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 :).