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One limit and integral

  1. Feb 22, 2008 #1
    What do we have to assume of the function f so that following limit is correct,

    [tex]
    \lim_{L\to\infty} \int\limits_{-L}^L f(\frac{x}{L}) \frac{\sin(x)}{x} dx = f(0)\int\limits_{-\infty}^{\infty} \frac{\sin(x)}{x} dx = \pi f(0)
    [/tex]

    If we first fix the integration domain like this

    [tex]
    \lim_{L\to\infty} \int\limits_{-\infty}^{\infty} f(\frac{x}{L})\frac{\sin(x)}{x} \chi_{[-L,L]}(x)dx
    [/tex]

    the problem is that the limit of the integrand is not Lebesgue integrable over [itex]\mathbb{R}[/itex], so the standard convergence results do not settle this immediately.
     
  2. jcsd
  3. Mar 4, 2008 #2
    I figured out one way to do this, but I had to use one inequality thats correctness is not clear to me. The triangle inequality

    [tex]
    |\int dx\; g(x)| \leq \int dx\;|g(x)|
    [/tex]

    is not suitable here. This instead,

    [tex]
    |\int dx\; f(x)g(x)| \leq \|f\|_{\textrm{sup}} |\int dx\; g(x)|
    [/tex]

    becomes useful. But is this correct? I don't know how to prove this. I wouldn't want to do much assumptions about g, but f can be assumed to be as nice as necessary.
     
  4. Mar 4, 2008 #3
    Argh! Not even correct!

    [tex]
    g(x) = \cos(x^2)
    [/tex]

    [tex]
    \int\limits_0^{\infty} \cos(x^2)dx = \frac{1}{2}\sqrt{\frac{\pi}{2}}
    [/tex]

    [tex]
    f(x)=\left\{\begin{array}{ll}
    1, \quad & \cos(x^2)\geq 0\\
    0,\quad &\cos(x^2) < 0 \\
    \end{array}\right.
    [/tex]

    [tex]
    |\int dx\; f(x)g(x)| = \infty > \frac{1}{2}\sqrt{\frac{\pi}{2}} = \|f\|_{\textrm{sup}} |\int dx\; g(x)|
    [/tex]




    hmhmh.... but here f is not integrable itself. It could be that the inequality is true if f's integral exists...
     
    Last edited: Mar 4, 2008
  5. Mar 4, 2008 #4
    Even this is not true. By fixing sufficiently large M, we obtain an integrable [itex]f\chi_{[0,M]}[/itex] such that

    [tex]
    |\int dx\; (f\chi_{[0,M]})(x) g(x)| > \|f\chi_{[0,M]}\|_{\textrm{sup}} |\int dx\; g(x)|
    [/tex]

    I'm back in the starting point :frown:
     
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