Existence of solution of $f'(x) = f(\phi(x))$ Let $\phi:[0,1]\to[0,1]$ be a none constant and continuous map. Prove that there exists a unique function $f\in C^1([0,1], \Bbb R)$ satisfying 
$$f'(x) = f(\phi(x))~~~f(0)=\alpha$$
where $\alpha\in\Bbb R$ is given. I attempted to use the Cauchy-Lipschitz Theorem but I couldn't go further. Any Hint?
 A: I think you can use the contraction principle.
Let us consider the map $T\colon C([0,1]) \to C([0,1])$ defined by
$$
Tf (x) := \alpha + \int_0^x f(\phi(s))\, ds, \qquad x\in [0,1].
$$
Here $C([0,1])$ is equipped with the uniform distance $d(f,g) := \max_{x\in [0,1]} |f(x) - g(x)|$, so that it is a complete metric space.
The original problem for $f$ is then equivalent to prove that $T$ admits a unique fixed point, i.e., $f\in C^1([0,1])$ is a solution to $f'(x) = f(\phi(x))$, $f(0) = \alpha$ if and only if $f\in C([0,1])$ is a fixed point of $T$.
If $f,g\in C([0,1])$, a simple estimate gives
$$
|Tf(x) - Tg(x)| \leq \int_0^x |f(\phi(s)) - g(\phi(s))|\, ds
\leq \|f-g\|_\infty\, x.
$$
Unfortunately the map $T$ is not a contraction, since the estimate above gives
$\|Tf - Tg\|_\infty \leq \|f-g\|_\infty$.
On the other hand, it is not difficult to prove that $T^2 := T\circ T$ is a contraction, since, by the previous estimate,
$$
|T^2 f(x) - T^2 g(x)| \leq \int_0^x |Tf(\phi(s) - Tg(\phi(s))|\, ds
\leq \int_0^x \|f - g\|_\infty \, \phi(s) \, ds
$$
so that
$$
\|T^2 f - T^2 g\|_\infty \leq L \|f - g\|_\infty,
\qquad \text{with}\ L := \int_0^1 \phi(s)\, ds.
$$
Since $\phi$ is continuous, not constant, and $0\leq\phi\leq 1$, we clearly have that $L < 1$, so that $T^2$ is a contraction and hence $T$ admits a unique fixed point.
