[proofplan] The regular point hypothesis first identifies the current locally with an integer multiple of integration over a smooth embedded submanifold. Local area-minimality then implies stationarity under every compactly supported interior variation, because the flow of such a vector field gives admissible competitors with the same boundary. Applying the smooth [first variation formula](/theorems/2728) to $q \llbracket M \rrbracket$ gives an integral identity against all compactly supported ambient vector fields. Finally, a localization argument with cutoff vector fields forces the smooth mean curvature vector $H_M$ to vanish pointwise. [/proofplan]

[step:Use regularity to represent the current by a smooth embedded submanifold] Since $x_0$ is an interior regular point of $T$, there exist $r > 0$, an integer $q \in \mathbb{N}$, and a smooth embedded oriented $m$-dimensional submanifold $M \subset B(x_0,r)$ such that $\overline{B}(x_0,r) \subset U$, $B(x_0,r) \cap \operatorname{spt}(\partial T) = \varnothing$, and \begin{align*} T \llcorner B(x_0,r) = q \llbracket M \rrbracket. \end{align*} Here $\llbracket M \rrbracket$ denotes the integral current obtained by integration over $M$ with its chosen orientation and multiplicity one. Shrink $r > 0$ if necessary so that every compactly supported vector field in $B(x_0,r)$ has support disjoint from $\operatorname{spt}(\partial T)$. This shrinking preserves the displayed representation because restriction of currents is compatible with restriction to smaller open balls. [/step]

[step:Show that local minimality gives stationarity for compactly supported interior variations]Let $X: B(x_0,r) \to \mathbb{R}^n$ be a smooth vector field with compact support in $B(x_0,r)$. Extend $X$ by zero to a smooth compactly supported vector field on $U$, still denoted $X$. Let $C:=\operatorname{spt}(X)\subset B(x_0,r)$, which is compact. Since $C$ has positive distance from $\partial B(x_0,r)$ and $X$ is compactly supported in $U$, the [flow existence theorem for smooth vector fields](/page/Flow) gives a flow $\Phi_t:U\to U$ of $X$ for all $|t|<\varepsilon$ after choosing $\varepsilon>0$ small enough; this flow remains inside $U$ and is the identity on $U\setminus C$. Choose a compact set $K\subset B(x_0,r)$ such that $C\cup\Phi_t(C)$ is contained in the interior of $K$ for all $|t|<\varepsilon$, decreasing $\varepsilon$ if necessary. This interior containment ensures that no mass crosses $\partial K$ during the variation. For each $t\in(-\varepsilon,\varepsilon)$, define the pushed-forward current \begin{align*} T_t := (\Phi_t)_\# T. \end{align*} Because $\Phi_t$ is the identity near $\operatorname{spt}(\partial T)$, the [boundary commutation formula for pushforwards](/page/Pushforward%20of%20Current) gives \begin{align*} \partial T_t = \partial ((\Phi_t)_\# T) = (\Phi_t)_\#(\partial T) = \partial T. \end{align*} Also $\operatorname{spt}(T_t - T)\subset K$. Hence $T_t$ is an admissible competitor for $T$ in the local minimizing property on the compact set $K$. Therefore the function $f:(-\varepsilon,\varepsilon)\to\mathbb{R}$ defined by $f(t):=\mathsf{M}_{K}(T_t)$ has a minimum at $t=0$. Define the first variation of mass of $T$ in the direction $X$ by \begin{align*} \delta \mathsf{M}(T)(X) := \frac{d}{dt}\Big|_{t=0}\mathsf{M}_{K}((\Phi_t)_\#T), \end{align*} where $K$ is any compact set whose interior contains the deformation support for all sufficiently small $|t|$. Since $\Phi_t$ is the identity outside $K$ and $X$ is supported in the interior of $K$, this derivative is independent of the chosen such $K$ and is precisely the first variation of mass in the direction $X$. Since the first variation exists for the smooth current representation on the support of $X$, this gives \begin{align*} 0 = f'(0) = \delta \mathsf{M}(T)(X). \end{align*}[/step]

