Proof. (a) $\Rightarrow$ (b):

This is implicitly proved in [Reference Liu and Pakzad43, Lemma 3.7]. We pursue a slightly different strategy. For any $f: \Omega \to {\mathbb {R}}$, let $C_{f}$ be the set on which $f$ is locally constant:

\[ C_{f} := \{ x\in \Omega; f \text{ is constant in a neighbourhood of } x\}. \]

Note that $f$ is constant on the connected components of the open set $C_{f}$. If $f$ satisfies condition (a) of the proposition, for each $x\in \Omega \setminus C_f$ we choose a line $L_x$ so that $f$ is constant on the connected component $x$ in $L_x\cap \Omega$, denoted by $l_x$.

We begin by the following simple but useful lemma.

Lemma 2.7 Let $f: \Omega \to {\mathbb {R}}^m$ be continuous and satisfy condition (a) in proposition 2.1. Let $\Delta \subset \Omega$ be a closed triangular domain. If $f$ is constant on two edges of $\Delta$, then $f$ is constant on $\Delta$.

Proof. Let $x,\,y,\,z$ be the vertices of $\Delta$ and assume that $f$ is constant on $[x,\,y]$ and $[x,\,z]$. For any point $w\in \Delta \setminus C_f$, $l_w$ must cross one of the two segments $[x,\,y]$ or $[x,\,z]$. This implies $f(w) =f(x)$. If $w\in \Delta \cap C_f$, and if the segment $[w,\,x] \subset C_f$, then once again $f(w)=f(x)$. Finally, if $[w,\,x] \not \subset C_f$, let $\tilde w$ be the closest element to $w$ on the segment $[w,\,x]$ not in $C_{f}$. We already proved that $f(\tilde w) = f(x)$. But since $\tilde w$ is in the closure of the connected component of $w$ in $C_{f}$, and $f$ is continuous, we have also $f(w) = f(\tilde w) = f(x)$. We conclude that $f|_\Delta \equiv f(x)$.

Proof. Let $x\in \Omega \setminus C_f$ and assume that for two lines $L \neq \Lambda$, $f$ is constant on the connected components $l$ and $\lambda$ of $x$ in respectively $L\cap \Omega$ and $\Lambda \cap \Omega$. Choose a disk $B(x,\,\rho ) \subset B(x,\, 2\rho ) \subset \Omega$ and note that $\partial B(x,\,\rho )$ intersects $l$ and $\lambda$ on 4 points $x_1,\, x_2 \in l$ and $\chi _1,\, \chi _2 \in \lambda$. Now the assumptions of lemma 2.7 is satisfied for the four closed triangular domains with vertices $x,\, x_i,\, \chi _j$, $i,\,j\in \{1,\,2\}$. This implies that $x\in C_f$, which contradicts the first assumption on $x$.

Proof. For showing the first conclusion, we argue by contradiction: Let $y \in C_{f} \cap l_x$ and let $V$ be an open neighbourhood of $y$ in $\Omega$ on which $f$ is constant. We observe that for $\delta >0$ small enough $B(x,\,\delta ) \setminus l_x$ cannot entirely lie in $C_{f}$, since otherwise, $B(x,\,\delta ) \setminus l_x$ having only two connected components on both sides of $l_x$, and $f$ being continuous, the value of $f$ would be constant on $B(x,\,\delta )$, contradicting $x \notin C_{f}$. Therefore, there is a sequence $x_k \in \Omega \setminus C_{f}$ converging to $x$ such that $f(x_k) \neq f(x)$. Since the value of $f$ on $x_k$ and $x$ differ, $l_{x_k} \cap l_x =\emptyset$ for all $k$. As $x_k\to x$, we deduce that $l_{x_k}$ must locally uniformly converge to $l_x$. This implies that $l_{x_k} \cap V$ is non-empty for $k$ large enough and hence $f (x_k) = f(y) = f(x)$, which is a contradiction.

To prove the second statement, also assume by contradiction that $l_y \neq l_z$ and that $x\in l_y \cap l_z \neq \emptyset$. Choose a disk $B(x,\,\rho ) \subset B(x,\, 2\rho ) \subset \Omega$ and note that $\partial B(x,\,\rho )$ intersects $l_y$ and $l_z$ on four points $y_1,\, y_2 \in l_y$ and $z_1,\, z_2 \in l_z$. Now the assumptions of lemma 2.7 are satisfied for the four closed triangular domains with vertices $x,\, y_i,\, z_j$, $i,\,j\in \{1,\,2\}$. This implies that $x\in C_f$, which contradicts the first statement, as $y\notin C_f$ but $x\in l_y \cap C_f$.

