[proofplan] We prove the equality pointwise. If no nonzero covector over $x_0$ lies in $WF(u)$, then each direction in the unit sphere has a conic neighbourhood on which a localized [Fourier transform](/page/Fourier%20Transform) decays rapidly; compactness of the unit sphere lets us pass from directional decay to rapid decay in all large frequencies. The Fourier smoothness criterion for compactly supported distributions then shows that a cutoff of $u$ near $x_0$ is smooth, so $x_0$ is outside the singular support. Conversely, if $u$ is already smooth near $x_0$, then every sufficiently small localization has rapidly decreasing Fourier transform in every cone, so no nonzero covector over $x_0$ belongs to the wave front set. [/proofplan]

[step:Define the projection and reduce the claim to two pointwise inclusions] Let \begin{align*} \pi: T^*X \setminus 0 \to X, \qquad (x,\xi) \mapsto x \end{align*} be the base projection, and use the stated identification of $T_x^*X$ with $\mathbb{R}^n$ through the standard coordinate covectors $dx_1,\dots,dx_n$. Define the unit sphere \begin{align*} S^{n-1} := \{\omega \in \mathbb{R}^n : |\omega|=1\}. \end{align*} We must prove that for each $x_0 \in X$, \begin{align*} x_0 \notin \pi(WF(u)) \iff x_0 \notin \operatorname{sing\,supp}(u). \end{align*} By the definition of the wave front set, $x_0 \notin \pi(WF(u))$ means that $(x_0,\xi_0) \notin WF(u)$ for every $\xi_0 \in \mathbb{R}^n \setminus \{0\}$. By the definition of singular support, $x_0 \notin \operatorname{sing\,supp}(u)$ means that there is an open neighbourhood $V \subset X$ of $x_0$ and a smooth function $f: V \to \mathbb{C}$ such that $u|_V$ is the [regular distribution](/page/Regular%20Distribution) induced by $f$ on $V$. [/step]

[step:Cover all cotangent directions by finitely many microlocally regular cones]Assume $x_0 \notin \pi(WF(u))$. For every $\omega \in S^{n-1}$, the point $(x_0,\omega)$ is not in $WF(u)$. By the definition of the wave front set, for each $\omega \in S^{n-1}$ there exist an open neighbourhood $V_\omega \subset X$ of $x_0$, a conic [open set](/page/Open%20Set) $\Gamma_\omega \subset \mathbb{R}^n \setminus \{0\}$ containing $\omega$, and a cutoff \begin{align*} \varphi_\omega: X \to \mathbb{C} \end{align*} with $\varphi_\omega \in C_c^\infty(V_\omega)$ and $\varphi_\omega(x_0) \ne 0$, such that the Fourier transform \begin{align*} \widehat{\varphi_\omega u}: \mathbb{R}^n \to \mathbb{C} \end{align*} is rapidly decreasing in $\Gamma_\omega$. Explicitly, for every integer $N \ge 0$ there is a constant $C_{\omega,N} > 0$ such that \begin{align*} |\widehat{\varphi_\omega u}(\xi)| \le C_{\omega,N}(1+|\xi|)^{-N} \end{align*} for every $\xi \in \Gamma_\omega$. The sets $\Gamma_\omega \cap S^{n-1}$ form an open cover of $S^{n-1}$. Since $S^{n-1}$ is compact, choose finitely many directions $\omega_1,\dots,\omega_m \in S^{n-1}$ such that \begin{align*} S^{n-1} \subset \bigcup_{j=1}^m \Gamma_{\omega_j}. \end{align*} Since $S^{n-1}$ is a compact [metric space](/page/Metric%20Space), hence