# Cauchy-Schwarz Inequality in $\mathbb{Z}$-modules

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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Since $V$ must be torsion free, why not tensor up to $\mathbb{Q}$? – Arturo Magidin Apr 15 '12 at 0:28
@ArturoMagidin I know what is torsion free means and what a tensor product of two modules is, but I don't understand what you mean by tensoring up to $\mathbb{Q}$ ? – Klaus Apr 15 '12 at 0:34
He means that since you have no torsion you may as well exchange $\mathbb{Z}^n$ for $\mathbb{Q}^n$. – Alex Youcis Apr 15 '12 at 0:39
@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. – Arturo Magidin Apr 15 '12 at 0:49

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.
How can you use the same proof if $X\in\mathbb{Z}$ ? There could be two roots $x_1<x_2$ such that $[x_1,x_2]\cap\mathbb{Z}=\emptyset$. – Klaus Apr 15 '12 at 0:59
Ok, thank you! How did you find that this case couldn't happen at the first time ? I couldn't find a way to get rid of it. If there are two such roots, then $4f(v,w)^2-4f(v,v)f(w,w)<f(v,v)$ and $f(v,v)f(w,w)<f(v,w)^2$, and then... ? [Oh, you edited your message meanwhile. Do what you said about this case still holds ?] – Klaus Apr 15 '12 at 1:10
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. – zyx Apr 15 '12 at 1:47