Proof. We choose a disk $B_x = B(x,\, 2\delta ) \subset \Omega$ according to condition (b) and we let $B=B(x,\, \delta )$. For any $p\in B$, there exists a line $L_p$ such that $f$ is constant on $\tilde l_p = L_p \cap B_x$, and no two such lines meet within $B_x$. We define the mapping ${\bf \Lambda }: B\to {\mathbb {R}}\mathbb {P}^1$ which associates to each point $p\in B$ the element of the real projective line ${\mathbb {R}}\mathbb {P}^1$ determined by the direction of $L_p$, and we note that it is constant along the segments $L_p\cap B$ and continuous. Since $B$ is simply connected, ${\bf \Lambda }$ can be lifted to a continuous mapping $\vec \eta : B \to \mathbb {S}^1$. By constancy of ${\bf \Lambda }$ along the segments $L_p\cap B$, $\vec \eta$ can only take two distinct values along them, and so its continuity implies that $\vec \eta$ is constant along these directions, which are now determined by $\vec \eta$ itself. Since two distinct lines $L_p$ and $L_q$, for $p,\,q\in B$, do not meet except possibly at an at least a $\delta$-distance from $\partial B$, we conclude that ${\rm Lip} \,\vec \eta \le 1/\delta$.

Proof. condition (b) states that the distributional derivative of $f$ in direction of the vector field $\vec \eta$ vanishes. The proof shows that a regular enough change of variable reduces the problem to the case when $\vec \eta$ is constant.

Let $\vec \xi = \vec \eta ^\perp$ and $x \in B$. Choose $T_0>0$ such that $\gamma : (-T_0,\,T_0) \to B$ is a solution to the ODE $\gamma ' (t) = \vec \xi (\gamma (t))$ in $B$ with initial value $\gamma (0) =x$. Let $L = {\rm Lip} \,\vec \eta$, $k(t)= -\gamma ''(t) \cdot \vec \eta (\gamma (t))$ and note that $\|k\|_{L^\infty (-T_0, T_0)} \le L$. It is straightforward to see that there exists $0< t_0\le T_0/2$ such that for all $t,\, t'\in [-t_0,\,t_0]$, and all $s,\, s' \in {\mathbb {R}}$,

\[ \gamma(t) + s \vec \eta (\gamma (t)), \gamma(t') + s' \vec \eta (\gamma (t'))\in B \implies \gamma(t) + s \vec \eta (\gamma (t)) \neq \gamma(t') + s' \vec \eta (\gamma (t')), \]

unless $t=t'$ and $s=s'$. Indeed, assume by contradiction that there exist sequences $t_k,\, t'_k\to 0$, $s_k,\, s'_k$, such that for all $k$, $(t_k,\, s_k) \neq (t'_k,\, s'_k)$, and

\[ \gamma(t_k) + s_k \vec \eta (\gamma (t_k)) = \gamma(t'_k) + s'_k \vec \eta (\gamma (t'_k)) \in B. \]

Note that $t_k = t'_k$ implies $s_k = s'_k$, and hence we must have $t_k \neq t'_k$ for all $k$, implying on its turn that $\gamma (t_k) \neq \gamma (t'_k)$ for all $k$. On the other hand, by the main property of $\vec \eta$ we obtain

\[ \vec \eta(\gamma (t_k)) = \vec \eta (\gamma (t_k) + s_k \vec \eta (\gamma (t_k)) = \vec \eta (\gamma (t'_k) + s'_k \vec \eta (\gamma (t'_k)) = \vec \eta(\gamma (t'_k)). \]

This yields

\[ \gamma(t_k) - \gamma(t'_k) = (s'_k - s_k) \vec \eta (\gamma (t_k)), \]

implying that

\[ \frac{\gamma(t_k) - \gamma(t'_k)}{{t_k - t'_k} } \cdot \vec\xi (\gamma (t_k)) =0. \]

Passing to the limit we obtain $|\vec \xi (x)|^2 = \gamma '(0) \cdot \vec \xi (\gamma (0)) =0$, which is a contradiction.

