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What is a (linear) functional?

  1. Apr 23, 2012 #1
    Hey all,

    I have been reading up on Green Functions and I stumbled upon the term "linear functional". I know the properties of the linear operator, but i can't really grasp what a functional does.

    In my notes it says that it indicates a linear function whose domain is a function space, and that is maps a function to its value at a point, such that:

    [tex]L_ξ=u(ξ)[/tex]

    Can someone clarify what this means? I don't understand the qualitative aspect, i.e. what it actually "does". Is it just a way to talk about the operation of going from the function domain space to its value space? What is the added value of using it instead of saying X->Y?
     
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  3. Apr 23, 2012 #2

    micromass

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    It's just some terminology that occurs a lot.

    For example, let [itex]\mathcal{C}(\mathbb{R},\mathbb{R})[/itex] be the continuous functions from [itex]\mathbb{R}[/itex] to [itex]\mathbb{R}[/itex].

    Then your functional is the same as saying (for example)

    [tex]\mathcal{C}(\mathbb{R},\mathbb{R})\rightarrow \mathbb{R}:f\rightarrow f(0)[/tex]

    (where f(0) can be replaced by f(2) or f(-10) or whatever).
     
  4. Apr 23, 2012 #3
    Thanks for your reply!

    I think I understand. I suppose though that by using the functional we can can actually discuss about the properties of the operation in a way not possible (?) by the notation you used. For instance, I am reading this in the context of generalized functions and my notes say that the functional:

    [tex]L_g:C^0[a,b]->R[/tex]

    where g is a fixed continuous function, is not always valid. It goes on to say that there is no actual delta function [itex]δ_ξ(x)[/itex] such that the identity ([itex]L^2[/itex] inner product):

    [tex]L_ξ=<δ_ξ;u>=\int_a^bδ_ξ(x)u(x)dx=u(ξ)[/tex]

    holds for every continuous function u(x), and that every (continuous) function defines a linear functional, but not conversely. To be honest I don't really understand why this is the case, but is it the use of a functional that enables us to discuss this issue?
     
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