# Arithmetic mean of $L^2$ function is $L^2$

I have found the following problem, to which I do not find the solution:

Consider $f(x), x > 0$ a function such as $$\int_0^\infty f^2(x) dx < \infty$$ and let $g(x) = \frac 1x \int_0^x f(t)dt$. Then prove that $$\int_0^\infty g^2(x) dx < \infty$$

Directly using the Cauchy Schwarz inequality does not seem to lead anywhere.

By Cauchy-Schwarz we obtain $|\int _0^x f(t)dt |\leq x^{1/2}\int _0^xf(t)^2dt=o_{x\to 0}(x^{1/2})$.
By using integration by parts : $\int _0^\infty \frac{1}{x^2}(\int _0^xf(t)dt)^2dx=2\int _0^\infty \frac{1}{x}f(x)\int _0^xf(t)dtdx-[\frac{1}{x}(\int _0^xf(t)dt )^2]_0^\infty$. But $[\frac{1}{x}(\int _0^xf(t)dt )^2]_0^\infty =0$ by Cauchy-Schwarz.
Now observe that $|\int _0^T \frac{1}{x}f(x)\int _0^xf(t)dtdx| \leq (\int f^2\int g^2)^{1/2}$. Then $\int g^2 \leq 2 (\int f^2\int g^2)^{1/2}$. If $\int g^2\neq 0$ you can divise both side by $(\int g^2)^{1/2}$ and obtain $(\int g^2)^{1/2}\leq 2 (\int f^2)^{1/2}$. Then let tend $T \to \infty$.
• It's true when $x\to 0$ and that's all we need. – Patissot Apr 17 '15 at 19:02