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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.

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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 $T \to \infty$.

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  • $\begingroup$ The little 'o' in your first line should be a big 'O'. $\endgroup$ – Frank Apr 17 '15 at 18:53
  • $\begingroup$ It's true when $x\to 0$ and that's all we need. $\endgroup$ – Patissot Apr 17 '15 at 19:02

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