Necessity of absolute value in Cauchy Schwarz inequality

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
SamRoss
Gold Member
Messages
256
Reaction score
36
TL;DR
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> ?
 
on Phys.org
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.