Necessity of absolute value in Cauchy Schwarz inequality

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Discussion Overview

The discussion revolves around the necessity of absolute value in the Cauchy-Schwarz inequality, particularly in the context of complex scalars. Participants explore the implications of squaring inequalities and the reasoning behind using absolute values in certain expressions.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions the introduction of absolute value in the expression derived from the triangle inequality, asking why it is not simply written without absolute values.
  • Another participant suggests that the absolute value is used to succinctly capture both cases of the real part being non-negative or non-positive.
  • Concerns are raised about the validity of the "triangle logic" used to prove the triangle inequality, with a suggestion that Cauchy-Schwarz is a standard way to prove it.
  • There is confusion regarding whether the expression involving the real part is equivalent to the product of the magnitudes of the vectors involved.
  • A later reply asserts that the two expressions are not the same and advises starting with the Cauchy-Schwarz inequality over real numbers before addressing the complex case.

Areas of Agreement / Disagreement

Participants express differing views on the necessity and reasoning behind the use of absolute values in the context of the Cauchy-Schwarz inequality. There is no consensus on the validity of the triangle logic employed or the equivalence of certain expressions.

Contextual Notes

Participants note potential issues with the assumptions made in the original text and the implications of proving the triangle inequality through different methods.

SamRoss
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In going from |X|+|Y|>=|X+Y| to 2|X||Y|>=|<X|Y>+<Y|X>|, I'm not sure why the last set of absolute value symbols are necessary.
Reading The Theoretical Minimum by Susskind and Friedman. They state the following...

$$\left|X\right|=\sqrt {\langle X|X \rangle}\\
\left|Y\right|=\sqrt {\langle Y|Y \rangle}\\
\left|X+Y\right|=\sqrt {\left({\left<X\right|+\left<Y\right|}\right)\left({\left|X\right>+\left|Y\right>}\right)}$$

Then they state (based on triangle logic described earlier)...

$$\left|X\right|+\left|Y\right|\geq\left|X+Y\right|$$

They then say that if we square the above inequality on both sides and simplify (which, as far as I can see, simply amounts to subtracting ##\left|X\right|^2+\left|Y\right|^2## from both sides), we get...

2|X||Y| >= |<X|Y>+<Y|X>| (Sorry for this rendering. I was having trouble with Latex.)

My question is - where did the absolute value symbols on the right side come from? Why isn't it just 2|X||Y| >= <X|Y>+<Y|X> ?
 
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SamRoss said:
They then say that if we square the above inequality on both sides and simplify (which, as far as I can see, simply amounts to subtracting ##\left|X\right|^2+\left|Y\right|^2## from both sides), we get...

2|X||Y| >= |<X|Y>+<Y|X>| (Sorry for this rendering. I was having trouble with Latex.)

My question is - where did the absolute value symbols on the right side come from? Why isn't it just 2|X||Y| >= <X|Y>+<Y|X> ?

I'm assuming we're working with scalars in ##\mathbb C##

##\langle X|Y\rangle+ \langle Y|X \rangle =2 \cdot re\big(\langle X|Y \rangle\big) \leq 2 \cdot \big \vert re\big(\langle X|Y\rangle \big)\big \vert##

If you step the the proof you should be able to see that there is an upper bound we can apply to the right hand side, and said upper bound holds whether or not

##re\big(\langle X|Y \rangle\big)##
is non-negative or non-positive, so why not be more succinct and capture both cases at the same time with an absolute value sign? Basically kill two bird with one stone is why they use the absolute value here.

- - - -
I have some concerns here though

SamRoss said:
Then they state (based on triangle logic described earlier)...

$$\left|X\right|+\left|Y\right|\geq\left|X+Y\right|$$

They then say that if we square the above inequality on both sides and simplify..

There isn't really "triangle logic", and in fact the standard way of proving triangle inequality is to use cauchy-schwarz. If you legitimately arrive at the triangle inequality by other means (e.g. convexity) then yes it implies cauchy-schwarz, but I rather doubt this was done here.
 
StoneTemplePython said:
##\langle X|Y\rangle+ \langle Y|X \rangle =2 \cdot re\big(\langle X|Y \rangle\big) \leq 2 \cdot \big \vert re\big(\langle X|Y\rangle \big)\big \vert##
I'm still confused. Is ##2 \cdot \big \vert re\big(\langle X|Y\rangle \big)\big \vert## the same thing as 2|X||Y|?
 
SamRoss said:
I'm still confused. Is ##2 \cdot \big \vert re\big(\langle X|Y\rangle \big)\big \vert## the same thing as 2|X||Y|?
No.

I'd advise first proving cauchy-schwarz over reals then returning to the complex case. I'd also advise first looking in a math text instead of this... the fact that your author seems to assume what he wants to prove (triangle inequality) is not a good sign
 
StoneTemplePython said:
No.

I'd advise first proving cauchy-schwarz over reals then returning to the complex case. I'd also advise first looking in a math text instead of this... the fact that your author seems to assume what he wants to prove (triangle inequality) is not a good sign

Good advice. I'll try that. Thanks for your help.
 

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