[proofplan] We prove the two inclusions pointwise. If $u$ is smooth near a point $x_0$, then every compactly supported localization of $u$ near $x_0$ is smooth, so its [Fourier transform](/page/Fourier%20Transform) is rapidly decreasing in every conic direction; hence no nonzero covector over $x_0$ lies in $WF(u)$. Conversely, if no nonzero covector over $x_0$ lies in $WF(u)$, compactness of the unit cosphere in one coordinate chart gives finitely many conic Fourier estimates. A single cutoff near $x_0$ is then rapidly decreasing in all frequency directions, so the local Fourier criterion for smoothness gives smoothness of $u$ near $x_0$. [/proofplan] [step:Reduce both inclusions to a fixed point of the manifold] Fix $x_0 \in X$. We shall prove \begin{align*} x_0 \notin \operatorname{sing\,supp}(u) \iff (x_0,\xi_0) \notin WF(u) \text{ for every } \xi_0 \in T_{x_0}^*X \setminus \{0\}. \end{align*} Since $\pi(WF(u))$ is precisely the set of all base points $x \in X$ for which there exists a nonzero covector $\xi \in T_x^*X$ with $(x,\xi)\in WF(u)$, this pointwise equivalence is exactly \begin{align*} \pi(WF(u))=\operatorname{sing\,supp}(u). \end{align*} [/step] [step:Exclude all covectors over a point where the distribution is smooth] Assume $x_0 \notin \operatorname{sing\,supp}(u)$. By the definition of singular support, there exists an open neighbourhood $O \subset X$ of $x_0$ such that the restriction of $u$ to $O$ is induced by a smooth function on $O$. Let $\xi_0 \in T_{x_0}^*X \setminus \{0\}$. Choose a coordinate chart $(U,\kappa)$ with $x_0 \in U \subset O$, where \begin{align*} \kappa:U \to \kappa(U) \subset \mathbb{R}^n \end{align*} is a diffeomorphism onto an open subset. Let $\rho \in C_c^\infty(U)$ satisfy $\rho(x_0)\ne 0$. The localized distribution $\rho u$ is represented in this chart by a compactly supported smooth function on $\kappa(U)$. Therefore its Fourier transform is rapidly decreasing in every conic subset of $\mathbb{R}^n_0$. This is exactly the local Fourier definition of the wave front set, so $(x_0,\xi_0)\notin WF(u)$. Since $\xi_0$ was arbitrary, \begin{align*} \{x_0\}\times (T_{x_0}^*X\setminus \{0\}) \cap WF(u)=\varnothing. \end{align*} [guided] Assume $x_0 \notin \operatorname{sing\,supp}(u)$. The definition of singular support says that this means more than pointwise regularity: there is an open neighbourhood $O \subset X$ of $x_0$ on which $u$ is a smooth distribution, meaning that the restriction of $u$ to $O$ is represented by a smooth function. Now fix an arbitrary nonzero covector $\xi_0 \in T_{x_0}^*X \setminus \{0\}$. To test whether $(x_0,\xi_0)$ belongs to the wave front set, we work in local coordinates. Choose a coordinate chart $(U,\kappa)$ with $x_0 \in U \subset O$, where \begin{align*} \kappa:U \to \kappa(U) \subset \mathbb{R}^n \end{align*} is a smooth diffeomorphism onto an open subset of Euclidean space. Let $\rho \in C_c^\infty(U)$ be a cutoff satisfying $\rho(x_0)\ne 0$. Since $u$ is smooth on $O$ and $\operatorname{supp}\rho \subset U \subset O$, the product $\rho u$ is represented in the chart by a compactly supported smooth function on $\kappa(U)$. The Fourier transform of a compactly supported smooth function is rapidly decreasing in every frequency direction. Thus, in every conic neighbourhood of the coordinate representative of $\xi_0$, the Fourier transform of the localized distribution $\rho u$ satisfies the rapid-decay estimates required in the definition of the wave front set. Hence $(x_0,\xi_0)\notin WF(u)$. Because the covector $\xi_0$ was arbitrary, no nonzero covector over $x_0$ belongs to $WF(u)$: \begin{align*} \{x_0\}\times (T_{x_0}^*X\setminus \{0\}) \cap WF(u)=\varnothing. \end{align*} [/guided] [/step] [step:Choose one coordinate chart and finitely many conic Fourier estimates] Assume conversely that \begin{align*} (x_0,\xi_0)\notin WF(u) \end{align*} for every $\xi_0 \in T_{x_0}^*X\setminus\{0\}$. Since the theorem assumes $n \ge 1$, define the punctured Euclidean space and unit sphere by \begin{align*} \mathbb{R}^n_0:=\mathbb{R}^n\setminus\{0\}, \qquad S^{n-1}:=\{\omega\in\mathbb{R}^n:|\omega|=1\}. \end{align*} Choose a coordinate chart $(U,\kappa)$ with $x_0\in U$, where \begin{align*} \kappa:U\to \kappa(U)\subset\mathbb{R}^n \end{align*} is a diffeomorphism, and write $y_0:=\kappa(x_0)$. For each $x \in U$, define the coordinate cotangent map \begin{align*} \tau_x:T_x^*X \to (\mathbb{R}^n)^*, \qquad \tau_x(\xi) := \xi\circ (d\kappa_x)^{-1}. \end{align*} Identify $(\mathbb{R}^n)^*$ with $\mathbb{R}^n$ via the standard Euclidean [inner product](/page/Inner%20Product), so $\tau_x(\xi)$ may be regarded as a vector in $\mathbb{R}^n$. This identifies each nonzero covector over $U$ with a point of $\kappa(U)\times \mathbb{R}^n_0$ via $(x,\xi)\mapsto (\kappa(x),\tau_x(\xi))$; in particular, $\tau_{x_0}$ identifies nonzero covectors in $T_{x_0}^*X$ with vectors in $\mathbb{R}^n_0$. For each $\omega\in S^{n-1}$, let $\xi_\omega\in T_{x_0}^*X\setminus\{0\}$ be the unique covector satisfying $\tau_{x_0}(\xi_\omega)=\omega$. The hypothesis says that $(x_0,\xi_\omega)\notin WF(u)$, equivalently $(x_0,\omega)$ is not in the coordinate representative of $WF(u)$. Hence there exist a cutoff $\phi_\omega\in C_c^\infty(U)$ with $\phi_\omega(x_0)\ne 0$ and an open conic neighbourhood $\Gamma_\omega\subset\mathbb{R}^n_0$ of $\omega$ such that the Fourier transform of the coordinate localization $\phi_\omega u$ is rapidly decreasing in $\Gamma_\omega$. Since $\phi_\omega(x_0)\ne 0$, after shrinking around $x_0$ there is an open neighbourhood $V_\omega\subset U$ of $x_0$ on which $\phi_\omega$ has no zeros. The open sets $\Gamma_\omega\cap S^{n-1}$ cover the compact unit sphere $S^{n-1}$. Choose finitely many directions $\omega_1,\dots,\omega_N\in S^{n-1}$ and closed subsets $L_j\subset S^{n-1}$ such that $L_j\subset \Gamma_{\omega_j}\cap S^{n-1}$ for every $j\in\{1,\dots,N\}$ and \begin{align*} S^{n-1}\subset \bigcup_{j=1}^N L_j. \end{align*} For each $j\in\{1,\dots,N\}$, define the