A well known representation of Dirac's delta-distribution is via the Fourier transform of distributions: \begin{equation} \delta[f]:=f(0)=\int_{\mathbb{R}}\int_{\mathbb{R}} e^{\mathrm{i}xk}f(k)\mathrm{d}k\mathrm{d}x. \end{equation} Can this be used to define the delta distribution composed with a function $\phi:\mathbb{R}\to\mathbb{R}$ via \begin{equation} (\delta\circ\phi)[f]:=\int_{\mathbb{R}}\int_{\mathbb{R}} e^{\mathrm{i}x\phi(k)}f(k)\mathrm{d}k\mathrm{d}x? \end{equation} Does this make sense? Heuristically and in a more physics-style notation, I would argue that \begin{equation} \int\int e^{\mathrm{i}x\phi(k)}f(k)\mathrm{d}k\mathrm{d}x "=" \int\left(\int e^{\mathrm{i}x\phi(k)}\mathrm{d}x\right)f(k)\mathrm{d}k "=" \int\delta(\phi(k))f(k)\mathrm{d}k. \end{equation} If it does not make sense, what can I say about the above expression for general functions $\phi$? E.g. I would expect the above integral to be positive if $f$ is point-wise positive.
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1$\begingroup$ What do you assume about $f$ ? That it is Schwartz, that $f$ and its Fourier transform are $L^1$ ? What do you assume about $\phi$ ? Let $T_n (x) = \int_{-n}^n e^{i xy}dy$, it converges to $\delta$ in the sense of distributions and $\int_0^x T_n(v)dv$ converges locally uniformly to $sign(x)$ away from $x =0$ and boundedly around $x=0$. Thus If $f\in C_c^0$ and $\phi $ is $C^1$ with no double zero and finitely zeros on each lnterval then $(T_n \circ \phi ,f ) \to( \delta\circ \phi ,f )$ $\endgroup$– reunsCommented Jan 14, 2019 at 20:11
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$\begingroup$ Composing the Dirac delta with functions is a common practice in physics. I related this trick to Laplace transforms to determine the Fuss-Catalan sequences in "Discriminating Deltas, Depressed Equations, and Generalized Catalan Numbers" tcjpn.files.wordpress.com/2013/04/discrdeltas9-6-20122.pdf. I think this might be extendable to general formal power series and a certain class of Laurent series, but that's a future project. $\endgroup$– Tom CopelandCommented Mar 16, 2023 at 15:37
3 Answers
What you want to do is the pull-back of distributions. And there is a theorem (cf. Hörmander 1, Theorem 8.2.4) that if the set $\{(\phi(x),\eta) \colon \phi'(x) \eta = 0\}$ and $\operatorname{WF}(\delta) = \{ (0,\eta) \colon \eta \not = 0\}$ have empty intersection, then the pull-back is well-defined.
If you are only interested in the delta-Distribution, then there is also Theorem 6.1.5 saying that for any smooth function $\phi : X \to \mathbb{R}$ with $|\phi'| \not = 0$ on $\phi = 0$, one has that $\delta^* \phi = \frac{dS}{|\phi'|}$, where $dS$ is the Euclidean surface measure on $\{\phi = 0\}$. Even though I mentioned no integrals, this has quite strong flavour of oscillatory integrals (cf. Shubin, Chapter 1) and FIOs (Hörmander 4) to it.
Literature:
L. Hörmander - The Analysis of Linear Partial Differential Operators 1-4
M. Shubin - Pseudodifferential Operators and Spectral Theory
If $\phi(k)$ vanishes at $k=k_n$, $n=1,2,\ldots$, and $\phi'(k_n)\neq 0$ for all $n$, then \begin{equation} \int\int e^{\mathrm{i}x\phi(k)}f(k)\,\mathrm{d}k\mathrm{d}x = 2 \pi \int\delta(\phi(k))f(k)\,\mathrm{d}k=2\pi\sum_{n}\frac{f(k_n)}{|\phi'(k_n)|}, \end{equation} so yes, the integral is positive for point-wise positive $f$.
The point of this coda to the above answers is to document the fact that it is possible to address this topic in a perfectly rigorous and elementary fashion. Let me start with the remarks that there is such a theory of parametrised integrals in a distributional sense (and has been for over 60 years) within which the fact that the delta distribution is the F.T. of the constant function is a perfectly valid statement (i.e. not in any sense merely formal--no quotation marks required). All you need to know is that if an integral $\int f(x,y) dy$ exists in the classical sense, then also in the distributional one. Further, the theorem of Euler is always valid--if $\int f(x,y) dy$ converges in the distributional sense, then so does $\int D_x f(x,y) dy$ and we have $$D_x \int f(x,y) dy=\int D_x f(x,y) dy.$$
This allows a simple proof of the required formula for the F.T. of the constant function. First consider the classical fact that the F.T. of $\frac 1 {1+y^2}$ is $e^{-|x|}$ (a simple exercise in the calculus of residues) and apply the operator $I-D_x^2$ to both sides.
Turning now to your suggestion that one can use this fact to define distributions of the type $\delta \circ \phi$ using the F.T., I think that this is just kicking the ball further down the park since one then has to find conditions on $\phi$ which ensure that your integral exists (in the distributional sense). Best of luck on that.
However, there is another path which is already well-trodden (and has been since the 60´s of the previous century). There are perfectly valid examples of situations where the composition of distributions can be defined in a simple and natural way: here are three which suffice for many examples:
We can define $f\circ \phi$
when $f$ and $g$ are functions, say continuous;
when $f$ is a distribution of finite order, i.e., of the form $D^n F$ as a higher distributional derivative of a continuous function , and $\phi$ is a diffeomorphism whose derivative never vanishes;
by a recollement des morceaux argument if we can cover the real line with open sets on each of which one of the above conditions holds.
In 2. the composition is defined as $(\frac 1{\phi´} D_x)^n (F\circ \phi)$.
Using this machinery, it is possible to give a positive answer to your second question in an elementary and rigorous manner. I woud be happy to supply references on request.
