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Non continuously differentiable but inner product finite

  1. Mar 21, 2008 #1

    I was trying to understand Green's function and I stumbled across the following statements which is confusing to me.

    I was referring to the following site

    http://www.math.ohio-state.edu/~gerlach/math/BVtypset/node79.html [Broken]

    Here the author says the following

    "What if $ u$ is not a continuously differentiable function? Then its image $ Lu$ is not square-integrable, but the inner product <v, Lu> is still well-defined because it is finite. For example, if u is a function which has a kink, then $ Lu$ would not be defined at that point and $ Lu$ would not be square-integrable. Nevertheless, the integral of $ \overline v Lu$ would be perfectly finite."

    I dont understand the fact is if Lu is not defined how can u define an inner product with v at any point, ie <v, Lu>. What does it mean physically at all, is it a mathematical jugglery to move the L operator to v and then say that look it is still defined? I am totally confused.

    Thanks a lot for any help in advance.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Mar 21, 2008 #2


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    Science Advisor

    Why do you say "Lu is not defined"? If Lu is not square-integrable, then it is not in L2 but it is in some larger space, of which L2 is a subspace. The innerproduct can be defined in that larger space.
  4. Mar 22, 2008 #3

    I say L is not defined because of the following. Lets give an example. Since L can be d^2/dx^2 + a(x) d/dx + b(x) and if u consider the function u s.t

    for say (a< x <b), a<0, b >0

    u(x) = 0 x<0,
    = x x>= 0

    The fn u is cont but is not differentiable at x = 0. So I am not sure how for such functions u can define the operator like this. This is my question.

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