[proofplan] We prove the two implications separately. If $u$ is smooth near $x_0$, we choose a cutoff supported inside a smoothness neighbourhood and reduce the problem to the standard rapid Fourier decay estimate for a compactly supported smooth function, obtained by repeated [integration by parts](/theorems/210). Conversely, if one cutoff has rapidly decreasing [Fourier transform](/page/Fourier%20Transform), Fourier inversion produces a smooth function representing the compactly supported distribution $\chi u$. Since $\chi=1$ near $x_0$, this smooth representative agrees with $u$ on a neighbourhood of $x_0$, which is exactly the assertion that $x_0$ is not in the singular support. [/proofplan]

[step:Choose a cutoff inside a neighbourhood where the distribution is smooth]Assume first that $x_0 \notin \operatorname{sing\,supp} u$. By the definition of singular support, there exist an [open set](/page/Open%20Set) $V \subset U$ with $x_0 \in V$ and a function $g: V \to \mathbb{C}$ with $g \in C^\infty(V)$ such that \begin{align*} u(\varphi)=\int_V g(x)\varphi(x)\,d\mathcal{L}^n(x) \end{align*} for every $\varphi \in C_c^\infty(V)$. Choose an open set $W \subset V$ such that $x_0 \in W$ and $\overline{W}$ is compact in $V$. By the smooth bump function construction in Euclidean space, choose $\chi \in C_c^\infty(V)$ such that $\chi=1$ on an open neighbourhood of $x_0$ and $\operatorname{supp}\chi \subset W$. Extend $\chi u$ by zero from $U$ to $\mathbb{R}^n$. Since $\operatorname{supp}\chi \subset V$ and $u$ is represented by $g$ on $V$, this extension is the [regular distribution](/page/Regular%20Distribution) associated to the compactly supported smooth function \begin{align*} h:\mathbb{R}^n &\to \mathbb{C} \end{align*} defined by $h(x)=\chi(x)g(x)$ for $x \in V$ and $h(x)=0$ for $x \notin V$.[/step]

[guided]The assumption $x_0 \notin \operatorname{sing\,supp} u$ means precisely that $u$ has no singularity in some open neighbourhood of $x_0$. Thus there are an open set $V \subset U$ with $x_0 \in V$ and a smooth function $g:V \to \mathbb{C}$ such that, on test functions supported in $V$, the distribution $u$ acts by integration against $g$: \begin{align*} u(\varphi)=\int_V g(x)\varphi(x)\,d\mathcal{L}^n(x) \end{align*} for every $\varphi \in C_c^\infty(V)$. We need a cutoff $\chi$ whose support stays inside this smoothness region, because then multiplying by $\chi$ removes every part of $u$ outside the region where we know it is a smooth function. Choose an open set $W \subset V$ with $x_0 \in W$ and $\overline{W}$ compact in $V$. The Euclidean bump function construction gives a function $\chi \in C_c^\infty(V)$ satisfying $\chi=1$ on an open neighbourhood of $x_0$ and $\operatorname{supp}\chi \subset W$. Now regard $\chi u$ as a compactly supported distribution on $\mathbb{R}^n$ by extension by zero. Because $\operatorname{supp}\chi \subset V$, this distribution is represented by the compactly supported smooth function $h:\mathbb{R}^n \to \mathbb{C}$ defined by $h(x)=\chi(x)g(x)$ for $x \in V$ and $h(x)=0$ for $x \notin V$. The compact support is inherited from $\chi$, and smoothness across the boundary of $V$ is harmless because $\chi$ vanishes outside the compact set $W \subset V$.[/guided]

[step:Integrate by parts to obtain rapid Fourier decay] We use the symmetric Fourier transform convention \begin{align*} \widehat{h}(\xi)=(2\pi)^{-n/2}\int_{\mathbb{R}^n}h(x)e^{-i\xi\cdot x}\,d\mathcal{L}^n(x). \end{align*} Let $M \in \mathbb{N}$, and define the differential operator $L_M:C^\infty(\mathbb{R}^n)\to C^\infty(\mathbb{R}^n)$ by \begin{align*} L_M f=(1-\Delta)^M f. \end{align*} Since \begin{align*} (1-\Delta_x)^M e^{-i\xi\cdot x}=(1+|\xi|^2)^M e^{-i\xi\cdot x}, \end{align*} [integration by parts](/theorems/2098) is justified without boundary terms because $h \in C_c^\infty(\mathbb{R}^n)$, and gives \begin{align*} (1+|\xi|^2)^M\widehat{h}(\xi)=(2\pi)^{-n/2}\int_{\mathbb{R}^n} L_Mh(x)e^{-i\xi\cdot x}\,d\mathcal{L}^n(x). \end{align*} Taking absolute values and using $|e^{-i\xi\cdot x}|=1$, we obtain \begin{align*} |\widehat{h}(\xi)|\leq (2\pi)^{-n/2}\|L_Mh\|_{L^1(\mathbb{R}^n)}(1+|\xi|^2)^{-M}. \end{align*} Given $N \in \mathbb{N}$, choose $M \in \mathbb{N}$ with $2M \geq N$. Since $1+|\xi|^2 \geq 2^{-1}(1+|\xi|)^2$, there is a constant \begin{align*} C_N=(2\pi)^{-n/2}2^M\|L_Mh\|_{L^1(\mathbb{R}^n)} \end{align*} such that \begin{align*} |\widehat{\chi u}(\xi)|=|\widehat{h}(\xi)|\leq C_N(1+|\xi|)^{-N} \end{align*} for every $\xi \in \mathbb{R}^n$. [/step]

