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Supremum of series difference question

  1. Nov 15, 2009 #1

    lom

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    [tex]f_n(x)=1,1\leq x\leq n\\[/tex]
    [tex]f_n(x)=0,1< n< \infty[/tex]
    f_n converges to f which is 1
    at the beggining f_n is 0 but when n goes to infinity its 1

    so why sup(f_n(x)-f(x))=1 ?

    f is allways 1

    but f_n is 0 and going to one

    in one case its 1-1
    in the other its 0-1

    the supremum is 0

    so the supremumum of their difference is 0 not 1



    ?
     
  2. jcsd
  3. Nov 15, 2009 #2

    LCKurtz

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    You obviously have typos there. Is the second one supposed to read:

    [tex]f_n(x)=0,n < x < \infty[/tex]?


    That's "beginning". What do you mean by "at the beginning fn = 0"? If you state things more precisely, it might help you understand the problem better.

    Given any n, can you find an x where |fn(x) - f(x)| = 1?
     
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