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Math Help - A question about step functions

  1. #1
    C.E
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    A question about step functions

    I am really stuck on the following, I don't know how to start. Can somebody please help?

    Suppose f:[a,b]->R is a weakly increasing function, i.e. u<v gives f(u)<=f(v). Show that f is the limit of a sequence of step functions.
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    Quote Originally Posted by C.E View Post
    I am really stuck on the following, I don't know how to start. Can somebody please help?

    Suppose f:[a,b]->R is a weakly increasing function, i.e. u<v gives f(u)<=f(v). Show that f is the limit of a sequence of step functions.
    For each n, divide the interval from a to b into n intervals. Define F_n(x) to be the step function that is equal to f(x_n) for all x between x_n and x_{n+1}.
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  3. #3
    C.E
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    I can imagine that the sequence of step functions f_n would converge to f. However I do not know how to prove it. Can you please help? Here is where I am stuck.

    I pick e>0, I now need to show that there exists N so that for all n>N,
    |f_(x)-f(x)|<e. However, as I don't know which specific function f is I am really struggling. How do I pick n?
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    You may find it a bit easier to divide the range of the function (the interval [f(a),f(b)]) into subintervals of length less than \epsilon. If J is one of those subintervals, use the weakly increasing condition to show that the set f^{-1}(J) =\{x\in [a,b]:f(x)\in J\} is an interval. Let y_J be the value that f takes at some point of J. Then the function that takes the constant value y_J on J differs from f by an amount less than \epsilon throughout J. Doing this for each of the subintervals of [f(a),f(b)], you get a step function that is always close to f.
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  5. #5
    C.E
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    I think I have got it. Is this right?
    pick n so that (f(b)-f(a))/n <e.
    let p_i= f(a) + i(f(b)-f(a))/n (i between 1 and n-1)
    let J=(p_i, p_i+1)
    pick any z,y in [a,b] such that f(z)=p_i, f(y)=p_i+1
    let g(x)=0.5*(p_i+1 +p_i)/2 for all x in [z,y].
    for any x in [p_i,p_i+1], |g(x)-f(x)|<=|p_i+1-f(x)|<=|p_i+1-p_i|<e
    Last edited by C.E; January 23rd 2010 at 10:16 AM.
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