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Can anyone come up with a Lebesgue-integrable function that

  1. Aug 2, 2010 #1
    ...satisfies the following conditions:

    (1) Is continuous on [itex][1,\infty)[/itex], and

    (2) Does not have a limit as [itex]x\to \infty[/itex].

    Apparently, such a function [itex]f(x)[/itex] exists, but I cannot think of an example for the life of me. Remember: The function must also satisfy

    [tex]
    \int_1^\infty |f(x)|dx < \infty,
    [/tex]

    where "[itex]\int[/itex]" is the Lebesgue integral.
     
  2. jcsd
  3. Aug 2, 2010 #2

    Hurkyl

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    Can you think of a Riemann-integrable function f such that:
    • Max(f) = 1
    • Min(f) = 0
    • The integral over all of R is a (a is a previously chosen positive real number
     
  4. Aug 2, 2010 #3
    Of course. How about

    [tex]
    f(x) = \begin{cases}
    1, & x \in [0,a],\\
    0, & \text{otherwise}.
    \end{cases}
    [/tex]
     
  5. Aug 2, 2010 #4

    disregardthat

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    On the line [tex][1,\infty)[/tex], at each integer point n, draw an isoceles of height [tex]2^n[/tex] and with base width [tex]4^{-n}[/tex]. Let the function be the curve of these isoceles when they occur, and 0 when they don't. This function is obviously continuous, and

    [tex]\int^{\infty}_1 |f(x)| dx = \sum_{n=1}^{\infty} \frac{2^{n}4^{-n}}{2}= \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1}{2}[/tex]

    However, the function has no limit as [tex]x \to \infty[/tex]. As an extra bonus it is not even bounded.
     
  6. Aug 2, 2010 #5

    Ben Niehoff

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    Jarle, some of your triangles overlap, but the basic principle still works.

    Is it possible to find a [itex]C^\infty[/itex] function that satisfies the OP's criteria? Yes, I see it is possible after just writing that...

    How about an analytic function that satisfies the criteria? For that, I'm not sure...
     
  7. Aug 2, 2010 #6
    I don't see why this is so. Can you give an example of two overlapping triangles?

    Also, what do you mean by "OP's criteria?"
     
    Last edited: Aug 2, 2010
  8. Aug 2, 2010 #7
    Perfect. I feel like a fool for not having thought of this myself. Thanks!
     
  9. Aug 2, 2010 #8

    Hurkyl

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    I feel like this is one of those questions that is either "obviously yes" or "obviously no", but I don't know which. :biggrin:

    My first inclination is a function like
    [tex]f(x) = \exp\left(\frac{x}{2} \log( \sin(x)^2 ) \right)[/tex]​
    which, on R, simplifies to
    [tex]|\sin x|^{x}[/tex]​
    or maybe replace x/2 with something even faster growing.


    I haven't ground through the analysis to see if this actually has a finite integral.


    However, http://www.wolframalpha.com/input/?i=Integrate[Exp[Exp[x]/2+Log[Sin[x]^2]],+x]


    Is that analytic? Well, this one is more obviously so: http://www.wolframalpha.com/input/?i=Integrate[Exp[Exp[x]/2+Log[(1/2)+++(1/2)Sin[x]^2]],+x]
     
    Last edited: Aug 2, 2010
  10. Aug 2, 2010 #9

    Mute

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    How about replacing the triangles in the given solution to the original problem with Gaussians?

    [tex]f(x) = \sum_{n=1}^\infty \frac{2^n}{\sqrt{2\pi}} \exp\left[-\frac{(x-n)^2}{2\sigma_n^2}\right][/tex]
    where [itex]\sigma_n = 4^{-n}[/itex]?

    It's possible this isn't analytic since we're summing an infinite number of terms, but it'd be the first thing I'd try (if I felt like trying to prove/disprove things like analyticity). (Of course, even if analyticity is proved doing the resulting integral wouldn't be too fun - at least on the [1,infinity) interval. (-infinity,infinity) wouldn't be so bad. ;))
     
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