[step:Use rapid decay to construct a smooth inverse Fourier transform]Conversely, assume that there exists $\chi \in C_c^\infty(U)$ with $\chi=1$ on an open neighbourhood of $x_0$ and such that $\widehat{\chi u}$ is rapidly decreasing in the stated sense. Let $T \in \mathcal{E}'(\mathbb{R}^n)$ denote the extension by zero of $\chi u$ from $U$ to $\mathbb{R}^n$, and define \begin{align*} F:\mathbb{R}^n &\to \mathbb{C} \end{align*} by $F(\xi)=\widehat{T}(\xi)$. For every multi-index $\alpha$, rapid decay with $N>n+|\alpha|$ implies that the function \begin{align*} \xi \mapsto \xi^\alpha F(\xi) \end{align*} belongs to $L^1(\mathbb{R}^n)$. Define \begin{align*} f:\mathbb{R}^n &\to \mathbb{C} \end{align*} by \begin{align*} f(x)=(2\pi)^{-n/2}\int_{\mathbb{R}^n}F(\xi)e^{ix\cdot\xi}\,d\mathcal{L}^n(\xi). \end{align*} By differentiating under the integral sign, justified by the integrability of $\xi^\alpha F(\xi)$ for every multi-index $\alpha$, we obtain \begin{align*} D^\alpha f(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 $f \in C^\infty(\mathbb{R}^n)$.[/step]

[guided]Now suppose the Fourier transform of the cutoff distribution decays faster than every power. Let $T \in \mathcal{E}'(\mathbb{R}^n)$ be the compactly supported distribution obtained by extending $\chi u$ by zero from $U$ to $\mathbb{R}^n$, and define the function $F:\mathbb{R}^n\to\mathbb{C}$ by $F(\xi)=\widehat{T}(\xi)$. The goal is to recover $T$ from its Fourier transform by Fourier inversion. The only point requiring verification is that the inverse Fourier integral and all its derivatives converge absolutely. Let $\alpha$ be a multi-index. The rapid decay assumption says that for every $N \in \mathbb{N}$ there is $C_N>0$ with \begin{align*} |F(\xi)|\leq C_N(1+|\xi|)^{-N}. \end{align*} Choose $N>n+|\alpha|$. Since $|\xi^\alpha|\leq |\xi|^{|\alpha|}\leq (1+|\xi|)^{|\alpha|}$, we have \begin{align*} |\xi^\alpha F(\xi)|\leq C_N(1+|\xi|)^{|\alpha|-N}. \end{align*} The exponent $|\alpha|-N$ is strictly less than $-n$, so the right-hand side is integrable over $\mathbb{R}^n$ with respect to $\mathcal{L}^n$. Hence $\xi \mapsto \xi^\alpha F(\xi)$ belongs to $L^1(\mathbb{R}^n)$. We may therefore define the inverse Fourier transform \begin{align*} f(x)=(2\pi)^{-n/2}\int_{\mathbb{R}^n}F(\xi)e^{ix\cdot\xi}\,d\mathcal{L}^n(\xi) \end{align*} for every $x \in \mathbb{R}^n$. Differentiating under the integral sign is justified because each derivative introduces multiplication by a polynomial factor $(i\xi)^\alpha$, and that factor remains integrable against $F$. Thus, for every multi-index $\alpha$, \begin{align*} D^\alpha f(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*} This proves $f \in C^\infty(\mathbb{R}^n)$.[/guided]

[step:Identify the inverse Fourier transform with the cutoff distribution] The [Fourier inversion theorem](/theorems/246) for rapidly decreasing functions and the injectivity of the Fourier transform on [tempered distributions](/page/Tempered%20Distributions) imply that the regular distribution $T_f$ associated to $f$ has Fourier transform $F$. Since $F=\widehat{T}$ and both $T_f$ and $T$ are tempered distributions, injectivity gives \begin{align*} T=T_f \end{align*} in $\mathcal{D}'(\mathbb{R}^n)$. Equivalently, for every $\varphi \in C_c^\infty(\mathbb{R}^n)$, \begin{align*} T(\varphi)=\int_{\mathbb{R}^n}f(x)\varphi(x)\,d\mathcal{L}^n(x). \end{align*} Here the cited input is the standard Fourier inversion and uniqueness theorem for tempered distributions; if this result is not yet in the wiki, it should be added as the prerequisite: Fourier inversion and injectivity for tempered distributions. [/step]

[step:Restrict to the neighbourhood where the cutoff equals one] Let $O \subset U$ be an open neighbourhood of $x_0$ such that $\chi=1$ on $O$. For every $\varphi \in C_c^\infty(O)$, multiplication by $\chi$ leaves $\varphi$ unchanged, so \begin{align*} u(\varphi)=(\chi u)(\varphi)=T(\varphi)=\int_{\mathbb{R}^n}f(x)\varphi(x)\,d\mathcal{L}^n(x)=\int_O f(x)\varphi(x)\,d\mathcal{L}^n(x). \end{align*} Thus $u$ is represented on $O$ by the smooth function $f|_O:O\to\mathbb{C}$. By the definition of singular support, this proves $x_0 \notin \operatorname{sing\,supp}u$. Combining this implication with the forward implication proves the equivalence. [/step]