Necessity of absolute value in Cauchy Schwarz inequality

  • #1
SamRoss
Gold Member
205
21

Summary:

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.

Main Question or Discussion Point

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> ?
 

Answers and Replies

  • #2
StoneTemplePython
Science Advisor
Gold Member
2019 Award
1,153
560
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

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.
 
  • #3
SamRoss
Gold Member
205
21
##\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|?
 
  • #4
StoneTemplePython
Science Advisor
Gold Member
2019 Award
1,153
560
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
 
  • #5
SamRoss
Gold Member
205
21
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.
 

Related Threads on Necessity of absolute value in Cauchy Schwarz inequality

  • Last Post
Replies
4
Views
747
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
10
Views
4K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
4
Views
3K
Replies
7
Views
657
Replies
6
Views
3K
Replies
3
Views
745
  • Last Post
Replies
3
Views
2K
Top