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Integrability and Lipschitz continuity

  1. Sep 20, 2011 #1
    (I've been lighting this board up recently; sorry about that. I've been thinking about a lot of things, and my professors all generally have better things to do or are out of town.)

    Is there an easy way to show that if [itex]f[/itex] is Lipschitz (on all of [itex]\mathbb R[/itex]), then

    [tex]
    \int_{-\infty}^\infty f^2(x) e^{-\frac{x^2}{2t}}dx < \infty \quad ?
    [/tex]

    I've tried a number of different approaches - boundedness of the derivative and integration by parts, bounded/dominated/monotone convergence theorem, approximation of [itex]f^2[/itex] by polynomials - but I haven't gotten any of them to work. Apparently, just because [itex]f[/itex] is Lipschitz doesn't mean [itex]f^2[/itex] is Lipschitz. (Just look at [itex]f(x) = x[/itex].)

    Can anyone think of a Lipschitz function for which the integral above is infinite?
     
  2. jcsd
  3. Sep 21, 2011 #2

    CompuChip

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    Science Advisor
    Homework Helper

    They only thing I can think of that you haven't tried yet is going back to definitions.
    For example, when you write out the integral as the limit of a Riemann sum, you will probably get summands like
    [tex]|f^2(x + \Delta x) e^{-(x + \Delta x)^2 / 2t} - f^2(x) e^{-x^2/2t}| \le \cdots \le |f(x + \Delta x)|^2 e^{-x^2/2t} \le (K {\Delta x})^2 e^{-x^2/2t}[/tex]
    and eventually, maybe, rewrite it to
    [tex]\cdots \le C \int_{-\infty}^\infty e^{-x^2/2t} \, dx[/tex]
    or something else for which you know it is finite.
     
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