Coordinate Invariance of the Wave Front Set (Theorem # 8169)
Theorem
Let $U,V \subset \mathbb{R}^n$ be open sets, let $\mathcal{L}^n$ denote $n$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}^n$, and let $\mathbb{R}^n_0:=\mathbb{R}^n\setminus\{0\}$. Let $\kappa:V \to U$ be a $C^\infty$ diffeomorphism, and let $u \in \mathcal{D}'(U)$. The pullback distribution $\kappa^*u\in\mathcal{D}'(V)$ is defined by
\begin{align*}
(\kappa^*u)(\varphi)=u\left(x\mapsto \varphi(\kappa^{-1}(x))\,|\det d\kappa^{-1}_x|\right)
\end{align*}
for every $\varphi\in C_c^\infty(V)$. Fourier transforms are taken with the convention
\begin{align*}
\widehat{f}(\xi)=\int_{\mathbb{R}^n} f(x)e^{-ix\cdot\xi}\,d\mathcal{L}^n(x)
\end{align*}
for $f\in C_c^\infty(\mathbb{R}^n)$, and by duality for compactly supported distributions. The wave front set $\operatorname{WF}(u)$ is defined by this Fourier-transform convention: $(x_0,\xi_0)\notin\operatorname{WF}(u)$ if and only if there are $\chi\in C_c^\infty(U)$ with $\chi(x_0)\neq0$ and an open conic neighbourhood $\Gamma\subset\mathbb{R}^n_0$ of $\xi_0$ such that $\widehat{\chi u}$ decays rapidly on $\Gamma$. For each $y \in V$, let
\begin{align*}
(d\kappa_y)^\top:(\mathbb{R}^n)^* \to (\mathbb{R}^n)^*
\end{align*}
denote the transpose of the differential $d\kappa_y:\mathbb{R}^n \to \mathbb{R}^n$ under the standard Euclidean coordinates. Then
\begin{align*}
\operatorname{WF}(\kappa^*u)=\{(y,(d\kappa_y)^\top\xi):(\kappa(y),\xi)\in \operatorname{WF}(u)\}.
\end{align*}
Knowledge Status
Analysis
Discussion
No discussion available for this theorem.
Proof
[proofplan]
We prove first that no new singular covectors appear after pulling back by $\kappa$. Fix a point $y_0 \in V$ and a covector $\eta_0 \neq 0$ which is not of the form $(d\kappa_{y_0})^\top \xi$ with $(\kappa(y_0),\xi)\in \operatorname{WF}(u)$; we then choose cutoffs and conic neighbourhoods so that the inverse transpose cotangent map sends the chosen $\eta$-cone into a microlocally smooth $\xi$-cone for $u$. The localized [Fourier transform](/page/Fourier%20Transform) of $\kappa^*u$ is rewritten as $u$ applied to an oscillatory [test function](/page/Test%20Function) on $U$, and this test function is split into a good frequency part and a nonstationary complementary part. The good part is controlled by the defining rapid-decay estimates for $\operatorname{WF}(u)$, while the complementary part is controlled by repeated [integration by parts](/theorems/210) in a phase with uniformly nonzero gradient. Applying the same argument to the inverse diffeomorphism $\kappa^{-1}:U\to V$ gives the reverse inclusion.
[/proofplan]
[step:Choose conic neighbourhoods compatible with the inverse transpose cotangent map]
Fix $y_0 \in V$ and $\eta_0 \in \mathbb{R}^n_0$ such that
\begin{align*}
(y_0,\eta_0)\notin \{(y,(d\kappa_y)^\top\xi):(\kappa(y),\xi)\in \operatorname{WF}(u)\}.
\end{align*}
Define $x_0:=\kappa(y_0)$ and
\begin{align*}
\xi_0:=(d\kappa_{y_0})^{-\top}\eta_0.
\end{align*}
Then $\xi_0\neq 0$ and $(x_0,\xi_0)\notin \operatorname{WF}(u)$.
By the definition of the wave front set, there exist $\chi_0\in C_c^\infty(U)$ with $\chi_0(x_0)\neq 0$ and an open conic neighbourhood $\Gamma_\xi\subset \mathbb{R}^n_0$ of $\xi_0$ such that for every integer $N\geq 0$ there is a constant $C_N>0$ satisfying
\begin{align*}
|\widehat{\chi_0u}(\xi)|\leq C_N(1+|\xi|)^{-N}
\end{align*}
for every $\xi\in \Gamma_\xi$. Here $\chi_0u\in \mathcal{E}'(U)$ denotes the compactly supported distribution $\phi\mapsto u(\chi_0\phi)$, and $\widehat{\chi_0u}(\xi)$ denotes its Fourier transform.
