Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 327 / 3

With $⟨\theta ⟩=(1+|\theta {|}^2{)}^{1/2}$, the symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)=S_{1,0}^N$ consists of complex-valued $a\in C^{\infty }(X\times {\mathbb{R}}^k)$ such that, for every compact $K⋐X$ and every pair of multi-indices $\alpha ,\beta$,An $x$ derivative preserves symbol order, while a $\theta$ derivative lowers it by one. The definition is local in $x$ and imposes estimates of every derivative order.

In the homogeneous convention, a phase function is real, smooth on $X\times ({\mathbb{R}}^k\setminus \{0\})$, positively homogeneous of degree one in $\theta$, and satisfiesOnly the large-frequency behavior is relevant to the regularization argument; any prescribed smooth modification near $\theta =0$ contributes an ordinary convergent integral. If the phase is presented on all of $X\times {\mathbb{R}}^k$, homogeneity can be required outside a bounded frequency region. Equivalently for this construction, its large-frequency symbol bounds and the following uniform estimate are sufficient:For a homogeneous phase function, this follows from compactness of $K\times S^{k-1}$. Nonvanishing of the total gradient is the condition; ${\nabla }_{\theta }\Phi$ may vanish. The displayed definition does not impose the stronger rank condition sometimes called a nondegenerate phase in Fourier-integral-operator theory.

We construct the oscillatory integral by integration by parts in both $x$ and $\theta$. For $|\theta |\ge 1$, setThen $L(e^{i\Phi })=e^{i\Phi }$. The coefficients of the $x$ derivatives have symbol order $-1$ and those of the $\theta$ derivatives have order zero. Use the bilinear formal transpose of a differential operator, not the Hermitian adjoint:The symbol order reduction by a phase integration operator follows from the product rule and the symbol class estimates: $L^t:S_{1,0}^s\to S_{1,0}^{s-1}$ locally in $x$. Differentiating the coefficients preserves precisely these orders because $q$ is elliptic of degree two on each compact $x$ set.

Let $\zeta \in C_c^{\infty }({\mathbb{R}}^k)$ equal one on $|\theta |\le 1$ and vanish on $|\theta |\ge 2$. The low-frequency part is an ordinary integral. For an integer $r>N+k$, define the high-frequency part byIf the homogeneous phase is only defined off zero, its value at the single point zero is immaterial to the low-frequency integral; its modulus remains one. In the high-frequency term all coefficients are used away from zero. For $\mathrm{supp}\phi \subset K$, repeated product rules giveSince $N-r<-k$, both terms converge absolutely. They are linear in the test function, with the continuity boundThus $I_{\Phi }(a)\in {\mathcal{D}}^{\prime }(X)$, with order of a distribution at most $r$ on each compact set.

To check that this construction is the intended oscillatory integral and does not depend on $r$, $\zeta$ or the large-frequency cutoff, let $\chi \in C_c^{\infty }({\mathbb{R}}^k)$ equal one near zero and formAfter $r$ applications of integration by parts, the term without a derivative on $\chi$ tends to the absolutely convergent high-frequency term by the dominated convergence theorem. Each extra term is supported in an annulus $|\theta |\asymp R$. A derivative of $\chi (\theta /R)$ contributes $R^{-1}$, so those terms have total absolute value at most $C_K{R}^{N-r+k}{\max }_{|\alpha |\le r}\Vert {\partial }^{\alpha }\phi {\Vert }_{\infty }\to 0$. The low-frequency part is unchanged for large $R$. Hence $J_R(\phi )$ tends to the displayed functional, independently of all cutoffs and of the permissible number of integrations. This proves the cutoff independence of an oscillatory integral as well as its continuity.

For the final distribution, the derivative convention $⟨{\delta }^{\prime },g⟩=-g^{\prime }(0)$ identifies it as $-x_2{\delta }^{\prime }(x_1)$. The delta derivatives from polynomial oscillatory amplitudes identity yieldsHere $k=1$, $a\in \mathrm{Sym}({\mathbb{R}}^2,\mathbb{R};1)$, and ${\nabla }_x\Phi =(\theta ,0)$ is nonzero for $\theta \ne 0$. Thus the phase is admissible even though ${\partial }_{\theta }\Phi =x_1$ vanishes on the distribution's support.

For a direct proof of the sign and normalization, put $g(s)={\int }_{\mathbb{R}}{x}_2\phi (s,x_2)d{x}_2\in C_c^{\infty }(\mathbb{R})$. The cutoff integral isThe Fourier transform $\hat{g}$ is rapidly decreasing, so the dominated convergence theorem and differentiated Fourier inversion giveThis is exactly the required pairing, so the oscillatory representation gives the prescribed distribution, not its negative or a multiple of it.