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Math Help - Infimum of integral of open set

  1. #1
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    Infimum of integral of open set

    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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  2. #2
    GJA
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    Re: Infimum of integral of open set

    Hi kanezila,

    In part b, to prove that F is continuous at f we need to show that for \epsilon>0 there is \delta>0 such that whenever \|f-g\|_{\infty}<\delta, we have |F(f)-F(g)|<\epsilon. Perhaps looking at an antiderivative will go somewhere, but this is probably best done by using the definition.

    To prove part c looking at the following sequence of functions \{f_{n}\} in A, where f_{n}(x)=0 for x\in [0,1/2], f_{n}(x)=(n+2)(x-1/2) for x\in [1/2, 1/2+1/(n+2)] and f_{n}(x)=1 for x\in [1/2+1/(n+2), 1]. Then we should be able to show that F(f_{n})=\frac{1}{2}(\frac{1}{n+2}). Since this goes to 0 as n goes to infinity the infimum must be 0, because each of the f_{n}'s belong to A.

    Does this clear things up? Good luck!
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  3. #3
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    Re: Infimum of integral of open set

    thank you very much for this, i have been trying to do this question for a whole day already.
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  4. #4
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    Re: Infimum of integral of open set

    could you tell me how to prove the first one???many thanks
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