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Minkowski's inequality!

  1. Jan 29, 2008 #1
    I need to proove the Minkowski's inequality for integrals.
    I am taking a course in analysis.

    [ int(f+g)^2 ] ^(1/2) =< [int(f^2)]^(1/2) + [int(g^2)]^(1/2)

    now we are given that both f and g are Riemann integrable on the interval.
    So by the properties of Riemann integrals, so is f^2,g^2 and fg.

    We are also given a hint to expand the integral on the left and then use the Cauchy-Bunyakovsky-Schwarz inequality (now this i've already prooved in a previous exercice using the discriminant).

    I was trying to expand the left side but i don't know what to do with the squared root, moreover i was trying to expand regardless the squared root and then at the end take a squared root but it still hasn't worked..

    I need help =)

  2. jcsd
  3. Jan 29, 2008 #2
    form what you wrote i assume you have an inner product given by

    <f,g> = int fg dx

    and the induced norm

    |f| = <f,f>^½

    so the Cauchy-Schwarz inequality is

    |<f,g>| <= |f||g|

    from what you get

    <f,g> <= |f||g|

    so starting with

    |f+g| = <f+g.f+g> = (int (f+g)^2)^½ = (int (f^2+g^2+2fg)^½
    = (int f^2 + int g^2 + int 2fg)^½ = (<f,f>+<g,g>+2<f,g>)^½
    = (|f|^2+|g|^2+2<f,g>)^½

    by cauchy-schwarz

    <= (|f|^2+|g|^2+2|f||g|)^½ = [(|f|+|g|)^2]^½ = |f|+|g|

  4. Jan 29, 2008 #3
    i forgot a ^½ int the line

    |f+g| = <f+g.f+g>

    it should be

    |f+g| = <f+g.f+g>^½
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