Since $\gamma ([-t_0,\, t_0])$ is compact in $B$, we choose $0< s_0< 1/{2L}$ such that the image of $[-t_0,\, t_0] \times [-s_0,\, s_0]$ under

\[ \Phi(t,s) := \gamma(t) + s \vec \eta (\gamma (t)), \,\, t\in [{-}t_0,t_0], s\in [{-}s_0, s_0], \]

lies compactly in $B$. We note that $\Phi$ is one-to-one and Lipschitz on $U:= (-t_0,\,t_0) \times (-s_0,\, s_0)$ and that we have

\[ \det \nabla \Phi (t,s) = 1+ sk(t) \ge \frac 12 \text{ a.e. in } U. \]

We conclude that $\Phi :U \to \Phi (U)$ is a bilipschitz change of variable.

We let $V_x= \Phi (U)$, which is an open neighbourhood of $x$ in $B$. We calculate for any $\psi \in C^\infty _c(V_x)$:

\[ {\rm div} (\psi \vec \eta) = \partial_{\vec \eta} (\psi \vec \eta) \cdot \vec \eta + \partial_{\vec \xi}(\psi \vec \eta) \cdot \vec \xi = \nabla \psi\cdot \vec \eta + \psi \partial_{\vec \xi} \vec \eta \cdot \vec \xi, \]

which gives for $\tilde \psi = \psi \circ \Phi$:

\[ {\rm div} (\psi \vec \eta) (\Phi (t,s))= \partial_s \tilde \psi + \tilde \psi \frac{k(t)}{1+sk(t)}. \]

Therefore, letting $\tilde f (t,\,s) = f \circ \Phi (s,\,t)= f(\gamma (t) + s \vec \eta (\gamma (t))$ we obtain

(2.1)\begin{equation} \begin{aligned} \int_{V_x} f {\rm div} (\psi \vec \eta) & = \int_U f{\rm div}(\psi \vec \eta) \circ \Phi (s,t) (1+ s k(t)) \,{\rm d}s{\rm d}t \\ & = \int_U \tilde f (s,t) \Bigg(\partial_s \tilde \psi + \tilde \psi \frac{k(t)}{1+sk(t)}\Bigg) (1+ sk(t)) \, {\rm d}s{\rm d}t \\ & =\int_U \tilde f(t,s) \Bigg( k(t) \tilde \psi + \partial_s \tilde \psi + s k(t) \partial_s \tilde \psi \Bigg) \, {\rm d}s{\rm d}t \\ & = \int_U \tilde f(t,s) \partial_s ((1+ sk(t))\tilde \psi)\, {\rm d}s{\rm d}t. \end{aligned} \end{equation}

If (a) is satisfied, then $\tilde f(t,\,s) = \tilde f(t)$ for all values of $s$ for which the function is defined, and hence, since $\tilde \psi$ is compactly supported in $U$, we obtain:

(2.2)\begin{equation} \forall \psi \in C^\infty_c(V_x) \quad \int_{V_x} f {\rm div} (\psi \vec \eta) =0. \end{equation}

Now for $\psi \in C^\infty _c(B)$, we let $K:= {\rm supp}(\psi ) \subset B$. $K$ is compact and so it admits a finite covering of open sets of the form $V_{x_i}$, for $i=1,\, \cdots,\, n$. We consider a partition of unity associated with this covering

\[ \displaystyle \theta_i \in C^\infty_c(V_{x_i}), \quad \sum_{i=1}^n \theta_i =1 \,\, \mbox{on}\,\, K, \]

and we conclude with (b) by (2.2):

\[ \int_{B} f {\rm div} (\psi \vec \eta) = \sum_{i=1}^n \int_B f {\rm div} (\theta_i \psi \vec \eta) = \sum_{i=1}^n \int_{V_{x_i}} f {\rm div} (\theta_i \psi \vec \eta) =0. \]

To prove the converse, assume (b) and let $\varphi \in L^1 (U)$, be such that

(2.3)\begin{equation} \int_{{-}s_0}^{s_0} \varphi(t,s)\,{\rm d}s =0, \end{equation}

and define

\[ \displaystyle \phi(t,s'):=\frac{1}{1+ s'k(t)} \int_{{-}s_0}^{s'} \varphi (t,s) \, {\rm d}s. \]

We can construct a sequence $\psi _k \in C^{0,1}_c(V_x)$ such that

\[ \psi_k \circ \Phi \to \phi \quad \mbox{in} \,\, L^1 (U) \quad \mbox {and}\quad \partial_s (\psi_k \circ \Phi) \to \partial_s \phi \quad \mbox{in} \,\, L^1 (U). \]

