# Cauchy-Schwarz for Multiple Integrals

Is there a generalization of the Cauchy-Schwarz Inequality for multiple integrals?

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You mean like $|\iint f\overline g|\leq\sqrt{\iint|f|^2}\sqrt{\iint|g|^2}$ on subsets of $\mathbb{R}^2$, and higher dimensional analogues? If so, then yes, these are all special cases of the Cauchy-Schwarz inequality: en.wikipedia.org/wiki/Cauchy-schwarz – Jonas Meyer Apr 4 '11 at 0:12

According to Michael Steele, one generalization is the following for double integrals: $S \subset \mathbb{R}^2, f: S \to \mathbb{R}$ and $g: S \to \mathbb{R}$, then $$A = \iint_{S} f^{2} \ dx \ dy, \ B = \iint_{S} fg \ dx \ dy, \ C = \iint_{S} g^{2} \ dx \ dy$$ satisfy $|B| \leq \sqrt{A} \cdot \sqrt{C}$.
There is an exact generalisation: for any space of functions over a measure space $(X,\Sigma,\mu)$ where $\int_X \lvert u \rvert^2 \, d\mu = 0$ if and only if $u =0$, then if we write $\langle u,v \rangle = \int_X \bar{u}v \, d\mu$, we have $$\langle u,v \rangle^2 \leqslant \langle u,u \rangle \langle v,v \rangle,$$ with equality if and only if $u=kv$ for some constant $k$.
If $\langle u,u \rangle=0$ or $\langle v,v \rangle=0$, this is clear. Suppose now neither is zero. Then $$\frac{\lvert \langle u,v \rangle \rvert}{\sqrt{\langle u,u \rangle \langle v,v \rangle}} = \left\lvert \int_X \frac{\bar{u}}{\sqrt{\langle u,u \rangle}} \frac{v}{\sqrt{\langle v,v \rangle}} \right\rvert \, d\mu \leqslant \left\lvert \frac{1}{2}\int_X \left( \frac{\lvert u \rvert^2}{\langle u,u \rangle} + \frac{\lvert v \rvert^2}{\langle v,v \rangle} \right) \right\rvert \, d\mu = 1,$$ using the elementary AM-GM inequality $2\lvert ab \rvert \leqslant \lvert a \rvert^2+ \lvert b \rvert^2$. It is clear that equality holds precisely when the functions are proportional, by the conditions of the AM-GM inequality.
A more direct way in this case is to note $$0 \leqslant \frac{1}{2}\iint \lvert u(x)v(y)-u(y)v(x) \rvert^2 \, d\mu(x) \, d\mu(y) = \langle u,u \rangle \langle v,v \rangle-\lvert \langle u,u \rangle\rvert^2,$$ obviously with equality only if $u(x)v(y)-u(y)v(x) = 0$, so $u(x)=kv(x)$.