[guided]The point of this step is to convert the minimizing hypothesis into the infinitesimal statement needed for mean curvature, while applying minimality only on compact subsets. Start with a smooth compactly supported vector field $X: B(x_0,r)\to\mathbb{R}^n$, and extend it by zero to a smooth compactly supported vector field on $U$, still denoted $X$. Let $C:=\operatorname{spt}(X)\subset B(x_0,r)$; this set is compact because $X$ is compactly supported. Since $C$ is compactly contained in $B(x_0,r)$, it has positive distance from $\partial B(x_0,r)$. Since the extended vector field is compactly supported in $U$, the standard flow existence theorem for smooth compactly supported vector fields gives a flow $\Phi_t:U\to U$ for all sufficiently small $|t|<\varepsilon$. By decreasing $\varepsilon>0$ if needed, the trajectories starting in $C$ remain inside the interior of a fixed compact set $K\subset B(x_0,r)$, and the flow is the identity on $U\setminus C$ because $X=0$ there. The interior containment of the deformation support in $K$ ensures that differentiating $\mathsf{M}_{K}((\Phi_t)_\#T)$ does not acquire an artificial contribution from mass crossing $\partial K$. For each $t\in(-\varepsilon,\varepsilon)$, define \begin{align*} T_t := (\Phi_t)_\#T. \end{align*} This is the natural competitor obtained by deforming the current through the flow. We must check that it is admissible for the local minimizing property on the compact set $K$, not on the noncompact ball. First, since $\Phi_t$ is the identity outside $C$ and $\Phi_t(C)\subset K$, the difference $T_t-T$ is supported in $K$. Second, since $B(x_0,r)$ was chosen disjoint from $\operatorname{spt}(\partial T)$ and $\Phi_t$ is the identity outside $C\subset B(x_0,r)$, the map $\Phi_t$ fixes $\partial T$. Using the [boundary commutation formula for pushforwards](/page/Pushforward%20of%20Current), \begin{align*} \partial T_t = \partial ((\Phi_t)_\#T) = (\Phi_t)_\#(\partial T) = \partial T. \end{align*} Thus $T_t$ has the same boundary as $T$ and differs from $T$ only in the compact subset $K\subset U$. Local area-minimality now says that the mass cannot decrease under this compactly supported deformation: \begin{align*} \mathsf{M}_{K}(T) \leq \mathsf{M}_{K}(T_t) \end{align*} for all sufficiently small $t$. Therefore the function $f:(-\varepsilon,\varepsilon)\to\mathbb{R}$ defined by $f(t):=\mathsf{M}_{K}(T_t)$ has a minimum at $t=0$. Define the first variation of mass of $T$ in the direction $X$ by \begin{align*} \delta \mathsf{M}(T)(X) := \frac{d}{dt}\Big|_{t=0}\mathsf{M}_{K}((\Phi_t)_\#T), \end{align*} where $K$ is any compact set whose interior contains the deformation support for all sufficiently small $|t|$. Because the deformation is the identity outside $K$ and $X$ is supported in the interior of $K$, differentiating this compact-set mass at $t=0$ gives exactly the first variation in the direction $X$, independent of the chosen such $K$. Since $T$ is represented by the smooth current $q\llbracket M\rrbracket$ on the support of $X$, the first derivative exists. Hence \begin{align*} 0 = f'(0) = \delta \mathsf{M}(T)(X). \end{align*} This is stationarity: every compactly supported interior variation has zero first variation.[/guided]

[step:Apply the smooth first variation formula to the local representative] Because $T \llcorner B(x_0,r)=q\llbracket M\rrbracket$ and $X$ is supported in $B(x_0,r)$, stationarity gives \begin{align*} 0 = \delta \mathsf{M}(T)(X) = \delta \mathsf{M}(q\llbracket M\rrbracket)(X). \end{align*} Let $\mathcal{H}^m$ denote $m$-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) restricted to $M$. For a smooth ambient vector field $Y:B(x_0,r)\to\mathbb{R}^n$, let $\operatorname{div}_M Y:M\to\mathbb{R}$ denote the tangential divergence of $Y$ along $M$, namely the trace over $T_xM$ of the tangential derivative of $Y$ at each $x\in M$. On the support of $X$, the local representative has no boundary contribution: $X$ is compactly supported in $B(x_0,r)$ and $B(x_0,r)\cap\operatorname{spt}(\partial T)=\varnothing$. Therefore the [smooth first variation formula for submanifolds](/page/First%20Variation%20of%20Area) gives \begin{align*} \delta \mathsf{M}(q\llbracket M\rrbracket)(X) = q\int_M \operatorname{div}_M X \, d\mathcal{H}^m(x) = -q\int_M H_M(x)\cdot X(x)\, d\mathcal{H}^m(x), \end{align*} where \begin{align*} H_M: M \to \mathbb{R}^n \end{align*} is the smooth mean curvature vector of $M$. Therefore \begin{align*} \int_M H_M(x)\cdot X(x)\, d\mathcal{H}^m(x)=0 \end{align*} for every smooth compactly supported vector field $X: B(x_0,r)\to\mathbb{R}^n$. [/step]

[step:Localize the variational identity to force the mean curvature vector to vanish] We prove that $H_M(p)=0$ for every $p\in M$. Suppose, toward a contradiction, that there exists $p\in M$ with $H_M(p)\neq 0$. Define $v:=H_M(p)\in\mathbb{R}^n$. Since $H_M$ is continuous, there exists an open neighbourhood $V\subset M$ of $p$ such that \begin{align*} H_M(x)\cdot v > \frac{1}{2}|v|^2 \end{align*} for every $x\in V$. Choose a nonnegative smooth function \begin{align*} \psi: M \to [0,\infty) \end{align*} with compact support in $V$ and with $\psi(p)>0$. Since $M$ is smoothly embedded in $B(x_0,r)$, after shrinking the support of $\psi$ inside a coordinate neighbourhood if necessary, there exists a smooth compactly supported ambient function \begin{align*} \widetilde{\psi}: B(x_0,r)\to[0,\infty) \end{align*} such that $\widetilde{\psi}|_M=\psi$. Define the ambient vector field $X:B(x_0,r)\to\mathbb{R}^n$ by $X(x):=\widetilde{\psi}(x)v$. Then $X$ is smooth and compactly supported in $B(x_0,r)$, so the variational identity applies: \begin{align*} 0 = \int_M H_M(x)\cdot X(x)\, d\mathcal{H}^m(x) = \int_M \psi(x)\, H_M(x)\cdot v\, d\mathcal{H}^m(x). \end{align*} The integrand is nonnegative on $M$, is supported in $V$, and satisfies \begin{align*} \psi(x)\,H_M(x)\cdot v > \frac{1}{2}\psi(x)|v|^2 \end{align*} on the [open set](/page/Open%20Set) where $\psi>0$. Since $\psi(p)>0$ and $M$ is a smooth embedded submanifold, this open set has positive $\mathcal{H}^m$-measure. Hence \begin{align*} \int_M \psi(x)\,H_M(x)\cdot v\, d\mathcal{H}^m(x)>0, \end{align*} contradicting the variational identity. Therefore $H_M(p)=0$ for every $p\in M$. Thus $H_M$ vanishes identically on the neighbourhood where $T=q\llbracket M\rrbracket$, completing the proof. [/step]