# What examples of distributions should I keep in mind?

I'm learning a bit about the theory of distributions. What examples of distributions will help me develop good intuition?

Definitions: Let $U$ be an open subset of $\mathbb{R}^n$. Write $C_c^\infty(U)$ for the complex vector space of infinitely differentiable functions $U \to \mathbb{C}$ with compact support. A distribution on $U$ is a linear map $C^\infty_c(U) \to \mathbb{C}$, continuous with respect to a certain topology on $C^\infty_c(U)$.

Examples: If $\mu$ is a signed measure on $U$, finite on compact subsets, then $f \mapsto \int_U f \mathrm{d}\mu$ is a distribution. (This covers, for instance, the Dirac distribution, $f \mapsto f(0)$.)

More generally, write $D_i = \partial/\partial x_i$. Then $$f \mapsto \int_U D_{i_1} D_{i_2} \cdots D_{i_r} f \mathrm{d}\mu$$ is a distribution, for any indices $i_1, \ldots, i_r$ and measure $\mu$.

Any linear combination of such things is again a distribution, since distributions form a vector space. E.g. if $n \geq 3$ then there's a distribution $$f \mapsto \int_U D_3 D_1 f \mathrm{d}\mu + \int_U D_2^2 D_3 f \mathrm{d}\nu$$ for any measures $\mu$ and $\nu$. I guess we can also take infinite linear combinations, subject to convergence conditions.

My question: Is it OK if I go round thinking of things like the last example as being a typical sort of distribution? Or is the concept of distribution much more general than I'm realizing? The texts I've seen are short on this kind of intuition.

• The measure has to be finite on compact sets, right? – MLevi Nov 9 '09 at 3:31
• We can see the Dirichlet integral in terms of distribution , see my paper from my high school arxiv.org/pdf/1004.2653.pdf accepted in Elemente der Mathematik- European Mathematical Society – user21574 Dec 1 '17 at 21:03
• $D'(\mathbb{R})$ is the completion of $C^\infty_c$ for the weak topology induced by $\varphi \mapsto \langle \phi,\varphi \rangle$. The definition of $C^\infty_c(\mathbb{R})$ as the intersection of the Banach spaces $C^k_c(\Omega)$ for each compact $\Omega$ means the "order of a distribution" must be finite for each $\Omega$, and that any distribution is infinitely differentiable and integrable. – reuns Dec 1 '17 at 21:57
• @reuns Topology of the space of distributions is Montel topology – user21574 Dec 1 '17 at 22:25

From the way your question is phrased, it seems as though you want to get a handle on particular distributions rather than the space of all distributions. In which case, the result cited by Debraj is probably the most comprehensive. Properly stated, the result is:

Theorem: If $T \in C^\infty(\mathbb{R},\mathbb{C})'$ (continuous dual) with $supp T \subseteq K$ ($K$ compact) then there are integers $n_1$, $n_2$, ..., $n_p$ and continuous functions $f_1$, $f_1$, ..., $f_p$ with supports in $K$, such that

$$\sum_{j=1}^p f_j^{(n_j)} = T$$

The references for this are: Schwartz Theorie des distributions (1965) and Vo Khac Khoan Distributions, Analyse de Fourier. Operateurs aux derivess partielles (1972).

Then, of course, any arbitrary distribution can be written as the sum of distributions with compact support in a "nice" way.

From this point of view, the best examples are ones that are close enough to continuous functions that they are accessible (sorry, I know you're a category theorist but read that in British not categorish) but far enough away that you see some weird behaviour that you wouldn't expect if everything was a nice, continuous function. The examples mentioned in other answers are all good from this point of view: delta functions, derivatives of delta functions, $L^p$ functions, derivatives thereof. I'd add a few things like the Dirac comb, $\Delta_{a} = \sum_{n \in \mathbb{Z}} \delta_{n a}$ for $a \in \mathbb{R}$, $a \ne 0$, which has a particularly nice Fourier transform. You could integrate this to get an infinite staircase function (the floor function, that is). Indeed, any piecewise continuous function is actually a limit of a sequence of variations on the theme of Dirac's comb (i.e. where the tines can vary in length and separation) so Dirac's comb and its derivatives are the "only" distributions you need to know about.

But for me, this is the wrong way to think about distributions. If you want to understand distributions by looking at specific examples then you should really say that distributions are just smooth functions with compact support but in a slightly different topology. Once you've grokked the topology, then there's no reason not to simply think about really nice smooth functions. And if you haven't grokked the topology, then none of the "examples" is going to give you a good intuition as to how distributions behave. Indeed, I'd say that most of the examples are designed to make you think about the topology and to "shock" you into realising that the topology isn't what you naturally assume it should be when thinking about smooth functions.