closed conic set \begin{align*} K_j:=\{r\omega\in\mathbb{R}^n_0:r>0,\ \omega\in L_j\}. \end{align*} Then $K_j\subset\Gamma_{\omega_j}$ and the finitely many closed cones $K_1,\dots,K_N$ cover $\mathbb{R}^n_0$. Define \begin{align*} V:=\bigcap_{j=1}^N V_{\omega_j}. \end{align*} Then $V$ is an open neighbourhood of $x_0$ in $U$. [/step] [step:Transfer the finitely many directional estimates to one cutoff] Choose $\rho\in C_c^\infty(V)$ with $\rho(x_0)\ne 0$. For each $j\in\{1,\dots,N\}$, choose an [open set](/page/Open%20Set) $W_j$ with $\operatorname{supp}\rho \Subset W_j \Subset V_{\omega_j}$ and a cutoff $\theta_j\in C_c^\infty(V_{\omega_j})$ satisfying $\theta_j=1$ on a neighbourhood of $\operatorname{supp}\rho$. Define \begin{align*} a_j:U\to\mathbb{C} \end{align*} by setting \begin{align*} a_j(x)=\frac{\theta_j(x)\rho(x)}{\phi_{\omega_j}(x)} \end{align*} for $x\in V_{\omega_j}$ and $a_j(x)=0$ for $x\notin V_{\omega_j}$. Since $\phi_{\omega_j}$ has no zeros on $V_{\omega_j}$ and $\theta_j\rho$ vanishes near $\partial V_{\omega_j}$, this gives $a_j\in C_c^\infty(U)$ and \begin{align*} \rho u=a_j\phi_{\omega_j}u \end{align*} as distributions on $U$. Fix $j\in\{1,\dots,N\}$, and write $v_j:=\phi_{\omega_j}u$ as a compactly supported coordinate distribution on $\mathbb{R}^n$. Since $a_j\in C_c^\infty(U)$, its coordinate representative has Fourier transform $\widehat{a_j}$ rapidly decreasing. Since $v_j$ is compactly supported, its Fourier transform has polynomial growth: there exist constants $A_j>0$, $B_j>0$, and an integer $m_j\ge 0$ such that \begin{align*} |\widehat{v_j}(\eta)|\le A_j(1+|\eta|)^{m_j} \end{align*} for every $\eta\in\mathbb{R}^n$. Because $K_j\subset\Gamma_{\omega_j}$ is a closed conic subset and $\Gamma_{\omega_j}$ is open and conic, there exists $\delta_j>0$ such that whenever $\xi\in K_j$, $\eta\notin\Gamma_{\omega_j}$, and $|\eta|\ge |\xi|/2$, one has $|\xi-\eta|\ge\delta_j(|\xi|+|\eta|)$. The convolution formula for multiplication by $a_j$ gives \begin{align*} \widehat{a_jv_j}(\xi)=(2\pi)^{-n}\int_{\mathbb{R}^n}\widehat{a_j}(\xi-\eta)\widehat{v_j}(\eta)\,d\mathcal{L}^n(\eta). \end{align*} Split this integral into the regions $\eta\in\Gamma_{\omega_j}$ and $\eta\notin\Gamma_{\omega_j}$. Fix an integer $M\ge 0$. On $\Gamma_{\omega_j}$, choose the rapid-decay order for $\widehat{v_j}$ larger than $M+n+1$; convolution against the rapidly decreasing function $\widehat{a_j}$ then gives a constant $C_{j,M}^{(1)}>0$ such that the contribution from $\eta\in\Gamma_{\omega_j}$ is bounded by $C_{j,M}^{(1)}(1+|\xi|)^{-M}$ for $\xi\in K_j$. On the complementary region, split further into $|\eta|<|\xi|/2$ and $|\eta|\ge |\xi|/2$. If $|\eta|<|\xi|/2$, then $|\xi-\eta|\ge |\xi|/2$; using the polynomial growth bound for $\widehat{v_j}$ and choosing a Schwartz seminorm order $Q>M+m_j+n$ for $\widehat{a_j}$ gives a bound $C_{j,M}^{(2)}(1+|\xi|)^{-M}$. If $|\eta|\ge |\xi|/2$ and $\eta\notin\Gamma_{\omega_j}$, the conic separation estimate gives $|\xi-\eta|\ge\delta_j(|\xi|+|\eta|)$. Choosing $Q>M+m_j+n+1$ makes the integrand bounded by a constant multiple of $(1+|\xi|)^{-M}(1+|\eta|)^{-n-1}$, which is integrable with respect to $\mathcal{L}^n$ on $\mathbb{R}^n$. Hence this part is bounded by $C_{j,M}^{(3)}(1+|\xi|)^{-M}$. Combining the three bounds, there is a constant $C_{j,M}>0$, depending on $M$, the polynomial-growth constants for $v_j$, finitely many Schwartz seminorms of $a_j$, and the separation constant $\delta_j$, such that \begin{align*} |\widehat{\rho u}(\xi)|=|\widehat{a_jv_j}(\xi)|\le C_{j,M}(1+|\xi|)^{-M} \end{align*} for every $\xi\in K_j$. Since $K_1,\dots,K_N$ cover $\mathbb{R}^n_0$, setting $C_M:=\max_{1\le j\le N}C_{j,M}$ gives rapid decay of $\widehat{\rho u}$ on all of $\mathbb{R}^n_0$. [guided] The goal is to replace the many localizing cutoffs $\phi_{\omega_j}$ by one cutoff $\rho$ near $x_0$. This is necessary because smoothness near $x_0$ will be detected by a single compactly supported localization whose Fourier transform decays rapidly in every direction. Choose $\rho\in C_c^\infty(V)$ with $\rho(x_0)\ne 0$. Since $V\subset V_{\omega_j}$ and $\phi_{\omega_j}$ has no zeros on $V_{\omega_j}$, the quotient $\rho/\phi_{\omega_j}$ is smooth on $V_{\omega_j}$. Define a compactly supported smooth function \begin{align*} a_j:U\to\mathbb{C} \end{align*} by \begin{align*} a_j(x)=\frac{\rho(x)}{\phi_{\omega_j}(x)} \end{align*} on a neighbourhood of $\operatorname{supp}\rho$ contained in $V_{\omega_j}$, and by extending smoothly as $0$ outside that neighbourhood. Then $a_j\in C_c^\infty(U)$ and, because $a_j\phi_{\omega_j}=\rho$ on $\operatorname{supp}\rho$, \begin{align*} \rho u=a_j\phi_{\omega_j}u \end{align*} as distributions. Now define $v_j:=\phi_{\omega_j}u$, regarded in the same coordinate chart as a compactly supported distribution on $\mathbb{R}^n$. The Fourier transform $\widehat{v_j}$ has polynomial growth, so there are constants $A_j>0$, $B_j>0$, and an integer $m_j\ge 0$ such that \begin{align*} |\widehat{v_j}(\eta)|\le A_j(1+|\eta|)^{m_j} \end{align*} for every $\eta\in\mathbb{R}^n$. Since $a_j$ is smooth with compact support, $\widehat{a_j}$ is rapidly decreasing. We use the convolution formula for multiplication by $a_j$: \begin{align*} \widehat{a_jv_j}(\xi)=(2\pi)^{-n}\int_{\mathbb{R}^n}\widehat{a_j}(\xi-\eta)\widehat{v_j}(\eta)\,d\mathcal{L}^n(\eta). \end{align*} The closed cone $K_j$ is contained in the open cone $\Gamma_{\omega_j}$, so there is a separation constant $\delta_j>0$ such that if $\xi\in K_j$, $\eta\notin\Gamma_{\omega_j}$, and $|\eta|\ge |\xi|/2$, then $|\xi-\eta|\ge\delta_j(|\xi|+|\eta|)$. Split the convolution integral into the part where $\eta\in\Gamma_{\omega_j}$ and the part where $\eta\notin\Gamma_{\omega_j}$. On $\Gamma_{\omega_j}$, the defining estimate for $(x_0,\xi_{\omega_j})\notin WF(u)$ gives rapid decay of $\widehat{v_j}(\eta)$. On the complement, the separation estimate makes $\widehat{a_j}(\xi-\eta)$ rapidly decreasing in both $|\xi|$ and $|\eta|$, which dominates the