1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cauchy-Bunyakovsky-Schwarz inequality

  1. Oct 9, 2008 #1
    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?
     
  2. jcsd
  3. Oct 9, 2008 #2

    morphism

    User Avatar
    Science Advisor
    Homework Helper

  4. Oct 16, 2008 #3

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    Bump? What more do you want?
     
  5. Oct 16, 2008 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

  6. Oct 17, 2008 #5
    I dont want to solve it in terms of a vector space, I would like to see it using integrals
     
  7. Oct 17, 2008 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Cauchy-Bunyakovsky-Schwarz inequality
  1. Umm, cauchy-schwarz (Replies: 8)

  2. Schwarz inequality (Replies: 6)

Loading...