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Essential supremum

  1. Jun 16, 2008 #1
    Problem: Show an example of a sequence of measurable positive functions on (0,1) so that
    [tex]\left\|\underline{lim} f_{n}\right\| < \underline{lim}\left\|f_{n}\right\| for n\rightarrow\infty[/tex]

    My work: I think its just the indicator function [tex]I_{[n,n+1]}[/tex]

    Since [tex]\left\|\underline{lim} I_{[n,n+1]}\right\|= 0 < \underline{lim}\left\|I_{[n,n+1]}\right\| =1 [/tex]

    For some reason I do not feel to confident in my answer, so any comments are welcome.
     
  2. jcsd
  3. Jun 16, 2008 #2
    correction

    Problem: Show an example of a sequence of measurable positive functions on (0,1) so that
    [tex]\left\|\underline{lim} f_{n}\right\|_{\infty} < \underline{lim}\left\|f_{n}\right\|_{\infty} for n\rightarrow\infty[/tex]

    My work: I think its just the indicator function [tex]I_{[n,n+1]}[/tex]

    Since [tex]\left\|\underline{lim} I_{[n,n+1]}\right\|_{\infty}= 0 < \underline{lim}\left\|I_{[n,n+1]}\right\|_{\infty} =1 [/tex]

    For some reason I do not feel to confident in my answer, so any comments are welcome.
     
  4. Jun 16, 2008 #3

    Dick

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    That's pretty good. But the domain of the indicator functions isn't (0,1). Can you build a very similar example using functions defined only on (0,1)?
     
  5. Jun 16, 2008 #4
    Thanks for the response Dick.

    If [tex]f_{n}=I_{(\frac{n-1}{n},1)}, then \left\|\underline{lim} I_{(\frac{n-1}{n},1)}\right\|_{\infty}= 0 < \underline{lim}\left\|I_{(\frac{n-1}{n},1)}\right\|_{\infty} =1[/tex]

    Please correct me if I am wrong.
     
  6. Jun 16, 2008 #5

    Dick

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    Sure. That's fine. I was thinking of I_(0,1/n), but you can put stuff on the other side of the interval as well.
     
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