii) Consider the equation
\begin{align}
    3u(x) = x + u^2(x) + \int_0^1 | x - u(y) |^{1/2} \,dy \qquad (\ast)
\end{align}
Using Krasnoselskii's fixed point theorem, show that $(\ast)$ has a continuous solution $u$ satisfying $0 \leq u(x) \leq 1$ for $x \in [0,1]$.
My attempt (haven't come very far):
Krasnoselskii’s fixed point theorem states, that if F is a closed bounded convex subset of a Banach space X and $T_1$ and $T_2$ are mappings from F into X, such that

*

*$T_1(x)+T_2(x) \in F \quad \forall x,y \in F$

*$T_1$ is a contraction and

*$T_2$ is continuous and compact,

then $T_1 + T_2$ has a fixed point in $F$.
Now set $F=[0,1]$ which is obviously closed, convex and bounded. I have to rewrite the equation into
$u=T_1 u + T_2 u$ such that I get one of the operators to be a contraction and one to be compact and continuous.
Now if I write
\begin{align*}
u(x) = \frac{1}{3} \left[ x + u^2(x) + \int_0^1 | x - u(y) |^{1/2} \,dy \right]
\end{align*}
and assume $|u| \leq 1, |x| \leq 1$ I would like to choose
\begin{align*}
T_1 u = \frac{1}{3} u^2
\end{align*}
to obtain the contraction, i.e.
\begin{align*}
|T_1 u  - T_1 v|= \frac{1}{3} |u^2 - v^2| \leq \frac{2}{3} |u-v|.
\end{align*}
For $T_2$ I then obtain
\begin{align*}
T_2 u(x) = \frac{1}{3} \left[ x + \int_0^1 | x - u(y) |^{1/2} \,dy \right] \leq 1,
\end{align*}
using the Cauchy- Schwartz inequality.
Could you give me a hint how to prove compactness of $T_2$?
 A: Note, really $F=\{ w\in C[0,1]: 0\leq w\leq 1\}$, which is closed, bounded and convex.
You can put the $x/3$ term in $T_1$ since it's just a constant, so I'll only address the integral operator, which I'll call $S$ for simplicity (so $3T_2=S$). It's a nonlinear operator, so you need to justify both the continuity and compactness.
First notice that by the triangle inequality
$$
0\leq S(u)(x) \leq \int_0^1 |x+|u(y)||^{1/2}\, dy \leq \int_0^1[ x^{1/2}+u(y)^{1/2}]\, dy \leq \dfrac{2}{3} + 1=\dfrac{5}{3},
$$
where we used that $(a+b)^{1/2}\leq a^{1/2}+b^{1/2}$ for $a,b\geq 0$. This shows $T_2:F\to F$.
Continuity: Consider $u,v\in F$, then
$$
|S(u)-S(v)|(x)= \left| \int_0^1[ |x-u(y)|^{1/2}-|x-v(y)|^{1/2}]\, dy \right| \leq \int_0^1||x-u(y)|-|x-v(y)||^{1/2}\, dy,
$$
where we have used that $|a^{1/2}-b^{1/2}|\leq |a-b|^{1/2}$ for any $a,b\geq 0$.
Now by the reverse triangle inequality we can further estimate,
$$
|S(u)-S(v)|(x)\leq \int_0^1 |u(y)-v(y)|^{1/2}\, dy \leq \| u-v\|^{1/2},
$$
where $\| \cdot\|$ denotes the supremum norm on $C[0,1]$. This implies that $S$ is (Hölder) continuous.
Compactness: Here we use Arzela-Ascoli. Since $\| Su\|\leq 5/3$ for any $u\in F$, it remains to show equicontinuity; this will be similar to the previous point. Fix $x,y\in [0,1]$ and $u\in F$, then
$$
|S u(x)-Su(y)|\leq \int_0^1||x-u(z)|-|y-u(z)||^{1/2}\, dz \leq |x-y|^{1/2}.
$$
This gives the uniform equicontinuity.
