Smoothness of distributions defined by oscillation integrals

Question

In M.A. Shubin's book Pseudodifferential Operators and Spectral Theory, we have the following statement.

Let $X\subset\mathbb{R}^n$ be an open set, and fix a symbol $a\in S_{\varrho,\delta}^m(X\times\mathbb{R}^N)$ for $m\in\mathbb{R}$, $0<\varrho\le 1$ and $0\le\delta<1$, i.e. $$\lvert\partial_\theta^\alpha\partial_x^\beta a(x,\theta)\rvert\le C_{\alpha,\beta,K}\langle\theta\rangle^{m-\varrho\lvert\alpha\rvert + \delta\lvert\beta\rvert}$$ for all $x\in K\Subset X$, $\theta\in\mathbb{R}^N$, and for all multiindices $\alpha,\beta$, for some constants $C_{\alpha,\beta,K}$. Let $\Phi:X\times\mathbb{R}^N\to\mathbb{R}$ be a phase function, i.e. $$\Phi\in\mathscr{C}^\infty(X\times(\mathbb{R}^N\setminus\{0\}))$$ such that $\Phi(x,t\theta)=t\Phi(x,\theta)$ for all $t>0$ and $(x,\theta)\in X\times(\mathbb{R}^N\setminus\{0\})$, and $$\nabla_{x,\theta}\Phi(x,\theta)\ne 0$$ for $(x,\theta)\in X\times(\mathbb{R}^N\setminus\{0\})$. Define $$ R_\Phi = \{ x\in X\ |\ \nabla_\theta\Phi(x,\theta)\ne 0\ \forall\ \theta\in\mathbb{R}^N\}.$$

If we define a distribution $A\in\mathcal{D}'(X)$ by setting $$\langle A,u\rangle = I_{\Phi}(au) := \int\!\!\!\!\int\!\! e^{i\Phi(x,\theta)}a(x,\theta)u(x)\,\mathrm{d}x\,\mathrm{d}\theta $$ defined as an oscillation integral, for example as $$ I_\Phi(au)=\lim_{\varepsilon\to0} \int\!\!\!\!\int\!\! \chi(\varepsilon \theta)e^{i\Phi(x,\theta)}a(x,\theta)u(x)\,\mathrm{d}x\,\mathrm{d}\theta $$ for some $\chi\in\mathcal{D}(\mathbb{R}^N)$ that is $1$ in a neighborhood of $0$.

Then we have that $A$ is smooth in $R_\Phi$, i.e. for $u\in\mathcal{D}(X)$ with support in $R_\Phi$, $$I_\Phi(au) = \int_X\! A(x)u(x)\,\mathrm{d}x$$ for some $A\in\mathscr{C}^\infty(R_\Phi)$.

In the course of the "proof", we are told that $$A(x) = \int\!\! e^{i\Phi(x,\theta)}a(x,\theta)\,\mathrm{d}\theta$$ exists, considered as an oscillatory integral, and that this is smooth. But why is that so?

Answer 1

For $x_0 \in R_\Phi$, the function $\theta \mapsto \Phi(x_0,\theta)$ has non-vanishing differential, so we can construct a differential operator (wrt $\theta$) $L$ such that $L e^{i\Phi} = e^{i\Phi}$ and use integration by parts (and the cut-off $\chi$) to regularize $A(x_0)$ as oscillatory integral (Shubin's second method).

Now, if we formally differentiate $A(x)$ we get the same type of oscillatory integral. The remaining problem is why one can change the limit $\varepsilon \to 0$ and differentiation (should be an application of the dominated convergence theorem, but i didn't check it). This problem is avoided when you regularize your integral by integration by parts (Shubin's first method).