Since the map
\begin{align*}
F:V\times \mathbb{R}^n_0&\to U\times \mathbb{R}^n_0
\end{align*}
\begin{align*}
(y,\eta)&\mapsto \bigl(\kappa(y),(d\kappa_y)^{-\top}\eta\bigr)
\end{align*}
is continuous and $F(y_0,\eta_0)=(x_0,\xi_0)$, we may choose a relatively compact open neighbourhood $W\subset V$ of $y_0$, an open conic neighbourhood $\Gamma_\eta\subset\mathbb{R}^n_0$ of $\eta_0$, and an open conic neighbourhood $\Gamma_\xi'\subset\mathbb{R}^n_0$ of $\xi_0$ with $\overline{\Gamma_\xi'}\cap S^{n-1}\subset\Gamma_\xi$ such that
\begin{align*}
\kappa(\overline{W})\subset \{x\in U:\chi_0(x)\neq 0\}
\end{align*}
and
\begin{align*}
(d\kappa_y)^{-\top}\eta\in \Gamma_\xi'
\end{align*}
for all $y\in W$ and all $\eta\in \Gamma_\eta$. Define the closed conic set
\begin{align*}
\Sigma_\xi:=\{r\xi:r>0,\xi\in \overline{\Gamma_\xi'}\cap S^{n-1}\}\subset \Gamma_\xi.
\end{align*}
Then the stationary frequency directions for the later Fourier variable satisfy
\begin{align*}
-(d\kappa_y)^{-\top}\eta\in -\Sigma_\xi
\end{align*}
for every $y\in W$ and every $\eta\in\Gamma_\eta$.
Choose $\psi\in C_c^\infty(W)$ with $\psi(y_0)\neq 0$. Choose $\chi\in C_c^\infty(U)$ such that $\chi=1$ on a neighbourhood of $\kappa(\operatorname{supp}\psi)$ and $\operatorname{supp}\chi\subset \{\chi_0\neq 0\}$. Define $q\in C_c^\infty(U)$ by $q=\chi/\chi_0$ on a neighbourhood of $\operatorname{supp}\chi$ and extend $q$ smoothly with compact support in $U$. Then $\chi u=q(\chi_0u)$.
We now record why the rapid-decay estimate transfers from $\chi_0u$ to $\chi u$. The Fourier transform of multiplication by $q$ is convolution with the Schwartz function $\widehat q$:
\begin{align*}
\widehat{\chi u}(\xi)=(2\pi)^{-n}\int_{\mathbb{R}^n}\widehat q(\xi-\zeta)\widehat{\chi_0u}(\zeta)\,d\mathcal{L}^n(\zeta).
\end{align*}
Choose a closed conic neighbourhood $\Gamma_\xi''$ of $\xi_0$ with $\Gamma_\xi''\cap S^{n-1}\subset\Gamma_\xi$ and shrink $\Gamma_\xi'$ so that $\Sigma_\xi\subset\Gamma_\xi''$. If $\xi\in\Gamma_\xi''$, split the integral into $\zeta\in\Gamma_\xi$ and $\zeta\notin\Gamma_\xi$. On $\zeta\in\Gamma_\xi$ the defining estimate for $\chi_0u$ gives arbitrary decay. On $\zeta\notin\Gamma_\xi$, the closed conic separation between $\Gamma_\xi''$ and $\mathbb{R}^n_0\setminus\Gamma_\xi$ gives $|\xi-\zeta|\geq c_1(|\xi|+|\zeta|)$ outside a fixed bounded set, for some constant $c_1>0$; rapid decay of $\widehat q$ then dominates the polynomial growth of $\widehat{\chi_0u}$. Hence, for every integer $N\geq0$, there is a constant $C_N'>0$ such that
\begin{align*}
|\widehat{\chi u}(\xi)|\leq C_N'(1+|\xi|)^{-N}
\end{align*}
for every $\xi\in\Gamma_\xi''$, and in particular for every $\xi\in\Sigma_\xi$.
[/step]
[step:Rewrite the localized Fourier transform as a distributional pairing on $U$]
Let $\widehat{\psi\,\kappa^*u}:\mathbb{R}^n\to \mathbb{C}$ denote the Fourier transform of the compactly supported distribution $\psi\,\kappa^*u$, namely
\begin{align*}
\widehat{\psi\,\kappa^*u}(\eta)=(\kappa^*u)(y\mapsto \psi(y)e^{-iy\cdot\eta}).
\end{align*}
The pullback of $u$ by the diffeomorphism $\kappa$ is defined by
\begin{align*}
(\kappa^*u)(\varphi)=u\left(x\mapsto \varphi(\kappa^{-1}(x))\,|\det d\kappa^{-1}_x|\right)
\end{align*}
for every $\varphi\in C_c^\infty(V)$.
