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Distributional derivative of one-parameter family of distributions

  1. Jun 16, 2014 #1
    Suppose, for a suitable class of real-valued test functions [itex]T(\mathbb{R}^n)[/itex], that [itex]\{G_x\}[/itex] is a one-parameter family of distributions. That is, [itex]\forall x \in \mathbb{R}^n, G_x: T(\mathbb{R}^n) \to \mathbb{R}[/itex].

    Now, suppose [itex]L[/itex] is a linear differential operator. That is, [itex]\forall g \in T(\mathbb{R}^n)[/itex] makes sense in terms of the normal definitions of derivates (assuming, of course, that [itex]g[/itex] is sufficiently smooth). [itex]L[/itex] also has meaning when acting on distributions by interpreting all derivatives as distributional derivatives. For example, the derivative of the distribution [itex] \frac{\partial}{\partial x_i} G_{x_0} [/itex] is defined by: [itex]\forall g \in T(\mathbb{R}^n), \frac{\partial}{\partial x_i} G_{x_0}(g) = G_{x_0}(- \frac{\partial}{\partial x_i}g)[/itex].

    Note that smooth functions can multiply distributions to form a new distribution in the following way. Suppose [itex]f:\mathbb{R}^n \to \mathbb{R}[/itex] is a smooth function. Then [itex]f G_x[/itex] is defined by: [itex]\forall g \in T(\mathbb{R}^n), (f G_x) (g) = G_x(f g) [/itex]

    These facts give [itex]L G_x[/itex] meaning.

    Also, for fixed [itex]g \in T(\mathbb{R}^n)[/itex], [itex] (x \mapsto G_x (g)) [/itex] is a possibly (non-smooth?) function. Define the function [itex]\psi_g : \mathbb{R}^n \to \mathbb{R}[/itex] by [itex]\psi_g (x) = G_x(g) [/itex]

    Now, here is my question: When is the following equality true?

    [tex] L (\psi_g) (x_0) = (L G_{x_0}) (g), \forall x_0 \in \mathbb{R}^n [/tex]
     
  2. jcsd
  3. Jul 10, 2014 #2
    I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
     
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