normal, the finite open cover admits a finite shrinking. Thus choose open sets $\Omega_j \subset S^{n-1}$ for $1 \le j \le m$ such that \begin{align*} S^{n-1} \subset \bigcup_{j=1}^m \Omega_j \end{align*} and \begin{align*} \overline{\Omega_j} \subset \Gamma_{\omega_j} \cap S^{n-1}. \end{align*} Define the smaller conic open sets \begin{align*} \Gamma_j' := \{r\theta : r>0 \text{ and } \theta \in \Omega_j\} \subset \mathbb{R}^n \setminus \{0\}. \end{align*} Then the cones $\Gamma_j'$ cover $\mathbb{R}^n \setminus \{0\}$, and each $\Gamma_j'$ has positive angular separation from $\mathbb{R}^n \setminus \Gamma_{\omega_j}$. Choose an open neighbourhood $V \subset X$ of $x_0$ such that $V \subset \bigcap_{j=1}^m V_{\omega_j}$ and choose \begin{align*} \psi: X \to \mathbb{C} \end{align*} with $\psi \in C_c^\infty(V)$ and $\psi=1$ on some open neighbourhood $W \subset V$ of $x_0$.[/step]

[guided]The hypothesis says that every nonzero covector over $x_0$ is microlocally regular. Since covectors are homogeneous in the definition of $WF(u)$, it is enough to look at their directions, namely points of the unit sphere $S^{n-1}$. Fix a direction $\omega \in S^{n-1}$. Because $(x_0,\omega) \notin WF(u)$, the definition of the wave front set gives three pieces of data: an open neighbourhood $V_\omega \subset X$ of $x_0$, a conic open set $\Gamma_\omega \subset \mathbb{R}^n \setminus \{0\}$ containing $\omega$, and a cutoff \begin{align*} \varphi_\omega: X \to \mathbb{C} \end{align*} with $\varphi_\omega \in C_c^\infty(V_\omega)$ and $\varphi_\omega(x_0) \ne 0$. The microlocal regularity condition is that the localized Fourier transform \begin{align*} \widehat{\varphi_\omega u}: \mathbb{R}^n \to \mathbb{C} \end{align*} decays faster than every inverse power of $|\xi|$ inside the cone $\Gamma_\omega$. That is, for every integer $N \ge 0$ there is a constant $C_{\omega,N} > 0$ satisfying \begin{align*} |\widehat{\varphi_\omega u}(\xi)| \le C_{\omega,N}(1+|\xi|)^{-N} \end{align*} for every $\xi \in \Gamma_\omega$. Why do we now use compactness? The definition gives one cone at a time, but smoothness requires rapid decay in every frequency direction simultaneously. The spherical pieces $\Gamma_\omega \cap S^{n-1}$ cover $S^{n-1}$. Since $S^{n-1}$ is compact, finitely many of them suffice: choose $\omega_1,\dots,\omega_m \in S^{n-1}$ with \begin{align*} S^{n-1} \subset \bigcup_{j=1}^m \Gamma_{\omega_j}. \end{align*} There is one technical point: after multiplying by another cutoff, Fourier transforms are convolved, so decay on the boundary of a cone need not remain in the same cone. We therefore shrink the cones before doing the multiplication. Because $S^{n-1}$ is a compact metric space, it is normal, so the finite open cover admits a shrinking. Choose open sets $\Omega_j \subset S^{n-1}$ with \begin{align*} S^{n-1} \subset \bigcup_{j=1}^m \Omega_j \end{align*} and \begin{align*} \overline{\Omega_j} \subset \Gamma_{\omega_j} \cap S^{n-1}. \end{align*} Define \begin{align*} \Gamma_j' := \{r\theta : r>0 \text{ and } \theta \in \Omega_j\}. \end{align*} These smaller cones still cover every nonzero frequency, but each of them lies with positive angular margin inside the cone where the original microlocal decay estimate is known. Now choose a single cutoff valid for all these finitely many cones. Since each $V_{\omega_j}$ is an open neighbourhood of $x_0$, their finite intersection is again an open neighbourhood of $x_0$. Choose an open neighbourhood $V \subset X$ of $x_0$ with \begin{align*} V \subset \bigcap_{j=1}^m V_{\omega_j} \end{align*} and choose \begin{align*} \psi: X \to \mathbb{C} \end{align*} with $\psi \in C_c^\infty(V)$ and $\psi=1$ on some open neighbourhood $W \subset V$ of $x_0$. This cutoff $\psi$ localizes $u$ inside every neighbourhood where the finitely many microlocal estimates are available.[/guided]

[step:Transfer the conic estimates to the smaller cones for the common cutoff]For each $j \in \{1,\dots,m\}$, because $\varphi_{\omega_j}(x_0)\ne 0$, after shrinking $V$ and choosing $\psi$ with sufficiently small support, we may assume $\varphi_{\omega_j}$ is nonzero on an open neighbourhood $U_j \subset V_{\omega_j}$ of $\operatorname{supp}\psi$. Define \begin{align*} a_j: U_j \to \mathbb{C}, \qquad x \mapsto \frac{\psi(x)}{\varphi_{\omega_j}(x)}. \end{align*} Extend $a_j$ by zero from a compactly supported smooth function on $U_j$ to a function \begin{align*} a_j: X \to \mathbb{C}. \end{align*} Then $a_j \in C_c^\infty(V_{\omega_j})$ and $\psi=a_j\varphi_{\omega_j}$ on $X$. Hence \begin{align*} \psi u = a_j(\varphi_{\omega_j}u) \end{align*} as compactly supported distributions on $X$. We prove that this multiplication transfers rapid decay from $\Gamma_{\omega_j}$ to the smaller cone $\Gamma_j'$. Let $T_j := \varphi_{\omega_j}u$. Since $T_j$ is compactly supported, its Fourier transform is polynomially bounded: there are constants $B_j>0$ and $M_j \ge 0$ such that \begin{align*} |\widehat{T_j}(\eta)| \le B_j(1+|\eta|)^{M_j} \end{align*} for every $\eta \in \mathbb{R}^n$. Since $a_j \in C_c^\infty(X)$, the Fourier transform $\widehat{a_j}$ is rapidly decreasing: for every integer $L \ge 0$ there is a constant $D_{j,L}>0$ such that \begin{align*} |\widehat{a_j}(\zeta)| \le D_{j,L}(1+|\zeta|)^{-L} \end{align*} for every $\zeta \in \mathbb{R}^n$. The angular separation of $\Gamma_j'$ from $\mathbb{R}^n \setminus \Gamma_{\omega_j}$ gives a number $c_j>0$ such that, whenever $\xi \in \Gamma_j'$ and $\eta \notin \Gamma_{\omega_j}$ with $|\eta| \ge |\xi|/2$, one has \begin{align*} |\xi-\eta| \ge c_j(|\xi|+|\eta|). \end{align*} Indeed, the directions $\xi/|\xi|$ and $\eta/|\eta|$ have a positive minimum angular separation on the compact set $\overline{\Omega_j} \times (S^{n-1} \setminus \Gamma_{\omega_j})$, and the ratio $|r\theta-s\rho|/(r+s)$ has a positive minimum for $\theta \in \overline{\Omega_j}$, $\rho \notin \Gamma_{\omega_j}$, and $s \ge r/2$. Using the convolution formula for the Fourier transform of a product, \begin{align*} \widehat{\psi