Schwarz inequality


by Rasalhague
Tags: inner product, inner product space, schwarz inequality
Rasalhague
Rasalhague is offline
#1
Aug2-09, 11:25 AM
P: 1,400
The couple of proofs that I've seen of Schwarz's inequality

[tex]\left|\mathbf{u} \cdot \mathbf{v} \right| \leq ||\mathbf{u}|| \: ||\mathbf{v}||[/tex]

both begin with statements to the effect that it's trivially true when u or v = 0. How so? In Bowen and Wang's Introduction to Vectors and Tensors, the inequality is said to be true for any inner product space, defined as a vector space (V,F,s), where V is an additive abelian group over a set of vectors, F a field, and s the function of scalar multiplication relating V and F, along with a function

[tex]f : V \rightarrow F[/tex]

called the inner product, defined by the following axioms:

[tex](1)\; f\left(\mathbf{u},\mathbf{v} \right) = \overline{ f\left(\mathbf{v},\mathbf{u} \right) }[/tex]

[tex](2)\; \lambda f\left(\mathbf{u},\mathbf{v} \right) = f\left(\lambda \mathbf{u},\mathbf{v} \right)[/tex]

[tex](3)\; f\left(\mathbf{u_{1}} + \mathbf{u_{2}},\mathbf{v} \right) = f\left(\mathbf{u_{1}},\mathbf{v} \right) + f\left(\mathbf{u_{2}},\mathbf{v} \right)[/tex]

[tex](4)\; f\left(\mathbf{u},\mathbf{u} \right) \geq 0 \; and \; f(\mathbf{u},\mathbf{u}) = 0 \Leftrightarrow \mathbf{u} = \mathbf{0}[/tex]

They give an example of the vector space [tex]\mathbb{C}^{n}[/tex], where elements of V are n-tuples of complex numbers. (Is n-tuple synonymous with "ordered n-tuple"?) And they defined an inner product for this space:

[tex]\mathbf{u} \cdot \mathbf{v} = \sum_{i=1}^{n} u_{i} \overline{v}_{i}[/tex]

No doubt the Schwarz inequality is true for such an inner product space when u or v = 0, since it just comes down to multiplication of complex numbers, but I don't know where to begin showing that it must be true generally for any inner product space as defined by those four axioms. Since the bar in the first axiom denotes the complex conjugate, presumable the field has to be the complex numbers, a subfield thereof, or something analogous? In that case, I can see that the right-hand side of the Schwarz inequality must equal 0. The left-hand sight must be positive, since it's the modulus of a complex number, so this trivial part of the proof amounts to showing that [tex]\left|\mathbf{u} \cdot \mathbf{v} \right|[/tex] = 0 if u or v = 0. Does it depend somehow on the first axiom?

[tex]f\left( \mathbf{0},\mathbf{v} \right) \; \overline{ f\left( \mathbf{0},\mathbf{v} \right)} = f\left( \mathbf{v},\mathbf{0} \right) \overline{ f\left( \mathbf{v},\mathbf{0} \right)}[/tex]

And then does it follow from some property of complex numbers, some basic algebra, or from something in the definition of a function, or something to do with the linearity of the dot product operation? It certainly feels as if it ought to be 0, but how to prove it only with these axioms. I'm new to the subject, and what may be trivial to the authors just isn't jumping out at me yet.
Phys.Org News Partner Science news on Phys.org
Better thermal-imaging lens from waste sulfur
Hackathon team's GoogolPlex gives Siri extra powers
Bright points in Sun's atmosphere mark patterns deep in its interior
jasonRF
jasonRF is offline
#2
Aug2-09, 12:51 PM
P: 654
I'm guessing your issue is with the LHS. Hint:
[tex]
\mathbf{0} = 0 \mathbf{0}
[/tex]

and use the axioms you have to show
[tex]
f(\mathbf{0},\mathbf{w})=0.
[/tex]
Rasalhague
Rasalhague is offline
#3
Aug2-09, 01:31 PM
P: 1,400
Quote Quote by jasonRF View Post
I'm guessing your issue is with the LHS. Hint:
[tex]
\mathbf{0} = 0 \mathbf{0}
[/tex]

and use the axioms you have to show
[tex]
f(\mathbf{0},\mathbf{w})=0.
[/tex]
Ooh, thanks, I think I've got it now! By Axiom 2 of the inner product space,

[tex]\mathbf{0} \cdot \mathbf{v} = 0 \left(\mathbf{0} \cdot \mathbf{v}\right) = 0[/tex]

and by Axioms 1 and 2,

[tex]\mathbf{v} \cdot \mathbf{0} = \overline{ \mathbf{0} \cdot \mathbf{v}} = \overline{ 0 \left(\mathbf{0} \cdot \mathbf{v}\right)} = 0 \overline{ \left(\mathbf{0} \cdot \mathbf{v}\right)} = 0[/tex]

because, by the distributive properties of a field,

[tex]0a = \left(b-b \right)a = ba - ba = 0[/tex]
[tex]a0 = a \left(b-b \right) = ab - ab = 0[/tex]

