Anisotropic gauge and unresolved uniqueness

On the common smooth core, the identity between the fixed-atlas and Euclidean $W^{1,2}$ norms is bounded in both directions. Indeed, each fixed chart representative is obtained by multiplying by a fixed smooth cutoff and composing with a smooth flattening chart, so [F3] bounds the finite fixed-atlas sum by the Euclidean norm. Conversely, the finite partition of unity writes $u$ as the sum of its localized chart pieces; pull each piece back and use [F3] to bound the Euclidean $W^{1,2}$ norm by the fixed-atlas sum. The constants are finite because the atlas and partition are finite. By the density and completeness in [F3], the identity extends to bounded inverse maps on the completions. Since the two norms are equivalent on $C_c^{\mathrm{\infty }}(\mathrm{\Omega })$, their closures give the same zero-boundary subspace.

For $u\in H_{\mathrm{f}\mathrm{i}\mathrm{x}}^1$, take its localized boundary-chart representatives in the fixed-atlas definition. Apply [F4] to each representative (reflecting $t<0$ to $t>0$ when required), and sum the finitely many resulting local traces. The fixed-atlas norm [F1] gives a bounded map $T:H_{\mathrm{f}\mathrm{i}\mathrm{x}}^1\to H_{\mathrm{f}\mathrm{i}\mathrm{x}}^{1\mathrm{/}2}$. For smooth $u$ this sum is the classical boundary restriction because the partition equals one near the boundary and the half-space traces agree with classical restriction. As an $L^2$-valued map, this trace agrees with the Euclidean trace [F5]: the maps are bounded into $L^2$ by [F2] and [F5], and the Euclidean and fixed $H^1$ norms are equivalent by step 1.1, so equality on the dense smooth core extends to all of $H^1$.

If $Tu=0$ in $H_{\mathrm{f}\mathrm{i}\mathrm{x}}^{1\mathrm{/}2}$, it is zero as an $L^2$ class by [F2]; the $L^2$ trace agreement of step 2.1 and [F6] imply $u\in H_{0,\mathrm{E}}^1$. Step 1.1 identifies this with $H_{0,\mathrm{f}\mathrm{i}\mathrm{x}}^1$. Conversely, if $u\in H_{0,\mathrm{f}\mathrm{i}\mathrm{x}}^1$, approximate it in the fixed $H^1$ norm by interior compactly supported smooth functions. Each has zero classical trace, so continuity of $T$ gives $Tu=0$. Therefore $\ker T=H_{0,\mathrm{f}\mathrm{i}\mathrm{x}}^1$.

By [F7] and the boundary-norm equivalence [F2], the Euclidean $L^2$ trace has a bounded right inverse $R:H_{\mathrm{f}\mathrm{i}\mathrm{x}}^{1\mathrm{/}2}(\mathrm{\partial }\mathrm{\Omega })\to H_{\mathrm{E}}^1(\mathrm{\Omega })$. The norm equivalence of step 1.1 makes $R$ bounded into $H_{\mathrm{f}\mathrm{i}\mathrm{x}}^1$ as well. For every boundary class $f$, the Euclidean trace of $Rf$ is $f$; by step 2.1 the fixed trace agrees with it in $L^2$, hence as an $H_{\mathrm{f}\mathrm{i}\mathrm{x}}^{1\mathrm{/}2}$ class. So $T$ is onto. The kernel identity of step 3.1 induces a bounded bijection $\bar{T}:H_{\mathrm{f}\mathrm{i}\mathrm{x}}^1\mathrm{/}{H}_{0,\mathrm{f}\mathrm{i}\mathrm{x}}^1\to H_{\mathrm{f}\mathrm{i}\mathrm{x}}^{1\mathrm{/}2}$. For any representative $u+v$ with $v\in H_{0,\mathrm{f}\mathrm{i}\mathrm{x}}^1$, step 2.1 gives $\mathrm{\parallel }Tu{\mathrm{\parallel }}_{H^{1\mathrm{/}2}}\le C\mathrm{\parallel }u+v{\mathrm{\parallel }}_{H_{\mathrm{f}\mathrm{i}\mathrm{x}}^1}$; taking the infimum shows $\bar{T}$ is bounded. Also $\mathrm{\parallel }{\bar{T}}^{-1}f\mathrm{\parallel }\le \mathrm{\parallel }Rf{\mathrm{\parallel }}_{H_{\mathrm{f}\mathrm{i}\mathrm{x}}^1}\le C_R\mathrm{\parallel }f{\mathrm{\parallel }}_{H^{1\mathrm{/}2}}$, so its inverse is bounded. Composing with the quotient identity isomorphism of step 1.1 proves the asserted Euclidean quotient identification.

