Let $f : [0, \infty) \rightarrow \mathbb{R}$ be a function such that $|f(x) - f(y)| \leq |x - y|, \forall x, y \geq 0$, and let $F$ be one of its antiderivatives such that $F(0) = 0$.
a) Prove that $|yF(x) - xF(y)| \leq \frac{xy|x-y|}{2}, \forall x, y \geq 0.$
b) Knowing that $f(x) \geq 0, \forall x \geq 0$, how many solutions does the equation $F(x) = x^2$ have?
My initial thought was observing that the function is Lipschitz with $L = 1$, thus continuous, thus Riemann integrable. Because we are told that $F(0) = 0$, I wanted to consider $F(x) = \int_0^x f(t)dt$ and then $F(x) - F(y) = \int_y^x f(t)dt$, but since we need $|yF(x) - xF(y)|$, I tried both integrating $\int_y^x tf(t)dt$ and seeing where that gets me, as well as integrating the first inequality with respect to x, using the fact that $|\int_a^b f(t)dt| \leq \int_a^b |f(t)|dt$, but to no avail.
For b), we know that $F'(x) = f(x) \geq 0$, so then $F$ is an increasing function. I believe we have to show somehow that $x_0 = 0$ is the only solution to this equation using a), but I am missing something.
By considering $f(x)-f(0)$ we can reduce the proof to the case$f(0)=0$.
Now $|yF(x)-xF(y)| \leq y|F(x)-F(y)|+|x-y||F(y)|$. Let $x<y$ and observe that $|f(x)|\leq x$ by the hypothesis and the condition $f(0)=0$. Thus, $y|F(x)-F(y)| \leq y|\int_x^{y} f(t)dt|\leq y\frac {y^{2}-x^{2}} 2$. A similar estoimation of he second term gives part a).
The derivative of $F(x)-x^{2}$ is $f(x)-2x$. Note that $|f(x)| \leq x<2x$ (for $x>0$) so $F(x)-x^{2}$ is strictly increasing. Since it vanishes at $0$ we see that $0$ is the only point where it vanishes.