u}(\xi)=\widehat{a_jT_j}(\xi)=C_F\int_{\mathbb{R}^n}\widehat{a_j}(\xi-\eta)\widehat{T_j}(\eta)\,d\mathcal{L}^n(\eta) \end{align*} with a Fourier-normalisation constant $C_F>0$, where the integral is absolutely convergent because $\widehat{a_j}$ is rapidly decreasing and $\widehat{T_j}$ is polynomially bounded, split the integral into the regions $E_1 := \Gamma_{\omega_j}$, $E_2 := \{\eta \in \mathbb{R}^n \setminus \Gamma_{\omega_j}: |\eta|<|\xi|/2\}$, and $E_3 := \{\eta \in \mathbb{R}^n \setminus \Gamma_{\omega_j}: |\eta|\ge |\xi|/2\}$. On $E_1$, choose an integer $K>N+n+1$. Let $C_{j,K}>0$ be the constant in the rapid-decay estimate for $\widehat{T_j}$ on $\Gamma_{\omega_j}$ with exponent $K$. The rapid decay of both factors gives \begin{align*} |\widehat{a_j}(\xi-\eta)|\,|\widehat{T_j}(\eta)| \le D_{j,K}C_{j,K}(1+|\xi-\eta|)^{-K}(1+|\eta|)^{-K}. \end{align*} The convolution estimate for the integrable weight $(1+|\cdot|)^{-K}$ gives a constant $C'_{j,N}>0$ such that \begin{align*} \int_{E_1}|\widehat{a_j}(\xi-\eta)|\,|\widehat{T_j}(\eta)|\,d\mathcal{L}^n(\eta) \le C'_{j,N}(1+|\xi|)^{-N}. \end{align*} For example, this follows by splitting $\mathbb{R}^n$ into the two sets $|\eta|\ge |\xi|/2$ and $|\xi-\eta|\ge |\xi|/2$, using one factor to supply $(1+|\xi|)^{-N}$ and the other factor to supply an integrable majorant. On $E_2$, the inequality $|\eta|<|\xi|/2$ gives $|\xi-\eta|\ge |\xi|/2$. Choose $L>N+n+M_j+1$. Then \begin{align*} \int_{E_2}|\widehat{a_j}(\xi-\eta)|\,|\widehat{T_j}(\eta)|\,d\mathcal{L}^n(\eta) \le C''_{j,N}(1+|\xi|)^{-N} \end{align*} because $\widehat{a_j}$ contributes $(1+|\xi|)^{-L}$ and the polynomial bound on $\widehat{T_j}$ contributes at most a factor of order $(1+|\xi|)^{M_j+n}$ over the ball $|\eta|<|\xi|/2$. On $E_3$, the stronger angular estimate gives \begin{align*} |\widehat{a_j}(\xi-\eta)|\,|\widehat{T_j}(\eta)| \le D_{j,L}B_j c_j^{-L}(1+|\xi|+|\eta|)^{-L}(1+|\eta|)^{M_j}. \end{align*} Since $L>N+n+M_j+1$, integration over $E_3$ gives a constant $C'''_{j,N}>0$ such that \begin{align*} \int_{E_3}|\widehat{a_j}(\xi-\eta)|\,|\widehat{T_j}(\eta)|\,d\mathcal{L}^n(\eta) \le C'''_{j,N}(1+|\xi|)^{-N}. \end{align*} Combining the three region estimates gives the desired rapid decay on $\Gamma_j'$. Therefore, for every integer $N \ge 0$ there is a constant $A_{j,N}>0$ such that \begin{align*} |\widehat{\psi u}(\xi)| \le A_{j,N}(1+|\xi|)^{-N} \end{align*} for every $\xi \in \Gamma_j'$. Since the finitely many smaller cones $\Gamma_j'$ cover every nonzero frequency direction, every $\xi \in \mathbb{R}^n \setminus \{0\}$ belongs to some $\Gamma_j'$. For each $N \ge 0$, define \begin{align*} A_N := \max_{1 \le j \le m} A_{j,N}. \end{align*} Then \begin{align*} |\widehat{\psi u}(\xi)| \le A_N(1+|\xi|)^{-N} \end{align*} for every $\xi \in \mathbb{R}^n \setminus \{0\}$. Enlarging $A_N$ if necessary to dominate the finite value $|\widehat{\psi u}(0)|$, the same estimate holds for every $\xi \in \mathbb{R}^n$. Thus $\widehat{\psi u}$ is rapidly decreasing on all of $\mathbb{R}^n$.[/step]