polynomial growth of $\widehat{v_j}(\eta)$. Therefore, for every integer $M\ge 0$, there is a constant $C_{j,M}>0$ such that \begin{align*} |\widehat{\rho u}(\xi)|=|\widehat{a_jv_j}(\xi)|\le C_{j,M}(1+|\xi|)^{-M} \end{align*} for every $\xi\in K_j$. The constant $C_{j,M}$ comes from the rapid-decay constants for $\widehat{v_j}$ on $\Gamma_{\omega_j}$, the polynomial-growth constants for $\widehat{v_j}$ on $\mathbb{R}^n$, finitely many Schwartz seminorms of $\widehat{a_j}$, and the separation constant $\delta_j$. The finite-cover step is now used. The closed cones $K_1,\dots,K_N$ cover $\mathbb{R}^n_0$. Since there are only finitely many cones, for each $M\ge 0$ we may define $C_M:=\max_{1\le j\le N}C_{j,M}$. Then \begin{align*} |\widehat{\rho u}(\xi)|\le C_M(1+|\xi|)^{-M} \end{align*} for every $\xi\in\mathbb{R}^n_0$. Hence the Fourier transform of the single localization $\rho u$ is rapidly decreasing in all nonzero frequency directions, equivalently on $\mathbb{R}^n_0$. [/guided] [/step] [step:Apply the local Fourier smoothness criterion] Regard $\rho u$ in the chart $(U,\kappa)$ as a compactly supported distribution on $\mathbb{R}^n$ by extension by zero from $\kappa(U)$. The previous step shows that its Fourier transform is rapidly decreasing on $\mathbb{R}^n_0$. Since the Fourier transform of a compactly supported distribution has at most polynomial growth, this rapid decay implies rapid decay on all of $\mathbb{R}^n$ after enlarging the estimate over the bounded set $\{\xi\in\mathbb{R}^n:|\xi|\le 1\}$. By [citetheorem:8166], applied in the coordinate open set $\kappa(U)\subset\mathbb{R}^n$ to the compactly supported coordinate representative of $\rho u$, the localization $\rho u$ is smooth near $x_0$. The hypotheses are satisfied because $\rho u$ is compactly supported in the coordinate chart and the preceding paragraph gives rapid decay of its Fourier transform on all of $\mathbb{R}^n$. Since $\rho(x_0)\ne 0$, there is an open neighbourhood $W\subset V$ of $x_0$ on which $\rho$ has no zeros. Define the reciprocal function $r:W\to\mathbb{C}$ by $r(x)=1/\rho(x)$. Then $r$ is smooth on $W$. On $W$, \begin{align*} u=r(\rho u) \end{align*} as distributions, and the right-hand side is smooth because it is the product of the smooth function $r$ and the smooth distribution $\rho u$. Therefore $u$ is smooth on $W$, so \begin{align*} x_0\notin \operatorname{sing\,supp}(u) \end{align*} [/step] [step:Conclude the equality of the two sets] We have proved that, for every $x_0\in X$, \begin{align*} x_0 \notin \operatorname{sing\,supp}(u) \iff (x_0,\xi_0)\notin WF(u) \text{ for every } \xi_0\in T_{x_0}^*X\setminus\{0\}. \end{align*} Taking complements in $X$, this is equivalent to \begin{align*} x_0 \in \operatorname{sing\,supp}(u) \iff \text{there exists } \xi_0\in T_{x_0}^*X\setminus\{0\} \text{ such that } (x_0,\xi_0)\in WF(u). \end{align*} The right-hand condition is exactly $x_0\in \pi(WF(u))$. Hence \begin{align*} \pi(WF(u))=\operatorname{sing\,supp}(u). \end{align*} [/step]