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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?

Irving Rabin
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    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} $ – Masacroso Apr 10 '22 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. – Irving Rabin Apr 10 '22 at 20:18
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    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 $r\in \mathbb{C}$. – Masacroso Apr 10 '22 at 20:20
  • They are still equal though right @Masacroso, if I actually write out $t = \alpha+\beta i$ and $\langle x,y \rangle = a+bi$ then they do agree unless my arithmetic is wrong. Also is there a way to just modify my approach? We proved CS without leveraging orthogonality. – Irving Rabin Apr 10 '22 at 20:21
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    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. – Masacroso Apr 10 '22 at 20:25
  • Yeah I just worked it out and saw my arithmetic was wrong. Is there any way to salvage this? Or I need to use orthogonality? – Irving Rabin Apr 10 '22 at 20:26
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    @AndreyYanyuk You may be interested in this. – Átila Correia Apr 10 '22 at 21:12

1 Answers1

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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 $\alpha = \frac{|(x, y)|}{(x, y)}$.

Mason
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  • So what is this choie of $\alpha$ doing? Are we forcing $\langle x,y \rangle $ to be real? – Irving Rabin Apr 11 '22 at 01:25
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    $x, y$ 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. – Mason Apr 11 '22 at 02:17
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    @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. – Mason Apr 11 '22 at 02:55
  • 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? – Irving Rabin Apr 11 '22 at 03:01
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    Yeah that's essentially what I did. You are just introducing $\alpha$ earlier than I am. Your $\alpha$ is the complex conjugate of mine. – Mason Apr 11 '22 at 03:08
  • Ah I hadn't put it together! Thank you very much! – Irving Rabin Apr 11 '22 at 03:10