[guided]The subtle point is that multiplying by $a_j$ becomes convolution after taking Fourier transforms, so frequency directions can mix. This is why we shrank from $\Gamma_{\omega_j}$ to $\Gamma_j'$: if $\xi \in \Gamma_j'$ and a frequency $\eta$ lies outside $\Gamma_{\omega_j}$, then either $\eta$ is small compared with $\xi$, or the angular separation forces $\xi-\eta$ to be large. Fix $j \in \{1,\dots,m\}$ and define $T_j := \varphi_{\omega_j}u$. Since $T_j$ is compactly supported, its Fourier transform has polynomial growth: there are constants $B_j>0$ and $M_j \ge 0$ such that \begin{align*} |\widehat{T_j}(\eta)| \le B_j(1+|\eta|)^{M_j} \end{align*} for every $\eta \in \mathbb{R}^n$. Since $a_j \in C_c^\infty(X)$, the Fourier transform $\widehat{a_j}$ is rapidly decreasing: for every integer $L \ge 0$ there is $D_{j,L}>0$ such that \begin{align*} |\widehat{a_j}(\zeta)| \le D_{j,L}(1+|\zeta|)^{-L} \end{align*} for every $\zeta \in \mathbb{R}^n$. The product identity $\psi u=a_jT_j$ gives the convolution formula \begin{align*} \widehat{\psi u}(\xi)=C_F\int_{\mathbb{R}^n}\widehat{a_j}(\xi-\eta)\widehat{T_j}(\eta)\,d\mathcal{L}^n(\eta) \end{align*} for a Fourier-normalisation constant $C_F>0$. This is an ordinary absolutely convergent integral: $\widehat{a_j}$ decays faster than every inverse power, while $\widehat{T_j}$ grows at most polynomially. For $\xi \in \Gamma_j'$, split the integration domain into $E_1 := \Gamma_{\omega_j}$, $E_2 := \{\eta \notin \Gamma_{\omega_j}: |\eta|<|\xi|/2\}$, and $E_3 := \{\eta \notin \Gamma_{\omega_j}: |\eta|\ge |\xi|/2\}$. On $E_1$, both factors are useful. The microlocal hypothesis gives rapid decay of $\widehat{T_j}(\eta)$ in $\eta$, and smooth compact support of $a_j$ gives rapid decay of $\widehat{a_j}(\xi-\eta)$ in $\xi-\eta$. Choose an integer $K>N+n+1$, and let $C_{j,K}>0$ be the constant in the rapid-decay estimate for $\widehat{T_j}$ on $\Gamma_{\omega_j}$ with exponent $K$. Then \begin{align*} |\widehat{a_j}(\xi-\eta)|\,|\widehat{T_j}(\eta)| \le D_{j,K}C_{j,K}(1+|\xi-\eta|)^{-K}(1+|\eta|)^{-K}. \end{align*} The convolution of these two integrable weights is bounded by a constant times $(1+|\xi|)^{-N}$. To see this, split the integration domain into $|\eta|\ge |\xi|/2$ and $|\xi-\eta|\ge |\xi|/2$; one factor then supplies $(1+|\xi|)^{-N}$ and the other remains integrable because $K>n$. On $E_2$, the condition $|\eta|<|\xi|/2$ implies $|\xi-\eta|\ge |\xi|/2$. Choose $L>N+n+M_j+1$. The rapid decay of $\widehat{a_j}(\xi-\eta)$ gives a factor of order $(1+|\xi|)^{-L}$, while the polynomial bound for $\widehat{T_j}(\eta)$ contributes at most order $(1+|\xi|)^{M_j+n}$ after integration over the ball $|\eta|<|\xi|/2$. Hence the $E_2$ contribution is bounded by a constant times $(1+|\xi|)^{-N}$. On $E_3$, we need the stronger form of angular separation. Since the directions in $\Gamma_j'$ have closure inside $\Gamma_{\omega_j}$, there is $c_j>0$ such that \begin{align*} |\xi-\eta| \ge c_j(|\xi|+|\eta|) \end{align*} whenever $\xi \in \Gamma_j'$, $\eta \notin \Gamma_{\omega_j}$, and $|\eta|\ge |\xi|/2$. Therefore \begin{align*} |\widehat{a_j}(\xi-\eta)|\,|\widehat{T_j}(\eta)| \le D_{j,L}B_jc_j^{-L}(1+|\xi|+|\eta|)^{-L}(1+|\eta|)^{M_j}. \end{align*} Because $L>N+n+M_j+1$, this last majorant is integrable in $\eta$ and its integral is bounded by a constant times $(1+|\xi|)^{-N}$. Thus for every integer $N\ge 0$ there is $A_{j,N}>0$ such that \begin{align*} |\widehat{\psi u}(\xi)| \le A_{j,N}(1+|\xi|)^{-N} \end{align*} for every $\xi \in \Gamma_j'$. The smaller cones $\Gamma_j'$ cover $\mathbb{R}^n\setminus\{0\}$. Taking the maximum of the finitely many constants $A_{j,N}$ gives rapid decay for every nonzero frequency, and increasing the constant to include $\xi=0$ gives rapid decay on all of $\mathbb{R}^n$.[/guided]

[step:Use Fourier smoothness to exclude $x_0$ from the singular support] The distribution $\psi u$ has compact support because $\psi \in C_c^\infty(X)$. We have proved that its Fourier transform is rapidly decreasing on $\mathbb{R}^n$. By the standard Fourier smoothness criterion for compactly supported distributions, a compactly supported distribution whose Fourier transform is rapidly decreasing is induced by a function in $C^\infty(\mathbb{R}^n)$. Hence there exists \begin{align*} g: \mathbb{R}^n \to \mathbb{C} \end{align*} with $g \in C^\infty(\mathbb{R}^n)$ such that $\psi u$ is the regular distribution induced by $g$. Since $\psi=1$ on the open neighbourhood $W$ of $x_0$, the restrictions of $u$ and $\psi u$ to $W$ agree. Therefore $u|_W$ is induced by the smooth function $g|_W: W \to \mathbb{C}$. Thus $x_0 \notin \operatorname{sing\,supp}(u)$. [/step]

[step:Show smoothness near $x_0$ rules out every covector over $x_0$] Conversely, assume $x_0 \notin \operatorname{sing\,supp}(u)$. Then there exist an open neighbourhood $V \subset X$ of $x_0$ and a smooth function \begin{align*} f: V \to \mathbb{C} \end{align*} with $f \in C^\infty(V)$ such that $u|_V$ is the regular distribution induced by $f$. Let $\xi_0 \in \mathbb{R}^n \setminus \{0\}$. Choose \begin{align*} \chi: X \to \mathbb{C} \end{align*} with $\chi \in C_c^\infty(V)$ and $\chi(x_0)\ne 0$. Then $\chi u$ is the regular distribution induced by the compactly supported smooth function $\chi f: X \to \mathbb{C}$. Since $\chi f \in C_c^\infty(X)$, its Fourier transform is rapidly decreasing on $\mathbb{R}^n$, and therefore it is rapidly decreasing in every conic neighbourhood of $\xi_0$. By the definition of the wave front set, $(x_0,\xi_0)\notin WF(u)$. Because $\xi_0 \in \mathbb{R}^n \setminus \{0\}$ was arbitrary, no nonzero covector over $x_0$ belongs to $WF(u)$. Hence $x_0 \notin \pi(WF(u))$. [/step]

[step:Conclude the equality of the two subsets of $X$] We have shown that for every $x_0 \in X$, \begin{align*} x_0 \notin \pi(WF(u)) \implies x_0 \notin \operatorname{sing\,supp}(u) \end{align*} and \begin{align*} x_0 \notin \operatorname{sing\,supp}(u) \implies x_0 \notin \pi(WF(u)). \end{align*} Taking complements in $X$ gives \begin{align*} \pi(WF(u)) = \operatorname{sing\,supp}(u). \end{align*} This is the desired statement. [/step]