Squarable delta function

Question

How much the Fourier analysis and the theory of Laplace transforms would be different if we assumed Dirac Delta to be a function $\overline{\delta}(x)$ rather than distribution $\delta(x)$, in other words, a function such that at zero it takes some infinitely-large value (say, $\omega/\pi$, equivalent to a divergent integral $\frac1{\pi}\int_0^\infty dx$) and otherwise $0$?

For instance, the following property would not hold: $\delta(a x)=\frac1{|a|}\delta(x)$. Instead, it would be $\overline{\delta}(a x)=\overline{\delta}(x)$.

Thus, $\int_{-\infty}^\infty \overline{\delta}(f(x))dx$ would represent the number of roots of the function $f(x)$.

It seems to me that in this case many things would be more convenient. For instance, we would be able to express powers of $\overline{\delta}(x)$, for any $p>0$:

$\overline{\delta}(x)^p={\begin{cases}\frac p{\pi^p}\int_0^\infty x^{p-1}dx,&{\text{if }}x=0\\ 0,&{\text{if }}x\ne 0.\end{cases}}$

For $p=1$ this coincides with the standard Fourier transform. (Notice though that we used trivial definition of multiplication of divergent integrals here, that is $(f(0)+\int_0^\infty f'(x)dx)(g(0)+\int_0^\infty g'(x)dx)=f(0)g(0)+\int_0^\infty (f(x)g(x))'dx$, that is, product of germs at infinity is the germ of the product at infinity, there can be a different, umbral definition).

On the other hand, $\overline{\delta}(x)$ would not be differentiable as a function.

Due to this aproach, we would be able to make continuous all functions with poles of even order or of the form $f(x)=1/|x|^n$. For instance, if we extend the function $f(x)=\frac1 {x^2}$ with $f(0)=\frac{\pi^2\overline{\delta}(0)^2}2$, the function becomes continuous. We even can make continuous the function $f(x)=\ln|x|$ by adding the value $f(0)=-\ln \overline{\delta}(0)-\ln \pi-\gamma$.

I consider a function to be continuous if the germs at right, at left and the value of the function at the point all coincide, in other words, if $f(a)+\int_a^x f'(t)dt=f(b)-\int_x^b f'(x)dx=f(x)$.

Like in the theory of hyperfunctions, we can differentiate discontinuous functions, for instance, for odd $p>0$,

$(\frac1{x^p})'={\begin{cases}\int_0^\infty \frac{p+1}{x^{p+2}} dx,&{\text{if }}x=0\\ -p/{x^{p+1}},&{\text{if }}x\ne 0\end{cases}}={\begin{cases}\frac1{p!}\int_0^\infty x^p dx,&{\text{if }}x=0\\ -p/{x^{p+1}},&{\text{if }}x\ne 0\end{cases}}={\begin{cases}\frac{\pi^{p+1}\overline{\delta}(0)^{p+1}}{(p+1)!},&{\text{if }}x=0\\ -p/{x^{p+1}},&{\text{if }}x\ne 0.\end{cases}}$

So, by sacrificing differentiability of Dirac Delta, it seems we can gain the ability to rise it to any positive power, find germs of functions at poles and make discontinuous functions continuous as well as differentiate them.

There are other interesting consequences in considering Dirac Delta a function, for instance, we would be able to generalize the $\operatorname{sign}$ function to dual numbers in such a way so it to preserve the property $\operatorname{sign} (u v)=\operatorname{sign} u\cdot \operatorname{sign} v$ still holds (in the linked question the formula 1 follows from delta distribution, formula 2 comes if Dirac Delta is a function).

We also would be able to write down expressions for $\ln \varepsilon$ and $\varepsilon^\varepsilon$ in duals.

But the main question is: how far can we develop the Laplace and Fourier transforms in such a world?

P.S.

A follow-up question, can we formally define this modified delta via usual delta as a functional $\overline{\delta}(f(x))=\delta(f(x))|f'(x)|$, because the later expression also represents the number of roots of the function $f(x)$? If so, the two deltas would be equal if their argument is the integration variable, and the difference will appear only when the argument is some function of integration variable.