Rasalhague
Rasalhague is offline
#4
Aug3-09, 09:58 AM
P: 1,400

Schwarz inequality


So, moving on to the actual proof, for the case where neither u nor v = 0, Bowen and Wang proceed by asking us to consider the vector

[tex](\mathbf{u\cdot \mathbf{v}})\mathbf{v} - (\mathbf{v}\cdot \mathbf{u})\mathbf{u}[/tex]

and employ Axiom 4, "which requires that every vector have a nonzero length, hence"

[tex]\left ( \left \| \mathbf{u}^{2} \right \| \left \| \mathbf{v}^{2} \right \| - (\mathbf{u\cdot v})( \overline{\mathbf{u\cdot v}})\right )\left \| \mathbf{u}^{2} \right \|\geq 0[/tex]

I can see how this inequality would imply the Schwarz inequality, but how is it established? Employing Axiom 4 tells me that the product of the squared lengths of u and v is positive, as is the complex norm of the inner product of these two vectors. Subtracting the latter from the former is therefore to subtract one positive real number from another. But the result of such an operation isnít necessarily positive, is it? And if the result wasnít positive, the inequality wouldnít hold. So what am I not taking into account? Of course, if the Schwarz inequality was true, this inequality would have to be true too, but itís the Schwarz inequality that this is meant to be a proof of.
jasonRF
jasonRF is offline
#5
Aug3-09, 11:59 AM
P: 654
Your book may have a typo (and it may not!). There are many ways to prove the Schwartz inequality, but to use the same basic approach as the book try looking at
[tex]
f(a\mathbf{u} + b\mathbf{v}, a\mathbf{u} + b\mathbf{v})
[/tex]
and make a clever choice for [tex]a[/tex] and [tex]b[/tex].
DrGreg
DrGreg is offline
#6
Aug3-09, 09:15 PM
Sci Advisor
PF Gold
DrGreg's Avatar
P: 1,806
There's a typo. You should consider
[tex]\mathbf{x} = \left\langle\mathbf{u} ,\mathbf{\color{red}{u}}}\right\rangle\mathbf{v} - \left\langle\mathbf{v} , \mathbf{u}\right\rangle\mathbf{u}[/tex]
and the fact that [itex]\left\langle\mathbf{x} , \mathbf{x}\right\rangle \geq 0[/itex].
Rasalhague
Rasalhague is offline
#7
Aug4-09, 09:38 AM
P: 1,400
Thanks to you both! I reckon I've got it at last - phew. The typo was my mistake. The book had it right.

Let x = [tex]\left(\mathbf{u \cdot u} \right) \mathbf{v} - \left(\mathbf{v \cdot u} \right) \mathbf{u}[/tex].

[tex]\mathbf{x \cdot x} \geq 0[/tex]

[tex]\mathbf{x \cdot x} = \left[ \left(\mathbf{u \cdot u} \right) \mathbf{v} - \left(\mathbf{v \cdot u} \right) \mathbf{u} \right] \mathbf{ \cdot x}[/tex]

[tex]= \left(\mathbf{u \cdot u} \right) \mathbf{v \cdot x} - \left(\mathbf{v \cdot u} \right) \mathbf{u \cdot x}[/tex]

[tex]= ||\mathbf{u}^{2}|| \; \left( \mathbf{v \cdot x}\right) - \left(\mathbf{v \cdot u} \right) \left( \mathbf{u \cdot x} \right)[/tex]

[tex]= ||\mathbf{u}^{2}|| \; \overline{\mathbf{x \cdot v}} - \left(\mathbf{v \cdot u} \right) \overline{ \mathbf{x \cdot u}}[/tex]

[tex]= ||\mathbf{u}^{2}|| \; \left[ \overline{||\mathbf{u}||^{2} \; ||\mathbf{v}||^{2} - \left(\mathbf{v \cdot u} \right) \left(\mathbf{u \cdot v} \right)}\right] - \left(\mathbf{v \cdot u} \right) \left[ \overline{||\mathbf{u}||^{2}\left(\mathbf{v \cdot u} \right) - ||\mathbf{u}||^{2}\left(\mathbf{v \cdot u} \right)}\right][/tex]

[tex]= ||\mathbf{u}||^{2}\left(||\mathbf{u}||^{2} \; ||\mathbf{v}||^{2} - |\mathbf{u \cdot v}|^{2}\right) \geq 0[/tex]

[tex]\Rightarrow ||\mathbf{u}||^{2} \; ||\mathbf{v}||^{2} - |\mathbf{u \cdot v}|^{2} \geq 0[/tex]

[tex]\Rightarrow ||\mathbf{u}||^{2} \; ||\mathbf{v}||^{2} \geq |\mathbf{u \cdot v}|^{2}[/tex]

Huzzah!


Register to reply

Related Discussions
Schwarz Inequality Proof Advanced Physics Homework 3
[SOLVED] Consequence of Schwarz Inequality Calculus & Beyond Homework 7
question about the Schwarz inequality Calculus & Beyond Homework 2
Schwarz Inequality Calculus 5
Cauchy Schwarz Inequality Calculus & Beyond Homework 4