Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question is perhaps a little vague; part of what I want to know is what question I should ask.

First, recall 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 $(v,w)\cdot(v,w) \le (v,v) \cdot (v,w)$.

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.


Here are some motivating examples.

As a fairly simple one, let $X$ be any set, and $W = \mathbb{R}^X$ the vector space of all real-valued functions on $X$. We can equip $W$ with the pointwise multiplication and ordering. Then let $V$ be any linear subspace of $W$, and let the bilinear form $V \times V \to W$ also be pointwise multiplication. Then of course Cauchy-Schwarz holds since we can just prove it pointwise.

For a slightly less trivial example, let $(X,\mu)$ be a measure space, and $W = L^0(X,\mu)$ be the vector space of all measurable functions on $X$, mod $\mu$-almost-everywhere equality (so an element of $W$ is in fact an equivalence class of functions). Again let $\cdot$ be pointwise multiplication (which is well defined), and the ordering $f \le g$ when $f(x) \le g(x)$ almost everywhere. Take again a linear subspace $V \subset W$, and pointwise multiplication as the bilinear form. Now Cauchy-Schwarz holds because we can prove it pointwise on a set of full measure.

A related but more complicated example from probability (and my original motivation) 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 $W$ 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.

In each of these examples, we are working on function spaces (or quotients thereof), and the proof essentially operates pointwise. I'm hoping for some kind of more abstract global argument.

share|improve this question
add comment

1 Answer

I think the space $W$ should be defined a partial order $\leq$ and zero element $0$ firstly and satisfy:

  1. If $a\leq b$ then $ca\leq cb$, $\forall a,b\in W$ and $0\leq c\in W$.
  2. If $a\leq b$ then $a-b\leq 0$.

Secondly, a multiply operator $\cdot$ should be defined in $W$ and satisfy $0\leq a\cdot a\stackrel{\triangle}{=}a^2$ for $\forall a\in W$. Also, the inverse operator of $\cdot$ should be defined in $W$ (Alternatively, the inverse element is defined in $W$). That is, if $ab=c$ then $c\stackrel{\triangle}{=}a/b$ for $\forall a,b,c\in W$ and $b\neq 0$ where $/$is the inverse operator of $\cdot$. What is more, these operators should be closed in $W$. Say, if $\forall a,b\in W$ then $a\cdot b\in W$ and $a/b\in W$ if $b\neq 0$. Finally, the operators $\cdot$ and /should satisfy commutative law.

Thirdly, there should have a multiply operator between the elements from $W$ and $V$ because we will define inner product by using this operator. What is more, to hold the Cauchy-Schwarz inequality, the properties of inner product is important. I believe the Cauchy-Schwarz inequality is valid in a space which define a inner product whose definition is classical. In another word, if a space $V$ have been defined a inner product $(*,*)$ (say, a bilinear form that $V\times V\rightarrow W$) satisfy the following conditions:

  1. Commutative: $(x,y)=(y,x)$, $\forall x,y\in V$ (If V is a complex space, the right hand side should be dual. But for the sake of simplicity, we ignore it here.)
  2. Linearity: $(\alpha x+\beta y, z)=\alpha(x,z)+\beta(y,z)$, $\forall x,y,z\in V$ and $\alpha,\beta\in W$.
  3. Positive define: $(x,x)\geq0$, $\forall x\in V$. The equal sign is valid iff $x=0$ is valid where $0$ donate the zero element in $W$.

Then, by this definition, the Cauchy-Schwarz inequality is valid. The proof are as follow:

For $\forall\lambda\in W$ and $\forall x,y \in V$, we have: \begin{equation} 0\leq (x+\lambda y,x+\lambda y)=(x,x)+2\lambda(x,y)+\lambda^2(y,y) \end{equation} If $y=0$, that is a trivial case and Cauchy-Schwarz inequality is valid obviously. If $y\neq 0$, let $\lambda=-(x,y)/(y,y)$ then we have: \begin{equation} 0\leq(x,x)-2(x,y)^2/(y,y)+(x,y)^2/(y,y)^2(y,y)\\ (x,y)^2\leq (x,x)(y,y) \end{equation} This is the Cauchy-Schwarz inequality.

In fact, Cauchy-Schwarz inequality imply that the inner product of two elements is less than the their product of length because there is an angle between them. And $W$ is a space to measure the inner product of $V$. So I think the conditions I assume at start is reasonable.

share|improve this answer
    
Thank you very much for your answer. I edited my question to add some more (simpler) examples to better explain what I'm looking for, and in light of them it seems that your assumptions are rather onerous. For instance, in my first example, the pointwise multiplication on $W = \mathbb{R}^X$ doesn't admit an inverse (because a function can be zero at some points), and we can't expect that it makes sense to multiply $wv$ where $w \in W$ and $v \in V$ (since $V$ could be a very small subspace of $W$). –  Nate Eldredge Feb 6 at 22:23
    
I am a few confused. I can not get the meaning of 'pointwise multiplication' you mentioned. Do you refer to the multiply two functions at same point? I think both inverse and the multiplication between $V$ and $W$ is reasonable even in your example. For one thing, the inverse element (alternatively, division operator) should be define on some nontrivial subspace on $W$, say nonzero. Just like in $R$, as we all know, $1/0$ is make no sense. –  Lion Feb 7 at 10:29
    
I think the inverse element is necessary because if there are multiplication and unit element '1' on $W$, and the problem $a\cdot b=1$ is reasonable if $0\neq a,b\in W$. What is more, the inverse element have been used both in my and your proof. If we can avoid it, I think we can cancel the condition of inverse element. For another, the multiplication between $W$ and $V$ is also reasonable. Because, if cancel it, how to define the linearity of inner product? Obviously, define it by using complex numbers is unreasonable because the measure of $V$ do not define on $\mathbb{C}$ while in $W$. –  Lion Feb 7 at 10:38
    
Personally, it is a new problem that what construction on $W$ which used to measure $V$ when it is not a number set and I am very interesting on in. But I think Cauchy-Schwarz inequality describe a property of a well defined inner product on $V$. So the validation of the inequality is largely depend on the definition of the inner product. Maybe part of my answer have something wrong and I am very glad to find the correct answer together. –  Lion Feb 7 at 10:45
1  
I agree with @NateEldredge. His pointwise multiplication means that $f.g$ is a function from $X$ to $\mathbb{R}$ maps $x$ to $f(x).g(x)$. This operator doesn't admit an inverse. –  Du Phan Feb 8 at 19:46
show 19 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.