Therefore by (b)

\[ \int_{V_x} f {\rm div} (\psi \vec \eta) =0 \,\, \forall \psi\in C^\infty_c(V_x) \implies \int_{V_x} f {\rm div} (\psi_k \vec \eta) =0, \]

which in view of (2.1) implies

\begin{align*} \int_U \tilde f \varphi \, {\rm d}s{\rm d}t & = \int_U \tilde f (t,s) \partial_s ((1+ sk(t) ) \phi) \, {\rm d}s{\rm d}t\\ & = \lim_{k\to \infty} \int_U \tilde f (t,s) \partial_s \Bigg((1+ sk(t)) \psi_k \circ \Phi \Bigg) \, {\rm d}s{\rm d}t = 0. \end{align*}

This fact being true for all $\varphi \in L^1(U)$ satisfying (2.3) implies that

\[ \forall t\in ({-}t_0, t_0)\quad \forall s \in ({-}s_0, s_0) \quad \tilde f(t, s) = {\unicode{x2A0F}}_{{-}s_0}^{s_0} \tilde f(t,s) \, {\rm d}s.\]

This means that for $|s|< s_0$:

\[ f(x+ s\vec \eta (x)) = \tilde f(0, s) = \tilde f(0,0) = f(x). \]

Global property (a) is a direct consequence of this local property around each $x\in B$, and the fact that $\vec \eta$ itself is constant on $x+ s\vec \eta (x)\in B$ for all $s$.

Proof. (i) $\Rightarrow$ (iv): Assuming $f\in c^{0,\alpha }(\overline U)$, let $f_k\in C^1 (\overline U)$ be a sequence of functions converging to $f$ in $C^{0,\alpha }(\overline U)$ as $k\to \infty$. We estimate:

\[ [f]_{0,\alpha|r} \le [f-f_k]_{0,\alpha|r} + [f_k]_{0,\alpha|r} \le [f-f_k]_{0,\alpha} + c(U) \|\nabla f_k\|_0 r^{1-\alpha}, \]

and a straightforward argument by the auxiliary argument $k$ implies the required convergence.

(iv) $\Rightarrow$ (iii): This is straightforward since for all $r>0$:

\[ \frac{1}{r^\alpha} \omega_{f;U}(r) \le [f]_{0,\alpha;U|r}. \]

(iii) $\Rightarrow$ (ii): By the assumption $f$ is uniformly continuous in $U$. We extend $f$ to $\tilde f: {\mathbb {R}}^n\to {\mathbb {R}}$ by applying proposition B.2, obtaining that $\omega _{\tilde f;{\mathbb {R}}^n}(r) \le C \omega _{f;U} (r)$ is $o(r^\alpha )$. We conclude that

\[ [\tilde f]_{0,\alpha;{\mathbb{R}}^n|r} = \sup_{0 \le s \le r} \frac{1}{s^\alpha} \omega_{\tilde f;{\mathbb{R}}^n} (s) \]

is $o(1)$ as a function of $r$. Consider any bounded weakly Lipschitz domain $V\subset {\mathbb {R}}^n$ and note that $\tilde f|_V \in C^{0,\alpha }(\overline V)$. Applying lemma B.1 to $V$ implies that $\tilde f_\varepsilon |_V \to \tilde f|_V$ in $C^{0,\alpha }(\overline V)$ as $\varepsilon \to 0$ and thus $\tilde f|_V \in c^{0,\alpha } (\overline V)$. Both conclusions in (ii) follow.

(ii) $\Rightarrow$ (i): This statement trivially follows from definition 3.3.

Proof. The first inclusion is trivial. If $f\in C^{1,\beta } (\overline U)$, then $\omega _{f;U}(r)$ is $O(r^\beta )$ and so the second inclusion follows by proposition 3.5.

Proof. Let $\psi \in W^{1,1}_0(V)$, and for $\varepsilon < {\rm dist}( \overline V,\, \partial U)$ we define:

\[ a^{(\varepsilon)} := \int_{V} f_\varepsilon \partial_j h_\varepsilon \psi(x) {\rm d}x. \]

We first prove that $a^{(\varepsilon )}$ is a Cauchy sequence. For any $0<\varepsilon '<\varepsilon$ we have

