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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    $\begingroup$ Since $V$ must be torsion free, why not tensor up to $\mathbb{Q}$? $\endgroup$ – Arturo Magidin Apr 15 '12 at 0:28
  • $\begingroup$ @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 $\mathbb{Q}$ ? $\endgroup$ – Klaus Apr 15 '12 at 0:34
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    $\begingroup$ He means that since you have no torsion you may as well exchange $\mathbb{Z}^n$ for $\mathbb{Q}^n$. $\endgroup$ – Alex Youcis Apr 15 '12 at 0:39
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    $\begingroup$ @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. $\endgroup$ – 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.

  • $\begingroup$ How can you use the same proof if $X\in\mathbb{Z}$ ? There could be two roots $x_1<x_2$ such that $[x_1,x_2]\cap\mathbb{Z}=\emptyset$. $\endgroup$ – Klaus Apr 15 '12 at 0:59
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    $\begingroup$ If you consider f(Xv + Yw, Xv + Yw) you get a homogeneous non-negative quadratic form on ZxZ. It could be argued that finishing the argument from this point is the same tensor product with Q, but the extension of scalars is for the more concrete space ZxZ and not the Z-module given in the problem. $\endgroup$ – zyx Apr 15 '12 at 1:09
  • $\begingroup$ 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 ?] $\endgroup$ – Klaus Apr 15 '12 at 1:10
  • $\begingroup$ 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. $\endgroup$ – zyx Apr 15 '12 at 1:47

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