For $\eta\in \mathbb{R}^n$, define the smooth compactly supported function
\begin{align*}
a_\eta:U&\to \mathbb{C}
\end{align*}
\begin{align*}
x&\mapsto \chi(x)\,\psi(\kappa^{-1}(x))\,|\det d\kappa^{-1}_x|\,e^{-i\kappa^{-1}(x)\cdot\eta}.
\end{align*}
Because $\chi=1$ on a neighbourhood of $\kappa(\operatorname{supp}\psi)$, the definition of pullback gives
\begin{align*}
\widehat{\psi\,\kappa^*u}(\eta)=u(a_\eta)=(\chi u)\left(x\mapsto \psi(\kappa^{-1}(x))|\det d\kappa^{-1}_x|e^{-i\kappa^{-1}(x)\cdot\eta}\right).
\end{align*}
[guided]
The point of this step is to put the localized Fourier transform of $\kappa^*u$ into a form where the known microlocal estimates for $u$ can be used. We start with the compactly supported distribution $\psi\,\kappa^*u$ on $V$. Its Fourier transform is, by definition,
\begin{align*}
\widehat{\psi\,\kappa^*u}(\eta)=(\kappa^*u)(y\mapsto \psi(y)e^{-iy\cdot\eta}).
\end{align*}
Now we use the definition of pullback by a diffeomorphism. If $\varphi\in C_c^\infty(V)$, then
\begin{align*}
(\kappa^*u)(\varphi)=u\left(x\mapsto \varphi(\kappa^{-1}(x))|\det d\kappa^{-1}_x|\right).
\end{align*}
Applying this with
\begin{align*}
\varphi:V&\to \mathbb{C}
\end{align*}
\begin{align*}
y&\mapsto \psi(y)e^{-iy\cdot\eta}
\end{align*}
gives
\begin{align*}
\widehat{\psi\,\kappa^*u}(\eta)
=
u\left(x\mapsto \psi(\kappa^{-1}(x))|\det d\kappa^{-1}_x|e^{-i\kappa^{-1}(x)\cdot\eta}\right).
\end{align*}
The cutoff $\chi$ may be inserted because $\chi=1$ on a neighbourhood of $\kappa(\operatorname{supp}\psi)$. The function
\begin{align*}
x\mapsto \psi(\kappa^{-1}(x))|\det d\kappa^{-1}_x|e^{-i\kappa^{-1}(x)\cdot\eta}
\end{align*}
is supported in $\kappa(\operatorname{supp}\psi)$, so multiplying it by $\chi$ does not change it. Therefore
\begin{align*}
\widehat{\psi\,\kappa^*u}(\eta)
=
(\chi u)\left(x\mapsto \psi(\kappa^{-1}(x))|\det d\kappa^{-1}_x|e^{-i\kappa^{-1}(x)\cdot\eta}\right).
\end{align*}
This is the desired reduction: the distribution is now the compactly supported distribution $\chi u$, whose Fourier transform has rapid decay in the good cone $\Gamma_\xi$.
[/guided]
[/step]
[step:Split the transformed test function into good and nonstationary frequency regions]
Choose a function $\rho\in C^\infty(\mathbb{R}^n;[0,1])$ such that $\rho$ is homogeneous of degree $0$ on $\{\xi\in\mathbb{R}^n:|\xi|\geq 1\}$, $\rho(\xi)=1$ for every $\xi\in -\Sigma_\xi$ with $|\xi|\geq 1$, and $\operatorname{supp}\rho\cap\{\xi:|\xi|\geq1\}\subset -\Gamma_\xi$. Let $1-\rho$ denote the complementary smooth multiplier. This low-frequency modification makes $\rho$ defined on all of $\mathbb{R}^n$, which is needed in the inverse Fourier integrals. The sign is forced by the Fourier convention: in the pairing below, the distributional Fourier transform is evaluated at $-\xi$, so the microlocally smooth covectors for $u$ correspond to $\xi\in-\Gamma_\xi$.
For each $\eta\in \Gamma_\eta$, write
\begin{align*}
b_\eta:U&\to \mathbb{C}
\end{align*}
\begin{align*}
x&\mapsto \psi(\kappa^{-1}(x))|\det d\kappa^{-1}_x|e^{-i\kappa^{-1}(x)\cdot\eta}.
\end{align*}
Since $b_\eta\in C_c^\infty(U)$, Fourier inversion gives
\begin{align*}
b_\eta(x)=(2\pi)^{-n}\int_{\mathbb{R}^n}\widehat{b_\eta}(\xi)e^{ix\cdot\xi}\,d\mathcal{L}^n(\xi).
