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Short question about L infinity

  1. Nov 15, 2009 #1
    I want to say that f(x) = |1/x| is in L-infinity(E) when m(E)<infinity becuase the function has and esssup on any measurable set, E. Even if E = (-1, 1) f(0) is not a problem since it is only one point...

    But wait... what *is* the esssup for this function on (-1, 1)? I think it might not have one. This is why I'm confused.

    :(
     
  2. jcsd
  3. Nov 15, 2009 #2

    LCKurtz

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    You don't seem too confused to me. It doesn't have an esssup on (-1,1).
     
  4. Nov 15, 2009 #3
    Ok. So then it's not in L-infinity when x=0 is in E.

    (slowly this starts to make more sense...)
     
  5. Nov 15, 2009 #4

    LCKurtz

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    Not exactly. E could be (-oo, -1) union [0] union (1,oo) and it would be in Loo.
     
  6. Nov 15, 2009 #5
    oh good point.
     
  7. Nov 15, 2009 #6

    LCKurtz

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    You could have f(x) = sin(x) except on the rationals where it is 1/x and anything at 0. This would have an esssup of 1 because the set of x where |f(x)| > 1 has measure 0.
     
  8. Nov 15, 2009 #7
    Thanks, that's a good example to think about.
     
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