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Holder's inequality for integrals

  1. Apr 18, 2012 #1
    Does anyone know a simple proof for holder's inequality?

    I would be more interested in seeing the case of
    |∫fg|≤ sqrt(∫f^2)*sqrt(∫g^2)
     
  2. jcsd
  3. Apr 18, 2012 #2


    Be sure you can prove the following:

    1) For any [itex]x,y\in\mathbb R\,\,,\,\,xy\leq\frac{1}{2}(x^2+y^2)[/itex]

    2) Now put [itex]\frac{f}{\sqrt{\int f^2}}:= x\,,\,\,\frac{g}{\sqrt{\int g^2 dx}}:=y\,\,[/itex] in the above, integrate both sides and voila!, there you have your proof.

    DonAntonio
     
  4. Apr 19, 2012 #3
    Well the first part makes sense, I don't see why the second step follows.
     
  5. Apr 19, 2012 #4
    Also I don't see the first part either....

    If max(x,y) = x.
    Then xy <= x^2.
    and y^2 <= xy
    But the other part doesn't follow. I think your proof is lacking....
     
  6. Apr 19, 2012 #5

    micromass

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    Think about [itex](x-y)^2[/itex] and [itex](x+y)^2[/itex].
     
  7. Apr 19, 2012 #6
    Actually I see the argument xy ≤ 1/2 (x^2 + y^2)
    because if we start with x=y then the result is equality.
    Now if we let min(x,y) = x. Such that x+ε=y as ε→0,
    then xy doesn't decrease as fast as x^2. I.e. (y-ε)y < (y-ε)^2 = x^2
    Okay, so the first part makes since, but I still don't see Holder's inequality.
     
  8. Apr 19, 2012 #7

    micromass

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    What do you get if you fill in the x and y that DonAntonio suggested?
     
  9. Apr 19, 2012 #8

    [itex]xy\leq \frac{1}{2}(x^2+y^2)\Longleftrightarrow 0 \leq (x-y)^2[/itex] . I think something different is lacking...;)

    DonAntonio
     
  10. Apr 19, 2012 #9
    For the first part: Let x + ε= y. : ε≥0
    y^2 - εy ≤ y^2 - εy + 2ε^2
    y^2 - εy ≤ .5(2^2 - 2yε+ε^2)
    (y-ε)y = xy ≤ .5[ (y-ε)^2 + y^2] = .5(x^2 + y^2)

    The furthest I get for the second part is:
    fg/[ (sqrt(∫g^2)*sqrt(∫f^2) ] ≤ .5 * [ (f^2/∫f^2) + (g^2/∫g^2) ]
    Sorry, I don't see it.
     
  11. Apr 19, 2012 #10

    micromass

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    Why are you making it so difficult?? It's just basic algebra. What is [itex](x-y)^2[/itex]??

    Now integrate both sides, what do you get??
     
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