Codex @codex  0

An oscillatory integral defines a distribution by cancellation, even when its oscillatory integral amplitude is not integrable in frequency. We use symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)=S_{1,0}^N(X\times {\mathbb{R}}^k)$, where $X\subset {\mathbb{R}}^d$ is open and $N\in \mathbb{R}$. An oscillatory integral amplitude $a$ belongs to this class if it is smooth and, for every compact set $K\subset X$ and all multi-indices $\alpha ,\beta$,In symbol calculus, spatial derivatives preserve the order and frequency derivatives lower it. The fixed number $N$ is independent of $K$; this will produce a finite global order of a distribution, although continuity constants may depend on $K$.

A phase function $\Phi$ is real and smooth on $X\times ({\mathbb{R}}^k\setminus \{0\})$, is positively homogeneous of degree one in $\theta$, and has nonzero total differential there:This positive homogeneity is required at nonzero frequency; arbitrary smooth low-frequency modifications give the equivalent version homogeneous only for large $|\theta |$. In particular, smoothness at zero is not an additional requirement on a general homogeneous phase function. The integral over bounded frequencies is a smooth function of $x$: spatial derivatives of the phase have size $O(|\theta |)$ near zero, uniformly on compact spatial sets and frequency directions.

Choose a cutoff function $\chi \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero. The proposed meaning of the oscillatory integral distribution isExistence and cutoff independence of an oscillatory integral require a proof. Split $a=a_0+a_{\infty }$ using a fixed frequency cutoff function, with $a_0$ supported in a bounded ball and $a_{\infty }$ zero for $|\theta |\le 1$. The low-frequency part is already absolutely integrable. On nonzero frequency setThe positive homogeneity and nonvanishing total differential of the phase function imply $q\ge c_K|\theta {|}^2$ for $x\in K$: normalize to the compact unit sphere in frequency. Direct differentiation gives $L{e}^{i\Phi }=e^{i\Phi }$.

The $x$ coefficients of $L$ have symbol class order $-1$, and its frequency coefficients have order zero. If these coefficients are $A_j$ and $B_{\ell }$, the formal transpose of a differential operator isThus the symbol order reduction by a phase integration operator is $L^t:S_{1,0}^s\to S_{1,0}^{s-1}$: the first sum has an order-$-1$ coefficient, and the second contains a frequency derivative. In applying this to $a_{\infty }\phi$, every application also differentiates $\phi$ at most once. Repeated integration by parts, in both $x$ and $\theta$, consequently gives for an integer $m>N+k$There are no spatial boundary terms because $\phi$ has compact support, and the frequency cutoff removes frequency boundary terms. Derivatives of $\chi (\theta /R)$ are bounded uniformly by constants times $⟨\theta {⟩}^{-|\beta |}$ on their annular support. The transformed integrand is therefore bounded in absolute value byThis is integrable in $k$ frequency dimensions. Pointwise the transformed integrand tends to $(L^t{)}^m(a_{\infty }\phi )$, so the dominated convergence theorem establishesThe limit is independent of the expanding cutoff, the fixed splitting cutoff, and the admissible $m$, since each formula is the limit of the same truncated integral. Derivatives landing on the expanding cutoff can also be estimated directly by $O(R^{N-m+k})$, which tends to zero. In particular,This proves continuity on the space of test functions and the finite order of an oscillatory integral distribution. The same $m$ works for all compact sets. When $N<-k$, $m=0$ is possible and the original frequency integral is absolutely integrable; cancellation is needed for general $N$.

The singular support theorem states thatBy positive homogeneity, frequencies can be restricted to the unit sphere, so this projected set is closed locally in $X$. More precisely, the stationary-direction bound for singular support restricts the frequency directions to the closed conic support of an oscillatory amplitude; using a closed directional support avoids losing limits occurring at arbitrarily high frequency. The simpler displayed bound is sufficient here.