Answer 1

Let me put a long comment, about the scaling of measures. As I explained in this post here, the scaling property of the Dirac delta $\delta(ax) = \delta(x)/a$ comes from identifying distributions with functions. That is we want to think of $\delta(x)$ as a function absolutely continuous with respect to the Lebesgue measure, i.e. we want to think that $$ \langle \delta,\varphi\rangle \simeq \int_{\Bbb R} \varphi(x)\,\delta(x)\,\mathrm d x $$

As I explained here, the right-hand side notation is indeed not intuitive if one now thinks as the Dirac delta not as a distribution extending a function but as a measure (let me call it $\delta_0$). Then one should rather write $$ \int_{\Bbb R} \varphi(x)\,\delta_0(\mathrm d x) = \varphi(0). $$ From this suggestive notation, one could think of $\delta_0(\omega)$ as acting on volume elements $\mathrm d x$ and define $\delta_0(a\omega)$ by $$ \int_{\Bbb R} \varphi(x)\,\delta_0(a\,\mathrm d x) = \int_{\Bbb R} \varphi(y/a)\,\delta_0(\mathrm d y) = \varphi(0). $$ That is, $\delta_0(a\,\omega) = \delta_0(\omega)$. And this is indeed what follows from integrating against the measure $\delta_0$ acting on sets $A\subset \Bbb R$ as $$ \delta_0(A) = 1 \text{ if } x\in A \\ \delta_0(A) = 0 \text{ if } x\notin A $$ using the classical Lebesgue integration theory for general measures. The notation $\delta_0(\mathrm d x)$ can then be seen as a remainder of the fact that a measure acts on set, and not on points. Now one can define the notation $\lambda A = \{\lambda x: x\in A\}$, and indeed, $\delta_0(\lambda A) = \delta_0(A)$.

Conclusion. the appearance of $|f'(x)|$ in the usual change of variable formula for the Dirac delta is due to the identification $\delta \simeq \delta(x)\,\mathrm d x$, which factors out the Lebesgue measure. This is the change of variable in the Lebesgue measure that is responsible of the stretching term $|f'(x)|$. These problems do not occur when thinking of the Dirac delta as a different measure.

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Additional remark: there is no problem to define the Fourier transform for measures, that is $$ \mathcal F(\mu)(x) = \int_{\Bbb R} e^{-2i\pi x y} \mu(\mathrm d y). $$ It actually gives the same definition as the one for distributions $\langle\mathcal F(\mu),\varphi\rangle = \langle\mu,\mathcal F(\varphi)\rangle$.

Answer 2

It seems to me now that the Fourier and Laplace transforms would not be affected at all. Those would be exactly the same as with ordinary delta distribution, as long as the argument of the delta is the integration variable. The change-of-variable rules would be different though.

Moreover, contrary to the question's title, the differentiability would not suffer. The modified delta function would be as much differentiable as usual delta distribution. Its derivatives though would not be functions (e.g, would not be able to be defined piecewise), but "distributions" with the following scaling rules: $\overline{\delta}^{(n)}(a x)=| a| a^n \delta^{(n)}(a x)$. As long as their argument is the integration variable, their Fourier and Laplace transforms would be the same as of usual derivatives of Delta distribution, for instance, the following would hold: $\int_{-\infty}^\infty \overline{\delta}'(x)f(x)dx=-f'(0)$.

For the modified delta function the following identities hold:

• $f(\pi \overline{\delta} (0))=f(0)+\int_0^{\infty } f'(x) \, dx,$ for functions satisfying the Hardy field requirements. Thus, the value of a function evaluated at $\pi \overline{\delta} (0)$ would be equal to limit/germ of the function at positive infinity.

• $\int_{-\infty }^{\infty } \left( \begin{cases} f(\overline{\delta} (0)) & x=0 \\ 0 & x\neq 0 \\ \end{cases} \right) \, dx=f'(0)+\frac1\pi\int_0^{\infty } f''(x) \, dx=f'( \overline{\delta} (0)).$ This is the rule of integrating a function, which has an infinite (divergent) value at a point. Unlike Riemann integration, having infinite value at just one point can make its antiderivative to jump. For instance, if a function has value of $\overline{\delta} (0)^2$ at a point, its antiderivative jumps by $\frac{2}{\pi}\overline{\delta} (0)$ there.

Thus, if we change the scaling rule of the delta disctribution and make it a function which takes an infinite (but exact) value at zero, we can both differentiate it and rise to an arbitrary power.

This variant of delta function looks more natural to me, so I wonder, why another scaling rule for delta distribution was historically adopted.