Cauchy-Schwartz Inequality Proof

  • #1
RJLiberator
Gold Member
1,095
63

Homework Statement


Show that |<v|w>|^2 ≤ <v|v><w|w>
for any |v>,|w> ∈ ℂ^2

Homework Equations




The Attempt at a Solution



The Cauchy-Schwartz inequality is extremely relevant for the math/physics that I am interested in.
I feel like I have a very good proof here, but I am interested in a few things here.
1) Is my proof correct in the right context (Complex numbers here)
2) What is your favorite proof of the Cauchy Schwartz Inequality? (I want to master this one)

Proof:
Case 1: We have to show that if |v> or |w> = 0, then the sides are equal to 0. Which is somewhat trivial to prove by simply plugging in the values and reaching the conclusion using inner space axioms.

Case 2: Where |v> and |w> are not equal to 0.
Here, I let:
a = <v|v>
b = 2<v|w>
c = <w|w>

By Inner product axiom 3 stating that the inner product of any vector with itself is nonnegative we see
0 ≤ <tv+w | tv+w >
0 ≤ <v|v>t^2+2<v|w>t+<w|w>
0 ≤ at^2+bt+c

The conclusion here is that the inequality implies that the quadratic polynomial has either no real roots or a repeated real root. Therefore the discriminant must satisfy: b^2-4ac ≤ 0.

By plugging in the necessary values we see:
4<v|w>^2-4<v|v><w|w> ≤ 0
<v|w>^2 ≤ <v|v><w|w>

Which is what we set out to prove.

So, my one concern is that this may not be complete for complex inner products, but rather only for real numbers? Is there any cause for that concern?
 

Answers and Replies

  • #2
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722

Homework Statement


Show that |<v|w>|^2 ≤ <v|v><w|w>
for any |v>,|w> ∈ ℂ^2

Homework Equations




The Attempt at a Solution



The Cauchy-Schwartz inequality is extremely relevant for the math/physics that I am interested in.
I feel like I have a very good proof here, but I am interested in a few things here.
1) Is my proof correct in the right context (Complex numbers here)
2) What is your favorite proof of the Cauchy Schwartz Inequality? (I want to master this one)

Proof:
Case 1: We have to show that if |v> or |w> = 0, then the sides are equal to 0. Which is somewhat trivial to prove by simply plugging in the values and reaching the conclusion using inner space axioms.

Case 2: Where |v> and |w> are not equal to 0.
Here, I let:
a = <v|v>
b = 2<v|w>
c = <w|w>

By Inner product axiom 3 stating that the inner product of any vector with itself is nonnegative we see
0 ≤ <tv+w | tv+w >
0 ≤ <v|v>t^2+2<v|w>t+<w|w>
0 ≤ at^2+bt+c

The conclusion here is that the inequality implies that the quadratic polynomial has either no real roots or a repeated real root. Therefore the discriminant must satisfy: b^2-4ac ≤ 0.

By plugging in the necessary values we see:
4<v|w>^2-4<v|v><w|w> ≤ 0
<v|w>^2 ≤ <v|v><w|w>

Which is what we set out to prove.

So, my one concern is that this may not be complete for complex inner products, but rather only for real numbers? Is there any cause for that concern?

You can answer that for yourself, by answering the following questions:
In line 1 of your development, is <tv + w | tv + w> real>? Is t real?
In line 2, are <v|v>, 2, <v|w> and <w|w> real?
So, in line 3 are a, b, c and t real?
 
  • Like
Likes RJLiberator
  • #3
RJLiberator
Gold Member
1,095
63
In line 1 of your development, is <tv + w | tv + w> real>? Is t real?
Ah, I forgot to define t in this post, Let t ∈ℂ.

In line 2, are <v|v>, 2, <v|w> and <w|w> real?
|v>, |w> are originally complex, but when they are an inner product with themselves they become greater than 0, but still complex. 2 can be real or complex, I suppose.

So, in line 3 are a, b, c and t real?

Hm, well, t may be complex due to my definition, and what I need for this proof.
But I'm inclined to believe that the results have to be real.
 
  • #4
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
Ah, I forgot to define t in this post, Let t ∈ℂ.


|v>, |w> are originally complex, but when they are an inner product with themselves they become greater than 0, but still complex. 2 can be real or complex, I suppose.



Hm, well, t may be complex due to my definition, and what I need for this proof.
But I'm inclined to believe that the results have to be real.

Yes, they do, but they might not be in the form you used. For real t we have
0 ≤ <t v + w | t v + w> = t^2 <v|v> + t ( <v|w> + <w|v> ) + <w|w>. Here, the coefficients of t^2, t^1 and t^0 = 1 are, indeed, real. However, 2<v|w> might not be real, and in that case <v|w> + <w|v> ≠ 2<v|w>.
 
  • Like
Likes RJLiberator
  • #5
RJLiberator
Gold Member
1,095
63
Hm.
Well, this is interesting.

So we see that in the part: 2<v|w>t where I am stating that b = 2<v|w>
This does NOT have to be real, it can be complex.

But the answer for this inequality to hold must be real.

So what we are really stating here is that b = 2<v|w> does not need to be real based on our definitions.

And therefore this implies that the discriminant proof stating that the polynomial either has no real roots or a repeated root is therefore junk, garbage, not true.

So this proof is entirely incorrect.

Right?
 
  • #6
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
Hm.
Well, this is interesting.

So we see that in the part: 2<v|w>t where I am stating that b = 2<v|w>
This does NOT have to be real, it can be complex.

