Cauchy-Bunyakovsky-Schwarz inequality

  • Thread starter phyguy321
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  • #1
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absolute value( [tex]\int{fg}[/tex]) [tex]\leq[/tex][([tex]\int{f}[/tex][tex]^{2}[/tex])([tex]\int{g}[/tex][tex]^{2}[/tex])][tex]^{1/2}[/tex]

all integrals are from a to b
does anyone have any idea of a proof for this?
 

Answers and Replies

  • #3
morphism
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Bump? What more do you want?
 
  • #5
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Bump? What more do you want?
I dont want to solve it in terms of a vector space, I would like to see it using integrals
 
  • #6
HallsofIvy
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Then use the "vector space" proof when the vector space in question is a function space!

The "vector space" proof looks at the inner product of [itex]\vec{u}- \lambda\vec{v}[/itex] with itself, getting [itex]|\vec{u}-\lambda\vec{v}|= <\vec{u}- \lambda\vec{v},\vec{u}= \lambda\vec{v}>= <\vec{u},\vec{u}>- 2\lambda<\vec{u},\vec{v}>+ \lambda^2<\vec{v},\vec{v}>\ge 0[/itex], for all [itex]\lambda[/itex] because it is a "square".

In particular, if we let let
[tex]\lambda= \frac{<\vec{u},\vec{v}>}{<\vec{v},\vec{v}>}[/tex]
that becomes
[tex]<\vec{u},\vec{u}>-2\frac{\vec{u},\vec{v}>^2}{<\vec{v},\vec{v}}+ \frac{\vec{u},\vec{v}>^2}{\vec{v},\vec{v}}[/tex]
[tex]= <\vec{u},\vec{u}>- \frac{<\vec{u},\vec{v}>^2}{<\vec{v},\vec{v}>}\ge 0[/tex]
so
[tex]<\vec{u},\vec{u}>\ge \frac{<\vec{u},\vec{v}>^2}{<\vec{v},\vec{v}>}[/tex]
and
[tex]<\vec{u},\vec{u}><\vec{v},\vec{v}>\ge<\vec{u},\vec{v}>^2[/tex]

That is precisely the proof for "vector spaces" at the site I linked to. Now, the set of all integrable functions IS an inner product space with innerproduct given by [itex]<f, g>= \int fg[/itex].

Replacing [itex]<\vec{u},\vec{v}>[/itex] by [itex]\int fg[/itex] , [itex]<\vec{u}, \vec{v}>[/itex] with [itex]\int f^2[/itex], and [itex]<\vec{v}, \vec{v}>[/itex] with [itex]\int g^2[/itex] gives exactly the inequality you post. That's the whole point of "abstraction" in mathematics- you don't have to prove a lot of different versions of the same thing.
 

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