The space $H_0^1(\mathrm{\Omega })$ is a closed linear subspace of the complete space $H^1(\mathrm{\Omega })$ by [F1] and [F3]. Since $\mathrm{\Omega }$ is bounded, it lies in a slab $a<x\cdot e<b$ for some unit vector $e$ and finite $a<b$; applying [F2] with $p=2$ gives $\mathrm{\parallel }v{\mathrm{\parallel }}_{L^2}\le C(2)(b-a)\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_{L^2}$ for every $v\in H_0^1(\mathrm{\Omega })$. Thus $\mathrm{\parallel }v{\mathrm{\parallel }}_H:=\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_{L^2}$ is a norm equivalent to the inherited $H^1$ norm, and $H_0^1(\mathrm{\Omega })$ is a real Hilbert space for $(v,w{)}_H={\int }_{\mathrm{\Omega }}\mathrm{\nabla }v\cdot \mathrm{\nabla }wdx$.

Fix any representative $U\in H^1(\mathrm{\Omega })$ of $f$; the existence argument below is independent of which representative is fixed. On $H_0^1(\mathrm{\Omega })$ define $a_{\gamma }(z,v)={\int }_{\mathrm{\Omega }}\gamma \mathrm{\nabla }z\cdot \mathrm{\nabla }vdx$ and $F_U(v)=-{\int }_{\mathrm{\Omega }}\gamma \mathrm{\nabla }U\cdot \mathrm{\nabla }vdx$. By [F6], $a_{\gamma }$ is bounded by $C\mathrm{\parallel }z{\mathrm{\parallel }}_H\mathrm{\parallel }v{\mathrm{\parallel }}_H$, coercive by $c\mathrm{\parallel }z{\mathrm{\parallel }}_H^2$, and $F_U$ is bounded by $C\mathrm{\parallel }\mathrm{\nabla }U{\mathrm{\parallel }}_{L^2}\mathrm{\parallel }v{\mathrm{\parallel }}_H$. Countable Choice follows from [F5], so [F4] gives a unique $z\in H_0^1(\mathrm{\Omega })$ with $a_{\gamma }(z,v)=F_U(v)$ for every $v\in H_0^1(\mathrm{\Omega })$. Then $u_{\gamma ,f}:=U+z$ has quotient class $f$ and satisfies the stated weak equation.

If $u_1,u_2$ are two weak solutions with boundary class $f$, then $w=u_1-u_2\in H_0^1(\mathrm{\Omega })$ and ${\int }_{\mathrm{\Omega }}\gamma \mathrm{\nabla }w\cdot \mathrm{\nabla }vdx=0$ for every $v\in H_0^1(\mathrm{\Omega })$. Taking $v=w$ and using [F6] gives $c\mathrm{\parallel }\mathrm{\nabla }w{\mathrm{\parallel }}_{L^2}^2\le 0$, so $\mathrm{\parallel }w{\mathrm{\parallel }}_H=0$ and hence $w=0$ by the norm property in step 1.1.

Let $h\in {\mathcal{B}}_{\mathrm{\Omega }}$ and let $V,V^{\mathrm{\prime }}$ be any two representatives of $h$. Their difference lies in $H_0^1(\mathrm{\Omega })$ by [F1], so the weak equation for $u_{\gamma ,f}$ gives ${\int }_{\mathrm{\Omega }}\gamma \mathrm{\nabla }{u}_{\gamma ,f}\cdot \mathrm{\nabla }(V-V^{\mathrm{\prime }})dx=0$. Therefore the energy pairing is independent of the representative of $h$.