\end{align*}
Define
\begin{align*}
b_\eta^{\mathrm{good}}:U&\to \mathbb{C}
\end{align*}
\begin{align*}
x&\mapsto (2\pi)^{-n}\int_{\mathbb{R}^n}\rho(\xi)\widehat{b_\eta}(\xi)e^{ix\cdot\xi}\,d\mathcal{L}^n(\xi)
\end{align*}
and
\begin{align*}
b_\eta^{\mathrm{bad}}:U&\to \mathbb{C}
\end{align*}
\begin{align*}
x&\mapsto (2\pi)^{-n}\int_{\mathbb{R}^n}(1-\rho(\xi))\widehat{b_\eta}(\xi)e^{ix\cdot\xi}\,d\mathcal{L}^n(\xi).
\end{align*}
Then $b_\eta=b_\eta^{\mathrm{good}}+b_\eta^{\mathrm{bad}}$, and hence
\begin{align*}
\widehat{\psi\,\kappa^*u}(\eta)=(\chi u)(b_\eta^{\mathrm{good}})+(\chi u)(b_\eta^{\mathrm{bad}}).
\end{align*}
[/step]
[step:Control the good frequency contribution by combining microlocal decay with a finite order distribution estimate]
The distribution $\chi u$ is compactly supported. Hence there are a compact set $K_\chi\subset U$, an integer $s\geq0$, and a constant $A_s>0$ such that $\operatorname{supp}(\chi u)\subset K_\chi$ and
\begin{align*}
|(\chi u)(f)|\leq A_s\sum_{|\alpha|\leq s}\sup_{x\in K_\chi}|D^\alpha f(x)|
\end{align*}
for every $f\in C^\infty(U)$. Its Fourier transform $\widehat{\chi u}:\mathbb{R}^n\to\mathbb{C}$ is therefore a smooth function of at most polynomial growth. Since $-\operatorname{supp}\rho\subset\Gamma_\xi$, the microlocal smoothness estimate gives: for every integer $S\geq0$ there is $A_S>0$ such that
\begin{align*}
|\widehat{\chi u}(-\xi)|\leq A_S(1+|\xi|)^{-S}
\end{align*}
whenever $\rho(\xi)\neq0$.
The function $b_\eta$ is smooth and compactly supported in the fixed compact set $K:=\kappa(\operatorname{supp}\psi)$. For each $\eta\in\Gamma_\eta$, define the smooth map
\begin{align*}
\zeta_\eta:K&\to\mathbb{R}^n_0
\end{align*}
\begin{align*}
x&\mapsto (d\kappa^{-1}_x)^\top\eta.
\end{align*}
Define the closed stationary-frequency set
\begin{align*}
Z_\eta:=\{-\zeta_\eta(x):x\in K\}\subset\mathbb{R}^n_0.
\end{align*}
The set $Z_\eta$ records the stationary frequencies of $\widehat{b_\eta}$. The cone construction gives $\zeta_\eta(x)\in\Sigma_\xi$, and compactness of $K$ gives constants $0<m_0\leq M_0<\infty$ such that
\begin{align*}
m_0|\eta|\leq |\zeta_\eta(x)|\leq M_0|\eta|
\end{align*}
for every $x\in K$ and $\eta\in\Gamma_\eta$.
We justify the Fourier representation in the distributional pairing by first inserting a radial cutoff. Let $\theta\in C_c^\infty(\mathbb{R}^n)$ satisfy $0\leq\theta\leq1$ and $\theta=1$ on $B(0,1)$, and define $\theta_R:\mathbb{R}^n\to[0,1]$ by $\theta_R(\xi)=\theta(\xi/R)$ for $R\geq1$. Define the smooth compactly supported Fourier-cutoff regularisation
\begin{align*}
b_{\eta,R}^{\mathrm{good}}:U&\to\mathbb{C}
\end{align*}
\begin{align*}
x&\mapsto (2\pi)^{-n}\int_{\mathbb{R}^n}\theta_R(\xi)\rho(\xi)\widehat{b_\eta}(\xi)e^{ix\cdot\xi}\,d\mathcal{L}^n(\xi).
\end{align*}
Fourier inversion for compactly supported smooth functions gives $b_{\eta,R}^{\mathrm{good}}\to b_\eta^{\mathrm{good}}$ in $C^s(K_\chi)$ as $R\to\infty$, because repeated [integration by parts](/theorems/2098) in the compact support of $b_\eta$ gives rapid decay of $\widehat{b_\eta}(\xi)$ in $\xi$ uniformly for $\eta$ in compact angular subsets of $\Gamma_\eta$. The finite-order estimate for $\chi u$ therefore permits passage to the limit. For each finite $R$, the definition of the Fourier transform of the compactly supported distribution $\chi u$ gives
\begin{align*}
(\chi u)(b_{\eta,R}^{\mathrm{good}})=(2\pi)^{-n}\int_{\mathbb{R}^n}\theta_R(\xi)\rho(\xi)\widehat{b_\eta}(\xi)\widehat{\chi u}(-\xi)\,d\mathcal{L}^n(\xi).
