[proofplan] We prove both directions by localizing the distribution near $x_0$ with a smooth cutoff. If $u$ is smooth near $x_0$, a cutoff supported in that smooth neighbourhood turns $\chi u$ into a compactly supported smooth function, whose [Fourier transform](/page/Fourier%20Transform) has rapid decay by repeated [integration by parts](/theorems/210). Conversely, if some localized Fourier transform is rapidly decreasing, Fourier inversion produces a smooth function representing $\chi u$ as a distribution on $\mathbb{R}^n$. Since $\chi=1$ near $x_0$, this smooth representative also represents $u$ near $x_0$, so $x_0$ is not in the singular support. [/proofplan] [step:Choose a cutoff inside a smooth neighbourhood] Assume first that $x_0 \notin \operatorname{sing\,supp} u$. By the definition of singular support, there are an [open set](/page/Open%20Set) $V \subset U$ with $x_0 \in V$ and a function $f \in C^\infty(V)$ such that $u|_{\mathcal{D}'(V)} = T_f$, where \begin{align*}T_f: C_c^\infty(V) \to \mathbb{C}, \qquad \phi \mapsto \int_V f(x)\phi(x)\, d\mathcal{L}^n(x).\end{align*} Choose $\chi \in C_c^\infty(V)$ such that $\chi=1$ on an open neighbourhood $W \subset V$ of $x_0$. This uses the standard Euclidean smooth cutoff lemma. Define \begin{align*}h: V \to \mathbb{C}, \qquad x \mapsto \chi(x)f(x).\end{align*} Since $\operatorname{supp}\chi$ is compact and contained in $V$, extend $h$ by zero to a function $g \in C_c^\infty(\mathbb{R}^n)$. Let $T_\chi \in \mathcal{D}'(\mathbb{R}^n)$ be the zero extension of $\chi u$, so that for every $\phi \in C_c^\infty(\mathbb{R}^n)$, \begin{align*} T_\chi(\phi) = u((\chi\phi)|_U) = \int_V f(x)\chi(x)\phi(x)\, d\mathcal{L}^n(x) = \int_{\mathbb{R}^n} g(x)\phi(x)\, d\mathcal{L}^n(x). \end{align*} Thus $T_\chi$ is the [regular distribution](/page/Regular%20Distribution) associated to $g$. [guided] The meaning of $x_0 \notin \operatorname{sing\,supp} u$ is precisely that $u$ is represented by a smooth function in some neighbourhood of $x_0$. Thus we choose an open set $V \subset U$ with $x_0 \in V$ and a smooth function $f \in C^\infty(V)$ such that, for every [test function](/page/Test%20Function) $\psi \in C_c^\infty(V)$, \begin{align*} u(\psi) = \int_V f(x)\psi(x)\, d\mathcal{L}^n(x). \end{align*} We now localize inside this smooth region. The standard Euclidean smooth cutoff lemma gives a function $\chi \in C_c^\infty(V)$ such that $\chi=1$ on some open neighbourhood $W \subset V$ of $x_0$. The compact-support condition is important: it lets us view $\chi u$ as a compactly supported distribution on all of $\mathbb{R}^n$, so its Fourier transform is globally defined. Define a function $g: \mathbb{R}^n \to \mathbb{C}$ by setting $g(x)=\chi(x)f(x)$ for $x \in V$ and $g(x)=0$ outside $\operatorname{supp}\chi$. Since $\operatorname{supp}\chi$ is compactly contained in $V$, no boundary mismatch occurs at $\partial V$; near every point outside $\operatorname{supp}\chi$, the function is identically zero. Hence $g \in C_c^\infty(\mathbb{R}^n)$. We verify that $g$ represents the localized distribution. Let $T_\chi \in \mathcal{D}'(\mathbb{R}^n)$ denote the zero extension of $\chi u$, defined by $T_\chi(\phi)=u((\chi\phi)|_U)$ for $\phi \in C_c^\infty(\mathbb{R}^n)$. Since $\operatorname{supp}\chi$ is compact and contained in $V$, the test function $(\chi\phi)|_U$ has compact support contained in $V$, so the smooth representative formula for $u$ on $V$ applies: \begin{align*} T_\chi(\phi) = u((\chi\phi)|_U) = \int_V f(x)\chi(x)\phi(x)\, d\mathcal{L}^n(x). \end{align*} Because $g$ is exactly $\chi f$ on $V$ and zero outside $\operatorname{supp}\chi$, the last integral is \begin{align*} \int_{\mathbb{R}^n} g(x)\phi(x)\, d\mathcal{L}^n(x). \end{align*} Therefore the zero extension of $\chi u$ is the regular distribution associated to the compactly supported smooth function $g$. [/guided] [/step] [step:Integrate by parts to prove rapid decay of the localized Fourier transform] For $\xi \in \mathbb{R}^n$, the Fourier transform of $g$ is \begin{align*} \widehat{g}(\xi) = (2\pi)^{-n/2}\int_{\mathbb{R}^n} g(x)e^{-ix\cdot \xi}\, d\mathcal{L}^n(x). \end{align*} Let $\alpha=(\alpha_1,\dots,\alpha_n) \in \mathbb{N}_0^n$ be a multi-index, with $D^\alpha=\partial_{x_1}^{\alpha_1}\cdots\partial_{x_n}^{\alpha_n}$ and $|\alpha|=\alpha_1+\cdots+\alpha_n$ as declared in the theorem statement. Since $g \in C_c^\infty(\mathbb{R}^n)$, [integration by parts](/theorems/2098) in each coordinate gives \begin{align*} \xi^\alpha \widehat{g}(\xi) = (2\pi)^{-n/2} (-i)^{|\alpha|}\int_{\mathbb{R}^n} D^\alpha g(x)e^{-ix\cdot \xi}\, d\mathcal{L}^n(x). \end{align*} Taking absolute values yields \begin{align*} |\xi^\alpha \widehat{g}(\xi)| \leq (2\pi)^{-n/2}\int_{\mathbb{R}^n} |D^\alpha g(x)|\, d\mathcal{L}^n(x). \end{align*} For each integer $N \geq 0$, the elementary polynomial comparison gives a constant $A_{N,n}>0$ such that \begin{align*} (1+|\xi|)^N \leq A_{N,n}\sum_{|\alpha|\leq N} |\xi^\alpha| \end{align*} for all $\xi \in \mathbb{R}^n$. Hence \begin{align*} (1+|\xi|)^N|\widehat{g}(\xi)| \leq A_{N,n}(2\pi)^{-n/2}\sum_{|\alpha|\leq N}\int_{\mathbb{R}^n} |D^\alpha g(x)|\, d\mathcal{L}^n(x). \end{align*} The right-hand side is finite because every $D^\alpha g$ is continuous with compact support. Defining \begin{align*} C_N := A_{N,n}(2\pi)^{-n/2}\sum_{|\alpha|\leq N}\int_{\mathbb{R}^n} |D^\alpha g(x)|\, d\mathcal{L}^n(x) \end{align*} proves \begin{align*} |\widehat{\chi u}(\xi)| = |\widehat{g}(\xi)| \leq C_N(1+|\xi|)^{-N}. \end{align*} Thus $\widehat{\chi u}$ is rapidly decreasing. [/step] [step:Use rapid decay to construct a smooth inverse Fourier transform] Conversely, assume there exists $\chi \in C_c^\infty(U)$ such that $\chi=1$ on an open neighbourhood $W \subset U$ of $x_0$ and $\widehat{\chi u}$ is rapidly decreasing. Let $T$ denote the zero extension of $\chi u$ to a compactly supported distribution on $\mathbb{R}^n$, and define \begin{align*}F: \mathbb{R}^n \to \mathbb{C}, \qquad \xi \mapsto \widehat{T}(\xi).\end{align*} By hypothesis, for every integer $N \geq 0$ there is $C_N>0$ such that \begin{align*} |F(\xi)| \leq C_N(1+|\xi|)^{-N} \end{align*} for all $\xi \in \mathbb{R}^n$. Define \begin{align*}v: \mathbb{R}^n \to \mathbb{C}, \qquad x \mapsto (2\pi)^{-n/2}\int_{\mathbb{R}^n} F(\xi)e^{ix\cdot \xi}\, d\mathcal{L}^n(\xi).\end{align*} For any multi-index $\alpha \in \mathbb{N}_0^n$, choose an integer $N_\alpha>n+|\alpha|$. Then \begin{align*} |\xi^\alpha F(\xi)| \leq C_{N_\alpha}|\xi|^{|\alpha|}(1+|\xi|)^{-N_\alpha} \leq C_{N_\alpha}(1+|\xi|)^{|\alpha|-N_\alpha}. \end{align*} The function $\xi \mapsto (1+|\xi|)^{|\alpha|-N_\alpha}$ is integrable on $\mathbb{R}^n$ with respect to $\mathcal{L}^n$ because $N_\alpha-|\alpha|>n$. Differentiation under the integral sign is therefore justified by dominated convergence, and \begin{align*} D^\alpha v(x) = (2\pi)^{-n/2}\int_{\mathbb{R}^n} (i\xi)^\alpha F(\xi)e^{ix\cdot \xi}\, d\mathcal{L}^n(\xi). \end{align*} Thus $v \in C^\infty(\mathbb{R}^n)$. [guided] The rapid decay hypothesis is designed to make the inverse Fourier integral converge absolutely after multiplying by any power of $\xi$. Let $T$ be the zero extension