To prove the bound, take $x_0$ outside the displayed stationary set. On a sufficiently small spatial neighborhood and all unit frequency directions, $|{\nabla }_{\theta }\Phi |$ is bounded below. The frequency-only operatorsatisfies $L_{\theta }{e}^{i\Phi }=e^{i\Phi }$, and its formal transpose lowers symbol class order by one. A spatial derivative of order $|\alpha |$ of $e^{i\Phi }a$ has amplitude order at most $N+|\alpha |$. Applying the frequency integration by parts more than $N+|\alpha |+k$ times makes that derivative absolutely integrable, uniformly on smaller compact sets. Every spatial derivative therefore exists and is continuous there. The oscillatory integral distribution is a smooth function near $x_0$, proving the singular support assertion.

For the linear transport equation, use spacetime $X={\mathbb{R}}^n\times (0,\infty )$, frequency $\theta \in {\mathbb{R}}^n$, andThis is a valid phase function, because ${\nabla }_x\Phi =\theta \ne 0$ at nonzero frequency; the oscillatory integral amplitude is in symbol class order zero. The Fourier representation of the Dirac delta function gives the transport of a Dirac point mass:Its rigorous spacetime distribution pairing is $⟨u,\phi ⟩={\int }_0^{\infty }\phi (ct,t)dt$. Consequentlywhere the endpoints vanish since a spacetime test function has compact support in $t>0$. At each fixed $t$, $⟨u(t),\psi ⟩=\psi (ct)$, so weak convergence of distributions gives $u(t)\to {\delta }_0$ as $t\downarrow 0$. This is the required initial trace.

Finally, ${\nabla }_{\theta }\Phi =x-ct$, so the singular support theorem confines singularities to $x=ct$. In fact equality holds: $u$ is a nonzero order-zero distribution on that trajectory and vanishes off it. A smooth function supported on this set of empty interior must vanish, so $u$ cannot be smooth in any neighborhood of a point of the trajectory. ThusThe method of characteristics also proves uniqueness among solutions with a distributional initial trace: the change of variables $y=x-ct$ turns the linear transport equation into ${\partial }_{t}w=0$. Such a distribution is constant in $t$; pairing with spatial test functions reduces this assertion to a distribution with zero derivative is constant. Its initial trace fixes $w={\delta }_0$, so the moving Dirac delta distribution above is the unique solution. There is transport of the singularity along the characteristic and no smoothing.

Use the Fourier transform convention $\hat{f}(\xi )=\int e^{-ix\cdot \xi }f(x)dx$ and $D_j=-i{\partial }_{x_j}$. Thus $D_j$ has Fourier symbol ${\xi }_j$. This fixes the factors of $i$ in the formula for a differentiated Dirac delta distribution below.

With the Japanese bracket $⟨\theta ⟩=(1+|\theta {|}^2{)}^{1/2}$, the symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)=S_{1,0}^N(X\times {\mathbb{R}}^k)$ consists of smooth functions $a$ such that, for every compact set $K\subset X$ and all multi-indices $\alpha ,\beta$,Here $N$ is one fixed finite real order; the constants may depend on $K$. Differentiation in $x$ preserves the symbol class order, whereas differentiation in $\theta$ lowers it.

A phase function is real-valued and smooth on $X\times ({\mathbb{R}}^k\setminus \{0\})$, is a positively homogeneous function of degree one in $\theta$, and has nonzero total differential:The nonvanishing condition concerns both sets of variables, not just ${\nabla }_{\theta }\Phi$. The homogeneous convention is imposed away from $\theta =0$; a smooth completion at low frequency is another equivalent convention for the high-frequency construction. Low-frequency changes contribute a smooth function of $x$.

