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?


1 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).

  • $\begingroup$ But how can we regularize the integral $$\int_{\mathbb{R}^N}\!\chi(\varepsilon\theta)e^{i\Phi(x,\theta)}a(x,\theta)\,\mathrm{d}\theta\ ?$$ In Shubin, we construct a differential operator $L$ in the variables $(x,\theta)$ such that $L^* e^{i\Phi} = e^{i\Phi}$, but in order to integrate by parts, we would need to also integrate with respect to $x$. I understand the general idea that this problem would be solved by regularizing the above integral, but I don't immediately see how Shubin's trick can be used here. $\endgroup$
    – D. Wynter
    Oct 29, 2017 at 22:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.