Necessity of absolute value in Cauchy Schwarz inequality

In summary, the authors of The Theoretical Minimum, Susskind and Friedman, state the triangle inequality and then proceed to prove it using the triangle logic described earlier. They use the absolute value symbols on the right side of the inequality to capture both positive and negative cases at the same time. However, this approach may not be the most rigorous and it is recommended to first prove Cauchy-Schwarz over reals before moving on to the complex case.
  • #1
SamRoss
Gold Member
254
36
TL;DR 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.
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> ?
 
Mathematics news on Phys.org
  • #2
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.
 
  • #3
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|?
 
  • #4
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
 
  • #5
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.
 

What is the Cauchy Schwarz inequality?

The Cauchy Schwarz inequality, also known as the Cauchy-Bunyakovsky-Schwarz inequality, is a mathematical inequality that states that the square of the inner product of two vectors is always less than or equal to the product of the squares of the individual vector lengths. It is commonly used in various areas of mathematics, including linear algebra, functional analysis, and probability theory.

Why is the Cauchy Schwarz inequality important?

The Cauchy Schwarz inequality is important because it provides a way to measure the relationship between two vectors in a vector space. It is a fundamental tool in many areas of mathematics and has numerous applications, such as in proving the convergence of series and in optimization problems.

What is the necessity of absolute value in the Cauchy Schwarz inequality?

The absolute value is necessary in the Cauchy Schwarz inequality because it ensures that the inequality holds for all possible values of the vectors. Without the absolute value, the inequality may not hold for certain combinations of vectors.

How is the Cauchy Schwarz inequality related to other mathematical concepts?

The Cauchy Schwarz inequality is related to other mathematical concepts, such as the triangle inequality and the concept of orthogonality. It can also be generalized to other types of inner products, such as complex inner products in a complex vector space.

Can the Cauchy Schwarz inequality be extended to higher dimensions?

Yes, the Cauchy Schwarz inequality can be extended to higher dimensions. In fact, it is a special case of the more general Hölder's inequality, which holds for any number of vectors in a vector space. The Cauchy Schwarz inequality can also be extended to infinite dimensions in the context of functional analysis.

Similar threads

Replies
1
Views
802
  • General Math
Replies
7
Views
1K
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
886
Replies
7
Views
1K
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
19
Views
829
  • Quantum Physics
Replies
9
Views
947
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
497
Back
Top