\end{align*}
On the support of $\rho$ with $|\xi|\geq1$ we have $-\xi\in\Gamma_\xi$, so the microlocal estimate gives rapid decay of $\widehat{\chi u}(-\xi)$. On $|\xi|<1$ the Fourier transform $\widehat{\chi u}$ is smooth and hence bounded. Combining these bounds with the rapid decay of $\widehat{b_\eta}$ in $\xi$ gives an integrable dominating function independent of $R$ once the decay order is chosen larger than $s+n+N+1$. The [dominated convergence theorem](/theorems/4) then yields
\begin{align*}
(\chi u)(b_\eta^{\mathrm{good}})=(2\pi)^{-n}\int_{\mathbb{R}^n}\rho(\xi)\widehat{b_\eta}(\xi)\widehat{\chi u}(-\xi)\,d\mathcal{L}^n(\xi).
\end{align*}
We derive the parameter-dependent integration-by-parts estimate for the phase $x\mapsto \kappa^{-1}(x)\cdot\eta+x\cdot\xi$. The Fourier transform of $b_\eta$ is
\begin{align*}
\widehat{b_\eta}(\xi)=\int_K \psi(\kappa^{-1}(x))|\det d\kappa^{-1}_x|e^{-i(\kappa^{-1}(x)\cdot\eta+x\cdot\xi)}\,d\mathcal{L}^n(x).
\end{align*}
For fixed $\eta\in\Gamma_\eta$ and $\xi\in\mathbb{R}^n$, define
\begin{align*}
G_{\eta,\xi}:K&\to\mathbb{R}^n
\end{align*}
\begin{align*}
x&\mapsto \zeta_\eta(x)+\xi.
\end{align*}
Since $K$ is compact and $\zeta_\eta$ depends linearly on $\eta$, there is a constant $c_2>0$ such that
\begin{align*}
\sup_{x\in K}|G_{\eta,\xi}(x)|\geq c_2\operatorname{dist}(\xi,Z_\eta).
\end{align*}
On a finite [partition of unity](/page/Partition%20of%20Unity) of $K$, choose a coordinate direction in which the corresponding component of $G_{\eta,\xi}$ has size comparable to $\operatorname{dist}(\xi,Z_\eta)$, and integrate by parts with the one-dimensional operator obtained by dividing by that component. Derivatives falling on $G_{\eta,\xi}$ and on the amplitude are bounded by powers of $1+|\eta|+|\xi|$ because all derivatives of $\kappa^{-1}$ are bounded on $K$. Therefore, for every pair of integers $M,L\geq0$, after increasing $L$ if necessary, there is $B_{M,L}>0$ such that
\begin{align*}
|\widehat{b_\eta}(\xi)|\leq B_{M,L}(1+|\eta|+|\xi|)^L(1+\operatorname{dist}(\xi,Z_\eta))^{-M}
\end{align*}
for all $\eta\in\Gamma_\eta$ and $\xi\in\mathbb{R}^n$.
Split the integral into $|\xi|\geq (m_0/2)|\eta|$ and $|\xi|<(m_0/2)|\eta|$. On the first region, the rapid decay of $\widehat{\chi u}(-\xi)$ gives a factor bounded by a constant multiple of $(1+|\eta|)^{-S}$, and choosing $S$ larger than $L+n+N+1$ gives a contribution bounded by $C_N(1+|\eta|)^{-N}$. On the complementary region, every point of $Z_\eta$ has norm at least $m_0|\eta|$, so $\operatorname{dist}(\xi,Z_\eta)\geq (m_0/2)|\eta|$; choosing $M$ larger than $L+n+N+1$ gives the same bound. Therefore, for every integer $N\geq0$, there exists $C_N^{\mathrm{good}}>0$ such that
\begin{align*}
|(\chi u)(b_\eta^{\mathrm{good}})|\leq C_N^{\mathrm{good}}(1+|\eta|)^{-N}
\end{align*}
for every $\eta\in\Gamma_\eta$.
[/step]
[step:Control the complementary contribution by uniform nonstationary phase]
For the complementary multiplier, the relevant phase is
\begin{align*}
\Phi_\eta: \kappa(\operatorname{supp}\psi)&\to \mathbb{R}
\end{align*}
\begin{align*}
x&\mapsto \kappa^{-1}(x)\cdot\eta.
