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Questions about generalised functions, and delta

  1. Aug 27, 2006 #1
    Hi everybody,

    I have just started reading some things about generalised functions, and i have some question. The source I am reading from is a book of partial differential equations so it's not a very formal introduction to generalized functions and functionals, but there are some basic things to help understand the Green function etc...

    First of all, it says that a functional of an integrable on any closed interval function f is a mapping between a set of continuous and infinite times derivable functions (h) and the real numbers. The mapping is defined by the following rule:

    [tex]D(h)=\int_{-\infty}^{\infty}f(x)h(x) dx[/tex]

    Then, it goes one to define the generalised function as a linear and continuous functional. I will use the symbol <f,h> for the generalised function [tex]T(h)=<f,h>=\int_{-\infty}^{\infty}f(x)h(x) dx[/tex]

    The derivative of T is defined like this: [tex]T^{(n)}(h)\equiv(-1)^{n} T(h^{(n)})[/tex]
    and it is always a generalised function. But in the book, it also mentions that to reach this definition, we thought of the functional of <f',h> with f' integrable on any interval,

    So, my basic question is(concerning delta, too): When we write <f,h> we mean a generalised function. So does this mean that f(x) exists and is a real function and we can write f(x), meaning the real function from which the functional occured? This actually relates to delta. I mean, delta is defined as the derivative of the functional of Heaviside. The Heaviside function (let's call it H(x)) is integrable on any interval [a,b] so we can define <H,h>=T(h). Its first derivative is T'(h)=delta(h). But H'(x) isn't integrable on any [a,b], a<0<b, right? So what do we mean by writing delta(x)??

    I am afraid that my writing is too messed up for anyone to understand, so I will restate my question in another way and then end the topic. Is there any real integrable function f so that <f,h>=delta?? If there isn't any, how can we say that delta is a generalised function??

    This is actually what troubles me also when I read delta(x)? How can we do the transition and define delta as a real function, too??

    That's all. Sorry for the size of the post and my bad English.


    Note: I have checked arildno's post on delta. The theory mentioned there is a bit different, and I don't really have much time to read and understand what he writes, because an exam approaches, so I would appreciate if any answer was based on the definitions I mention.
    Last edited: Aug 27, 2006
  2. jcsd
  3. Aug 27, 2006 #2


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    More generally, a "functional" maps functions in some function space to numbers. It does not correspond to any function f. This formula shows how to associate a functional with any function and is defining a particular class of functionals. The "distributions" include functionals created in that way but also much more. The functionals created in that way are, trivially, "linear and continuous" as you say below. More generally, if we have a "linear and continuous functional", F(h), that can not be created from a function like that, we can talk about the "distribution" (or "generalized function"), f, such that [itex]F(h)= \int_{-\infty}^\infty f(x)h(x)dx[/itex].
    For example, the functional F(h)= h(0) that, to every function, associates its value at x= 0 is clearly both linear (if v(x)= ah(x)+ bk(x) then v(0)= ah(0)+ bk(0) so F(v)= ah(0)+ bk(0)= aF(h)+ bF(k)) and continuous (of course, all constant functions are continuous!). Then we say that [itex]F(h)= \int_{-\infty}^\infty h(x)\delta(x)dx[/itex] so that this functional "defines" the delta function used by P. A. M. Dirac. Of course, that is not a function. Physicists can get away with that but mathematicians have to make it precise.

    More correctly, that is the nth derivative.

    Careful: that is saying that, for each generalized function that can be defined by a "true" differentiable function, we want the derivative of the functional to correspond to what we would get from the derivative of the function. Of course, even those generalized functions that do correspond to functions may not correspond to differentiable functions.

    More correctly, <f, h> means the result of applying the functional corresponding to f, whether f is a true function or not, to the function h.
    No. Some generalized functions correspond to "true" functions using the integral formula above but most don't- the "true" functions are a (small) subset of all generalized functions.

    Not exactly true. [itex]\delta(x)[/itex] is defined as the functional that, to every function h(x), assigns the numeric value h(0). It can then be shown that the derivative of distribution defined from the Heaviside function, in the sense of distributions, is the [itex]\delta[/b] "function".

  4. Aug 27, 2006 #3
    First of all, thank you for your thorough answer.

    From what you say in your post, I assume that I have misunderstood some things about functionals and generalised functions, partially due to the fact that I have read only a very brief introduction to functionals, and this coming from a book of differential equations and not one about functionals.
    You are probably right. I have checked the definition of functional on mathworld (-> http://mathworld.wolfram.com/Functional.html) and it seems that I got it wrong that each functional is related to ("corresponds to" as you mention) any function. This false ideal actually created the most problems in my head, because I thought that every functional was based on a real integrable function.
    If the "function" f in the integral is not a real function, like in the example with the delta function, then why do we think of it as a generalised function that belongs in the, as you mention, class of functionals that are given by the integral-formula and we use integrals with delta in it? We should just consider it a linear, continuous functional, having nothing to do with integrals. I say that, because in the case of delta and many others, the integral definition must change if we are to accept integrals with delta for example. Or maybe it is just a symbolism, although I don't think that's the case, because I think I have seen examples that we use properties of the Riemmann Integral to calculate integrals with delta in them.
    I mentioned that the delta was defined as the derivative of Heaviside, because with the misconceptions that i had I thought that this proved that actually delta existed, and that if we gave the definition you give, we wouldn't be sure that such a generalised function exists. But of course, this thought was totally wrong. You are right.

    That's all for now. I am still somewhat confused, but you helped me clear many things up. Thanks again
  5. Aug 27, 2006 #4


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    For a little perspective... integrals are rarely interesting things. The things we really want to know are (multi)linear functionals. For example, we might want to know the electrostatic force that one charged body exerts on another. (this is bilinear in the two charge distributions)

    It just so happens that the idea of electrostatic force is differentiable, which allows us to looking at things like charge density and electrostatic field strength. And then, we integrate to undo the differentiation and recover the thing we're really interested in.

    However, distributions like that describing a point charge are not differentiable, so we can't approach a problem using those methods.

    (I'm using differentiable in a slightly different sense than you're used to, but I think it's still the right word to use)
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