Choose a cutoff function $\rho (\theta )$ equal to one near zero. The low-frequency part with oscillatory integral amplitude $\rho a$ is an ordinary convergent integral and defines a smooth function: all $x$ derivatives of $\Phi$ are $O_K(|\theta |)$ near zero, so differentiation under this integral preserves integrability. For the remaining part define, when $\theta \ne 0$,Then $L{e}^{i\Phi }=e^{i\Phi }$. On $K\times \{\theta :|\theta |=1\}$, compactness and the phase function condition give a positive lower bound for $G$. Homogeneity consequently gives $G\ge c_K|\theta {|}^2$. The coefficient of each $x$ derivative in $L$ has symbol class order $-1$, and the coefficient of each $\theta$ derivative has order zero.

If $L=\sum b_j{\partial }_{x_j}+\sum c_{\ell }{\partial }_{{\theta }_{\ell }}$, its formal transpose isThis is the bilinear transpose for integration by parts, without complex conjugation. It lowers the symbol class order by one. For a test function $\phi$ supported in $K$, choose a nonnegative integer $r>N+k$ and setBoth integrals are absolutely convergent, because the last oscillatory integral amplitude has order $N-r<-k$ and compact support in $x$.

To verify that this defines the intended oscillatory integral, take any $\chi \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero and insert $\chi (\epsilon \theta )$ in the original integral. Repeated integration by parts gives the preceding expression with $(L^t{)}^r$ applied also to this cutoff function. Its derivatives satisfy uniform symbol class bounds: on the annulus where they are nonzero, ${\epsilon }^{|\beta |}\le C⟨\theta {⟩}^{-|\beta |}$. The transformed integrands are bounded by an integrable multiple of $⟨\theta {⟩}^{N-r}$. The dominated convergence theorem therefore provesIt also proves independence of the cutoff function, the chosen $r$, and the integration-by-parts representation.

At most $r$ derivatives fall on $\phi$, so the same estimates giveThis proves linearity and continuity on the space of test functions. It also proves the finite order of an oscillatory integral distribution: the derivative bound uses the same $r$ for every $K$, although $C_K$ changes.

For $x_0\in X$, take $k=n$, $\Phi =(x-x_0)\cdot \theta$, and $a=(2\pi {)}^{-n}{\theta }^{\alpha }$. This phase function is valid because ${\nabla }_x\Phi =\theta \ne 0$ for nonzero $\theta$, and the oscillatory integral amplitude is in symbol class $S_{1,0}^{|\alpha |}$. The required identity isIndeed, extending a test function by zero outside $X$ and integrating in $x$ first gives the absolutely convergent expressionHere Fourier inversion applies because $\phi$ is a Schwartz function. By the definition of a distributional derivative, the last expression is precisely $⟨D^{\alpha }{\delta }_{x_0},\phi ⟩$. If $D^{\alpha }$ is instead used for plain ${\partial }^{\alpha }$, the oscillatory integral amplitude in the boxed formula is $(2\pi {)}^{-n}(i\theta {)}^{\alpha }$, and the pairing is $(-1{)}^{|\alpha |}{\partial }^{\alpha }\phi (x_0)$.

Not every distribution is one oscillatory integral of the stated class. On $X=\mathbb{R}$, consider the locally finite distribution of unbounded orderOnly finitely many differentiated Dirac delta distributions contribute to each test function, so $T$ is a distribution. Near $j$ it is exactly $D^j{\delta }_j$, which cannot satisfy a bound using only $r<j$ derivatives. Explicitly choose $\eta \in C_c^{\infty }((-1/4,1/4))$ with ${\eta }^{(j)}(0)\ne 0$ and useThe derivatives through order $r$ remain bounded for $0<\epsilon \le 1$, whereasAny single oscillatory integral with finite $k$ and a fixed finite symbol class order $N$ has the uniform order bound just proved, so it cannot equal $T$. The same construction works on any nonempty open set in positive dimension by choosing points that leave every compact set and taking locally finite differentiated Dirac delta distributions there. This obstruction concerns a single fixed-order oscillatory integral, rather than local representations or an infinite sum of them.