# 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
Right. Thanks. – Tom Leinster Nov 9 '09 at 4:12
Yeah... gets confusing sometimes when so general ;) – MLevi Nov 9 '09 at 4:17

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?! – Loop Space 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. – Loop Space 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

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.

-

(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

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. – Loop Space 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. – Loop Space Nov 19 '09 at 8:55

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.

-

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 $L^2({\mathbb R})$!).

-