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all integrals are from a to b

does anyone have any idea of a proof for this?

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all integrals are from a to b

does anyone have any idea of a proof for this?

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morphism

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It will help if you know before hand that [itex]\langle f,g \rangle = \int_a^b fg[/itex] (with f and g coming from an appropriate vector space) is an inner product.

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morphism

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Bump? What more do you want?

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HallsofIvy

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[tex]\int (f(x)+ \lambda g(x))^2 dx[/tex]

where [itex]\lambda[/itex] is a complex number.

The details are given at

http://en.wikipedia.org/wiki/Cauchy–Schwarz_inequality

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I dont want to solve it in terms of a vector space, I would like to see it using integralsBump? What more do you want?

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HallsofIvy

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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

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