The weak solution minimizes the quadratic energy in its boundary class: if $W$ is any representative of $f$, then $W-u_{\gamma ,f}\in H_0^1(\mathrm{\Omega })$, and expanding the symmetric quadratic form with the weak equation gives ${\int }_{\mathrm{\Omega }}(\gamma \mathrm{\nabla }W)\cdot \mathrm{\nabla }Wdx={\int }_{\mathrm{\Omega }}(\gamma \mathrm{\nabla }{u}_{\gamma ,f})\cdot \mathrm{\nabla }{u}_{\gamma ,f}dx+{\int }_{\mathrm{\Omega }}(\gamma \mathrm{\nabla }(W-u_{\gamma ,f}))\cdot \mathrm{\nabla }(W-u_{\gamma ,f})dx$. Using [F6] and taking the infimum over representatives yields $\mathrm{\parallel }\mathrm{\nabla }{u}_{\gamma ,f}{\mathrm{\parallel }}_{L^2}\le \sqrt{C\mathrm{/}c}\mathrm{\parallel }f{\mathrm{\parallel }}_{{\mathcal{B}}_{\mathrm{\Omega }}}$. The same infimum argument gives representatives $V$ of $h$ with $\mathrm{\parallel }V{\mathrm{\parallel }}_{H^1}\le \mathrm{\parallel }h{\mathrm{\parallel }}_{{\mathcal{B}}_{\mathrm{\Omega }}}+\epsilon$ for every $\epsilon >0$. Cauchy–Schwarz and the upper bound in [F6] now give $\mathrm{\mid }⟨{\mathrm{\Lambda }}_{\gamma }f,h⟩\mathrm{\mid }\le C\sqrt{C\mathrm{/}c}\mathrm{\parallel }f{\mathrm{\parallel }}_{{\mathcal{B}}_{\mathrm{\Omega }}}(\mathrm{\parallel }h{\mathrm{\parallel }}_{{\mathcal{B}}_{\mathrm{\Omega }}}+\epsilon )$; letting $\epsilon \downarrow 0$ proves boundedness. To prove bilinearity, for $f_1,f_2\in {\mathcal{B}}_{\mathrm{\Omega }}$ and $a,b\in \mathbb{R}$, the function $a{u}_{\gamma ,f_1}+b{u}_{\gamma ,f_2}$ has boundary class $a{f}_1+b{f}_2$ and satisfies the weak equation by linearity of the quotient map and of the integral. Uniqueness from step 3.1 therefore gives $u_{\gamma ,a{f}_1+b{f}_2}=a{u}_{\gamma ,f_1}+b{u}_{\gamma ,f_2}$. For $h_1,h_2\in {\mathcal{B}}_{\mathrm{\Omega }}$, representatives $V_{h_1},V_{h_2}$ give $a{V}_{h_1}+b{V}_{h_2}$ as a representative of $a{h}_1+b{h}_2$; representative independence from step 3.2 and linearity of the integral show that the energy pairing is linear in its second argument as well. Thus the energy form is bilinear, and the bound just proved makes it a bounded bilinear form on ${\mathcal{B}}_{\mathrm{\Omega }}\times {\mathcal{B}}_{\mathrm{\Omega }}$.

The pushforward coefficient is admissible. Since $F$ and $F^{-1}$ are smooth on the compact closure, $DF$, $D{F}^{-1}$, $J_F$, and $J_F^{-1}$ are bounded, with $J_F$ positive. The formula in the statement is smooth. For each $\xi \in {\mathbb{R}}^n$ and $x=F(y)$, ${\xi }^T(F_{\ast }\gamma )(x)\xi =(DF(y{)}^T\xi {)}^T\gamma (y)(DF(y{)}^T\xi )\mathrm{/}{J}_F(y)$; the ellipticity of $\gamma$ and the uniform upper and lower singular-value bounds for $DF$ and $J_F$ give positive finite uniform ellipticity constants for $F_{\ast }\gamma$.

In the Riemannian setting of the statement, the chain rule [F3] gives $d(u\circ \mathrm{\Phi })=D{\mathrm{\Phi }}^T(du\circ \mathrm{\Phi })$ in a coordinate chart, while [F8] gives $({\mathrm{\Phi }}^{\ast }g)=D{\mathrm{\Phi }}^T(g\circ \mathrm{\Phi })D\mathrm{\Phi }$. Thus $({\mathrm{\Phi }}^{\ast }g{)}^{-1}=D{\mathrm{\Phi }}^{-1}(g^{-1}\circ \mathrm{\Phi })D{\mathrm{\Phi }}^{-T}$, and multiplication shows $⟨d(u\circ \mathrm{\Phi }),d(v\circ \mathrm{\Phi }){⟩}_{{\mathrm{\Phi }}^{\ast }g}=(⟨du,dv{⟩}_g)\circ \mathrm{\Phi }$. The determinant formula [F9] gives $d{V}_{{\mathrm{\Phi }}^{\ast }g}=\mathrm{\mid }\det D\mathrm{\Phi }\mathrm{\mid }(d{V}_g\circ \mathrm{\Phi })$, with the absolute Jacobian and no orientation choice.

