Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 327 / 3

A phase function is a real smooth function $\Phi$ on $X\times ({\mathbb{R}}^k\setminus \{0\})$, positively homogeneous of degree one in $\theta$, with $d_{x,\theta }\Phi \ne 0$. The symbol classconsists of smooth amplitudes satisfying, on each compact $K\subset X$,

To define the oscillatory integral, insert a cutoff $\chi (\epsilon \theta )$ equal to one near zero and setRepeated integration by parts with an operator $L$ satisfying $L{e}^{i\Phi }=e^{i\Phi }$ makes the integral absolutely convergent after enough iterations and shows that the limit defines a distribution.

Applying $D_x^{\alpha }$ changes only the constants in the defining estimates, while each $D_{\theta }$ lowers the power of $⟨\theta ⟩$ by one. Hence

The Leibniz rule writes every derivative of $a_1{a}_2$ as a finite sum of products of derivatives of the factors. Multiplying their symbol bounds adds the orders, so

Differentiating a function positively homogeneous of degree $M$ in $\theta$ makes $D_{\theta }^{\beta }b$ homogeneous of degree $M-|\beta |$, while $x$-derivatives preserve the degree. On the unit sphere these derivatives are bounded uniformly over compact subsets of $X$. Scaling then givesfor large $\theta$; smoothness controls the remaining compact frequency region. Thus

The singular support of $u\in {\mathcal{D}}^{\prime }(X)$ is the complement of the largest open subset of $X$ on which $u$ is represented by a smooth function.

Suppose $x_0$ has a neighborhood on which ${\nabla }_{\theta }\Phi (x,\theta )\ne 0$ for every $\theta \ne 0$. There one may integrate by parts repeatedly withEach adjoint application lowers the effective symbol order. After enough repetitions, the integral and all its $x$-derivatives converge absolutely and define a smooth function near $x_0$. Therefore

The claim is false because vanishing of the amplitude at one spatial point need not control its derivatives nearby. Take $X=\mathbb{R}$, $k=1$,Then $0\in Z(a)$, but distributionallyup to the Fourier-transform sign convention. Thus $0$ lies in the singular support of $I_{\Phi }(a)$ despite belonging to $Z(a)$.