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Cauchy-Schwarz inequality for inner products

If $V$ is a real vector space and $f: V\times V\to \mathbb{R}$ is a symmetric bilinear positive map, then we have the Cauchy-Schwarz inequality $f(v,w)^2\le f(v,v)f(w,w)\text{ for all }v,w\in V,$ which is proved for example by examining the discriminant of the quadratic function $f(Xv+w,Xv+w)=f(v,v)X^2+2f(v,w)X+f(w,w).$

A generalization ?

Now let $V$ is a $\mathbb{Z}$-module and $f: V\times V\to \mathbb{R}$ a symmetric bilinear positive function such that $f(nv,mw)=nmf(v,w)$ for $v,w\in V$ and $n,m\in\mathbb{Z}$.

The question is: Do we still have a Cauchy-Schwarz inequality on $f$ ?

The idea of the proof above can be used to prove that $f(v,w)^2\le f(v,v)(f(v,v)/4+f(w,w))$, but we can't seem to do better with this idea since $\mathbb{Z}$ itself is not a field.

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    @Klau: View $\mathbb{Q}$ as a $\mathbb{Z}$-module, and consider $W=V\otimes_{\mathbb{Z}}\mathbb{Q}$. Then $W$ has a natural structure as a $\mathbb{Q}$-module (i.e. a vector space). Essentially, you embed $V$ into a $\mathbb{Q}$-vector space and work there.2012-04-15

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The same proof gives $4 \times$ (Cauchy Schwarz inequality). Because the values of $f$ considered in the inequality are real numbers, not elements of the $Z$-module, division by $4$ is possible and the factor of $4$ can be removed.

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    On second thought, no extension to Q is ever required. If you diagonalize the non-negative quadratic form $aX^2 + bXY + cY^2$ by "completing the square" the result is $a(X+ {b/{2a}}Y)^2 - ${(b^2 - 4ac)/4a} and this calculation can be done over Z if one multiplies by $4a$. It is easy to see that $a$ and $b$ are positive. So the exact statement is that the usual proof, avoiding denominators, says (4a)x(Cauchy-Schwarz) is positive in R where one can cancel the $4a$. It could be said that any use of homogeneity is the same as extension to Q but no reference to tensor product is needed.2012-04-15