# Suppose $f$ is continuous and verifies $f(x) = o(x)$. Does it follow that $\int_{0}^xf(y)\,dy = o(x^2)$?

I know $F(x) = \int_{0}^xf(y) dy$ means $F^\prime(x) = f(x)$, but I have no ideas how to relate it to little-oh test.

The only method in my mind is to find example $f(y)$ and $F(x)$, but I don't think that is the right approach.

Any suggestions?

• The notation $\int_{0}^x f(x)\,dx$ should be avoided. Use different variables: $$\int_{0}^x f(t)\,dt.$$ Jan 12, 2016 at 1:28
• If you know l'Hopital's rule you could use it for $\lim F(x) / x^2$. Also note that $o(x^2)$ may not necessarily be the best constraint in all cases. For example, while $f(x)=sin(x)$ is $o(x)$ its antiderivative $F(x)$ is also $o(x)$.
– dxiv
Jan 12, 2016 at 1:57

More generally, if $f$ is Riemann (or Lebesgue) integrable on some $[0,a], a >0,$ and $f(x) = o(x)$ as $x\to 0,$ then $\int_0^x f = o(x^2).$ Proof: Let $\epsilon > 0.$ Then there is $x_0>0$ such that $0<x<x_0$ implies $|f(x)/x| < \epsilon.$ Thus, for the same $x,$
$$|\int_0^x f| \le \int_0^x (|f(t)|/t)t\,dt \le \epsilon \int_0^x t\, dt = \epsilon(x^2/2).$$
This gives the $o(x^2)$ estimate.
Yes: By the mean theorem since $F$ is continue and its derivative is $f$, $F(x)-F(0)=f(c_x)(x-0)$ where $c_x\in [0,x]$ $F(x)=xf(c_x)$ where $c_x\in [0,x]$. $xf(c_x)$ is $o(x^2)$ since $\mid xf(c_x)\mid\leq \mid x\mid \mid o(x)\mid$.
• Sorry, could you please explain a little bit more? What does $h_x$ mean in this case. Is it a constant? Jan 12, 2016 at 1:21