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There is a proof of the real case of Cauchy-Schwarz inequality that expands $\|\lambda v - w\|^2 \geq 0 $, gets a quadratic in $\lambda$, and takes the discriminant to get the Cauchy-Schwarz inequality. In trying to do the same thing in the complex case, I ran into some trouble. First, there are proofs here, here, here, and here, but none of them do it the way I'm thinking of.

If I similarly expand $\|\lambda v - w\|^2_{\mathbb{C}},$ I get $|\lambda|^2\|v\|^2 - 2 \text{Re}(\lambda \langle v,w\rangle) + \|w\|^2$. How can I manipulate this to get Cauchy-Schwarz using the discriminant? My problem is that $$|\lambda|^2\|v\|^2 - 2 \text{Re}(\lambda \langle v,w\rangle) + \|w\|^2 \geq |\lambda|^2\|v\|^2 - 2 | \lambda ||\langle v,w\rangle | + \|w\|^2,$$ so I can't be sure that the latter term is $\geq 0$.

Community
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Eric Auld
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3 Answers3

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I just thought of $$|\lambda|^2\|v\|^2 - 2 \text{Re}(\lambda \langle v,w\rangle) + \|w\|^2 \leq |\lambda|^2\|v\|^2 + 2 | \lambda ||\langle v,w\rangle | + \|w\|^2,$$ which yields the same discriminant. This seems to work!

Eric Auld
  • 28,127
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If I similarly expand $\|\lambda v - w\|^2_{\mathbb{C}},$ I get $|\lambda|^2\|v\|^2 - 2 \text{Re}(\lambda \langle v,w\rangle) + \|w\|^2$.

Your idea can work almost the same way.

Consider values $\lambda(r) = r \exp i\theta$, taking $\theta$ such as $\lambda(r) \langle v,w\rangle \in \mathbb R$. You get $$ 0\le|\lambda|^2\|v\|^2 - 2 \text{Re}(\lambda \langle v,w\rangle) + \|w\|^2 = r^2\|v\|^2 \pm 2 r |\langle v,w\rangle| + \|w\|^2 $$

the $\pm$ depends on the sign of the real part.

mookid
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I recently revisited this problem and stumbled into another way around this, using a standard linear analysis trick. For the sake of completeness for future readers, I will add it in this answer.

It actually suffices to show that $$| \text{Re} \langle x,y \rangle | \leq \lVert x \rVert \lVert y \rVert$$ for all $x,y$, which one gets from using the discriminant trick on the expansion of $\lVert \lambda x + y \rVert^2$. To see this, consider vectors $x,y$. If $\langle x, y \rangle = 0$, there is nothing to show. Hence, suppose otherwise and let $\alpha = \frac{\overline{ \langle x, y \rangle}}{| \langle x, y \rangle| }$. Then $| \alpha | = 1$ and we have

\begin{align*} 0 \leq | \langle x, y \rangle | &= \alpha \langle x, y \rangle = \langle \alpha x, y \rangle \\ &= |\text{Re} \langle \alpha x, y \rangle| \leq \lVert \alpha x \rVert \lVert y \rVert \\ &= \lVert x \rVert \lVert y \rVert. \end{align*}

Thusle Gadelankz
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