Cauchy Schwarz proof with alternative dot product definition

[dustbin]

Homework Statement

Does the Cauchy Schwarz inequality hold if we define the dot product of two vectors A,B \in V_n by \sum_{k=1}^n |a_ib_i|? If so, prove it.

Homework Equations

The Cauchy-Schwarz inequality: (A\cdot B)^2 \leq (A\cdot A)(B\cdot B). Equality holds iff one of the vectors is a scalar multiple of the other.

The Attempt at a Solution

The result is trivial if either vector is the zero vector. Assume A,B\neq O.

Let A',B'\in V_n s.t. a'_i = |a_i| and b'_i = |b_i| for each i=1,2,...,n. Let \theta be the angle between A' and B'. Notice that A\cdot B = A'\cdot B', ||A|| = ||A'||, ||B|| = ||B'||, and A\cdot B > 0. Then A\cdot B = ||A||\,||B||\,\cos\theta \implies A\cdot B = ||A||\,||B||\,|\cos\theta| \leq ||A||\,||B|| since 0\leq |\cos\theta|\leq 1. Then (A\cdot B)^2 \leq ||A||^2||B||^2 = (A\cdot A)(B\cdot B). This proves the inequality.

We now show that equality holds iff B = kA for some k\in\mathbb{R}.

(\Longrightarrow) We prove this direction by the contrapositive. If B\neq k A for any k\in\mathbb{R}, then B' \neq |k| A'. Hence \theta \neq \alpha \pi for any \alpha \in \mathbb{Z}. Thus 0 < |\cos\theta| < 1 and therefore A\cdot B < ||A||\,||B||. In other words, equality does not hold.

(\Longleftarrow) If A = kB for some k\in\mathbb{R}, then B' = |k| A'. Hence \theta = \alpha \pi for some \alpha \in \mathbb{Z}. Therefore \cos\theta = \pm 1. Thus equality holds.

Does this look okay?

 

[verty]

I may have missed something but I think you redefine $A \cdot B$ then try to use the fact that $A \cdot B = ||A||\,||B||\,\cos\theta$, but this is negative if the angle between them is obtuse.

 

[dustbin]

But don't A',B' still satisfy A'\cdot B' = ||A'||\,||B'||\,\cos\theta, where \theta is the angle between A', B'? As well, A\cdot B = A'\cdot B', ||A|| = ||A'||, and ||B|| = ||B'||.

 

[verty]

Oh, you should call it $\theta'$ then, it is the angle between $A'$ and $B'$, not between $A$ and $B$.

 

[dustbin]

Okay. I thought it was clear. Thanks for pointing that out!

Does the proof look okay, then?

 

[verty]

Well, you say the inequality is proved, but the original inequality has a square on the left. Something is amiss.

 

[dustbin]

Edited. What do you tink?

 

[verty]

I think you're done. You were asked to prove the equality and I see no gaps. My work here is done :).