I think of distributions simply as dual to smooth functions. The fact that we can think of functions as distributions is simply down to the fact that we have a pairing

$$(f,g) \mapsto \int_{\mathbb{R}} f(t) g(t) d t$$

between many of the different function spaces that we can define. (Note the lack of conjugation.) This pairing defines a map from the one function space into the dual of the other and we can ask how much of the dual we can see in this way. That's essentially what the results about representing distributions try to answer. But this doesn't give much intuition as to what the dual space looks like as a whole because it tries to build it up piece by piece, each time saying "have we got it all yet"?

For example, many of the answers you got talk about differentiation of distributions. How do we know that we can differentiate these? In one answer, you got the formula $\partial \phi (f) = - \phi( \partial f)$. Where did that minus sign come from? After all, if I'm in tempered distributions then I can define the Fourier transform of a distribution and then the formula is $\mathcal{F}(\phi)(f) = \phi(\mathcal{F}(f))$. Why a minus sign on the one and not on the other? And I can multiply smooth functions, so why can't I multiply distributions? What's going on?

The truth is that by simply embedding functions into distributions you miss out on the whole duality story and the difference between defining a dual operator versus an extension operator.

But I've already written up this part on the n-lab so I'll simply refer you to there for the next chapter. Take a look over there. And while you're there, add your favourite of the above examples and correct the statement of the theorem.

• Hey! Not fair! My first answer that gets more than 10 votes, and it's a community wiki question. Where's the referee when you need them?! – Andrew Stacey Nov 10 '09 at 7:55
• Andrew, you'll just have to bask in the warm glow of appreciation. Your answer was very helpful, especially to me. – Tom Leinster Nov 11 '09 at 23:26
• Bask. Bask. Given that it's below freezing here, then a warm glow of appreciation is more use than the reputation points so I'll content myself with that. – Andrew Stacey Nov 12 '09 at 8:25

A rather crazy (and very useful) example is a fundamental solution of an arbitrary differential equation with constant coefficients, i.e., a distribution $u$ satisfying $P(D)u=\delta_0$ where $P$ is a polynomial and $D$ is the differentiation operator. The construction can be found in many decent PDE textbooks. It is as far from the standard "take a non-smooth function, differentiate a few times" idea of how to get distributions as possible.

Another thing to understand is that, like with everything else, it is even more important to learn what you can and what you cannot do with distributions than what they can be.

• This is a good family of examples, but it seems more algebraic than analytic. The construction with P yields a D-module supported on the zero set of P. – S. Carnahan Nov 9 '09 at 15:09

(1) Every distribution can be locally represented as a (distributional) partial derivative of a continuous function. For example, for the dirac delta at 0, we can start from the function which is 0 for negative x, and equal to x for positive x and take two derivatives. Therefore, it is important to understand that not all distributions are made equal -- the more complicated ones are made by taking more derivatives of continuous functions.

(2) Some examples to definitely keep in mind (to emphasize the subtleness of the notion) while thinking about distributions are the principal value p.v $\frac{1}{x}$ and the pseudofunctions p.f. $\frac{1}{x^n}$

• I thought p.f. was pronounced as "finite part", but "pseudofunction" makes sense too. – timur Jun 27 '11 at 1:47
• @timur, it is Hadamard's "partie finie", en francais, hence "p.f." – paul garrett Dec 28 '17 at 21:46

(Not so much a response to the OP, but to the question in title.)

One of the motivations for Schwartz's distributions was to clarify the solution to the (free) wave equation (the Cauchy problem), after the earlier works of Hadamard and M. Riesz. Unlike the elliptic and parabolic counterparts, the fundamental solution to the wave equation, for spatial dimension = 3, 5, 7 ... etc., is truly singular; more precisely its support is contained in the lightcone, a phenomenon known as the Huygens' principle.

