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Infimum of integral of open set

  1. Dec 1, 2012 #1
    I have already done part a and b. Part a is easy, for part b, i let the anti-derivative of f to be k(t)+c and arrive at the answer that F(f)= 1/2+ 2*k(1/2) - k(1). But i don't know how to do the next part, can anyone give me a hint? the question c ask me to show that the infimum of F is 0 and it is never attained on A.
     

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  3. Dec 1, 2012 #2

    LCKurtz

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    Hint: For F(f) to be near zero then f needs to be near 0 on [0,1/2) and near 1 on (1/2,1].
     
  4. Dec 1, 2012 #3

    Dick

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    It's conceptually pretty easy. Both integrands are nonnegative. You can make it as small as you want by jumping from f=0 on most of [0,1/2] to f=1 on most of [1/2,1]. Is that enough of a hint?
     
  5. Dec 2, 2012 #4
    but in that case, F(f) will be -1/2,right? im sorry but i don't really understand the hint,can you give a more specific "hint"?(i know,im bad at this)
     
  6. Dec 2, 2012 #5

    HallsofIvy

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    I don't know where you got that since "f" was not given explicitely. It sounds to me like you are integrating f from 0 to 1 rather than f from 0 to 1/2 and then 1- f from 1/2 to 1.

    Suppose f(x)= 0 for [itex]0\le x\le .4[/itex], f(x)= 5(x- .4) for [itex].4\le x\le .6[/itex], f(x)= 1 for [itex].6\le x\le 1[/itex]. That is, f is 0 from 0 to .4, then the straight line from (.4, 0) to (.6, 1), then is 1 from .6 to 1. What is the integral of that?

    Now, do the same thing for , say .45 instead of .4 and .55 instead of .6- f(x)= 0 for [itex]0\le x\le .45[/itex], is the straight line from (.45, 0) to (.55, 1), then 1 from .55 to 1. What is the integral of that? (You don't even need to write out the formula for the function- the integral is the area of a right triangle.)

    What happens as you keep moving those two points toward x= .5?
     
  7. Dec 2, 2012 #6
    thank you very much for this, i have been trying to do this question for a whole day already.
     
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