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A few weeks ago, I looked into the theory of locally convex spaces. Now I want to continue with the theory of distributions. The following description is somewhat confusing to me:

(*)

The basic idea in the theory of distributions is, to comprehend a function $f$ as a functional $T_f$ via $T_f(\phi)=\int f(x) \phi(x) dx$ on a suitable function space.


Here is what I know: $(**)$

For $\Omega \subseteq \mathbb{R}^n$, we have the vector space of smooth functions of $\Omega$, $\mathcal{E}(\Omega):=C^{\infty}(\Omega):=\{\phi: \Omega \rightarrow \mathbb{R}| \phi \text{ is a } C^{\infty} \text{ function}\}$.

Further,for $C \subset \mathbb{R}^n$ compact denote $\mathcal{D}_C(\Omega):= \{\phi \in C^{\infty}| \text{supp}(\phi) \subseteq C\}$.

Then one can define the space of test functions as $$\mathcal{D}(\Omega):=\underset{C \subset \Omega, C \text{ compact} }{\bigcup} \!\!\mathcal{D}_C(\Omega).$$

On $\mathcal{D}_C(\Omega)$ one can define a locally convex space via the seminorms $$p_m(\phi):=\underset{|\alpha| \leq m}{\max} {\sup_{x \in \Omega}|\partial^{\alpha \phi(x)}|},$$ where $\phi \in \mathcal{D}_C(\Omega)$.

On $\mathcal{D}(\Omega)$ the set of seminorms $p$ on $\mathcal{D}(\Omega)$ such that $p_{|\mathcal{D}_C(\Omega)}$ is continuous w.r.t $\tau_C $for every compact $C \subset \Omega$ generates a locally convex topology on $\mathcal{D}_C(\Omega)$.


My question is the following:

In detail, what is the basic idea of distributions?

Especially, since the description of the idea in $(*)$ sounds a lot like some (Riesz-) Representation theorem.

For context, I do know the basics about locally convex spaces. I haven't encounter distributions thus far.

Denis
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    What is your background in Laplace and Fourier transforms? How about differential equations? – J W Aug 04 '25 at 14:06
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    My own experience was that I began to understand distributions when I began to understand the Dirac delta function. This might or might not work for you. – Jayanth R Varma Aug 04 '25 at 14:07
  • If $f\in L^\infty$, then for all $g$ with compact support ("test functions"), you can look at $\int_{\mathbb R} fg$. The map $g\mapsto \inte_\RR fg$ is a distribution. It faithfully represents $f$. So every function like $f$ is a distribution. So distributions are generalizations of functions. They are useful for understanding PDE. – Wulfhartus Aug 04 '25 at 14:38
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    Have you ever heard of the Dirac delta function? That is the most basic distribution - it is a misnomer to call it a function (although it is commonly described and used as one.) – Thomas Andrews Aug 04 '25 at 14:57
  • In general, a lot of functions do not have a Fourier transform function, but they do have Fourier transform distributions. So it gives us an idea of what space the Fourier transform and other transforms "really" act on - not function spaces, but distribution spaces. – Thomas Andrews Aug 04 '25 at 14:59
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    My two comments above come together in realizing that the Dirac $\delta$ "function" is the distribution you get taking the Fourier transform of the constant function $f(x)=1.$ – Thomas Andrews Aug 04 '25 at 15:04
  • @JW I do know about Fourier Transformation and I have some basic knowledge about ordinary differential equations. But no knowledge on Laplace Transform or pde. – Denis Aug 04 '25 at 20:18
  • @DenisGerhardt, it's often in the study of partial differential equations that the motivation for much of functional analysis becomes clearer. – J W Aug 05 '25 at 07:59
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    Tao has good notes explaining this topic https://terrytao.wordpress.com/2009/04/19/245c-notes-3-distributions/ – D.R. Aug 05 '25 at 18:59

1 Answers1

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The main idea behind distributions is the generalization of the notion of functions and their derivatives.

Weakly Differentiable Functions

There are many functions in mathematics, calculus, and engineering that are not differentiable at every point.
A simple example is
$$ F(x) = |x|. $$ Its derivative is $-1$ for $x<0$ and $1$ for $x>0$, but it is not defined at $x=0$.
However, if we consider
$$ f(x) = \operatorname{sgn}(x) := \begin{cases} -1, & x<0,\\ 0, & x=0,\\ 1, & x>0, \end{cases} $$ we see that
$$ F(x)-F(y) = \int_x^y f(t)\, dt, $$ for every $x,y \in \mathbb{R}$.

Thus, $F$ behaves like a primitive of $f$ on $\mathbb{R}$, except that $f$ is not regular enough to apply the classical rules of calculus, where primitives are defined for continuous functions. In particular, we cannot conclude that $F'(0) = f(0)$, since $f$ is discontinuous at $0$.

Nevertheless, in some weak sense, $f$ is the derivative of $F$.

This might look trivial, but there are natural situations where weak derivatives are convenient. A classical example arises in the earliest partial differential equations in the literature: the wave equation and the heat equation.