Riesz essentially constructed the fundamental solution by means of analytic continuation: putting aside some (important) factors, we may start with

$$R^s(x) = \begin{cases} \frac{1}{\Gamma_n(s)} (x_1^2 - x_2^2 - \cdots - x_n^2)^{(s-n)/2} & x \in C_+ \\ 0 & x \not\in C_+ \end{cases}$$

which defines a distribution for $\operatorname{Re} s>n-2$. Here, $C_+$ is the future cone: $C_+=\{x\in\mathbb R^n | \sqrt{x_2^2+\cdots+x_n^2}\leq x_1\}$. If we let the wave operator $\square = \partial_1^2-\partial_2^2-\cdots-\partial_n^2$ act on $R^s$ (on the $x$ variables), we see by direct calculation that $\square R^s = c R^{s-2}$ for some constant $c$ (depending on $s$). So we may define $\Gamma_n(s)$ appropriately so that it goes away: $\square R^s = R^{s-2}$. This, then, allows one to analytically continue $R^s$ to all $s\in\mathbb C$ (after one notes that $\Gamma_n(s)$, essentially with two factors of the usual Gamma function, is analytic in $s$).

Note what happens to $R^{n-2}$ (which was just shy of being defined by the original expression). It is (by definition) a derivative of $R^n$, which is constant inside (and outside) the cone, so must vanish there upon differentiation: $\operatorname{supp} R^{n-2} \subseteq \partial C_+$. If $n=2$ (i.e., one-dimensional wave equation, and the "lightcone" $\partial C_+$ consists of two rays), one can compute directly that $R^0 = \delta$ (up to some constant; we could define $\Gamma_n(s)$ appropriately to make that go away).

For $n=4$ (our space-time), the distribution $R^{n-2} = R^2$ has a rather simple description: $$R^2(x) = \frac{\delta(x_1 - r)}{4\pi r}, \qquad r=\sqrt{x_2^2+x_3^2+ x_4^2}$$ and, by applying $\square$ on it again, we magically have $R^0=\delta$. Thus, the fundamental solution is $R^2$. (If I had to pick, that is the one example of distribution I'd keep in mind.)

For higher even dimensions ($n=6, 8,\ldots$), we get succesively $$R^{n-2}, R^{n-4}, \ldots, R^2, R^0$$ all but the last one supported on the lightcone -- but more and more "singular" (or "twisted" in some sense) -- and bingo, $R^0=\delta$! This again makes $R^2$ the fundamental solution. (One may of course continue applying $\square$ to obtain $R^{-2}, R^{-4}, \ldots$, all supported at the origin.)

For odd dimensions, $R^{n-2},R^{n-4}, \ldots$ do have the same support structure, but it skips over $R^0$ (which incidentally is still $\delta)$. The true fundamental solution $R^2$ is the repeated derivative of some function that does not vanish inside the cone, so $\operatorname{supp} R^2 = C_+$ (and Huygens' principle fails). The dichotomy (with the parity of $n$) could be made more clear by stating the $\operatorname{supp} R^s$ for all $s\in\mathbb C$.

The crucial fact that $R^0=\delta$ is a surprisingly simple consequence of the fact that the factor $\Gamma_n(s)$ has a pole at $s=0$, or that $\frac{1}{\Gamma_n(s)}$ has a zero. I suspect it was essentially in Riesz, but is perhaps more clear in the language of distributions in Duistermaat [91] and Kolk-Varadarajan [91]. It probably appeared somewhere in Vol. I of Gelfand-Shilov, if not in Schwartz. (I'd welcome opinions on this from experts.)

So, it may be true that any distribution is a derivative of some measure, but it can have some unexpected behavior when the support suddenly jumps -- or rather, shrinks. Of course it happens very rarely, but they are the most interesting cases since all fundamental solutions are just like that.

I believe you should start with the theory of tempered distributions, which are the linear functionals $\phi:\mathcal S(\mathbb R^n) \to \mathbb C$ where $\mathcal S(\mathbb R^n)$ is the Schwartz space on $\mathbb R^n$, i. e. the $C^\infty$ functions on $\mathbb R^n$ which are bounded together with all their derivatives.

You can get more intuition in $\mathcal S'$, since the tempered distributions behave pretty much as functions. In fact, every $f\in L^p$ is a distribution, via $$f(g) = \int fg$$ for every $g\in\mathcal S$. You can take a derivative $\partial$ of a distribution $\phi$ via $$\partial \phi(f) = -\phi(\partial f),$$ or the Fourier transform via $$\hat\phi(f) = \phi(\hat f\ ).$$ A good reference is Folland's Real Analysis book, Chapter 9.

Although $L^1_{loc}$ does not contain the Dirac distribution, it may be useful to distinguish $L^1_{loc}$-distributions from say distributions represented by Radon measures. Your question is interesting, because it is definitely important to understand examples of distributions. That said, perhaps the motivation for distributions is equally important. Distributions help us take weak derivatives. The definition of a derivative of a distribution is motivated by Integration by Parts.