\end{align*}
Its differential is
\begin{align*}
\nabla_x\Phi_\eta(x)=(d\kappa^{-1}_x)^\top\eta.
\end{align*}
If $x=\kappa(y)$ with $y\in\operatorname{supp}\psi\subset W$, then the refined cone construction gives
\begin{align*}
\nabla_x\Phi_\eta(x)=(d\kappa_y)^{-\top}\eta\in\Sigma_\xi
\end{align*}
for every $\eta\in\Gamma_\eta$. Hence the stationary frequencies for the Fourier variable $\xi$ lie in $-\Sigma_\xi$. Since $\rho=1$ on $-\Sigma_\xi$ for $|\xi|\geq1$, the support of $1-\rho$ is separated from this stationary cone at high frequency. More precisely, compactness of the closed subsets
\begin{align*}
\{\omega\in S^{n-1}:r\omega\in -\Sigma_\xi\text{ for some }r>0\}
\end{align*}
and
\begin{align*}
\{\omega\in S^{n-1}:r\omega\in \operatorname{supp}(1-\rho)\text{ for some }r\geq1\}
\end{align*}
gives a number $\delta>0$ such that $|\omega-\omega'|\geq\delta$ whenever the first unit vector lies in the stationary set and the second lies in the high-frequency complementary multiplier set. Since $d\kappa_y^{-\top}$ is uniformly invertible for $y\in\overline{W}$, there are constants $c>0$ and $R>0$ such that
\begin{align*}
|\nabla_x(\Phi_\eta(x)+x\cdot\xi)|\geq c(|\eta|+|\xi|)
\end{align*}
whenever $x\in \kappa(\operatorname{supp}\psi)$, $\eta\in\Gamma_\eta$, $(1-\rho)(\xi)\neq 0$, and $|\xi|\geq1$.
It remains to handle the low-frequency part $|\xi|<1$. The same uniform invertibility gives a constant $c_0>0$ such that
\begin{align*}
|\nabla_x\Phi_\eta(x)|\geq c_0|\eta|
\end{align*}
for all $x\in\kappa(\operatorname{supp}\psi)$ and $\eta\in\Gamma_\eta$. Hence, after increasing $R$, if $|\xi|<1$ and $|\eta|\geq R$, then
\begin{align*}
|\nabla_x(\Phi_\eta(x)+x\cdot\xi)|\geq \frac{c_0}{2}|\eta|.
\end{align*}
The remaining set $|\xi|<1$ and $|\eta|<R$ is bounded and is absorbed into the constants.
Define the integration-by-parts operator
\begin{align*}
L_{\eta,\xi}:C^\infty(\kappa(\operatorname{supp}\psi))&\to C^\infty(\kappa(\operatorname{supp}\psi))
\end{align*}
\begin{align*}
f&\mapsto -\frac{1}{i|\nabla_x(\Phi_\eta+x\cdot\xi)|^2}\nabla_x(\Phi_\eta+x\cdot\xi)\cdot\nabla f.
\end{align*}
Then $L_{\eta,\xi}(e^{-i(\Phi_\eta(x)+x\cdot\xi)})=e^{-i(\Phi_\eta(x)+x\cdot\xi)}$. Repeatedly transferring $L_{\eta,\xi}$ to the smooth compactly supported amplitude in the Fourier integral defining $(1-\rho)(\xi)\widehat{b_\eta}(\xi)$ gives rapid decay in the nonstationary parameter. Derivatives falling on the multiplier $1-\rho$ are controlled because $1-\rho$ is smooth on all of $\mathbb{R}^n$ and is a symbol of order $0$ outside $B(0,1)$. On $|\xi|\geq1$ the lower bound gives powers of $(1+|\eta|+|\xi|)^{-1}$; on $|\xi|<1$ and $|\eta|\geq R$ it gives powers of $(1+|\eta|)^{-1}$. Choosing the number of integrations larger than $n+s+Q+1$ makes the differentiated inverse Fourier integrals absolutely convergent. Thus, for every pair of integers $Q,s\geq0$ and every compact set $K_\chi\subset U$, there is a constant $E_{Q,s,K_\chi}>0$ such that
\begin{align*}
\sup_{x\in K_\chi}|D^\alpha b_\eta^{\mathrm{bad}}(x)|\leq E_{Q,s,K_\chi}(1+|\eta|)^{-Q}
\end{align*}
for every multi-index $\alpha$ with $|\alpha|\leq s$ and every $\eta\in\Gamma_\eta$. The bounded region $|\xi|<1$ and $|\eta|<R$ is included by increasing $E_{Q,s,K_\chi}$.
