0
$\begingroup$

I want to proof $|\langle x, y \rangle| \leq \|x\| \, \|y\|$ for all $x,y \in \mathbb{R}^n$ or $\mathbb{C}^n $. I know there exist a ton of proofs for this inequality, but it want to proof it through a specific schematic.

  1. We know that the length of a Vector is $\geq 0$, hence for an arbitrary $t \in \mathbb{R}$ it follows that $0 \leq \| tx + y \|$.
  2. I think its possible to rearrange this inequality such that $0 \leq at^2 + bt +c$.
  3. Know we can look at the discriminant $b^2 -4ac$ formulate a new inequality such that the Cauchy-Schwarz inequality follows.
$\endgroup$
2
$\begingroup$

Start by squaring your inequality in (1), giving $$ 0 \le \|tx+y\|^2 = \def\<#1>{\left<#1\right>}\<tx+y,tx+y> = t^2\|x\|^2 + 2t\<x,y> + \|y\|^2 $$ The discriminant is therefore given by $$ 4\<x,y>^2 - 4\|x\|^2\|y\|^2 $$ As the discriminant cannot be positive (note that a non-negative quadratic real polynomial has at most one root), we have $$ 4\<x,y>^2 - 4\|x\|^2\|y\|^2 \le 0 \iff \<x,y>^2 \le \|x\|^2\|y\|^2. $$ Taking square roots gives Cauchy-Schwarz.

$\endgroup$
  • $\begingroup$ Thank you! I have two questions. 1) For $\mathbb{R}$ i understand the rearrangement, but for $\mathbb{C}$ this is only sesquilinear, isn't it? Why is it also for $\mathbb{C}$ correct ? 2) Why exactly ist the discriminant not positiv? $\endgroup$ – faoeoe Oct 16 '18 at 19:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.