Consider a functional acting on two functions of the form $f,g\to\int_{-\infty}^\infty \delta(f(x))|f'(x)|g(x)dx$, where $\delta$ is Dirac Delta distribution.
If we define a symbol $\overline{\delta}$ the following way: $\int_{-\infty}^\infty \overline{\delta}(f(x))g(x)dx=\int_{-\infty}^\infty \d...
@reuns function in a sense that it canbe assigned a value at all points unlike the Dirac Delta distribution. The space should be understood in general sense (not necessarily real numbers).
@MoisheKohan I have missed $dx$ in all integrals and now it's fixed. But otherwise, it looks the formula makes sense, why it should not?
Why don't you see that if $\bar{\delta}(x)$ means $\delta(f(x))|f'(x)|$ with $f(x)=x$ then $\bar{\delta}(x)=\delta(x)$ and your "is it a function" question doesn't make sense?
I found that the OP made the following comment to LL3.14: "Well, I was not talkning about real functions, I was talking about function $\overline{\delta}(x)={\begin{cases}\frac1\pi\int_0^\infty xdx,&{\text{if }}x=0\\ 0,&{\text{if }}x\ne 0.\end{cases}}$, a piece-wise defined one, which also can be defined via delta distribution: $\overline{\delta}(f(x))=\delta(f(x))|f'(x)|$". Voting to close as I believe OP is using some unusual implicit definition for 'function'
@reuns as i see it, $\overline{\delta}(x)$ behaves as a pointwise-defined function, while $\delta$ is not a function, thus equivalence is impossible, for instance $\overline{\delta}(ax)=\overline{\delta}(x)$, like in the case of the sign function, thus equivalence between a function and a non-function distribution is impossible outside the integral sign. The equivalence $\overline{\delta}(f(x))=\delta(f(x))|f'(x)|$ holds only when x is integration variable, that is, under integral.
@reuns from reals to reals plus divergent integrals (infinitely large germs). But this question intentionally avoids the nature of the sets. What it puts emphasis on, is that symbol seems to define a function (something which is defined at all points and behaves differently from non-function distribution or generalized function under change of variable).
And so the definition finally comes out. Also re:"behaves differently from non-function distribution or generalized function under change of variable", this assumption of yours that took so many comments to come out is wrong. The definition of CoV for generalised functions was chosen specifically to generalise CoV for functions, something LL3.14 tried very hard to convey to you. If your $\bar\delta$ behaves differently, then you have in fact proved it is not even a function.
@CalvinKhor hmm, why? Can you elaborate on this last statement? Do you claim that a function cannot have the property $f(ax)=f(x)$? But a counterexample is $0^{|x|}$.
If it was a function, then it would also be a generalised function, and the scaling for the function would induce precisely the same scaling for a distribution...? Perhaps you should consult LL3.14's answer again. PS a much much simpler function solving $f(ax) = f(x)$ is $f=0$...
@CalvinKhor I think, the term "generalized function" is used in the theory of distributions generalizes the functions $\mathbb R\to\mathbb R$. In other words, any such function is also a "generalized function" in this sense. But if the function is not $\mathbb R\to\mathbb R$, it may not belong to "generalized functions" in the sense of theory of distributions.
Sorry, but I will completely ignore all mentions of other definitions of generalised functions and divergent integrals, especially because it is not in the question body
@CalvinKhor by "function" I mean a map from one set to another. This is a quite common definition and I never seen this kind of functions to be called "generalized". I think, $\overline{\delta}$ is a one-to-one map and has a value at all points of real axis. Thus, it is simply a function, not "generalized function".
Last comment on the issue: no, because if you are talking about functionals and distribution theory (see tags) then the domain and codomain of the functions (which are part of the definition of the function) need to belong to a certain list of sets, or else you should throw the whole theory out. So if you want to add your own definition of function, then you need to write a few books of definitions before this question makes any sense. And definitely give it its own name. Otherwise, you have fun confusing everyone and getting nowhere
@CalvinKhor I see the problems, but the very purpose of this question was to verify the idea that such object $\overline{\delta}$ can be expressed via conventional delta-distribution. Why do I think this relation holds? Simply because $\int_{-\infty}^\infty \overline{\delta}(f(x))dx$ is equal to the number of roots of $f(x)$ (by definition), and also the expression $\int_{-\infty}^\infty \delta(f(x))|f'(x)|dx$ (when defined) gives the number of roots of $f(x)$ in the theory of distributions.
@CalvinKhor Thus, my conjecture was that the expressions are equal when the both are defined, in other words, they generalize each other.
Discussion on question by Anixx: Does…
Discussion on question by Anixx: Does…