Cauchy-Schwarz Hermitian Inner Product Remainder

Question

A couple weeks ago, someone showed me a proof of Cauchy-Schwarz where he ended up deriving something of the form $$|\langle a,b\rangle|^2=|\langle a,a\rangle||\langle b,b\rangle| +f(a,b)$$ Where $f(a,b)$ was positive definite about $a=b$. Recall that a Hermitian inner product satisfies $\langle a,b\rangle=\overline{\langle b,a\rangle}$ and $\langle a,cb\rangle=\overline{c}\langle a,b\rangle$ for $c\in\mathbb{C}$. I've forgotten both how he did it and what $f(a,b)$ turned out to be and it's borderline driving me bonkers.

Answer 1

Instead of posting more and more comments (I've deleted most of them) it's better to formulate sort of an answer.
Let's first get straight the derivation of Schwarz inequality for the complex inner product. The most concise proof I could find on the internet is in the (Dutch) Wikipedia page "Ongelijkheid van Cauchy-Schwarz". The inequality reads, with $a$ and $b$ complex vectors: $$ |\langle a,b \rangle|^2 \le \langle a,a \rangle \langle b,b \rangle $$ Equality is when $b = \lambda a$ ($\lambda$ complex). Then trivialiter: $$ |\langle a,b \rangle|^2 = |\lambda|^2 \langle a,a \rangle = \langle a,a \rangle \langle \lambda a, \lambda a \rangle = \langle a,a \rangle \langle b,b \rangle $$ Assume trivial cases done and assume that $b \ne 0$ . Then for each complex number $\lambda$ : $$ 0 \le \langle a-\lambda b, a-\lambda b \rangle = \langle a,a \rangle - \lambda \langle b,a \rangle - \overline{\lambda} \langle a,b \rangle + |\lambda|^2 \langle b,b \rangle $$ Where $\overline{\lambda}$ is the complex conjugate of $\lambda$. Now take: $$ \lambda = \frac{\langle a,b \rangle}{\langle b,b \rangle} $$ Then the Schwarz inequality easily follows from: $$ 0 \le \langle a,a \rangle - \frac{\langle a,b \rangle}{\langle b,b \rangle}\langle b,a \rangle - \frac{\langle b,a \rangle}{\langle b,b \rangle} \langle a,b \rangle + \left|\frac{\langle a,b \rangle}{\langle b,b \rangle}\right|^2 \langle b,b \rangle\\ \Longleftrightarrow \qquad 0 \le \langle a,a \rangle - \frac{|\langle a,b \rangle|^2}{\langle b,b \rangle} $$ Of course $f(a,b)$ must be a positive function if we write instead: $$ |\langle a,b \rangle|^2 = \langle a,a \rangle \langle b,b \rangle - f(a,b) \quad \Longrightarrow \quad f(a,b) = \langle a,a \rangle \langle b,b \rangle - |\langle a,b \rangle|^2 $$ If that is what's meant by the OP ..

Answer 2

It's not too challenging.

Consider for some positive $t$ and some unit scalar $\alpha$, $\langle x-\alpha ty, x-\alpha ty\rangle$:

This gives us:

$$\begin{align*} \langle x-\alpha ty, x-\alpha ty\rangle &= \langle x, x-\alpha ty\rangle - \alpha t \langle y, x-\alpha ty\rangle \\ &= \langle x,x\rangle - \overline{\alpha t}\langle x,y\rangle - \alpha t \langle y,x\rangle + \alpha t (\overline{\alpha t})\langle y,y\rangle \\ &= \|x\|^2 -2t \operatorname{Re} \langle x,\alpha y\rangle + \|y\|^2 \\ &= \|x\|^2-2t|\langle x,y\rangle | + \underbrace{\alpha \overline{\alpha}}_{=|\alpha|^2 = 1} t^2\|y\|^2. \end{align*} $$ Therefore, $$\langle x-\alpha ty,x-\alpha ty\rangle = \|x\|^2-2t|\langle x, y\rangle|+t^2\|y^2\|.$$ (I had an errant squared on the left-hand side before, but it wasn't used; it was just a typo).

This is a quadratic real-valued polynomial in $t$ that is non-negative. Therefore, the discriminant is such that $b^2-4ac \le 0$, where $a=\|y^2\|$, $b = -2|\langle x,y\rangle|$, and $c = \|x^2\|$.

This gives us $$4|\langle x,y\rangle|^2 \le 4 \|x\|^2\|y\|^2 = 4\langle x,x\rangle \langle y,y\rangle.$$

From this, we obtain

$$\langle x,y\rangle^2 \le \langle x,x\rangle \langle y,y\rangle$$ which is the standard Cauchy-Schwarz inequality. From basic analysis, this means that for all $x,y$, we have

$$\langle x,y\rangle^2 = \langle x,x\rangle \langle y,y\rangle + f(x,y).$$

Now, we know that equality must hold when $x=y$. Therefore, $f(x,y) = 0$ when $x=y$ and $f$ cannot be a non-zero constant. Now just show that $f$ must always have the same sign ($+1$).

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Another approach would be to use the Polarization identity, which I will attempt when I return from the market.