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Properties of L1, lebesgue integreable functions

  1. Dec 3, 2008 #1
    1. The problem statement, all variables and given/known data
    (X,S,u) is measur space, f,g are in L1. Prove that:
    1.min(f,g) is in L1
    2.[tex]min(\int(fdu),\int(gdu))\geq\int(min(f,g)du)[/tex]
    3. when does equality hold?

    The attempt at a solution
    1.Since both f,g are in L1, and min must be one of the f or g, both of which are in L1, thus min (f,g) is in L1.
    2.Star with simple functions and extend to limit?
    3. Guess: equality hold if f=g
     
    Last edited: Dec 3, 2008
  2. jcsd
  3. Dec 3, 2008 #2

    Office_Shredder

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    Question 1 isn't asking if min(f,g) is in L1... actually, I'm not really sure what it's asking, since you didn't post a function. was it supposed to be min(f,g)?

    For 2, notice that f>=min(f,g) so
    [tex] \int fdu \geq \int min(f,g) du[/tex]

    and the same for g.

    For 3, equality holds under a much broader class of conditions... try drawing some functions on R like x, x2, 8, and sin(x) and try some comparisons. Or you can try to glean it from the proof of 2
     
  4. Dec 3, 2008 #3
    sorry about that, I've corrected question 1.

    For3. Are you saying that monotonic functions satisfy equality?
     
  5. Dec 3, 2008 #4

    Office_Shredder

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    Only if sin(x) is monotonic :p

    Try drawing the graphs y=sin(x) and y=8 and see if it satisfies equality. Then try the same for y=sin(x) and y=1/2. What's the difference?
     
  6. Dec 3, 2008 #5
    #1 Is a little more complicated than your proof, because which function is the minimum may change from point to point. Think about f(x)=x and g(x)=x^2.
     
  7. Dec 3, 2008 #6
    I see, so the equality holds when there doesn't exist any x such that f(x)=g(x). , ie, they never cross, but how show rigorously that this is the condition?


    Regarding question 1. I think the proof is a one-liner using the comparison theorem, since if we can show|min(f,g)|<|f|, then since f is in L1, by comparison min(f,g) must also be in L1. The only problem is that in order to show |min(f,g)|<|f|, don't we need both f and g to be nonnegative?
     
  8. Dec 3, 2008 #7
    Yes, for that you would need f, g >= 0. Try this: let [tex]E = \{ x | f(x) \leq g(x) \}[/tex],
    so
    [tex]\begin{align*}
    \int |\min ( f, g )| d\mu = \int_E |\min ( f, g )| d\mu + \int_{ E^c } |\min ( f, g )| d\mu = \int_E |f| d\mu + \int_{ E^c } |g| d\mu,
    \end{align*}[/tex]
    and see where you can go from there.
     
  9. Dec 5, 2008 #8
    E is a measurable set, let x be the characteristic function, then
    |f|xE is in L1
    |g|xEc is in L1
    and |f|xE+|g|xEc=|min(f,g)| is in L1
    thus min (f,g) is in L1
    QED
    please tell me this is right, exam is next week, thanks
     
  10. Dec 5, 2008 #9
    That's exactly right!
     
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