An asymptotic integral inequality

Question

Suppose $f:\mathbb{R}\to\mathbb{R}$ is a continuous function, $g(x)=xf(x)-\int_0^xf(t)\ dt$, and we have $f(0)=0$ and $g(x)=O(x^2)$ as $x\to0$.

Is it true that $f(x)=O(x)$ as $x\to0$ ?

Answer 1

Since people seem to care a little, answering myself here, positively.

First of all, it is easy to see that, $g$ being given, there exists only one $f$ that satisfies the condition. If there were two such functions, say $f_1$ and $f_2$, and $\varphi(x)=\int_0^xf_1(t)-f_2(t)\ dt$, then $x\varphi'(x)=\varphi(x)$, hence $\varphi(x)=Cx$, and $f_1-f_2$ is constant; and since $f_1(0)=f_2(0)=0$, we have $f_1=f_2$.

Then we need to notice that such $f$ can be given by this formula:

$$f(x)=\frac{g(x)}{x}+\int_0^x\frac{g(t)}{t^2}\ dt.$$

Even though the easiest way to obtain this is to integrate by parts in $f(x)=\int_0^x\frac{g'(t)}{t}\,dt$, this formula can also be easily verified for non-smooth functions directly. Also, Riemann integral is sufficient here, since $g(t)/t^2$ is bounded and continuous almost everywhere.

Having this representation of $f$ in terms of $g$, we easily see that $f(x)=O(x)$ .