A well known representation of Dirac's deltadistribution 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 physicsstyle 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 pointwise positive.

$\begingroup$ 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 $(T_n \circ \phi ,f ) \to( \delta\circ \phi ,f )$ $\endgroup$– reunsJan 14, 2019 at 20:11
2 Answers
What you want to do is the pullback 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 pullback is welldefined.
If you are only interested in the deltaDistribution, 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 14
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 pointwise positive $f$.