of $\chi u$ to $\mathbb{R}^n$, and write \begin{align*} F: \mathbb{R}^n \to \mathbb{C}, \qquad \xi \mapsto \widehat{T}(\xi). \end{align*} The hypothesis says that for every integer $N \geq 0$ there is a constant $C_N>0$ such that \begin{align*} |F(\xi)| \leq C_N(1+|\xi|)^{-N} \end{align*} for all $\xi \in \mathbb{R}^n$. We define the inverse Fourier candidate \begin{align*} v: \mathbb{R}^n \to \mathbb{C}, \qquad x \mapsto (2\pi)^{-n/2}\int_{\mathbb{R}^n} F(\xi)e^{ix\cdot \xi}\, d\mathcal{L}^n(\xi). \end{align*} This integral is absolutely convergent: take $N>n$, and compare $|F(\xi)|$ with $C_N(1+|\xi|)^{-N}$, which is integrable on $\mathbb{R}^n$ with respect to $\mathcal{L}^n$. To prove smoothness, we must justify differentiating under the integral sign. Fix a multi-index $\alpha \in \mathbb{N}_0^n$. Formally differentiating $e^{ix\cdot \xi}$ produces the factor $(i\xi)^\alpha$, so we need $\xi \mapsto \xi^\alpha F(\xi)$ to be integrable. Choose an integer $N_\alpha>n+|\alpha|$. The rapid decay bound gives \begin{align*} |\xi^\alpha F(\xi)| \leq C_{N_\alpha}|\xi|^{|\alpha|}(1+|\xi|)^{-N_\alpha}. \end{align*} Since $|\xi|^{|\alpha|}\leq (1+|\xi|)^{|\alpha|}$, we have \begin{align*} |\xi^\alpha F(\xi)| \leq C_{N_\alpha}(1+|\xi|)^{|\alpha|-N_\alpha}. \end{align*} The exponent satisfies $|\alpha|-N_\alpha<-n$, so this majorant is integrable on $\mathbb{R}^n$. Dominated convergence therefore permits differentiation under the integral sign and gives \begin{align*} D^\alpha v(x) = (2\pi)^{-n/2}\int_{\mathbb{R}^n} (i\xi)^\alpha F(\xi)e^{ix\cdot \xi}\, d\mathcal{L}^n(\xi). \end{align*} Because this argument works for every multi-index $\alpha$, the function $v$ belongs to $C^\infty(\mathbb{R}^n)$. [/guided] [/step] [step:Identify the inverse Fourier transform with the localized distribution] The spaces $\mathcal{S}(\mathbb{R}^n)$ and $\mathcal{S}'(\mathbb{R}^n)$, together with the operators $\mathcal{F}$ and $\mathcal{F}^{-1}$, are as declared in the theorem statement. The distribution $T$ has compact support, hence it extends continuously to $\mathcal{S}(\mathbb{R}^n)$ and defines an element of $\mathcal{S}'(\mathbb{R}^n)$. By definition of $F$, the distributional Fourier transform of $T$ is the tempered distribution represented by the function $F$. Fourier inversion for [tempered distributions](/page/Tempered%20Distributions) with the symmetric normalization gives \begin{align*} \mathcal{F}^{-1}(F)=T \end{align*} in $\mathcal{S}'(\mathbb{R}^n)$. We now verify that the tempered distribution $\mathcal{F}^{-1}(F)$ is the regular distribution induced by $v$. Let $\phi \in \mathcal{S}(\mathbb{R}^n)$. Since $F$ is rapidly decreasing and $\phi$ is a Schwartz function, the product $F(\xi)\widehat{\phi}(\xi)$ is integrable with respect to $\mathcal{L}^n$. Using the symmetric Fourier normalization and the defining inverse Fourier integral for $v$, [Fubini's theorem](/theorems/2961) applies to the absolutely integrable function $(x,\xi)\mapsto F(\xi)e^{ix\cdot\xi}\phi(x)$ after the standard Schwartz approximation argument, and gives \begin{align*} \int_{\mathbb{R}^n} v(x)\phi(x)\, d\mathcal{L}^n(x)=(\mathcal{F}^{-1}F)(\phi). \end{align*} Thus $\mathcal{F}^{-1}(F)$ is represented by $v$. Combining this with distributional Fourier inversion, for every $\phi \in C_c^\infty(\mathbb{R}^n)$, \begin{align*} T(\phi) = \int_{\mathbb{R}^n} v(x)\phi(x)\, d\mathcal{L}^n(x). \end{align*} Thus the zero extension of $\chi u$ is the regular distribution associated to $v$ on $\mathbb{R}^n$, with $v \in C^\infty(\mathbb{R}^n)$. [guided] The point of this step is to identify