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Smooth and L^2 on R^n. Will it be bounded?

  1. Oct 31, 2009 #1
    smooth and L^2 on R^n. Will it be bounded??

    If a function, say u, is smooth and L^2 on R^n. Will it be bounded??

    In the case of n=1 I would say that it obviously is so. Because if it were unbounded then it wouldn't be L^2.

    But in the case of n=2 (or higher). I can imagine a function with a kind of ridge that gets thinner and thinner but higher and higher, the further away from the origin we get. So that it would be unbounded but still L^2.

    I guess I was just wondering if this line of thinking is correct? Thankful for any feedback.
  2. jcsd
  3. Oct 31, 2009 #2
    Re: smooth and L^2 on R^n. Will it be bounded??

    You have to define "smooth" ... For n=1 do you disallow a sequence of bumps, getting thinner but taller, as you go to infinity. Such a thing could be infinitely differentiable. But the derivative won't be bounded.
  4. Oct 31, 2009 #3
    Re: smooth and L^2 on R^n. Will it be bounded??

    As I understand it the question is about one single function, not about a sequence of functions, Any function in L^2, be it smooth or not, needs to be bounded, in the sense that the essential supremum of |f| is finite, for otherwise neither f nor f^2 would be integrable...

    Of course you can always change the values of f on set of measure zero so as to make f unbounded and without changing f (as an element of L^2).
  5. Oct 31, 2009 #4
    Re: smooth and L^2 on R^n. Will it be bounded??

    Thanks for the quick answers. With smooth I just meant that all the derivatives exist and are continuous. So with smooth I meant to disallow for the above mentioned example regarding sets of measure zero.

    And yes, the question was about one single function, but I think the sequence of bumps were meant as a series of bumps along the x-axis (n=1), i.e. one single function.

    So, this bump-function, let's call it f, is unbounded and smooth. Meaning that for every M>0 there is an x and and an epsilon such that f>M on the ball around x of radius epsilon. Couldn't this function still be L^2 if the epsilon went to zero sufficiently fast as M got larger?

    The problem for me, I think, is that I haven't studied measure theory. But am I correct if I say that smooth and unbounded implies that for every M>0 there exists a set A, where f>M on A,such that [tex]\int_A dS >0[/tex] . In the same way as above couldn't the L^2 norm of f be finite even if ess sup f isn't? Providing that the measure of A goes to zero as M gets larger??
    Last edited: Oct 31, 2009
  6. Nov 1, 2009 #5
    Re: smooth and L^2 on R^n. Will it be bounded??

    To make things clearer, consider this example(I think this is the function g_edgar mentioned).

    Let [tex]\psi(x)[/tex] be a smooth non-negative function such that its support is contained in the interval (-1,1) and such that [tex]\int_R\psi(x)dx=1[/tex].

    Now consider the series [tex]\sum_{k=1}^\infty k\psi ((x -2k)k^3 )[/tex]

    I believe this series converges pointwise to a smooth unbounded function f(x). But is [tex]f\in L^1 (R)[/tex]? If we were allowed to integrate the series term by term we would get

    [tex]\int_R\sum_{k=1}^\infty k\psi ((x -2k)k^3)dx = \sum_{k=1}^\infty 1/k^2dx[/tex]

    But the series doesn't converge uniformly so I guess we aren't allowed to do this??
    Last edited: Nov 1, 2009
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