[proofplan] The proof is an immediate consequence of the [first variation formula for area](/theorems/5660), together with the fact that every compactly supported section of the pulled-back tangent bundle $F^*TN$ occurs as the variational vector field of a compactly supported smooth variation. If $H=0$, the [first variation formula](/theorems/2728) gives zero first variation for every variation. Conversely, stationarity implies that the $L^2$ pairing of $H$ with every compactly supported normal variation field is zero; choosing the variation field to be a nonnegative cutoff multiple of $H$ forces $|H|_g^2$ to vanish locally, hence pointwise by continuity. [/proofplan] [step:Use the first variation formula to prove stationarity when $H$ vanishes] Assume $H(p)=0$ for every $p \in M$. Let $K \subset M$ be a compact domain, and let $\mathcal{F}:(-\varepsilon,\varepsilon)\times M \to N$ be a smooth compactly supported variation of $F$ whose support is contained in the interior of $K$, written $\mathcal{F}(t,p)=F_t(p)$. Let $X:M \to F^*TN$ be its variational vector field, defined by \begin{align*} X(p)=\left.\frac{\partial F_t(p)}{\partial t}\right|_{t=0}. \end{align*} We use the first variation formula for area in the following form: if $K\subset M$ is a compact domain, $F_t:M\to N$ is a smooth compactly supported variation through immersions with support contained in $\operatorname{int}K$, and $X$ is its variational vector field, then \begin{align*} \left.\frac{d}{dt}\right|_{t=0}\mu_{F_t^*g}(K) = -\int_K g_{F(p)}(H(p),X(p))\,d\mu_{F^*g}(p). \end{align*} The hypotheses of this formula are exactly the compact-domain condition on $K$, the smoothness of the variation, the immersion condition for $F_t$, and the compact support assumption stated above. Since $H(p)=0$ for every $p \in M$, the integrand is identically zero, and therefore \begin{align*} \left.\frac{d}{dt}\right|_{t=0}\mu_{F_t^*g}(K)=0. \end{align*} Thus $F$ is stationary for area. [/step] [step:Realize every compactly supported section as a variational vector field] Assume now that $F$ is stationary for area. Let $X:M \to F^*TN$ be a smooth compactly supported section. We show that $X$ is the variational vector field of a compactly supported smooth variation. Let $C:=\operatorname{supp}X \subset M$, which is compact by hypothesis. The ambient exponential map $\exp: \mathcal{U} \subset TN \to N$ is defined on an open neighbourhood $\mathcal{U}$ of the zero section of $TN$. Since $F(C)\subset N$ is compact, there exists $\varepsilon>0$ such that \begin{align*} tX(p)\in \mathcal{U}_{F(p)} \end{align*} for every $p\in C$ and every $t\in(-\varepsilon,\varepsilon)$, where $\mathcal{U}_{F(p)}:=\mathcal{U}\cap T_{F(p)}N$. Define the smooth map $F_t:M\to N$ by \begin{align*} F_t(p)=\exp_{F(p)}(tX(p)). \end{align*} For $p\notin C$, we have $X(p)=0$, so $F_t(p)=\exp_{F(p)}(0)=F(p)$. Thus the variation is compactly supported in $C$. Since $F_0=F$ is an immersion and the immersion condition is open on compact subsets, after decreasing $\varepsilon>0$ if necessary, each $F_t$ is an immersion for $|t|<\varepsilon$. Finally, \begin{align*} \left.\frac{\partial F_t(p)}{\partial t}\right|_{t=0}=X(p) \end{align*} by the defining property of the exponential map at the zero vector. Hence every smooth compactly supported section of $F^*TN$ is an admissible variational vector field. [guided] The goal of this step is to justify that stationarity can be tested against arbitrary compactly supported vector fields along the immersion, not only against vector fields that happen to come from some pre-existing variation. Let $X:M \to F^*TN$ be a smooth compactly supported section, so $X(p)\in T_{F(p)}N$ for each $p\in M$. Its support \begin{align*} C:=\operatorname{supp}X \end{align*} is compact. We use the exponential map of the ambient Riemannian manifold. The map $\exp:\mathcal{U}\subset TN \to N$ is defined on an open neighbourhood $\mathcal{U}$ of the zero section. Because $F(C)$ is compact and $X$ is continuous, the set of vectors $\{X(p):p\in C\}$ lies in a compact