Confused about the concept of distributions and functions

Question

I just learned the concept of distributions, but I'm confused about the concept, so my question is simple and may looks a bit strange (sorry about that..)

Firstly, I know that a distribution is a continuous linear functional, mapping from $D(\Omega)$ to a complex number.
So if $f\in D'(\Omega)$, then the parameter for $f$ should be a function, right? (which I mean $f$ should be like $f(g)$ or $f(\varphi)$, ($g$ and $\varphi$ are functions), but not $f(x)$ where $x$ is a real number)

And is this the main difference between a function and a functional?

But I also see examples like "let $f\in D'(\Omega)$ be a distribution defined by $f(x)=2x$ when $x>0$ and $f(x)=0$ when $x\leq 0$)". What does this mean? Is $f$ here a distribution or a function after all?

Secondly, a distribution $f$ operating on a function $\varphi$ should be defined by $<f,\varphi>$ equals something, but why both in my note and some textbooks, when discussing the multiplication by $C^{\infty}$ functions and differentiation, they all began with
$<af,\varphi>=\int_\Omega af\varphi dx=<f,a\varphi> \forall \varphi \in D(\Omega)$. Here suppose $f\in C$ (or $f\in L_{loc}^{1}(\Omega)$) and $a\in C^{\infty}(\Omega)$
and $<\partial^\alpha f,\varphi>=\int_{\mathbb R^N}(\partial ^\alpha f)\varphi dx=...$

Why here the distributions are set as integral automatically? Or shall we think of distribution as integral like above when doing operations on distributions?

Thanks so much!

Answer 1

If $f$ is ordinary function on an open set $\Omega\subseteq\mathbb{R}^n$, then you can define a linear functional by $$ \Phi_{f}(g) = \int_{\Omega}fg dx,\;\;\;g \in \mathcal{C}_{c}^{\infty}(\Omega), $$ where $\mathcal{C}_{c}^{\infty}(\Omega)$ consisting of compactly supported $\mathcal{C}^{\infty}$ functions. If $f$ were differentiable and $\alpha$ were an $n$-index, then the derivative $f^{(\alpha)}$ would satisfy the following identity obtained by integration by parts in each variable: $$ \Phi_{f^{(\alpha)}}(g)=\int_{\Omega}f^{(\alpha)}gdx=(-1)^{|\alpha|}\int_{\Omega}fg^{(\alpha)}dx,\;\;\; g\in\mathcal{C}^{\infty}_c(\Omega) $$ So, it is natural to define $\Phi_{f}^{(a)}$ to be the functional defined by $$ \Phi_{f}^{(\alpha)}(g)=(-1)^{|\alpha|}\int_{\Omega}fg^{(\alpha)}dx. $$ The advantage of this type of formalism is that every functional has all orders of derivatives because the derivatives fall on the test functions. And, if $f$ is classically differentiable, then $\Phi^{(\alpha)}_{f}=\Phi_{f^{(\alpha)}}$. So this is an actual extension of the notion of derivative, and not just an arbitrary definition. Even without that, it makes sense to define the following for general "distributions": $$ \Phi^{(\alpha)}(g) = (-1)^{|\alpha|}\Phi(g^{(\alpha)}). $$ Of course the main point is that this definition reduces to the ordinary one when the distributions are represented as $\Phi_{f}$ for a nice enough function $f$. And the reason that distributions are so useful is that they allow you to find "solutions" of partial differential equations. The tricky part is showing that the solution is an actual solution, which is not always true.

When you're working on $\Omega=\mathbb{R}^{n}$, the better topology on the functions involves looking at conditions that are preserved under the Fourier transform, which is how Schwartz space came about https://en.wikipedia.org/wiki/Schwartz_space . In this setting every distribution has a Fourier transform. The idea comes from the observation that $$ \int_{\mathbb{R}^n}\hat{f}g dx = \int_{\mathbb{R}^{n}}f\hat{g}dx,\;\;\; f,g\in L^1(\mathbb{R}^n). $$ So the Fourier transform of Schwartz distribution is defined as $$ \hat{\Phi}(f) = \Phi(\hat{f}). $$ The Dirac delta lives in the space of Schwartz distributions, because the following is a distribution: $$ \delta(f)=f(0) $$ And, as expected, $$ \hat{\delta}(f) = \delta(\hat{f})= \hat{f}(0)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n} fdx = \frac{1}{(2\pi)^{n/2}}1(f) \\ \hat{\delta} = \frac{1}{(2\pi)^{n/2}}1. $$ The usual hand-waving arguments that the $\delta$ function makes sense and has a Fourier transform give way to rigorous Mathematics in this context. The delta function(al) is a nice thing to have in Mathematics.

Answer 2

Note: I just started learning this myself, so please correct me if I'm wrong.

There's an association made between the functional $F[\varphi] = \langle F,\varphi\rangle$ and the (generalized) function $f$ such that $$F[\varphi] = \int_\Omega f(x)\varphi(x)dx$$

Both $F$ and $f$ are often called the distribution.

A functional is a linear mapping from a vector space (in this case the space of test functions) to the base field (the complex or real numbers).

Note that not every distribution $F$ can be written as $$F[\varphi] = \int_\Omega f(x)\varphi(x)dx$$ for an actual function $f:\Omega \to \Bbb C$. But I guess people like the notation so much that we just extend the notion of function so that no matter what functional $F$ is, we write it in that integral form. Then if there's no such function that'll do the job we want of $f$, we just define the properties of $f$ such that it matches the prescription of $F$ and call it a generalized function. But again, the nomenclature is pretty ambiguous as both $F$ and $f$ may be called a generalized function or a distribution. But, taking the lead of physicist Bernard Jancewicz, I like to separate the ideas ($F$ is the distribution and $f$ is the generalized function).

Answer 3

All "reasonable" functions are distributions, but not all distributions are functions. When you hear talk of a function $f$ as a distribution, the talkers usually mean the functional $\varphi \to \int_\Omega f\varphi.$ If for example $f$ is Lebesgue integrable on each compact subset of $\Omega,$ like your $2x,0$ example, then $f$ is reasonable. Even some "unreasonable" functions can define distributions, but that is a discusssion for another rainy day.