Lemma: Well-posedness and continuity

The extension exists uniquely and depends linearly and continuously on $f$. The form ${B_{g,A}}^{\Gamma }$ is bounded, Hermitian, and nonnegative on ${\mathcal{H}}_{\Gamma }$.

Proof. On zero-trace sections, the scalar Poincaré inequality and $|d|u||\le |{\nabla }_{A}u|$ bound $\Vert u{\Vert }_2$ by $\Vert {\nabla }_{A}u{\Vert }_2$. Since $A$ is bounded, the identity $du={\nabla }_{A}u-Au$ then gives an equivalent ${H^1}_0$ norm. Consequently the connection energy is coercive there.

Choose a bounded trace extension $F$ of $f$, and solve for $v\in {H^1}_0$ the equation

The Hilbert-space variational theorem gives a unique $v$; $u_f=F+v$ is independent of the chosen extension. The coercive estimate and bounded trace extension give continuous dependence. The stated properties of $B$ follow from its integral definition.