# Proof of Cauchy-Schwarz in a Complex Inner Product space.

Setup: Given a complex inner product space $$V$$ the Cauchy-Schwarz inequality is $$|\langle x,y \rangle|\leq \sqrt{\langle x,x \rangle} \sqrt{\langle y,y \rangle}.$$ I know that the RHS is just the product of the induced norms that come from our inner product, but I don't want to pass to any properties of a norm function. I took the standard proof approach (I think?) but I'm not very comfortable with complex variables so I don't know if I'm doing any arithmetic that is invalid over $$\mathbb{C}$$. Any help would be appreciated!

Proof Attempt: Define a function $$f(t) = \langle x+ty,x+ty \rangle$$. We can expand this inner product as \begin{align*} \langle x+ty,x+ty \rangle &= \langle x,x+ty\rangle + \langle ty,x+ty \rangle, \textrm{ linear in first arg.} \\ &=\langle x,x\rangle +\langle x,ty \rangle +\langle ty,x \rangle +\langle ty,ty \rangle, \textrm{ linear in second argument.}\\ &=\langle x,x \rangle + \overline{t} \langle x, y \rangle + t\langle y,x \rangle +t\overline{t} \langle y,y\rangle, \textrm{ conjugate linearity in second argument.} \end{align*} Now because $$t \in \mathbb{C}$$ we know that $$t\overline{t} = |t|^2$$. We also have by conjugate symmetry that $$\overline{t}\langle x,y \rangle= t \langle y,x \rangle.$$ Now because the inner product is positive definite, we can conclude that $$0 \leq \langle x,x \rangle + 2\overline{t}\langle x,y \rangle +|t|^2 \langle y,y\rangle.$$ Now just like in the case where we are over the reals, I would like to conclude by making the claim that this is a quadratic in $$t$$ that opens upwards, meaning it has no real roots and hence the discriminant $$b^2-4ac \leq 0$$ but I don't know if that's valid. It seems fair-ish because $$\langle x,x \rangle$$ and $$\langle y,y \rangle$$ are both real numbers, but is there a way to conclude that $$\langle x,y \rangle$$ is also a real number? I think I need $$\langle x,y \rangle$$ to be real to apply the quadratic formula right? I suppose intuitively $$\langle x,y \rangle$$ must be real because there's no ordering on $$\mathbb{C}$$ so for the inequality to make sense it has to be real?

Maybe more importantly, does it even make sense to think of it as a quadratic because I'm using $$|t|^2$$ and $$\overline{t}$$ as the parameters?

• this $\overline{t}\langle x,y \rangle = \overline{\overline{t}\langle x,y \rangle}$ is false for a sesquilinear product. What you have is that $\overline{t}\langle x,y \rangle = \bar t\overline{\langle y,x \rangle}$ Apr 10, 2022 at 20:18
• Yeah i just realized as I was writing over but I can't figure out what the right approach is, trying to fix it now. Apr 10, 2022 at 20:18
• the usual proof decomposes $w$ in the product $\langle v,w \rangle$ as a sum of two orthogonal vectors, one of the form $rv$ for some $r\in \mathbb{C}$. Apr 10, 2022 at 20:20
• this $\overline{t}\langle x,y \rangle= t \langle y,x \rangle$ is still wrong. There you are assuming that $\bar zw=z\bar w$ for arbitrary complex numbers $z,w\in \mathbb{C}$, however if you took $z=1$ and $w=i$ you see that it cannot be true. Apr 10, 2022 at 20:25
• @AndreyYanyuk You may be interested in this. Apr 10, 2022 at 21:12

## 1 Answer

The correct formula is $$(x + ty, x + ty) = (x, x) + (ty, ty) + 2\Re((x, ty)) = \|x\|^2 + 2\Re(\bar{t}(x, y)) + |t|^2\|y\|^2.$$ If you want to apply calculus wrt $$t$$, then let's assume that $$t \in \mathbb{R}$$. Then we get $$0 \leq (x + ty, x + ty) = \|x\|^2 + 2t\Re((x, y)) + t^2\|y\|^2.$$ Then, since this is a parabola that opens upward and is above the $$x$$-axis, it has at most one real zero, so $$4\Re((x, y))^2 -4\|y\|^2\|x\|^2\leq 0,$$ i.e. $$|\Re((x, y))| \leq \|x\|\|y\|.$$ Now for any $$\alpha \in \mathbb{C}$$ with $$|\alpha| = 1$$, we can replace $$x$$ with $$\alpha x$$ to get $$|\Re(\alpha(x, y))| \leq \|x\|\|y\|.$$ Cauchy's inequality comes from $$\alpha = \frac{|(x, y)|}{(x, y)}$$.

• So what is this choie of $\alpha$ doing? Are we forcing $\langle x,y \rangle$ to be real? Apr 11, 2022 at 1:25
• $x, y$ are arbitrary elements of $V$. We choose $\alpha$ because it yields Cauchy's inequality when you plug it in. In fact, our choice of $\alpha$ maximizes the lhs. Apr 11, 2022 at 2:17
• @AndreyYanyuk You can also view the parabola trick as minimizing the rhs of the first inequality over $t$. So the entire proof consisted of introducing auxiliary parameters, getting inequalities involving the parameters, and then choosing the parameters to get the best inequality. Apr 11, 2022 at 2:55
• Thanks its a lot clearer now. What do you think about assuming that $\langle x,y \rangle$ is strictly positive without loss of generality? Because if pick an $\alpha \in \mathbb{C}$ for which $\langle x ,\alpha y \rangle$ is strictly positive, we have $|\langle x, \alpha y \rangle | = |\alpha| \langle x,y \rangle$ and $\sqrt{\langle \alpha y, \alpha y \rangle} = |\alpha| \sqrt{\langle y,y \rangle}$ so the addition of this $\alpha$ term has no bearing on the CS inequality? Does that make sense? Apr 11, 2022 at 3:01
• Yeah that's essentially what I did. You are just introducing $\alpha$ earlier than I am. Your $\alpha$ is the complex conjugate of mine. Apr 11, 2022 at 3:08