# Cauchy-Schwarz Hermitian Inner Product Remainder

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.

• You are correct, of course. Silly mistake. – Archaick Apr 18 '15 at 11:25
• Aside from the errant squared signs on the right, the statement is basically the same. (Note, I implicitly ignored them, likely by ignorance). – Emily Apr 18 '15 at 11:27
• Perhaps some clarifications need to be made. I want to find $f(a,b)$. Of course we know it exists, but I though it was obvious I wanted to know what it was. BTW, I'm still a bit confused about how you can ignore the fact that when you conjugate $<a+b,a+b>=||a||^2+2\mathfrak{R}(<a,b>)+||b||^2$. Meaning, only the real component of the cross term survives. Can you enlighten me? – Archaick Apr 18 '15 at 11:31
• @Archaick I have expanded slightly. – Emily Apr 18 '15 at 11:59
• Indeed, indeed. – Archaick Apr 18 '15 at 18:07

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 ..

• Oh, how often the simplest of solutions will elude us. Although I recognize at this point that we've effectively proved a tautology, I'd appreciate it if you could edit your answer to include this answer. – Archaick Apr 19 '15 at 0:08
• For the op: this answer is literally identical term by term to what I wrote. If you understand this response, it is worth comparing the structure. – Emily Apr 19 '15 at 1:41
• @Archaick: Many of us (including myself) have calculated $0=0$ before gaining some competence. – Han de Bruijn Apr 19 '15 at 10:29
• @Arkamis: Yes, maybe, but I think that Occam's razor makes a difference here. – Han de Bruijn Apr 19 '15 at 10:34
• @HandeBruijn Agreed. Besides, it's never bad to see something from a few different angles. – Emily Apr 19 '15 at 19:39

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 ($+1$).

--

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

• It seems to me that you forgot that I'm asking about an inner product which is conjugate bilinear. Moreover, forgive me if I'm just thick-skulled, but I don't see where the remainder comes in here. From where I'm sitting, this is a proof of the inequality, but I'm looking for the remainder term. – Archaick Apr 17 '15 at 15:34
• I am not forgetting that it's conjugate bilinear, I'm just whitewashing away that by going directly to the induced norm. The remainder comes in from the inequality, e.g. if $a \le b$ then $a = b + r$. Consider the quadratic in $t$ and complete the square -- I think that will get you what you want. Let me see if I can't quickly do the algebra. – Emily Apr 17 '15 at 15:36
• What is a "unit scalar" and what use has it here? – Han de Bruijn Apr 18 '15 at 10:14
• @hahn quite not. The unit scalar normalizes the expression for $y$. See for instance Folland chapter 5.5, this proof is copied almost verbatim from it. If you cannot see how to get the expression $x \le y \implies x < y + r$ for some $r>0$, I am not sure what to say. Note, I am not deriving the form of the $f$; I am simply proving that there is one. – Emily Apr 18 '15 at 11:12
• The down voter should also explain the down vote or ask for clarification. Nothing in this post is incorrect, except possibly an immaterial algebra error. – Emily Apr 18 '15 at 11:14