Since $\chi u$ is compactly supported, there are a compact set $K_\chi\subset U$, an integer $s\geq0$, and a constant $A_s>0$ such that $\operatorname{supp}(\chi u)\subset K_\chi$ and
\begin{align*}
|(\chi u)(f)|\leq A_s\sum_{|\alpha|\leq s}\sup_{x\in K_\chi}|D^\alpha f(x)|
\end{align*}
for every $f\in C^\infty(U)$. This estimate does not require $f$ to have compact support; it only uses the values of $f$ on the fixed compact set $K_\chi$. Applying this bound to $f=b_\eta^{\mathrm{bad}}$ gives, for every integer $Q\geq0$, a constant $D_Q>0$ such that
\begin{align*}
|(\chi u)(b_\eta^{\mathrm{bad}})|\leq D_Q(1+|\eta|)^{-Q}
\end{align*}
for every $\eta\in\Gamma_\eta$. The constant is uniform in $\eta$ because the lower bound for $|\nabla_x(\Phi_\eta+x\cdot\xi)|$, the symbol estimates for the multiplier $1-\rho$, and all derivatives of $\kappa^{-1}$ on the compact sets $K_\chi$ and $\kappa(\operatorname{supp}\psi)$ are uniform.
[guided]
The complementary term is the part where the Fourier variable $\xi$ is kept away from every possible stationary frequency of the transformed oscillation. Why should it decay rapidly? Because the phase then has a uniform nonzero gradient.
The phase generated by the coordinate change is
\begin{align*}
\Phi_\eta:\kappa(\operatorname{supp}\psi)&\to \mathbb{R}
\end{align*}
\begin{align*}
x&\mapsto \kappa^{-1}(x)\cdot\eta.
\end{align*}
Differentiating by the chain rule gives
\begin{align*}
\nabla_x\Phi_\eta(x)=(d\kappa^{-1}_x)^\top\eta.
\end{align*}
Since $x=\kappa(y)$ with $y\in\operatorname{supp}\psi\subset W$, the refined cone construction gives
\begin{align*}
(d\kappa^{-1}_x)^\top\eta=(d\kappa_y)^{-\top}\eta\in \Sigma_\xi
\end{align*}
for every $\eta\in\Gamma_\eta$. The stationary relation in the Fourier transform of $b_\eta$ is $\xi=-(d\kappa^{-1}_x)^\top\eta$, so the relevant stationary cone for $\xi$ is the closed cone $-\Sigma_\xi$. The cutoff $\rho$ was chosen to equal $1$ on $-\Sigma_\xi$ at high frequency. Therefore, on the support of $1-\rho$ and for $|\xi|\geq1$, the direction of $\xi$ is separated from every stationary direction. Compactness of the two closed unit-direction sets gives a positive angular separation, and the uniform invertibility of $d\kappa_y$ on $\overline W$ converts that angular separation into constants $c>0$ and $R>0$ such that
\begin{align*}
|\nabla_x(\Phi_\eta(x)+x\cdot\xi)|\geq c(|\eta|+|\xi|)
\end{align*}
for all $x\in\kappa(\operatorname{supp}\psi)$, all $\eta\in\Gamma_\eta$, and all $\xi$ with $(1-\rho)(\xi)\neq 0$ and $|\eta|+|\xi|\geq R$. For $|\xi|<1$ there is a separate estimate. Uniform invertibility of $d\kappa_y^{-\top}$ on $\overline W$ gives $c_0>0$ such that
\begin{align*}
|\nabla_x\Phi_\eta(x)|\geq c_0|\eta|
\end{align*}
for all $x\in\kappa(\operatorname{supp}\psi)$ and $\eta\in\Gamma_\eta$. Therefore, when $|\xi|<1$ and $|\eta|$ is large enough,
\begin{align*}
|\nabla_x(\Phi_\eta(x)+x\cdot\xi)|\geq \frac{c_0}{2}|\eta|.
\end{align*}
Only the set $|\xi|<1$ with $|\eta|$ bounded contributes directly to the constants.
Now define
\begin{align*}
L_{\eta,\xi}:C^\infty(\kappa(\operatorname{supp}\psi))&\to C^\infty(\kappa(\operatorname{supp}\psi))
\end{align*}
\begin{align*}
f&\mapsto -\frac{1}{i|\nabla_x(\Phi_\eta+x\cdot\xi)|^2}\nabla_x(\Phi_\eta+x\cdot\xi)\cdot\nabla f.
\end{align*}
This operator is chosen so that it reproduces the oscillatory exponential:
\begin{align*}
L_{\eta,\xi}(e^{-i(\Phi_\eta(x)+x\cdot\xi)})=e^{-i(\Phi_\eta(x)+x\cdot\xi)}.
