# How to prove Godunova's inequality?

Let $\phi$ be a positive and convex function on $(0,\infty)$. Then $$\int_0^\infty \phi\left(\frac{1}{x}\int_0^x g(t)\,dt\right)\frac{dx}{x} \leq \int_0^\infty \phi(g(x))\frac{dx}{x}$$

The application of this inequality is this :

$(1)$ Hardy's inequality.

With $\phi(u)=u^p$, we obtain that $$\int_0^\infty \left(\frac{1}{x} \int_0^x g(t) \, dt\right)^p\frac{dx}{x} \leq \int_0^\infty g^p(x)\frac{dx}{x}$$ (2) Polya-Knopp's inequality

By using it with $\phi(u)=u^p$, replacing $g(x)$ by $\log g(x)$ and making the substitution $h(x)=\frac{g(x)}{x}$ we obtain that

$$\int_0^\infty \exp\left(\frac{1}{x} \int_0^x \log h(t)\right) \, dt \leq e\int_0^\infty h(x) \, dx$$

How to prove Godunova's inequality? Is there any reference?

• If you're fluent in Russian, you can read the details just from the original article www.mathnet.ru/eng/rm6128. It is a combination of the Jensen inequality with the Fubini theorem. Jul 5 '14 at 15:17
• @Jack D'Aurizio Thank you! I'am not fluent in Russian, but I have a good friend, Google Translator, who is fluent in Russian. It is just the combination of the Jensen and the Fubini. Jul 5 '14 at 16:22

This is just an explicit execution (as a community wiki answer) of the hint given in the comments:

By Jensen's inequality, we get (because $(0,x)$ with the measure $\frac{dt}{x}$ is a probability space)

\begin{eqnarray*} \int_{0}^{\infty}\phi\left(\int_{0}^{x}g\left(t\right)\frac{dt}{x}\right)\frac{dx}{x} & \leq & \int_{0}^{\infty}\int_{0}^{x}\phi\left(g\left(t\right)\right)\frac{dt}{x}\,\frac{dx}{x}\\ & \overset{\text{Fubini}}{=} & \int_{0}^{\infty}\phi\left(g\left(t\right)\right)\,\int_{t}^{\infty}x^{-2}\, dx\, dt\\ & = & \int_{0}^{\infty}\phi\left(g\left(t\right)\right)\,\frac{dt}{t} \end{eqnarray*}

The application of Fubini's theorem is legitimate as the integrand is non-negative.