From Applications of Green's Functions in Science and Engineering by Michael D. Greenberg…
The author introduces generalized functions by considering the functional
$$\int_{-\infty}^\infty g(x)h(x) \, dx = \mathscr{F}(h) \tag{A}$$
That is for each function $h$ within some prescribed class
$\mathscr{D}$ of function, called the domain of $\mathscr{F}$, the
left side assigns a numerical value $\mathscr{F}(h)$. The choice of
the domain $\mathscr{D}$ rests with us.
He then goes on to supposing that if we specified the functional on the right side of (A) like
$$\int_{-\infty}^\infty g(x)h(x) \, dx = \int_{\xi}^\infty h(x) \, dx \tag{B}$$
then it follows that the kernel $g(x)$ must be the Heaviside step function. Agreed.
But then he claims that there is no such $g(x)$, in terms of "ordinary" functions that can be found for
$$\int_{-\infty}^\infty g(x)h(x) \, dx = h(0) \tag{C}$$
such that (C) holds for all $h$'s in $\mathscr{D}$
My questions is what does he mean by class of functions? Because obviously I could find a single $h$ for which we can find an "ordinary" $g(x)$, for example if $h(x) = e^{-\pi x^2} $, then $h(0) = 1$ and $g(x) \equiv 1$ and now, noting that I don't feel like doing the calculus but I am sure this is also true, for the function $h(x) = e^{-a x^4} $, where $a$ is some scaling factor for this gaussian type function.
So what I'm proposing is that there is a class of functions $\mathscr{D}$ comprised of
$$h_n(x) = e^{-a_n x^{2n}} \tag{$\star$}$$
where it is clear that $h_n(0) = 1 = \int_{-\infty}^\infty h_n(x) \, dx$. That is $g(x) \equiv 1$ would work for this entire class.
But I'm guessing I don't know what a class is, or that there's some other crucial part of the definition I'm not understanding.
Best Answer
(note: the end of this post contains a precise statement of what is what is meant by the statement quoted in the OP)
There are lots of strange functions; many operations you'd like to do to functions in analysis aren't really well-behaved (or even well-defined) for all of them.
For example, in a lot of settings the operation $\mathscr{F}(h) = h(0)$ tends to be badly behaved unless we restrict its domain to functions that are continuous at zero.
Now, generally one isn't interested in a single functional, but instead a whole collection of them that we can do algebra and analysis with; for example, we'd like to be able to talk about limits of functionals. This sort of idea is easiest to carry out when all functionals have the same domain.
So that's what $\mathscr{D}$ is — it is a class1 of functions we've agreed to take as the common domain of the functionals we're about to study. For example, we might take $\mathscr{D}$ to be all continuous functions.
Often, we have some recipe for taking $\mathscr{D}$ and producing a definition of what functionals we will study. One such recipe (called the "continuous dual) is to require that $\mathscr{D}$ be topological vector space, and then define "functional" to mean "continuous linear function $\mathscr{D} \to \mathbb{R}$".
Often, we can find nice analytic representations for functions. If $\mathscr{D}$ is a Hilbert space, for example, then the Riesz representation theorem says that for every continuous linear function $\mathscr{F} : \mathscr{D} \to \mathbb{R}$, there is an associated $f_{\mathscr{F}} \in \mathscr{D}$ such that $\mathscr{F}(d) = \langle f_{\mathscr{D}}, d \rangle$ is true for all $d$.
Often, Hilbert spaces are constructed whose elements are functions and whose inner product is a (Lebesgue) integral; e.g. $\langle f, g \rangle = \int_{-\infty}^{\infty} f(x) g(x) \, \mathrm{d}x $ is common. So for such examples, by the Riesz representation theorem, "functional" on this class means the same thing as "integration against a function in the Hilbert space".
Another important example is to take $\mathscr{D}$ to be the class of all continuous functions with compact support. In this case, the continuous dual construction produces the class of Radon measures, by the Riesz–Markov–Kakutani representation theorem.
The functional $\mathscr{F}(h) = h(0)$, for example, is a Radon measure. The statement you are asking about is the observation that it can't be written as integration against a function:
Explicitly, the negation of this false claim is:
1: Or collection or set or aggregation or type or whatever noun you like to use for this basic notion