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I know an ordinary function $f(x)$ which is locally Lebesgue integrable can map a regular generalized function with $\int f(x)φ(x)dx$ and the Dirac delta is a singular generalized function. Also, I know we can use derivatives, multiplication, and so on to generate new singular generalized function from Dirac delta.

Is there a singular generalized function that is not related with Dirac delta? Asked another way: is there a singular generalized function that is essentially not a Dirac delta $\delta$? If not, why is the Dirac delta is so special?

ps the definition of the "Singular generalized functions" is all generalized functions(linear continuous functionals) that is not regular generalized functions(generated by locally Lebesgue integrable function f(x) through $\int f(x)φ(x)dx$, which φ(x) is test function)

thanks so many answers, but i'm sorry to find my descriptions/terminology are different from your math-major's, the following link is the material of generalized function i referred(https://www.cs.odu.edu/~mln/ltrs-pdfs/tp3428.pdf), the material defines "regular/singular generalized function" as the following picture showsenter image description here and the ordinary function's definition is like the following picture showing enter image description here

In the before, I mistakenly thought these descriptions/terminology are used by every math-majored, so I didn't explain much at the time. Now please allowing me to rephrase my question: is there a distribution(linear&continuous functional) is not from $\int f(x)φ(x)dx$ or Dirac delta? ———— The reason why I ask this question is because, as far as I know, there are only two types of generalized functions, one is obtained by using the "classic" functions through the formal formula $\int f(x)φ(x)dx$, and the other is obtained by the definition of the Dirac function like δ[φ]=φ(0), so whether the two this way covers all generalized functions?

Aerterliusi
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  • So what do you mean with singular? – reuns Feb 15 '23 at 16:37
  • A distribution with compact support is a derivative (in the sense of distributions, of course) of a continuous function. Dirac distributions arise from computing the derivative of very mildly non-deriable continuous functions. – Mariano Suárez-Álvarez Feb 15 '23 at 16:50
  • @renus , singular mean not regular – Aerterliusi Feb 15 '23 at 23:34
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    @Aerterliusi, you are really making an effort not to be helped, huh – Mariano Suárez-Álvarez Feb 15 '23 at 23:34
  • I find myself wondering whether there is a distribution which would look something like "take the average value of $f$ on the Cantor set"; and if so, if that would be something like what the OP might be looking for. – Daniel Schepler Feb 15 '23 at 23:56
  • We don"t use the words regular and singular in distribution theory (we say $L^1_{loc}$ and not $L^1_{loc}$). So md2perpe 's answer is what you want, the distributional derivative of $\log |x|$ is $1/x$ together with some strange regularization at $0$: $\langle (\log |x|)',\phi \rangle = \int_{-\infty}^\infty (\phi(x)-\phi(0) 1_{|x|<1})\frac{dx}x$. Of course $1/x$ can't be said to be locally integrable around $0$. – reuns Feb 16 '23 at 00:51
  • @DanielSchepler, since the area of the cantor set is zero, that does not really work. – Mariano Suárez-Álvarez Feb 16 '23 at 10:10
  • @MarianoSuárez-Álvarez, I am pretty sure Daniel is talking about Cantor distribution. One way to realize it is to consider the sequence $(C_n)$ of sets appearing in the construction of the Cantor set $C$ (i.e., $I_0=[0,1]$ and $I_{n+1}$ is obtained by removing the middle third of each interval in $I_n$) and then taking limit of the probability measures $\mu_n(\mathrm{d}x)=(3/2)^n \mathbf{1}_{C_n}(x),\mathrm{d}x$ in the space of generalized functions. – Sangchul Lee Feb 16 '23 at 10:29
  • The way I was thinking about it was taking the canonical map from the Cantor set to $[0,1]$ (map ternary representations to binary representations), which has only countably many points with double inverse, and then use that map to transfer Lebesgue measure on $[0,1]$ to a measure on the Cantor set. Or equivalently, treat the Cantor set as being ${ 0, 1 }^\omega$ with the product topology, and transfer the usual "infinitely many coin flips" probability measure on that space to the Cantor set. – Daniel Schepler Feb 16 '23 at 15:55

3 Answers3

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A way to classify distributions is by talking about order, that is the number of derivatives that are needed on the test function to make sense of the distribution.

Functions. As you notice, the more "regular" distributions are just locally integrable functions. They define a distribution by the formula $$ \langle f,\varphi\rangle = \int f(x)\,\varphi(x)\,\mathrm d x $$ Here it is sufficient to have bounded test functions. The functions $f$ do not need to be bounded, one can take $f(x)= \ln(|x|)$ for instance.

Order 0: measures. More generally, any measure $\mu$ that gives a value to measurable sets defines an integral with respect to it, and the associated distribution is defined by $$ \langle \mu,\varphi\rangle = \int \varphi(x)\,\mu(\mathrm d x) $$ and it is sufficient to have continuous test functions (so still $0$ derivatives involved). This is the group where you will find the Dirac delta. However, there are also a lot of other measures that are singular in the sense that they are not absolutely continuous with respect tothe Lebesgue measure. For example, one I like is the following. Take $g$ to be the devil staircase, an look at its derivative in the sense of distributions $g'$. Since $g$ has bounded variations, $g'$ is a measure. Actually, one can also write the action of $g'$ as a distribution as a Stieljes integral $$ \langle g',\varphi\rangle = \int \varphi(x)\,\mathrm d g(x). $$

Order 1: derivatives of measures. To get higher order distributions, one can just take distributional derivatives of the previous ones. The example of md2perpe is a good one, the principal value of $1/x$ defines a distribution of order $1$, and is the distributional derivative of the function $\ln|x|$. You can have also Hadamard finite parts, such as $f = \mathrm{pf}(1/|x|)$ that is the distributional derivative of $x\ln(|x|)/|x|$ and can also be defined by $$ \langle f,\varphi\rangle = \int_{|x|\leq 1} \frac{\varphi(x)-\varphi(0)}{|x|}\,\mathrm{d}x + \int_{|x|> 1} \frac{\varphi(x)}{|x|} \,\mathrm{d}x $$ (See here for the derivation, take $d=1$ in dimension $1$). In all these examples, the test functions need to have at least one bounded derivative.

Higher order. Taking more derivatives will give higher order distributions, that is you will need more smooth test functions. You will get here the higher derivatives of the Dirac Delta, but also for example Hadamard finite parts of higher order, see for example the answers by Mark Viola such as here. You could also take higher derivatives of the devil's staircase, but I have no idea how to picture that in my mind. These distributions are useful when taking Fourier transforms. For example, the Fourier transform of $|x|^\alpha$ is a constant multiple of $h= \mathrm{pf}(1/|x|^{1+\alpha})$, that can be defined by $$ \langle h,\varphi\rangle = \int_{\Bbb R} \frac{\varphi(x)-\varphi(0)}{|x|^{1+\alpha}}\,\mathrm{d}x $$ when $\alpha\in(0,1)$, and this gives the formula for the fractional Laplacian $\Delta^{\alpha/2}f = C \,h* f$ for some constant $C$.

LL 3.14
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  • (+1) As always, you've written a beautifully constructed summary. And grateful for the reference. Thank you. – Mark Viola Feb 18 '23 at 18:52
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If by "singular" we understand "with support equal to $\{0\}$", then the answer is no. All distributions(a.k.a. generalized functions) with such support are finite linear combinations of $\delta_0$ and its derivatives.

TZakrevskiy
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A couple of distributions that are singular at $0$ but have support on all of $\mathbb{R}$ are $\ln|x|$ and its distributional derivative $\operatorname{pv}\frac{1}{x}.$

md2perpe
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