I previously posted this question on Math.SE but didn't receive an answer. It is perhaps a little vague; part of what I want to know is what question I should ask.
First, consider the following form of the Cauchy-Schwarz inequality: let $V$ be a real vector space, and suppose $(\cdot, \cdot) : V \times V \to \mathbb{R}$ is a symmetric bilinear form which is positive semidefinite, that is, $(x,x) \ge 0$ for all $x$. Then for any $x,y \in V$ we have $|(x,y)|^2 \le (x,x) (y,y)$.
I'd like to know what happens if we replace $\mathbb{R}$ by some other space $W$. Suppose at first that $W$ is a real vector space, equipped with a partial order $\le$ that makes it an an ordered vector space, as well as a multiplication operation $\cdot$ that makes it an algebra. Then it makes sense to speak of a positive semidefinite symmetric bilinear form $(\cdot, \cdot) : V \times V \to W$, and ask whether it satisfies the Cauchy-Schwarz inequality $(v,w)\cdot(v,w) \le (v,v) \cdot (w,w)$.
Under what conditions on $W$ does this "generalized Cauchy-Schwarz inequality" hold?
At a minimum I expect we will need some more structure on $W$; in particular I assume we would like the multiplication and the partial ordering in $W$ to interact in some reasonable way, so that for instance $w\cdot w \ge 0$ for all $w \in W$. Are there other properties that $W$ should have?
There are lots of proofs of the classical Cauchy-Schwarz inequality; presumably one should try to find one of them which generalizes. But I couldn't immediately see how to do this.
My motivating example is the quadratic variation form from probability. For instance, we could take $V$ to be the vector space of continuous $L^2$ martingales on some filtered probability space over some time interval $[0,T]$, and $W$ to be the vector space of continuous adapted processes of bounded variation, mod indistinguishability, with pointwise multiplication and the partial order $X \le Y$ iff $X_t \le Y_t$ for all $t$ almost surely. Then the quadratic variation $\langle M,N \rangle$ is a symmetric positive semidefinite bilinear form from $V \times V$ to $W$.
In this case I can prove the Cauchy-Schwarz inequality pointwise: fix $M,N \in V$. For almost every $\omega$, for all $t \in [0,T]$ and all $q \in \mathbb{Q}$ I can say
$$q^2 \langle M,M \rangle_t(\omega) \pm 2 \langle M,N \rangle_t(\omega) + \frac{1}{q^2} \langle N,N \rangle_t(\omega) = \langle q M \pm \frac{1}{q} N \rangle_t(\omega) \ge 0$$
and then letting $q$ be a rational very close to $\sqrt{\langle N,N \rangle_t(\omega) / \langle M,M \rangle_t(\omega)}$ shows that $$|\langle M,N \rangle_t(\omega)| \le \sqrt{\langle M,M \rangle_t(\omega) \langle N,N \rangle_t(\omega)}$$
which is what we want. But I have used in an essential way the fact that $W$ is a function space, and it would be nice to see if this can be avoided.
Best Answer
I'm not sure if this is the direction you are after, but there is a beautiful generalisation of the Cauchy-Schwarz inequality to C*-algebras, called the Kadison-Schwarz inequality. Instead of vector spaces $V$ and $\mathbb{R}$, you take C*-algebras $A$ and $B$. The positive semidefiniteness of the bilinear form translates into positivity of a map $A \to B$. That is, it has to send positive elements to positive elements (which are those of the form $a^\ast a$ for $a \in A$). If you like, you can restrict to self-adjoint ($a^\ast=a$) elements, which form an ordered vector space, and a so-called Jordan algebra. The beautiful thing is that the inequality still holds for noncommutative C*-algebras.