But the answer for this inequality to hold must be real.

So what we are really stating here is that b = 2<v|w> does not need to be real based on our definitions.

And therefore this implies that the discriminant proof stating that the polynomial either has no real roots or a repeated root is therefore junk, garbage, not true.

So this proof is entirely incorrect.

Right?

If your vector space is over the real field, you are done; your proof is correct and complete in that case. However, if it is over the complex field you have more work to do. In that case it is, indeed, important that t be allowed to assume complex values.
 
  • Like
Likes RJLiberator
  • #7
RJLiberator
Gold Member
1,095
63
Case 2: Where |v> and |w> are not equal to 0.
Here, I let:
a = <v|v>
b = 2<v|w>
c = <w|w>

By Inner product axiom 3 stating that the inner product of any vector with itself is nonnegative we see
0 ≤ <tv+w | tv+w >
0 ≤ <v|v>t^2+<tv|w>+<w|tv>+<w|w>
0 ≤ <v|v>t^2+t*<v|w>+t<w|v>+<w|w>


So here is where my initial proof failed for complex numbers.

So I have to adjust this part.
I'm just no longer sure where to go? There must be a property of inner products I need to use.
Is there a property that will allow me to switch the components? So that I can make this work?

Does this method even work this way anymore? 0_o
 
Last edited:
  • #8
RJLiberator
Gold Member
1,095
63
Screen Shot 2015-10-11 at 5.47.15 PM.png


In this proof, at the end, how do you go from

|v|^2 |w|^2 ≤ | |<v|w>|^2

to |<v|w>|^2 ≤ <v|v> <w|w>

I'm struggling to find which properties allow this.
 
  • #9
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
View attachment 90051

In this proof, at the end, how do you go from

|v|^2 |w|^2 ≤ | |<v|w>|^2

to |<v|w>|^2 ≤ <v|v> <w|w>

I'm struggling to find which properties allow this.

The inequality ##0 \leq |v|^2 - |\langle v,w \rangle |^2/|w|^2## is the same as ##|\langle v,w \rangle |^2/|w|^2 \leq |v|^2##. Do you see it now?

Anyway, since you have now obtained the solution already, I feel I can now legally show you how to complete the type of argument you were using before. For complex scalar ##t## we have
[tex] 0 \leq |v|^2 + |t|^2 |w|^2 + t \langle v,w \rangle + \overline{ t \langle v,w \rangle} = |v|^2 + |t|^2 |w|^2 + 2 \text{Re} \, t \langle v,w \rangle [/tex]
Here, ##\overline{z}## is the complex conjugate of a complex number ##z## and ##\text{Re} \,z## is the real part of ##z##. Let's use the polar representations:
[tex] t = r e^{i \theta}, \; \langle v,w \rangle = s e^{i \phi}, [/tex]
where ##r , s \geq 0##. Here, ##s, \phi ## are fixed but ##r, \theta ## are variable. We have that
[tex] \text{Re} \, t \langle v,w \rangle = r s \cos(\theta + \phi) [/tex]
We thus have
[tex] r^2 |w|^2 + 2 r s \cos(\theta + \phi) + |v|^2 \geq 0 [/tex]
for all ##r \geq 0## and all ##\theta##. However, if we replace ##\theta## by ##\theta + \pi## we will have that
[tex] r^2 |w|^2 - 2 r s \cos(\theta + \phi) + |v|^2 \geq 0 [/tex]
for all ##r \geq 0 ## and all ##\theta##. Taken together, these two results say that
[tex] |w|^2 x^2 + 2 s \cos(\theta + \phi) x + |v|^2 \geq 0 \; \text{for all real } \;\; x [/tex]
Now we can use the "discriminant" argument, to conclude that
[tex] 4 s^2 \cos^2 (\theta + \phi) \leq 4 |v|^2 |w|^2 [/tex]
for all angles ##\theta##. In particular, when ##\theta + \phi = 0 ## or when ##\theta + \phi = \pi## we get
[tex] s^2 \leq |v|^2 |w|^2 [/tex]
Since ##s = |\langle v,w \rangle|##, we are done.
 
  • Like
Likes RJLiberator
  • #10
RJLiberator
Gold Member
1,095
63
Bravo. Highly impressive and I thoroughly enjoyed reading over that proof.

There are parts involved in it there are amazingly tricky, but make poetic sense. Ah, it is truly a thing of art.

:DD:DD
 
  • #11
RJLiberator
Gold Member
1,095
63
I must ask, why did you change r to x prior to taking the discriminat?
Was this to make it easier to visualize in quadratic form, or is there a more significant meaning?
 
  • #12
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
I must ask, why did you change r to x prior to taking the discriminat?
Was this to make it easier to visualize in quadratic form, or is there a more significant meaning?

Initially, I said ##r \geq 0##. I could have changed that to "for all real ##r##"), but I thought it might be less confusing to change ##r## to ##x## and allow both signs of ##x##. Anyway, who cares if we call it ##r## or ##x##? It is just a variable. The main issue is the region over which that variable is allowed to roam---which is ultimately the whole real line in the end.
 
  • Like
Likes RJLiberator

Related Threads on Cauchy-Schwartz Inequality Proof

  • Last Post
Replies
16
Views
7K
Replies
1
Views
2K
Replies
4
Views
633
Replies
6
Views
3K
  • Last Post
Replies
1
Views
10K
  • Last Post
Replies
3
Views
6K
  • Last Post
Replies
1
Views
6K
Replies
2
Views
10K
Replies
4
Views
1K
Replies
14
Views
6K
Top