Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What does test mean in distribution theory

  1. Oct 20, 2009 #1
    What does "test" mean in distribution theory

    Dear all,

    I am recently exposed to the distribution theory. Why are "test" functions called "test" functions? What are they testing for? or What are they tested for?

    Why do we need to introduce the idea of distribution? It is merely for explaining the application of delta function in a more rigorous manner?

    Thank you for clarifying this.

    kzhu
     
  2. jcsd
  3. Oct 21, 2009 #2
    Re: What does "test" mean in distribution theory

    A test function in distribution theory is simply an infinitely smooth function, which is zero outside some interval. This enables us to differentiate any function an infinite number of times in the distributional sense. This in turn allows us to obtain weak solutions for a singular or any other ugly differential equation.
     
  4. Oct 21, 2009 #3
    Re: What does "test" mean in distribution theory

    Thank you for the clarification.
     
  5. Oct 21, 2009 #4
    Re: What does "test" mean in distribution theory

    On a second thought, would "distribution basis function" be a more descriptive name? I have also find the name "tempered function" difficult to grasp.
     
  6. Oct 22, 2009 #5

    Landau

    User Avatar
    Science Advisor

    Re: What does "test" mean in distribution theory

    Not just some interval, but a compact interval. In other words, test functions are smooth functions with compact support.
     
  7. Oct 24, 2009 #6
    Re: What does "test" mean in distribution theory

    Is there 'something' stronger than a distribution ??, i mean perhaps there are expressions that can not be handled even with distribution theory
     
  8. Oct 24, 2009 #7

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: What does "test" mean in distribution theory

    Why do we want a bigger space?

    There are various reasons why, when working with an infinite-dimensional Hilbert space, one might desire to have "more" vectors available. I can think of two off the top of my head:

    . You want differentiation to be an operator. (In common cases, it's only a partial operator)
    . You want every operator to have enough eigenvectors

    Let me clarify that last point -- the spectrum of an operator T is the set of all [itex]\lambda[/itex] such that the operator [itex]T - \lambda[/itex] is not invertible.

    In finite-dimensional linear algebra, this means [itex]\lambda[/itex] is an eigenvalue of T, and you can find associated eigenvectors v such that [itex]Tv = \lambda v[/itex]. If T is nice enough (or if we use generalized eigenvectors), we can find a set of T's eigenvectors that form a basis for our vector space. This is very convenient.

    However, infinite-dimensional linear algebra doesn't have this feature. Operators still have a spectrum, but many don't have eigenvalues or eigenvectors. The position (partial) operator X of quantum mechanics is a typical example.

    So, it is convenient to find a vector space larger than our Hilbert space that contains eigenvectors for our favorite operators.

    How do we build a bigger space?

    The easiest way to build a bigger space is via test functions. One of the key features of a Hilbert space H is that it is isomorphic to its dual space H*. If we pick a subspace T of H, then H* will be a subspace of T*.

    Furthermore, many of the properties of T* are determined by the properties of T -- in particular, if an operator acts on T, it also acts on T*. For example:

    . If T contains only differentiable functions, and is closed under differentiation, then differentiation acts on T. Therefore, differentiation acts on T* too -- every vector now has a derivative. (Even those from H; their derivatives aren't in H, but they are in T*)

    . If T is the space of tempered distributions, then Fourier transform acts on T. Therefore, Fourier transform acts on T* too. This is very useful if you want to use Fourier analysis!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: What does test mean in distribution theory
  1. What does this mean? (Replies: 2)

  2. What does this mean? (Replies: 8)

  3. What does ye mean? (Replies: 6)

Loading...