# Deriving Cauchy's inequality with Fubini's theorem

The exercise is the following:

Let $f,g: X \to \mathbb K$ be two measurable functions such that $|f|^2, |g|^2 \in \mathcal L^1$. Making use of Fubini's theorem and by considering the function $$(x,y) \mapsto |f(x)g(x) f(y) g(y)|$$ derive Cauchy's inequality: $$\left(\int_X |fg|\, d\mu \right)^2 \le \left(\int_X |f|^2 \, d\mu \right)\left(\int_X |g|^2\, d\mu\right)$$

I can prove that $\int_{X\times X} |f(x)g(x)f(y)g(y)| \, d(\mu\otimes \mu)= \left(\int_X |fg|\, d\mu\right)^2$ by an application of Fubini with $A = \{f\ne 0\}\cup \{g\ne 0\}$, which is $\sigma$-finite, since it can be written as a union $$A = \bigcup_{n = 1}^\infty (\{|f|^2>1/n\}\cup\{|g|^2\ge 1/n\})$$

where all sets on the RHS have to have finite measure, since $f,g \in L^2$. So

\begin{align*} \int_{X\times X} |f(x)g(x)f(y)g(y)| \, d(\mu\otimes \mu) &= \int_{A\times A} |f(x)g(x)f(y)g(y)| \, d(\mu\otimes \mu) \\ &= \left(\int_A |fg|\, d\mu\right)^2 \\ &= \left(\int_X |fg|\, d\mu\right)^2 \end{align*}

But I don't really see how to prove $$\int_{X\times X} |f(x)g(x)f(y)g(y)| \, d(\mu\otimes \mu)\le \left(\int_X |f|^2 \, d\mu\right)\left( \int_X |g|^2\, d\mu\right)$$ without making use of Young's inequality (or AM-GM). I think one should be able to see this last inequality directly somehow.

Why I don't want to use AM-GM: The derivation in this exercise should probably be an alternative to the usual one, where one integrates $$\frac{|fg|}{\Vert f\Vert_2 \Vert g\Vert_2} \le \frac12 \left(\frac{|f|^2}{\Vert f\Vert_2^2}+\frac{|g|^2}{\Vert g \Vert_2^2}\right)$$

Some help would be very much appreciated, thanks! =)

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Note that the inequality $(a-b)^2 \ge 0$ used in the first line is really a special case of the AM-GM inequality. – Nate Eldredge Feb 25 '12 at 16:38