Functional equation $f(x-F(x))=f(x)$

Question

Let $f:[0,1]\rightarrow(0,1]$ be a continuous function and $F$ its primitive such that $F(0)=0$. Find $f$ so that $$f(x-F(x))=f(x), \forall x \in [0,1]$$ Let $F:[0,1]\rightarrow\mathbb{R}$, $F(x)=\int_0^xf(t)dt$, so we have $F'(x)=f(x), \forall x \in [0,1]$. I have managed to show that $x-F(x) \in [0,1], \forall x \in [0,1]$. Does the continuity of $f$ imply that it is constant?

Answer 1

Since $f$ is continuous, we find that $\delta = \min\{f(x) : x\in[0,1] \} > 0$ by the assumption on $f$. So if we define $\phi(x) = x - F(x)$, then

$$ 0 \leq \phi(x) = \int_{0}^{x} (1-f(t)) \, dt \leq (1-\delta)x$$

and thus $0 \leq \phi^{\circ n}(x) \leq (1-\delta)^n x \to 0$ as $n\to\infty$. So

$$ f(x) = f(\phi(x)) = \cdots = f(\phi^{\circ n}(x)) \to f(0) $$

as $n\to\infty$ and therefore $f$ is a constant function.