Consider someone playing a guitar. Under some (idealized) assumptions, the motion of the string is described by the wave equation
$$ u_{tt}(x,t) = c^2\, u_{xx}(x,t), $$ where $u(x,t)$ is the height of the string at position $x$ and time $t$. Here, $u_{tt}$ and $u_{xx}$ are second-order partial derivatives in $t$ and $x$.

A simple difficulty occurs if the musician plucks the string: the initial shape of the string at $t=0$ may resemble the graph of $|x|$. Then $u(x,0)$ is not differentiable with respect to $x$ at some point. How can we use a differential equation if the initial shape is not differentiable?

The idea is to differentiate without ever differentiating directly, using the principle of integration by parts.

Integration by Parts

The classical Integration by Parts formula states that if $f$ and $\phi$ are smooth (say, $C^1$) then
$$ \int_a^b f'(x)\phi(x)\, dx = - \int_a^b f(x)\phi'(x)\, dx + f(b)\phi(b) - f(a)\phi(a). $$

If $\phi$ vanishes outside $(a,b)$, this simplifies to
$$ \int_\mathbb{R} f'(x)\phi(x)\, dx = - \int_\mathbb{R} f(x)\phi'(x)\, dx. $$

The surprising fact is that the right-hand side still makes sense even if $f$ is not classically differentiable, provided $\phi$ is smooth enough. Thus we can compute
$$ \int f'(x)\phi(x)\, dx $$ without ever differentiating $f$ directly.

Very Regular Functions as Probes

In practice, you never observe a function $f(x)$ directly. For example, if $f(x)$ is the temperature of a metal bar, you only measure temperature averages near a point. You are probing $f$ using a smoother function $\phi$. Formally, define

$$ T_f(\phi) := \int_\mathbb{R} f(x)\,\phi(x)\, dx. $$

An interesting example is when $\phi$ is a $C^\infty$ function with compact support so that $\phi\ge0$ and $\int \phi \, dx = 1$. Then $T_f(\phi)$ is a kind of weighted average of $f$ over the support of $\phi$.

Importantly, if $f$ and $g$ are continuous and
$$ T_f(\phi) = T_g(\phi) $$ for every $\phi\in C_c^\infty(\mathbb{R})$, then $f=g$. Here $C_c^\infty(\mathbb{R})$ denotes the vector space of all $C^\infty$ functions with compact support. Thus, the functional $T_f$ contains all the information about $f$, so we can identify $f$ with this linear functional on the space of test functions.

Moreover, if $f\in C^\infty$, integration by parts yields
$$ T_{f'}(\phi) = -\,T_f(\phi'), \tag{1} $$ so we can recover the derivative of $f$ in this weak sense.

We define $g$ to be the weak derivative of $f$ if
$$ T_g(\phi) = -\, T_f(\phi') $$ for every test function $\phi$.

The Breakthrough

Once we recognize that a function $f$ can be identified with its associated functional $T_f$ on $C^\infty_c$ and approach differentiation of $f$ in terms of such functionals, the next step is to consider arbitrary linear functionals.
We can then extend the notion of the derivative of a function to every such functional on $C^\infty_c$, even those not arising from a function.

Define a distribution as any linear functional
$$ T : C^\infty_c(\mathbb{R}) \to \mathbb{R}, $$ and its derivative by
$$ T'(\phi) := -\,T(\phi'). $$

This allows us to differentiate objects that are not functions and to obtain derivatives that are also not functions.

For example, differentiating $F(x)=|x|$ in the sense of distributions gives
$$ T'_F = T_{\operatorname{sgn}}, $$ where $\operatorname{sgn}$ is the sign function defined above.

Differentiating again,
$$ T''_F = T'_{\operatorname{sgn}} = 2\,\delta_0, $$ where $\delta_0$ is Dirac’s delta at $0$, defined by
$$ \delta_0(\phi) := \phi(0). $$

This distribution does not come from any classical function. It comes from a measure assigning weight $1$ to the point $0$. And we can keep differentiating ad infinitum.

The space $C_c^\infty$ is a classical space of test functions. We can also considered other test function spaces (such as $C^\infty$ functions with rapid decay), which we omit here.

That is a major advantage of distributions: they allow us to place many disparate objects in the same space: $L^p$ functions, wildly discontinuous functions, continuous functions, $C^k$ functions, regular measures, and even more exotic objects, are all distributions, functionals on a suitable space of test functions.

Within this framework, we can with great freedom take linear combinations, form series, and differentiate these otherwise unrelated objects. Distributions allow us to treat all of them on equal footing.

Why Distributions Are Useful

Distributions are extremely useful because they allow you to break down the proof of the existence of solutions to problems involving functions and its derivatives (such as partial differential equations) into two steps:

  1. Existence as a distribution: First, prove that a solution exists in the sense of distributions. At this stage, one can freely manipulate derivatives of arbitrary order without worrying too much about its existence in a strong sense and its regularity.

  2. Regularity improvement: Then, step by step, establish that this distribution is more regular, eventually showing that it corresponds to an actual function with the desired regularity.

