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Extremely hard concept

  1. May 4, 2008 #1
    I am trying to prove that the Limit as p approaches infinity of {integral from 0 to 1[|f(t)|^p dt]}^(1/p) is in fact equal to the max of |f(x)| between [0,1].

    Any suggestions im sure I need to set the limit to less than or equal to and greater than or equal to the max but i dont quite know how
     
  2. jcsd
  3. May 4, 2008 #2
    One hint.... What is the most contributing term when you take the n th power of each element and sum them up,

    keep it simple and start with taking two elements a,b and take the power of 20th...
     
  4. May 4, 2008 #3
    im sorry im having a hard time folowing yor terminology is there any way to rephrase
     
  5. May 5, 2008 #4
    Why is the following true?

    [tex]
    \left(\int_0^1{|f(t)|^p dt}\right)^{1/p} \leq \left(\int_0^1{\underbrace{(\max{|f(t)|})^p}_{const} dt}\right)^{1/p}=\max|f(t)|\int_0^1{dt}=\max|f(t)|
    [/tex]
    This proof lacks only one limiting argument, can you find it?
     
  6. May 5, 2008 #5
    ok so how do I show the opposite or that the function is greater than or equal to the max
     
  7. May 5, 2008 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What function and the max of what?
     
  8. May 5, 2008 #7
    the function is just vague f(t) and the max is the maximum of |f(t)| between [0,1]
     
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