Composition by $F^{-1}$ and by $F$ is bounded on $H^1(\mathrm{\Omega })$. For ambient smooth $u$, the chain rule [F3] and change of variables [F4], whose choice hypothesis follows from [A1] by [F11], yield ${\int }_{\mathrm{\Omega }}(\mathrm{\mid }u\circ F^{-1}{\mathrm{\mid }}^2+\mathrm{\mid }\mathrm{\nabla }(u\circ F^{-1}){\mathrm{\mid }}^2)dx\le C_F{\int }_{\mathrm{\Omega }}(\mathrm{\mid }u{\mathrm{\mid }}^2+\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^2)dy$, because $D{F}^{\pm 1}$ and $J_F^{\pm 1}$ are bounded; the nonnegative change-of-variables formula applies to the displayed squared terms. By [F5], smooth functions are dense in $H^1$; extending this estimate to limits defines the bounded compositions for every $H^1$ class. For smooth approximants $u_k$, the pointwise boundary identity $(u_k\circ F^{-1}){\mathrm{\mid }}_{\mathrm{\partial }\mathrm{\Omega }}=u_k{\mathrm{\mid }}_{\mathrm{\partial }\mathrm{\Omega }}$ holds since $F^{-1}$ fixes $\mathrm{\partial }\mathrm{\Omega }$. Continuity of the trace [F6] and convergence of both compositions in $H^1$ therefore give $T(u\circ F^{-1})=Tu$. The same argument with $F$ in place of $F^{-1}$ gives $T(u\circ F)=Tu$. Since $\ker T=H_0^1(\mathrm{\Omega })$ by [F6], both trace equalities imply $[u\circ F^{-1}]=[u\circ F]=[u]$ in the quotient. Finally, if $\varphi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$, then $\varphi \circ F^{-1}$ and $\varphi \circ F$ are smooth and compactly supported in $\mathrm{\Omega }$, because $F$ maps the interior to itself and maps compact sets to compact sets. Applying bounded composition to sequences converging in [F7] proves that both compositions preserve $H_0^1(\mathrm{\Omega })$. The weak-gradient chain rule and the energy change-of-variables identity also extend to all $H^1$ classes. Given $u\in H^1(\mathrm{\Omega })$, choose ambient smooth restrictions $u_k\to u$ in $H^1$ by [F5]. For each $k$, the classical chain rule gives $\mathrm{\nabla }(u_k\circ F^{-1})(F(y))=DF(y{)}^{-T}\mathrm{\nabla }{u}_k(y),\mathrm{\nabla }(u_k\circ F)(y)=DF(y{)}^T\mathrm{\nabla }{u}_k(F(y))\mathrm{.}$ The composition estimates just established give convergence of the left sides in $L^2$; the uniform bounds on $D{F}^{\pm 1}$ and the Jacobian, together with [F4], give convergence of the right sides in $L^2$. Passing to the limits identifies the weak gradients of $u\circ F^{-1}$ and $u\circ F$ by the same formulas. Applying these identities to $u,v\in H^1(\mathrm{\Omega })$ and using [F4] on the positive and negative parts of the integrable energy density yields ${\int }_{\mathrm{\Omega }}(F_{\ast }\gamma )(x)\mathrm{\nabla }(u\circ F^{-1})(x)\cdot \mathrm{\nabla }(v\circ F^{-1})(x)dx={\int }_{\mathrm{\Omega }}\gamma (y)\mathrm{\nabla }u(y)\cdot \mathrm{\nabla }v(y)dy\mathrm{.}$ Integrability follows from boundedness of the coefficients and Cauchy--Schwarz.

