The couple of proofs that I've seen of Schwarz's inequality(adsbygoogle = window.adsbygoogle || []).push({});

[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 whenuorv=0. How so? In Bowen and Wang'sIntroduction to Vectors and Tensors, the inequality is said to be true for any inner product space, defined as a vector space (V,F,s), whereVis an additive abelian group over a set of vectors,Fa field, andsthe function of scalar multiplication relatingVandF, 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 ofVare 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 whenuorv=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 ifuorv=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.

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# Schwarz inequality

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