A final example

I have also encountered distributions in my own work (and with my students), not in the classical PDE setting, but in the study of functions similar to the Weierstrass function $$ f(x)=\sum_{n=0}^\infty \frac{1}{2^n} \cos(2^n x). $$

This belongs to the famous family of examples by Weierstrass of nowhere differentiable functions. However, one is tempted to formally differentiate term by term:
$$ f'(x)=\sum_{n=0}^\infty -\,\sin(2^n x), \tag{2} $$ which does not make sense in the classical sense. Nevertheless, the right-hand side does define a distribution. Indeed, for a $2\pi$-periodic $C^\infty$ function $\phi$,
$$ T'_f(\phi) = - \int_0^{2\pi} f(x)\,\phi'(x)\, dx = \sum_{n=0}^\infty \int_0^{2\pi} -\,\sin(2^n x)\,\phi(x)\, dx. $$

Thus, the formal derivative in (2) makes sense in the sense of distributions, and this approach can be used to study the regularity of the Weierstrass function and similar examples.

Historical references

The main contributors to the development of the theory of distributions are Sobolev and L. Schwartz. For readers interested in the history of distribution theory, see

Jesper Lützen. The Prehistory of the Theory of Distributions. Studies in the History of Mathematics and Physical Sciences (HISTORY, volume 7). Springer.

Also of interest is the autobiography of Schwartz

Laurent Schwartz. A Mathematician Grappling with His Century. Birkhauser, 2001.

and

Jean-Michel Kantor. Mathematics east and west, theory and practice: the example of distributions. The Mathematical Intelligencer. Volume 26, pages 39–50, (2004).

Daniel Smania
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    This explanation is on par with ones seen in advanced PDE classes (and in some sense more intuitive). Would you mind demonstrating how $T'F=T{\text{sgn}}$ for $F(x)=|x|$? :) – Man-I-Fold Aug 04 '25 at 23:55
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    @Man-I-Fold Write $\int_{-\infty}^\infty |x|\phi'(x),dx = -\int_{-\infty}^0 x\phi'(x),dx + \int_0^\infty x\phi'(x),dx$ and integrate by parts to get $-\int_{-\infty}^\infty \text{sgn}(x)\phi(x),dx$. – Ted Shifrin Aug 05 '25 at 00:07
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    Here's another example that's easy on the eyes. Define $u,v:(0,2)\to\mathbb{R}$ by $u(x):=\begin{cases}x&\text{ if }0<x\leq 1\1&\text{ if }1\leq x<2\end{cases}$, $v(x):=\begin{cases}1&\text{ if }0<x\leq 1\0&\text{ if }1<x<2\end{cases}$. We claim that $v$ is a derivative of $u$ on $(0,2)$ in the weak sense from Evans' 2nd ed of PDEs. To see this, fix any $\phi\in C^\infty_c((0,2))$. We calculate \begin{align}\int_0^2u\phi'dx&=\int_0^1x\phi'dx+\int_1^2\phi'dx\&=-\int_0^1\phi dx+\phi(1)-\phi(1)\&=-\int_0^11\phi dx-\int_1^20\phi dx=-\int_0^2v\phi dx.\end{align} – Man-I-Fold Aug 05 '25 at 01:12
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    In the IBP section you say “The surprising fact is that the right-hand side still makes sense even if…”, but that’s not really surprising is it. It is very trivial that under very mild assumptions on $f$, and one regularity assumption on $\phi$, the quantity $\int f\phi’$ makes sense. Instead I think the ingenuity comes in the fact that we can use the RHS pretty much as a (generalized) definition for the LHS. And this entire idea is predicated on the principle that integrals are “better” (easier to handle/manipulate) than derivatives. – peek-a-boo Aug 05 '25 at 01:15
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    Also, it should be emphasized that generalizing definitions purely for the sake of generality is a silly idea. But the utility of distributions comes, as you’ve explained, because of how effectively they solve several problems (especially when dealing with PDEs, even ‘simple’ ones like Laplace’s equation and the flat linear wave equation). And thinking ahead, the introduction of the Schwartz space is a perfect intermediate between $C^{\infty}_c$ and $C^{\infty}$ :) – peek-a-boo Aug 05 '25 at 01:17
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    @peek-a-boo Well, it was truly surprising to me the first time I saw it.
    Perhaps less so now, with my tired old eyes. :)
    – Daniel Smania Aug 05 '25 at 01:18
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    @peek-a-boo I understand that Schwartz was sometimes criticized because this idea was already “in the air” and seemed almost obvious. But truly great ideas often look obvious a posteriori. Schwartz systematized ideas from many predecessors, including Sobolev.
    There is a certain beauty in this achievement that, in my view,
    not only includes but surpasses its utility. It became a language that permeates almost everything, and it is now practically impossible to escape it.
    – Daniel Smania Aug 05 '25 at 01:28
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    I love this answer. I would put emphasis on the "functions as probes" and "all these objects on equal footing" concepts. Objects like the Dirac delta are often treated "as functions" by physicists, which mathematically is inconsistent for a function (a function that is zero almost-everywhere but has integral equal to 1?) but works perfectly for a distribution. – Stef Aug 05 '25 at 08:54