\end{align*}
Thus every integration by parts moves one derivative from the exponential onto the smooth amplitude in the Fourier integral defining $(1-\rho)(\xi)\widehat{b_\eta}(\xi)$ and gains one factor comparable to $(|\eta|+|\xi|)^{-1}$. Since the amplitude is supported in the fixed compact set $\kappa(\operatorname{supp}\psi)$ and all derivatives of $\kappa^{-1}$ are bounded there, the constants in these estimates are uniform for $\eta\in\Gamma_\eta$. After enough integrations by parts, the derivatives falling on the multiplier $1-\rho$ are controlled by its symbol estimates, and the resulting inverse Fourier integrals are absolutely convergent after choosing the number of integrations larger than $n+s+Q+1$. Thus, on every fixed compact set $K_\chi\subset U$, differentiating the inverse Fourier representation of $b_\eta^{\mathrm{bad}}$ under the integral gives, for every pair of integers $Q,s\geq0$, a constant $E_{Q,s,K_\chi}>0$ such that
\begin{align*}
\sup_{x\in K_\chi}|D^\alpha b_\eta^{\mathrm{bad}}(x)|\leq E_{Q,s,K_\chi}(1+|\eta|)^{-Q}
\end{align*}
for all $|\alpha|\leq s$.
Finally, we use the continuity of the compactly supported distribution $\chi u$. There are a compact set $K_\chi\subset U$, an integer $s\geq0$, and a constant $A_s>0$ such that $\operatorname{supp}(\chi u)\subset K_\chi$ and
\begin{align*}
|(\chi u)(f)|\leq A_s\sum_{|\alpha|\leq s}\sup_{x\in K_\chi}|D^\alpha f(x)|
\end{align*}
for every $f\in C^\infty(U)$. This is the correct form of the estimate because $b_\eta^{\mathrm{bad}}$ need not be compactly supported after the Fourier multiplier $1-\rho$ is applied. Applying it to $f=b_\eta^{\mathrm{bad}}$ gives
\begin{align*}
|(\chi u)(b_\eta^{\mathrm{bad}})|\leq D_Q(1+|\eta|)^{-Q}
\end{align*}
for every $\eta\in\Gamma_\eta$, where $D_Q>0$ depends on $Q$, the fixed cutoffs, the compact set $K_\chi$, and finitely many derivatives of $\kappa^{-1}$, but not on $\eta$.
[/guided]
[/step]
[step:Conclude one inclusion from the rapid decay estimate]
Combining the good and bad estimates, for every integer $N\geq 0$ there exists $C_N>0$ such that
\begin{align*}
|\widehat{\psi\,\kappa^*u}(\eta)|\leq C_N(1+|\eta|)^{-N}
\end{align*}
for all $\eta\in\Gamma_\eta$. Since $\psi(y_0)\neq 0$, the definition of the wave front set gives
\begin{align*}
(y_0,\eta_0)\notin \operatorname{WF}(\kappa^*u).
\end{align*}
Therefore
\begin{align*}
\operatorname{WF}(\kappa^*u)\subseteq \{(y,(d\kappa_y)^\top\xi):(\kappa(y),\xi)\in \operatorname{WF}(u)\}.
\end{align*}
[/step]
[step:Apply the same argument to the inverse diffeomorphism]
Let $\lambda:=\kappa^{-1}:U\to V$. Applying the inclusion already proved to the diffeomorphism $\lambda$ and the distribution $\kappa^*u\in\mathcal{D}'(V)$ gives
\begin{align*}
\operatorname{WF}(u)=\operatorname{WF}(\lambda^*(\kappa^*u))
\subseteq
\{(x,(d\lambda_x)^\top\eta):(\lambda(x),\eta)\in \operatorname{WF}(\kappa^*u)\}.
\end{align*}
Set $x=\kappa(y)$. Since $d\lambda_{\kappa(y)}=(d\kappa_y)^{-1}$, the covector relation
\begin{align*}
\xi=(d\lambda_{\kappa(y)})^\top\eta
\end{align*}
is equivalent to
\begin{align*}
\eta=(d\kappa_y)^\top\xi.
\end{align*}
Thus the preceding inclusion is exactly the reverse containment
\begin{align*}
\{(y,(d\kappa_y)^\top\xi):(\kappa(y),\xi)\in \operatorname{WF}(u)\}
\subseteq
\operatorname{WF}(\kappa^*u).
\end{align*}
Together with the first inclusion, this proves
\begin{align*}
\operatorname{WF}(\kappa^*u)=\{(y,(d\kappa_y)^\top\xi):(\kappa(y),\xi)\in \operatorname{WF}(u)\}.
\end{align*}
[/step]
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