As you may know, often many mathematicians are more interested in working with specific distributions such as those in Sobolev spaces such as $W^{k,p}(U)$ ($1\leq p\leq \infty$), $BV(U)$ (integrable functions whose first order (weak) derivatives are signed measures with finite variation), or even tempered distributions. Then there are distributions like $$T(\phi):=\sum_{k=1}^{\infty} \int_{0}^{1}\frac{\phi(x)\sin(k\pi x)}{x}\, dx$$ for $\phi\in C_c^{\infty}((0,1))$. I guess the point is, be careful not to think that all distributions somehow behaving the same way.

• Sequences or nets of distributions have interesting properties : for example $\frac{\sin n\pi x}{\pi x} \to \delta$ implies a largest test function space such that $\varphi \ast \frac{\sin n\pi x}{\pi x} \to \varphi$ pointwise, uniformly, or for some norm, and this is one of the main problem in Fourier analysis. – reuns Dec 1 '17 at 22:01

Every distribution is, locally, a finite number of derivatives of measures. You can prove this with Hahn Banach: restricted to an open set $\Omega$ with compact closure, your distribution belongs to the dual of $C^k(\Omega)$ for some $k$. Note that $C^k$ embeds into $(C^0)^N$ for some large $N$ by taking $f \mapsto (f, D f, D^2 f, \ldots, D^k f)$. By Hahn Banach your distribution is the restriction of some linear function on the dual of $(C^0)^N$, which has the form $u(f) = \sum_{|\alpha| \leq k} \int \partial^\alpha f(x) d\mu_\alpha$. This characterization can then be used to deduce the one mentioned in other responses where you replace $\mu_\alpha$ with some continuous functions, but you have to take more than $k$ derivatives so the latter characterization can be a bit misleading.

So your intuition is right, all a distribution does is take a few derivatives and integrate. In this sense, distribution theory is the natural setting to combine differential calculus with measure theory. Once you understand how crazy measures can be (e.g. measures on hypersurfaces; the derivative of the Cantor function is a measure supported on the Cantor set), you basically have the extent of the pathologies of distributions. But a lot of distributions don't come given to you as derivatives of measures ($p.v. \frac{1}{x}$ is a good example; it requires one derivative to define, but the Hilbert transform $f \mapsto \frac{1}{\pi} p.v. \int \frac{f(x-y)}{y} dy$ is a bounded operator on $L^2({\mathbb R})$!).

Stepping back from the problem a little bit, I'd say that focusing on distributions is not the right approach. It's obvious from the way you've written your question that you understand the basics of distribution theory. Distributions are meant to fade into the background once you've established their theory. I'd say concentrate on Sobolev spaces, their embedding theorems, and their applications.

• Absolutely not! Distributions are fascinating in their own right and deserve to be centre stage, not simply as a background for Sobolev spaces and other such "constructed" spaces. – Andrew Stacey Nov 9 '09 at 9:35
• Andrew, maybe instead of implying that distributions should fade into the background, I should have said you may let distributions fade into the background if their applications are your main interest. – John D. Cook Nov 9 '09 at 22:17
• Yes, that's better. I agree with that. – Andrew Stacey Nov 19 '09 at 8:55

Let $(\mu_{\alpha})_{\alpha\in\mathbb{N}_0^d}$ be a collection of signed/complex Radon measures on $U\subset \mathbb{R}^d$. Let's call it locally finite if for every compact $K\subset U$, $|\mu_{\alpha}|(K)=0$ except for finitely many multiindices $\alpha$'s. Now given such a collection and for a test function in $f\in \mathcal{D}(U)$ define $$\phi(f)=\sum_{\alpha\in\mathbb{N}_0^d} \int_U \partial^{\alpha}f(x) \ d\mu_{\alpha}(x)\ .$$ This $\phi$ is in $\mathcal{D}'(U)$ and I am pretty sure every element on $\mathcal{D}$ looks like this. The writing is of course very non unique because of the redundancy between derivatives of the same function. This is just an elaboration on Isset's answer. I think it is nice to think of all these "layers" index by $\alpha$ simultaneously. Take the embedding $U\rightarrow \mathcal{D}(U)$ which sends a point to the corresponding Dirac unit mass. It is in the "top layer" $\alpha=0$. What about making this embedding smooth? By taking Gateau derivatives, you see that you start cascading down to deeper and deeper alphas. This is also in accordance with the physical intuition behind dipoles,...multipolar charge distributions which were a source of inspiration for L. Schwartz. BTW Hahn-Banach is not needed. One can use truncation/convolution to reduce to $\mathcal{S}'$ and the Hermite function basis (see Barry Simon's JMP article).