# Delta-distribution composed with a function from the Fourier representation

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.

• 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 $(T_n \circ \phi ,f ) \to( \delta\circ \phi ,f )$ – reuns Jan 14 '19 at 20:11

## 2 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 $$f$$.