Distributions more complicated than the Dirac δ and derivatives - MathOverflow

Distributions more complicated than the Dirac δ and derivatives

2010-05-21T22:41:50Z

The responses to another question clarifies that the best known examples of distributions that are not measures, are the derivatives of the delta and such. What I want to know is: Is that the only way a distribution is not a measure?

Are there distributions that are not measures, their derivatives, and their derivatives, and so on?

Answer by S. Carnahan for Distributions more complicated than the Dirac δ and derivatives

2010-05-21T22:48:59Z

You can take distributional derivatives of the devil's staircase. See Wikipedia for an explanation.

Edit: Sorry I was so curt. The answer to your revised question is no. The Wikipedia link above gives references to the fact that all distributions on the real line (and finite dimensional Euclidean spaces) are distributional derivatives of continuous functions, and continuous functions can be lifted canonically to measures.

As Rekalo mentioned, there is a notion of hyperfunction due to Sato that arises from studying boundary values of holomorphic functions, and hyperfunctions form a strictly larger space than distributions (in particular, essential singularities are allowed). I've heard of other spaces of generalized functions such as modules over sheaves of pseudodifferential operators, but I don't know how they relate to each other, or how one proves theorems with them.

Answer by Andy Putman for Distributions more complicated than the Dirac δ and derivatives

2010-05-21T23:33:37Z

There is a structure theorem for distributions that shows that they are all (possibly infinite, but locally finite) sums of derivatives.

Here's a proper statement. Let $T$ be a distribution on $\mathbb{R}^n$. Then there exists continuous functions $f_{\alpha}$ such that $T = \sum_{\alpha} (\frac{\partial}{\partial x})^{\alpha} f_{\alpha}$, where for each bounded open set $\Omega$, all but a finite number of the distributions $(\frac{\partial}{\partial x})^{\alpha} f_{\alpha}$ vanish identically on $\Omega$. Here $\alpha$ is a multiindex.

I recommend perusing the beautiful little book "A Guide to Distribution Theory and Fourier Transforms" by Strichartz. The above theorem is discussed in section 6.2.

Answer by Andrey Rekalo for Distributions more complicated than the Dirac δ and derivatives

2010-05-22T01:04:20Z

One can construct various "exotic" classes of distributions by playing with spaces of test functions. For example, if we choose a suitable space of real analytic functions, the dual space will include "distributions" which are not finite sums of derivatives of measures.

Let $$T(x)=\sum\limits_{k=0}^{\infty}c_k \delta^{(k)}(x),\quad c_k\in \mathbb C, \quad \limsup\limits_{k\to\infty}\left(k!|c_k|\right)^{1/k}=0.\qquad\qquad(*)$$ $T(.)$ is not a distribution in the classical sense. Its support is localized at $x=0$ but $T(.)$ cannot be written as a finite linear combination of $\delta^{(k)}(x)$. Yet $(*)$ defines a continuous linear functional on a subspace of the space of entire analytic functions on $\mathbb C$. This is a simple example of a hyperfunction (a.k.a. analytic functional).

The analogue of the structure theorem for hyperfunctions basically says that every hyperfunction can be written as a convolution $T*\mu$, where $T$ is of the form $(*)$ and $\mu$ is a Borel measure on $\mathbb R$.

Answer by Bazin for Distributions more complicated than the Dirac δ and derivatives

2012-04-19T18:07:34Z

Maybe a simple example to demonstrate that, although a distribution is of finite order on each compact, it could be of infinite order, i.e. not of finite order. Consider simply the following distribution on the real line $$ \langle T,\phi\rangle=\sum_{k\in \mathbb N}\phi^{(k)}(k). $$ Bazin.