The boundary fundamental lemma of the calculus of variations

Local chart at a boundary point. Fix $x_0\in \mathrm{\partial }\mathrm{\Omega }$. By [F1] we may, after translating and applying a rigid motion, assume $x_0=0$ and find $\rho >0$, $h\in C^1(B(0,\rho ))$ and a neighbourhood $U\ni 0$ with the boundary in $U$ is the graph of $h$ and the domain in $U$ is its subgraph, with both sets intersected with $U$; write $X(y):=(y,h(y))$ and $J:=\sqrt{1+\mathrm{\mid }Dh{\mathrm{\mid }}^2}\ge 1$. Any function on $\mathrm{\partial }\mathrm{\Omega }$ whose support lies in this patch has surface integral equal to the chart integral against $J$, by [F1].

A cutoff and suitable test functions. Choose $0<r<R$ with $B(0,R)\subseteq U$ and let $\eta$ be the smooth bump of [F2] with $\eta =1$ on ${\bar{B}}_r(0)$ and $\mathrm{supp}\eta \subseteq B_R(0)$. Choose $\delta >0$ with $\delta <\rho$ and $\mathrm{\mid }(y,h(y))\mathrm{\mid }<r$ for $\mathrm{\mid }y\mathrm{\mid }<\delta$. For every $\psi \in C_c^{\mathrm{\infty }}(\{\mathrm{\mid }y\mathrm{\mid }<\delta \})$ define $\phi (z):=\psi (z^{\mathrm{\prime }})\eta (z)$, where $z^{\mathrm{\prime }}=(z_1,\dots ,z_{n-1})$; then $\phi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$, hence $\phi \in C^{\mathrm{\infty }}(\bar{\mathrm{\Omega }})$, and for $\mathrm{\mid }y\mathrm{\mid }<\delta$ one has $\phi (X(y))=\psi (y)\eta (X(y))=\psi (y)$ because $X(y)\in {\bar{B}}_r(0)$ there.

The local integral identity. The hypothesis gives ${\int }_{\mathrm{\partial }\mathrm{\Omega }}g\phi d\sigma =0$ for the test function $\phi$ of step 2.1, whose boundary support lies in the patch of step 1.1; the chart formula therefore yields $0={\int }_{B(0,\rho )}g(X(y))\phi (X(y))J(y)dy={\int }_{B(0,\rho )}G(y)\psi (y)dy$, where $G:=g\circ X\cdot J$ is continuous because $g$ is continuous on $\bar{\mathrm{\Omega }}$ and $h$ is $C^1$. As $\psi \in C_c^{\mathrm{\infty }}(\{\mathrm{\mid }y\mathrm{\mid }<\delta \})$ was arbitrary, $G\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$ satisfies $\int G\psi =0$ for every test function supported in that ball.

The fundamental lemma at $x_0$. Applying [F3] to $G$ on the ball $\{\mathrm{\mid }y\mathrm{\mid }<\delta \}$ gives $G=0$ almost everywhere; since $J\ge 1$, this implies $g\circ X=0$ almost everywhere, and since $y\mapsto g(X(y))$ is continuous, $g(X(y))=0$ for every $\mathrm{\mid }y\mathrm{\mid }<\delta$. In particular $g(x_0)=g(X(0))=0$.

Conclusion on the boundary. The point $x_0\in \mathrm{\partial }\mathrm{\Omega }$ was arbitrary, so $g=0$ on $\mathrm{\partial }\mathrm{\Omega }$.

The vector-valued formulation. Let $h\in C(\bar{\mathrm{\Omega }};{\mathbb{R}}^n)$ satisfy ${\int }_{\mathrm{\partial }\mathrm{\Omega }}(h\cdot \nu )\phi d\sigma =0$ for every $\phi \in C^{\mathrm{\infty }}(\bar{\mathrm{\Omega }})$. The boundary function $\gamma :=(h\cdot \nu ){\mathrm{\mid }}_{\mathrm{\partial }\mathrm{\Omega }}$ is continuous, because $h$ is continuous on $\bar{\mathrm{\Omega }}$ and the normal field $\nu$ is continuous on the $C^1$ boundary [F1]; the argument of steps 1.1–5.1 uses only the boundary values of the continuous integrand and the linearity of the integral in it, so it applies with $g$ replaced by $\gamma$ and gives $\gamma =0$ on $\mathrm{\partial }\mathrm{\Omega }$, that is $h\cdot \nu =0$ on $\mathrm{\partial }\mathrm{\Omega }$.