Does Cauchy-Schwarz Inequality depend on positive definiteness?

Question

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.

Answer 1

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.

Addendum

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, $\lvert\langle u, v \rangle\rvert \leqslant \|u\|\|v\|$.

Answer 2

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 $v=(1,-1)$.

Answer 3

The question as well as the answer by Calum Gilhooley address the case over a real vector space. Here I briefly mention that the result carries over to the complex case and mention some proofs that carry over to this setting.

Let $X$ be a vector space over $\mathbb{K}$, where $\mathbb{K}$ is fixed as one of $\mathbb{R}$ or $\mathbb{C}$. We say a map $u\colon X\times X \to \mathbb{K}$ is a semi-inner product on $X$ if it satisfies the following three conditions.

1. For all $x, y, z\in X$ and $\lambda \in \mathbb{K}$ we have \begin{align*} u(x + y, z) &= u(x, z) + u(y, z) , \\ u(x, y + z) &= u(x, y) + u(x, z) , \\ u(\lambda x, y) &= \alpha u(x, y) \text{ and } \\ u(x, \alpha y) &= \overline{\lambda} u(x, y) . \end{align*}

2. We have $u(y, x) = \overline{u(x, y)}$ for all $x, y\in X$.

3. We have $u(x, x) \geq 0$ for all $x\in X$.

Observe that a semi-inner product $u$ on a vector space $X$ (over $\mathbb{K}$) is an inner product if and only if for each $x\in X$ the condition $u(x, x) = 0$ implies $x = 0$. We now show the following.

Proposition 1. Let $X$ be a vector space over $\mathbb{K}$ and $u$ a semi-inner product on $X$. Then for every $x, y\in X$ we have \begin{equation} |u(x, y)|^{2} \leq u(x, x) u(y, y) . \tag{1} \end{equation}

Proof 1. Let $x, y\in X$. If $u(y, y) = 0$ the result is clear. So we assume $u(y, y) \neq 0$. Let $\lambda \in \mathbb{K}$. We have \begin{equation} \begin{aligned} 0 \leq u(x - \lambda y, x - \lambda y) &= u(x, x) - \lambda u(y, x) - \overline{\lambda} u(x, y) + |\lambda |^{2} u(y, y) \\ &= u(x, x) - 2 \, {\rm Re}(\overline{\lambda} u(x, y) ) + |\lambda |^{2} u(y, y) . \end{aligned} \tag{2} \end{equation} By taking $\lambda = \frac{u(x, y) }{u(y, y) }$ in $(2)$ we get \begin{equation} 0 \leq u(x, x) - 2\frac{|u(x, y)|^{2} }{u(y, y) } + \frac{|u(x, y)|^{2} }{u(y, y) } = \frac{u(x, x) u(y, y) - |u(x, y)|^{2} }{u(y, y) } . \tag{3} \end{equation} From $(3)$ we deduce $0 \leq u(x, x) u(y, y) - |u(x, y)|^{2}$ and the result follows.

Proof 2. Let $x, y\in X$. If $u(y, y) = 0$ the result is clear. So we assume $u(y, y) \neq 0$. Choose $\alpha \in \mathbb{K}$ such that $u(x, y) = \alpha |u(x, y)|$. Define $z = \alpha y$. For each $t\in\mathbb{R}$ we have \begin{equation} \begin{aligned} 0 \leq u(x - tz, x - tz) &= u(x, x) - tu(z, x) - tu(x, z) + t^{2} u(z, z) \\ &= u(x, x) - 2t |u(x, y)| + t^{2} u(y, y) . \end{aligned} \tag{4} \end{equation} Define $p\colon \mathbb{R}\to\mathbb{R}$ by $p(t) := u(y, y) t^{2} - 2|u(x, y)| t + u(x, x)$. Then $p$ is a quadratic polynomial satisfying $p(t) \geq 0$ for all $t\in\mathbb{R}$. We readily see that the polynomial $p$ attains its minimum at the value $t_{0} = \frac{|u(x, y)| }{u(y, y) }$. Then by evaluating $(4)$ at this value, we get \begin{equation} \begin{aligned} 0 \leq u(x - t_{0}z, x - t_{0}z) &= u(x, x) - 2\frac{|u(x, y)|^{2} }{u(y, y) } + \frac{|u(x, y)|^{2} }{u(y, y) } \\ &= \frac{u(x, x) u(y, y) - |u(x, y)|^{2} }{u(y, y) } . \end{aligned} \tag{5} \end{equation} Hence we deduce from $(5)$ that $0 \leq u(x, x) u(y, y) - |u(x, y)|^{2}$ and the result follows.

Note that we can also characterise when equality in $(1)$ occurs.

Proposition 2. Let $X$ be a vector space over $\mathbb{K}$ and $u$ a semi-inner product on $X$. Then for every $x, y\in X$ we have \begin{equation} |u(x, y)|^{2} = u(x, x) u(y, y) \end{equation} if and only if we have one of $u(y, y) = 0$ or otherwise $u(x - \lambda y, x - \lambda y) = 0$ for some $\lambda\in\mathbb{K}$.

Proof. See here.

A reference for the result from Proposition 1 is Proposition 3.3 in Functional Analysis by van Neerven (see here).