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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.

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I suspect you may be describing a pre-Hilbert module, where the essential assumptions are that $W$ is a pre-C*-algebra and $V$ is a $W$-module with a $W$-valued inner product. However, I'm embarrassed to say that I'm not familiar enough with probabilistic terminology to understand your example. Would it be hard to restate the example in straight functional-analytic terms? –  Nik Weaver Jun 17 '13 at 16:32
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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.

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