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Chapter 9 in Folland's Real Analysis starts with the following sentence:

"At least as far back as Heaviside in the $1890$s, engineers and physicists have found it convenient to consider mathematical objects which, roughly speaking, resemble functions but are more singular than functions."

Even though I've never studied distribution theory in depth, it is a topic that's been arisen occasionally in my Analysis classes, and so I've come to feel almost comfortable with "the basics" of distributions. Now I'm trying to study the subject from Folland's book but this first sentence has me puzzled:

Q: How is it possible for a distribution to have singularities in a sense comparable to how ordinary functions (let's think from $\mathbb R$ to $\mathbb R$ to keep things simple) have singularities?

I've given some thought to this question but couldn't find a satisfactory explanation yet. As I think about it there are two possibilities:

1) A singularity of a distribution $T$ is a test function $\varphi$ for which $T\varphi=\infty$. This would be the most natural meaning to me, but it doesn't make much sense because by definition $T$ is a map from the space of test functions to $\mathbb R$, and so it has to have finite values at each test function. Is it maybe that we keep "singular test functions" out of the domain on definition of $T$, similar to when we define $1/x$ as a function on $\mathbb R\setminus\{0\}$ instead of as a function on $\mathbb R$? Also, if anything like this "definition" is the case, how would one compare how singular a distribution and a function are considering that the singularities themselves would be different objects?

2) A singularity of a distribution $T$ is a point $x$ in $\mathbb{R}$. In this case we can compare how singular $T$ is with respect to a function $f:\mathbb R\to\mathbb R$. Furthermore, we can regard locally integrable functions as distributions, so this definition would make sense at least when $T$ is actually a locally integrable function. However, even if this can be given precise sense in general, for any given subset $S\subset\mathbb R$ it's easy to define functions that are singular on $S$ like $f(x)=\big[\inf_{s\in S}|x-s|\big]^{-1}$ so how could distributions be more singular than functions?

Any insight is most welcome, thank you!

Edit: It seems I have misinterpreted the meaning of 'singular' in the above sentence. I'd still appreciate if someone could enumerate the ways in which distributions are more singular than functions.

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    $\begingroup$ I'm not sure Folland intends to mean that they are comparable - instead that distributions can have even less pleasant properties than functions. For instance, any function can be interpreted as a distribution, but the dirac delta cannot be interpreted as a total function. $\endgroup$ – B. Mehta Apr 16 '18 at 3:47
  • $\begingroup$ @B.Mehta So 'singular' in that sentence refers to unpleasant properties like not being differentiable in the classic sense and not to the usual notion of singularity? $\endgroup$ – Jonatan B. Bastos Apr 16 '18 at 4:07
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    $\begingroup$ Far more unpleasant - the dirac delta isn't even a well defined function, let alone being continuous or even differentiable! $\endgroup$ – B. Mehta Apr 16 '18 at 4:08
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    $\begingroup$ Exactly. The thing is, distributions, or rather the space of distributions, is an "unpleasant space". There's no neat and clean "representation" of these linear operators, like there is for functions (with the help of functional notation). SImilarly, one can embed smooth functions/ measures into distributions, but then not every distribution is of that form either. Hence, this is what I think "singular" means : it contains a large class of objects, but you sort of lose track of how to "represent" such an object, and in that sense it is "singular", or tricky. $\endgroup$ – астон вілла олоф мэллбэрг Apr 16 '18 at 4:40
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Although there are various ways to measure regularity, a natural and elementary one to use is the degree of differentiability of a function. If a function $f$ can be differentiated $k$ times it will be more regular if $k$ is high, and less regular if $k$ is low. In the context of classical functions this scale stops at functions which are not differentiable at all.

Now the theory of distributions extends this scale to something akin to negative degrees of differentiability. In fact, the structure theorem for distributions states that locally you can write every distribution as the derivative of a certain order of a continuous function. The more derivatives you need to take, the more singular such a distribution will be.

For example, the delta distribution $\delta$ can be obtained by differentiating the kink function $x_+$, defined by $x_+(x) = 0$ for $x<0$ and $x_+(x)=x$ for $x \ge 0$, twice: the first derivative gives the Heaviside function, whose derivative is the delta distribution.

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  • $\begingroup$ I didn't know about this structure theorem for distributions. This was not the answer I was expecting when I asked the question, but it actually helped me understand the use of 'singular' in the context of distributions, thank you! $\endgroup$ – Jonatan B. Bastos Apr 17 '18 at 23:24

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