The pointwise density identity in step 1.2 makes the left density the pullback of $⟨du,dv{⟩}_{g}d{V}_g$. The boundary has zero Riemannian volume: in each boundary chart it is contained in a coordinate hyperplane, which has Lebesgue measure zero by Fubini, and the smooth volume density is locally bounded. Cover $M$ by finitely many source neighborhoods each lying in a chart and mapped by $\mathrm{\Phi }$ into another chart; such neighborhoods exist by continuity and compactness. Partition their interiors into finitely many disjoint measurable pieces subordinate to this cover. On each piece apply the Euclidean change-of-variables formula [F4], whose choice hypothesis follows from [A1] by [F11], to the positive and negative parts of the smooth integrand, extended by zero within the corresponding chart. Summing the finitely many equalities and ignoring the null boundary gives ${\int }_M⟨d(u\circ \mathrm{\Phi }),d(v\circ \mathrm{\Phi }){⟩}_{{\mathrm{\Phi }}^{\ast }g}d{V}_{{\mathrm{\Phi }}^{\ast }g}={\int }_M⟨du,dv{⟩}_{g}d{V}_g$.

Let $f\in {\mathcal{B}}_{\mathrm{\Omega }}$ and let $u=u_{\gamma ,f}$. Set $v=u\circ F^{-1}$. By step 2.1 its boundary class is $f$. For any $\phi \in H_0^1(\mathrm{\Omega })$, $\phi \circ F\in H_0^1(\mathrm{\Omega })$ by step 2.1. The energy identity of step 2.1, with second input $\phi \circ F$, gives ${\int }_{\mathrm{\Omega }}(F_{\ast }\gamma )(x)\mathrm{\nabla }v(x)\cdot \mathrm{\nabla }\phi (x)dx={\int }_{\mathrm{\Omega }}\gamma (y)\mathrm{\nabla }u(y)\cdot \mathrm{\nabla }(\phi \circ F)(y)dy=0$. The matrix factors cancel because $D{F}^{T}D{F}^{-T}=I$, and the Jacobian $J_F$ in $dx$ cancels the denominator in $F_{\ast }\gamma$. Thus $v$ is the weak solution with coefficient $F_{\ast }\gamma$ and boundary class $f$; uniqueness [F2] identifies $v=u_{F_{\ast }\gamma ,f}$.

Fix any $h\in {\mathcal{B}}_{\mathrm{\Omega }}$ and a representative $V_h\in H^1(\mathrm{\Omega })$. Since $F$ fixes the boundary, step 2.1 gives $V_h\circ F$ the same boundary class $h$. The energy identity of step 2.1, with second input $V_h\circ F$, and the solution identification in step 3.1 give $⟨{\mathrm{\Lambda }}_{F_{\ast }\gamma }f,h⟩={\int }_{\mathrm{\Omega }}(F_{\ast }\gamma )\mathrm{\nabla }(u\circ F^{-1})\cdot \mathrm{\nabla }{V}_{h}dx={\int }_{\mathrm{\Omega }}\gamma \mathrm{\nabla }u\cdot \mathrm{\nabla }(V_h\circ F)dy=⟨{\mathrm{\Lambda }}_{\gamma }f,h⟩$. This holds for every $f,h$, so the bounded bilinear maps are equal.

If $u$ is smooth and $g$-harmonic, then for any $\psi \in C_c^{\mathrm{\infty }}(M^{\circ })$, $\psi \circ {\mathrm{\Phi }}^{-1}$ is also smooth and compactly supported in $M^{\circ }$. Applying the identity of step 2.2 to $u$ and $\psi \circ {\mathrm{\Phi }}^{-1}$ yields ${\int }_M⟨d(u\circ \mathrm{\Phi }),d\psi {⟩}_{{\mathrm{\Phi }}^{\ast }g}d{V}_{{\mathrm{\Phi }}^{\ast }g}=0$, so $u\circ \mathrm{\Phi }$ is ${\mathrm{\Phi }}^{\ast }g$-harmonic. Because $\mathrm{\Phi }$ fixes $\mathrm{\Gamma }$ and is bijective, it maps $\mathrm{\partial }M\setminus \mathrm{\Gamma }$ to itself; therefore it leaves unchanged every boundary function supported in $\mathrm{\Gamma }$. Step 2.2 gives equality of the local measurements for every such pair. When the stated Dirichlet solutions are unique, these harmonicity and trace facts identify the solution for ${\mathrm{\Phi }}^{\ast }g$ with the pullback of the solution for $g$, so their local energy maps agree.