subset of $TN$ over $F(C)$. Therefore, after choosing $\varepsilon>0$ sufficiently small, the scaled vectors $tX(p)$ all lie in the domain of the exponential map for every $p\in C$ and every $t\in(-\varepsilon,\varepsilon)$. Define $F_t:M \to N$ by \begin{align*} F_t(p)=\exp_{F(p)}(tX(p)). \end{align*} This is smooth because $F$, $X$, scalar multiplication in the vector bundle $TN$, and the exponential map are smooth on their domains. If $p\notin C$, then $X(p)=0$, and hence \begin{align*} F_t(p)=\exp_{F(p)}(0)=F(p). \end{align*} Thus the variation is compactly supported in $C$. It remains to ensure that $F_t$ is still an immersion for small $|t|$. On the compact set $C$, the differentials $dF_t$ depend continuously on $t$, and $dF_0=dF$ is injective at every point because $F$ is an immersion. Injectivity of a [linear map](/page/Linear%20Map) is an open condition in finite-dimensional operator space, so after decreasing $\varepsilon>0$ if necessary, $dF_t$ remains injective on $C$ for every $|t|<\varepsilon$. Outside $C$, $F_t=F$, so the immersion condition is unchanged. Hence $F_t$ is a smooth compactly supported variation through immersions. Finally, the derivative of the exponential map at the zero vector in each fibre is the identity on that fibre. Therefore \begin{align*} \left.\frac{\partial F_t(p)}{\partial t}\right|_{t=0}=X(p) \end{align*} for every $p\in M$. This proves that any compactly supported section of $F^*TN$ can be used as a variational vector field in the first variation formula. [/guided] [/step] [step:Test stationarity against compactly supported multiples of $H$] Let $q\in M$. Choose a coordinate neighbourhood $U\subset M$ of $q$ whose closure is compact in $M$. Let \begin{align*} \varphi:M &\to [0,\infty) \end{align*} be a smooth compactly supported function with $\operatorname{supp}\varphi\subset U$. Define the compactly supported section $X_\varphi:M \to F^*TN$ by \begin{align*} X_\varphi(p)=\varphi(p)H(p). \end{align*} By the previous step, $X_\varphi$ is the variational vector field of a compactly supported smooth variation. Since $F$ is stationary, the first variation vanishes. Applying the first variation formula on any compact domain $K\subset M$ with $\operatorname{supp}\varphi\subset \operatorname{int}K$ gives \begin{align*} 0=-\int_K g_{F(p)}(H(p),X_\varphi(p))\,d\mu_{F^*g}(p). \end{align*} Substituting $X_\varphi(p)=\varphi(p)H(p)$ into the integrand gives \begin{align*} 0=-\int_K \varphi(p)\, g_{F(p)}(H(p),H(p))\,d\mu_{F^*g}(p). \end{align*} Using the definition $|H(p)|_g^2=g_{F(p)}(H(p),H(p))$, we obtain \begin{align*} 0=-\int_K \varphi(p)|H(p)|_g^2\,d\mu_{F^*g}(p). \end{align*} Since $\operatorname{supp}\varphi\subset K$, this is equivalent to \begin{align*} \int_U \varphi(p)|H(p)|_g^2\,d\mu_{F^*g}(p)=0 \end{align*} for every smooth nonnegative compactly supported function $\varphi:M\to[0,\infty)$ with support contained in $U$. [/step] [step:Use nonnegative cutoffs to force $H$ to vanish pointwise] We prove that $H(q)=0$. Suppose instead that $|H(q)|_g^2>0$. Since $H$ is smooth, the function $h:U \to [0,\infty)$ defined by \begin{align*} h(p)=|H(p)|_g^2 \end{align*} is continuous. Hence there exists an open neighbourhood $V\subset U$ of $q$ and a constant $a>0$ such that \begin{align*} h(p)\geq a \end{align*} for every $p\in V$. Choose a smooth function \begin{align*} \varphi:M &\to [0,\infty) \end{align*} with compact support contained in $V$ and not identically zero. Since $\varphi$ is nonnegative and continuous, and since it is positive on some nonempty open subset of $V$, we have \begin{align*} \int_U \varphi(p)h(p)\,d\mu_{F^*g}(p) \geq a\int_U \varphi(p)\,d\mu_{F^*g}(p) > 0. \end{align*} This contradicts the identity \begin{align*} \int_U \varphi(p)|H(p)|_g^2\,d\mu_{F^*g}(p)=0 \end{align*} from the previous step. Therefore $|H(q)|_g^2=0$, so $H(q)=0$. Since $q\in M$ was arbitrary, $H=0$ on $M$. Combining this implication with the first step proves that $F$ is stationary for area if and only if its mean curvature vector field vanishes identically. [/step]