Does Cauchy-Schwarz Inequality depend on positive definiteness?

Let $V$ be a vector space over $\mathbb{R}$. Suppose we have a product $\langle \cdot,\cdot\rangle:V^2\to \mathbb{R}$ that satisfies all the inner product axioms except the second part of positive-definiteness: $$\langle x,x\rangle=0\iff x=0\tag{1}$$

So far, every proof I've seen that the Cauchy-Schwarz Inequality holds for all inner product spaces uses $(1)$. But does a proof of the Cauchy-Schwarz Inequality necessarily depend on $(1)$? Specifically, I'm looking for one of the following:

1. A proof that the Cauchy-Schwarz Inequality holds for all "inner product spaces" where $(1)$ does not necessarily hold.
2. A counterexample of a product $\langle \cdot,\cdot\rangle$ that follows symmetry, linearity in the first parameter, and $\langle u,u\rangle\ge 0$ for all $u\in V$, but where $\lvert \langle u,v\rangle\rvert\le \lvert\lvert u\rvert\rvert\ \lvert\lvert v\rvert\rvert$ does not always hold.

I suspect that there is a counterexample, but it's hard for me to come up with one.

• If $\langle .,. \rangle$ is bilinear, symmetric and semi-definite positive, then $\|x\| = \sqrt{\langle x,x\rangle}$ and $\|x\|=0 \iff x=0$ is the definition of the underlying inner product space. Dec 3 '17 at 8:24

Yes, Cauchy-Schwartz definitely depends on definiteness. It fails for indefinite forms: one can have $\left<u,u\right>=0$ but $\left<u,v\right>>0$. For instance consider $\left<(a,b),(c,d)\right>=ac-bd$ and $u=(1,1)$ and $v=(1,-1)$.

• It appears your inner product doesn't satisfy the first part of positive definiteness. I was asking about if you only get rid of the second part. In your example, $\langle (1,2), (1,2)\rangle=-3$ Dec 3 '17 at 6:21
• It satisfies "symmetry and linearity in the first parameter". Dec 3 '17 at 6:23
• Oh, sorry. I worded it correctly in the beginning but forgot about that in the list. I'll edit my question. Dec 3 '17 at 6:25

If $\|a\| = \|b\| = 0$, then \begin{align*} 0 & \leqslant \|a + b\|^2 = \|a\|^2 + \|b\|^2 + 2\langle a, b \rangle = +2\langle a, b \rangle, \\ 0 & \leqslant \|a - b\|^2 = \|a\|^2 + \|b\|^2 - 2\langle a, b \rangle = -2\langle a, b \rangle, \end{align*} therefore $\lvert\langle a, b \rangle\rvert = 0 \leqslant \|a\|\|b\|$.

Suppose, on the other hand, that $\|b\| > 0$. Then a fairly standard argument applies. Define $\lambda = \langle a, b \rangle/\|b\|^2$. Then $\langle a - \lambda b, b \rangle = \langle a, b \rangle - \lambda \|b\|^2 = 0$, therefore \begin{align*} 0 & \leqslant \langle a - \lambda b, a - \lambda b \rangle \\ & = \langle a, a - \lambda b \rangle - \lambda\langle b, a - \lambda b \rangle \\ & = \langle a, a - \lambda b \rangle - \lambda\langle a - \lambda b, b \rangle \\ & = \langle a, a - \lambda b \rangle \\ & = \|a\|^2 - \lambda \langle a, b \rangle, \end{align*} therefore $$\langle a, b \rangle^2 = \lambda\|b\|^2\langle a, b \rangle \leqslant \|a\|^2\|b\|^2,$$ therefore $$\lvert\langle a, b \rangle\rvert \leqslant \|a\|\|b\|.$$ The argument is similar when $\|a\| > 0$. So the Cauchy-Schwarz inequality holds in all cases.

It appears (see my series of shame-faced comments below for details) that this argument is merely an obfuscation of what is surely the most "standard" of all proofs of the Cauchy-Schwarz inequality. It is the one that is essentially due to Schwarz himself, and he had good reasons for using it, quite probably including the fact that it makes no use of the postulate that $\langle x, x \rangle = 0 \implies x = 0$! In a modern abstract formulation, it goes as follows (assuming, of course, that I haven't messed it up again). For all real $\lambda$, we have $\|u\|^2 - 2\lambda\langle u, v \rangle + \lambda^2\|v\|^2 = \|u - \lambda v\|^2 \geqslant 0$. Therefore, the discriminant of this quadratic function of $\lambda$ must be $\leqslant 0$. That is, $\langle u, v \rangle^2 \leqslant \|u\|^2\|v\|^2$; equivalently, $\lvert\langle u, v \rangle\rvert \leqslant \|u\|\|v\|$.
• I don't understand the first case. How does $0\le -2\langle a, b\rangle$ imply $\langle a,b \rangle=0$? What if $\langle a,b\rangle =-1$? Dec 3 '17 at 16:16
• I remember now "what I was thinking". It accounts both for the mistake and how easy it was to fix. My first thought was that if $\|a\|=\|b\|=0$, then for all real $\lambda$ we have $0 \leqslant \|a-\lambda b\|^2 = \|a\|^2-2\lambda\langle a,b\rangle+\lambda^2 \|b\|^2 = -2\lambda\langle a, b\rangle$, and this is only possible if $\langle a, b\rangle=0$. I mistakenly thought I could simplify this argument (simple enough though it is already) by considering only the case $\lambda = 1$. Of course that was wrong, but it somehow slipped through several anxious checks, which is why I was shocked. Dec 4 '17 at 11:59
• It was Schwarz who originally came up with the "standard" proof; it was specifically for the case where $\langle f,g\rangle$ is defined as an integral; and indeed in this case, the postulate $\langle f,f\rangle=0 \implies f=0$ fails to hold. So there may have been a very good reason why he chose the proof he did (in addition to the reason cited by Steele in his book The Cauchy-Schwarz Master Class, which I started to read recently - the reason cited is that passage to the limit from the discrete Cauchy inequality fails to preserve strictness, but Steele does say there are "several reasons"). Dec 4 '17 at 12:24