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Integral of open set

  1. Dec 3, 2012 #1
    1. The problem statement, all variables and given/known data

    2.png


    3. The attempt at a solution
    for part A i assume that f is in norm space C[0,1],||.|| , then choose a sequence fn in C[o,1] s.t fn->f then for 0<fn<1 so 0<f<1 i.e. A is closed i am not sure my answer here

    for part B i assume the anti-derivatice of f(t) to be K(t)+c therefore, by F(f)=2K(1/2)+1/2-K(1)-K(0) then how should i prove it is cts ???
     
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  3. Dec 3, 2012 #2

    micromass

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    I don't see how that proves anything. You need to prove that if [itex]f_n[/itex] is a convergent sequence in A such that [itex]f_n\rightarrow f[/itex], that then [itex]f[/itex] is an element of A.

    OK, but that doesn't really help. What definition of continuous would you like to use here? Can you state it?

    Also, it would be a great help for us if you would type your posts in LaTeX: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3
     
  4. Dec 3, 2012 #3
    δ
    if [itex]f_n[/itex] is a convergent sequence in A such that [itex]f_n\rightarrow f[/itex], then for [itex]f_n[/itex] is in A , A is a subset of C[0,1] so fn is bounded by 0 and 1 and continuous on [0,1] so fn is uniformly continuous to f ,then f is bouded by 0,1 which is in A????

    and secound part if i use the definition of continuous , then exists a δ,s.t. |f(x)-f(y)|<δ implies that |F(f(x))-F(g(x))|<esillope . and just fixed the esillope . and i can solve it out now many thanks
     
    Last edited: Dec 3, 2012
  5. Dec 3, 2012 #4

    micromass

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    Sure. But can you actually prove that f is bounded by 0 and 1? Once you proven that, I agree that the above shows that A is closed.

    Protip: we usually talk about epsilon instead of esillope

    OK, so you you need to make [itex]|F(f)-F(g)|[/itex] small somehow by making [itex]\|f-g\|_\infty[/itex] small. So, let us start by writing

    [tex]|F(f)-F(g)|=\left|\int_0^{1/2}f(t)dt + \int_{1/2}^1 (1-f(t))dt - \int_0^{1/2} g(t)dt - \int_{1/2}^1 (1-g(t))dt\right|[/tex]

    Try to find a good estimation for that.
     
  6. Dec 3, 2012 #5
    i not sure how to prove that 0<f<1 ,which way should i begin with??
     
  7. Dec 3, 2012 #6

    micromass

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    You should not prove 0<f<1, you should prove [itex]0\leq f\leq 1[/itex].

    Start by fixing an [itex]x\in [0,1][/itex]. Then try to deduce from [itex]f_n(x)\rightarrow f(x)[/itex] that [itex]0\leq f(x)\leq 1[/itex].
     
  8. Dec 3, 2012 #7
    for me ,it is obviouse. [itex]0\leq f_n\leq 1[/itex]. then max fn =1, min fn =0 and limfn =f so by sanwich rule [itex]0\leq f(x)\leq 1[/itex] i just dont know how should i prove that f is bounded by [0,1] here
     
  9. Dec 3, 2012 #8

    micromass

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    Why should [itex]\max f_n=1[/itex] and [itex]\min f_n=0[/itex]?? That's really weird.
    For example, take the map

    [tex]f_n:[0,1]\rightarrow \mathbb{R}:x\rightarrow x\rightarrow \frac{1}{n}[/tex]

    then the [itex]\max f_n=\min f_n=\frac{1}{n}[/itex].

    You're really making it harder than it is. I'm just asking you that if [itex](y_n)_n[/itex] is a convergent sequence with [itex]y_n\rightarrow y[/itex], and if [itex]0\leq y_n\leq 1[/itex], then [itex]0\leq y\leq 1[/itex]. It doesn't even have anything to do with functions.
     
  10. Dec 3, 2012 #9
    omg! thats what i thought at the beginning!!! many thanks !!!
     
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