the smooth function constructed from the inverse Fourier integral with the original localized distribution. Let $\mathcal{S}(\mathbb{R}^n)$ denote the [Schwartz space](/page/Schwartz%20Space) and let $\mathcal{S}'(\mathbb{R}^n)$ denote its tempered-distribution dual. Since $T$ has compact support, its action on compactly supported test functions extends continuously to Schwartz functions; hence $T \in \mathcal{S}'(\mathbb{R}^n)$. By definition, $F: \mathbb{R}^n \to \mathbb{C}$ is the function representing the distributional Fourier transform $\widehat{T}$. Fourier inversion for tempered distributions, with the symmetric normalization declared in the theorem statement, states that applying $\mathcal{F}^{-1}$ to $\widehat{T}$ recovers $T$. Therefore \begin{align*} \mathcal{F}^{-1}(F)=T \end{align*} as tempered distributions. Why is the rapid decay assumption enough here, even though no derivative bounds for $F$ were assumed? The previous step used only rapid decay of $F$ after multiplication by powers of $\xi$ to prove that the inverse Fourier integral \begin{align*} v(x)=(2\pi)^{-n/2}\int_{\mathbb{R}^n}F(\xi)e^{ix\cdot \xi}\,d\mathcal{L}^n(\xi) \end{align*} defines a smooth function. To identify this function with the inverse Fourier transform as a distribution, test against an arbitrary $\phi \in \mathcal{S}(\mathbb{R}^n)$. The rapid decay of $F$ and the Schwartz decay of $\phi$ make the relevant products integrable, so Fubini's theorem justifies interchanging the $x$- and $\xi$-integrals in the pairing of $v$ with $\phi$. This gives \begin{align*} \int_{\mathbb{R}^n}v(x)\phi(x)\,d\mathcal{L}^n(x)=(\mathcal{F}^{-1}F)(\phi). \end{align*} Thus $v$ represents $\mathcal{F}^{-1}F$ as a tempered distribution. Combining this identification with distributional Fourier inversion gives that $T$ is the regular distribution induced by $v$. Hence, for every $\phi \in C_c^\infty(\mathbb{R}^n)$, \begin{align*} T(\phi)=\int_{\mathbb{R}^n}v(x)\phi(x)\,d\mathcal{L}^n(x). \end{align*} [/guided] [/step] [step:Remove the cutoff near $x_0$] Since $\chi=1$ on the open neighbourhood $W$ of $x_0$, for every test function $\phi \in C_c^\infty(W)$ we have $\chi\phi=\phi$. Therefore \begin{align*} u(\phi) = u(\chi\phi) = T(\phi) = \int_{\mathbb{R}^n} v(x)\phi(x)\, d\mathcal{L}^n(x). \end{align*} Because $\operatorname{supp}\phi \subset W$, this is the same as \begin{align*} u(\phi) = \int_W v|_W(x)\phi(x)\, d\mathcal{L}^n(x). \end{align*} The restriction $v|_W$ lies in $C^\infty(W)$, so $u$ is represented by a smooth function on an open neighbourhood of $x_0$. Hence $x_0 \notin \operatorname{sing\,supp}u$. [guided] We now use the fact that the cutoff is identically one near the point under consideration. Let $\phi \in C_c^\infty(W)$ be an arbitrary test function. Since $\chi=1$ on $W$ and $\operatorname{supp}\phi \subset W$, the product satisfies $\chi\phi=\phi$. By the definition of the zero extension $T$ of $\chi u$ and by the previous step, \begin{align*} u(\phi)=u(\chi\phi)=T(\phi)=\int_{\mathbb{R}^n}v(x)\phi(x)\,d\mathcal{L}^n(x). \end{align*} Because $\phi$ is supported in $W$, the integral over $\mathbb{R}^n$ reduces to the integral over $W$: \begin{align*} u(\phi)=\int_W v|_W(x)\phi(x)\,d\mathcal{L}^n(x). \end{align*} The function $v|_W: W \to \mathbb{C}$ is smooth because $v \in C^\infty(\mathbb{R}^n)$. Thus $u$ agrees on $W$ with the regular distribution induced by a smooth function. This is exactly the assertion that $x_0 \notin \operatorname{sing\,supp}u$. [/guided] Combining this implication with the forward implication proves the equivalence. [/step]