[proofplan] The proof inserts the compactly supported normal variation field $\phi \nu$ into the [second variation formula](/theorems/2729) for area. Because $\Sigma$ is stable, this second variation is nonnegative. The second variation formula expresses the resulting quadratic form as the gradient energy of $\phi$ minus the curvature potential $\left(|A|_g^2+\operatorname{Ric}_h(\nu,\nu)\right)\phi^2$, and rearranging gives the desired coercive inequality. [/proofplan]

[step:Build the compactly supported normal variation generated by $\phi\nu$] Fix $\phi \in C_c^\infty(\Sigma)$. Define the normal variation field $V: \Sigma \to TN$ by $V(p)=\phi(p)\nu(p)$ for each $p \in \Sigma$. Since $\phi$ is smooth with compact support and $\nu$ is a smooth unit normal field along $\Sigma$, the map $V$ is a smooth compactly supported section of the normal bundle of $\Sigma$ in $N$. For each $p \in N$, let $\exp_p^N: \mathcal{D}_p \subset T_pN \to N$ denote the Riemannian exponential map of the ambient Riemannian manifold $(N,h)$ at $p$, defined on its maximal star-shaped domain $\mathcal{D}_p$. Choose $\varepsilon>0$ small enough that $t\phi(p)\nu(p) \in \mathcal{D}_p$ for every $p \in \operatorname{supp}\phi$ and every $t \in (-\varepsilon,\varepsilon)$. Define the variation $F: \Sigma \times (-\varepsilon,\varepsilon) \to N$ by $F(p,t)=\exp_p^N(t\phi(p)\nu(p))$. Then $F(\cdot,0)$ is the inclusion of $\Sigma$ into $N$, and its variational vector field is $V=\phi\nu$. [/step]

[step:Evaluate the second variation quadratic form]Let \begin{align*} \mathcal{A}: (-\varepsilon,\varepsilon) &\to \mathbb{R} \end{align*} denote the area of the varied hypersurface $F(\Sigma,t)$, computed only over a compact set containing $\operatorname{supp}\phi$, so that the variation is stationary outside that compact set. By the second variation formula for two-sided minimal hypersurfaces applied to the normal variation field $V=\phi\nu$ (citing a result not yet in the wiki: [Second Variation Formula for Minimal Hypersurfaces](/theorems/5663)), \begin{align*} \mathcal{A}''(0) = \int_\Sigma \left( |\nabla^\Sigma \phi|_g^2 - \left(|A|_g^2+\operatorname{Ric}_h(\nu,\nu)\right)\phi^2 \right) \,d\mu_\Sigma. \end{align*} The compact support of $\phi$ removes all boundary terms in the integration-by-parts form of the second variation formula.[/step]

[guided]We choose the variation field to be exactly $V=\phi\nu$ because stability tests the second variation against compactly supported normal variations. The function $\phi$ supplies the scalar amplitude, while the globally defined normal field $\nu$ converts that scalar function into a normal vector field. The second variation formula for a two-sided minimal hypersurface says that, for the normal variation field $V: \Sigma \to TN$ defined by $V(p)=\phi(p)\nu(p)$ for each $p \in \Sigma$, the second derivative of area is \begin{align*} \mathcal{A}''(0) = \int_\Sigma \left( |\nabla^\Sigma \phi|_g^2 - \left(|A|_g^2+\operatorname{Ric}_h(\nu,\nu)\right)\phi^2 \right) \,d\mu_\Sigma. \end{align*} The hypotheses required for this formula are satisfied: $\Sigma$ is minimal, so the first variation term vanishes; $\Sigma$ is two-sided, so the normal variation can be written globally as $\phi\nu$; and $\phi$ has compact support, so the integration-by-parts step in the Jacobi operator form produces no boundary contribution. The term $|\nabla^\Sigma \phi|_g^2$ is the tangential gradient energy on $\Sigma$, while $\left(|A|_g^2+\operatorname{Ric}_h(\nu,\nu)\right)\phi^2$ is the curvature potential term.[/guided]

[step:Use stability to force nonnegativity of the quadratic form] Since $\Sigma$ is stable, the second variation of area is nonnegative for every compactly supported smooth normal variation. Applying this to the variation generated by $V=\phi\nu$ gives \begin{align*} 0 \le \mathcal{A}''(0) = \int_\Sigma \left( |\nabla^\Sigma \phi|_g^2 - \left(|A|_g^2+\operatorname{Ric}_h(\nu,\nu)\right)\phi^2 \right) \,d\mu_\Sigma. \end{align*} Rearranging this inequality gives \begin{align*} \int_\Sigma \left(|A|_g^2+\operatorname{Ric}_h(\nu,\nu)\right)\phi^2\,d\mu_\Sigma \le \int_\Sigma |\nabla^\Sigma \phi|_g^2\,d\mu_\Sigma. \end{align*} Because $\phi \in C_c^\infty(\Sigma)$ was arbitrary, the inequality holds for every compactly supported smooth [test function](/page/Test%20Function) on $\Sigma$. [/step]