The distinction between "ordinary" functions and generalized functions.

Question

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.

Answer 1

Given a set $\mathscr{D}$ of test-functions (say $\mathscr{D}\subset C^0$) one defines a corresponding set of generalized-functions as the corresponding dual space, i.e. $$ \mathscr{D}'=\{\mathscr{F}\colon\mathscr{D}\rightarrow \mathbb{R} ~\text{linear}\}. $$ (That is only half the truth, as one usually also specifies a topology on $\mathscr{D}$ and only studies the continuous dual, i.e. functionals which are continuous with respect to that topology. But for the point I want to make we can ignore that.)

Common choices for $\mathscr{D}$ are $C_c^\infty$,$\mathcal{S}$ or $C^\infty$, giving rise to distributions, tempered distributions and compactly supported distributions. Why are those choices particularly good?

Well, you want to generalize ordinary functions, in this context we usually mean (classes of) functions in $L^1_{loc}$. That is you want to assure that for a large subset $F\subset L^1_{loc}$ you have $$ F \subset \mathscr{D}' $$ in some sense (namely as an embedding). And indeed to a function $f\in F$ you can oftenassociate a generalized function $\mathscr{F}_f$, by intergating against $f$. (For this you need products of the form $f \phi$ to be integrable for $\phi \in \mathscr{D}$.)

In order to speak of $F$ as a subset of $\mathscr{D}'$, you must not lose any information, i.e. $f \mapsto \mathscr{F}_f$ has to be injective. The fundamental lemma of calculus of variations tells you that this is the case for $\mathscr{D} = C_c^\infty$ and $F = L^1_{loc}$. But for your choice of $\mathscr{D}$ (consisting of Gaussians of some sort) the class $F$ on which $f \mapsto \mathscr{F}_f$ is injective is rather small. For example all odd functions with compact support get mapped to zero.

Note that $C_c^\infty\subset \mathcal{S} \subset C^\infty$, dualizing changes the order of inclusion and thus $\mathscr{D}=C_c^\infty$ gives rise to the larges class of distributions among those three and in most cases it is the common choice when wants to deal with generalized functions. In this case we have $L^1_{loc} \subset \mathscr{D}'$ as mentioned above, and what the author is referring to is that no $L^1_{loc}$ gives rise to the Dirac $\delta$. This fact follows directly from the fundamental lemma mentioned above.

Answer 2

(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:

Falsehood: There exists a measurable function $g$ on the reals such that, for every continuous function with compact support $h$ on $\mathbb{R}$, you have $h(0) = \int_{-\infty}^{\infty} g(x) h(x) \, \mathrm{d}x $.

Explicitly, the negation of this false claim is:

Theorem: For every measurable function $g$ on the reals, there exists a continuous function $h$ with compact support on the reals such that $h(0) \neq \int_{-\infty}^{\infty} g(x) h(x) \, \mathrm{d}x$.

_{1: Or collection or set or aggregation or type or whatever noun you like to use for this basic notion}