(3.3)\begin{equation} \begin{aligned} |a^{(\varepsilon)} - a^{(\varepsilon')}| & = \Bigg| \int_{V} (f_\varepsilon \partial_j h_\varepsilon - f_{\varepsilon'} \partial_j h_{\varepsilon'} ) \psi \Bigg| \\ & \le \Bigg|\int_{V} f_\varepsilon (\partial_j h_\varepsilon - \partial_j h_{\varepsilon'} ) \psi \Bigg| + \Bigg|\int_{V} (f_\varepsilon - f_{\varepsilon'}) \partial_j h_{\varepsilon'} \psi \Bigg| \\ & = \Bigg|- \int_{V} \partial_j f_\varepsilon (h_\varepsilon \!-\! h_{\varepsilon'} ) \psi - \!\int_{V} f_\varepsilon (h_\varepsilon - h_{\varepsilon'} ) \partial_j \psi \Bigg| \!+\! \Bigg|\int_{V} (f_\varepsilon - f_{\varepsilon'}) \partial_j h_{\varepsilon'} \psi \Bigg| \\ & \le \Bigg| \int_{V} \partial_j f_\varepsilon (h_\varepsilon - h_{\varepsilon'} ) \psi \Bigg| \!+\! \Bigg|\int_{V} f_\varepsilon (h_\varepsilon - h_{\varepsilon'} ) \partial_j \psi \Bigg| \!+\! \Bigg|\int_{V} (f_\varepsilon - f_{\varepsilon'}) \partial_j h_{\varepsilon'} \psi \Bigg| \\ & = I_1 (\varepsilon, \varepsilon') + I_2 (\varepsilon, \varepsilon') + I_3(\varepsilon, \varepsilon')\\ \end{aligned} \end{equation}

We estimate, using $\varepsilon '<\varepsilon$:

\begin{align*} I_1 (\varepsilon, \varepsilon') & = \Bigg|\int_{V} \psi(x) \Bigg( \int_{{\mathbb{R}}^n} (f(x-y) \partial_j \phi_\varepsilon(y)) {\rm d}y \Bigg)\\ & \quad \Bigg(\int_{{\mathbb{R}}^n} (h(x-z) (\phi_\varepsilon(z) - \phi_{\varepsilon'}(z)) dz \Bigg) {\rm d}x \Bigg| \\ & = \Bigg|\int_{V} \psi(x) \int_{{\mathbb{R}}^n} (f(x-y) -f(x)) \partial_j \phi_\varepsilon(y) {\rm d}y\\ & \quad \int_{{\mathbb{R}}^n} (h(x-z) -h(x) ) (\phi_\varepsilon(z) - \phi_{\varepsilon'}(z)) dz \, {\rm d}x \Bigg| \\ & \le \|\psi\|_{L^1} (c[f]_{0,\alpha} \varepsilon^{\alpha-1} )(c[h]_{0,\alpha|\varepsilon} \varepsilon^{\alpha}) \le \|\psi\|_{L^1} (c[f]_{0,\alpha} \varepsilon^{\alpha-1} )([h]_{0,\alpha} o(\varepsilon^{\alpha})) \\ & \le \|\psi\|_{L^1} [h]_{0,\alpha} [f]_{0,\alpha} o(\varepsilon^{2\alpha-1}), \end{align*}

where we used lemma 3.1-(i) and proposition 3.5 in the third line. For $I_3(\varepsilon,\, \varepsilon ')$, we note that $\varepsilon \not \le \varepsilon '$ and hence the same estimate as for $I_1$ is not obtained, but through the same calculations as for $I_1$, we obtain

\[ I_3 (\varepsilon, \varepsilon') \le \|\psi\|_{L^1} [h]_{0,\alpha} [f]_{0,\alpha} {\varepsilon'}^{{-}1} o({\varepsilon'}^\alpha) \varepsilon^\alpha. \]

In the same manner, we can also obtain an estimate for $I_2$:

\begin{align*} I_2 (\varepsilon, \varepsilon') & = \Bigg|\int_{V} f_\varepsilon (h_\varepsilon - h_{\varepsilon'} ) \partial_j \psi \Bigg| \\ & \le \|\partial_j \psi\|_{L^1} \|f\|_{0} \sup_{x\in {V}} \Bigg|\int_{{\mathbb{R}}^n} h(x-y) (\phi_\varepsilon(y) - \phi_{\varepsilon'}(y)) {\rm d}y \Bigg| \\ & = \|\partial_j \psi\|_{L^1} \|f\|_{0} \sup_{x\in {V}} \Bigg|\int_{{\mathbb{R}}^n} (h(x-y) - h(x)) (\phi_\varepsilon(y) - \phi_{\varepsilon'}(y)) {\rm d}y \Bigg| \\ & \le \|\partial_j \psi\|_{L^1} \|f\|_{0} [h]_{0,\alpha} o( \varepsilon^{\alpha}). \end{align*}

Putting the three estimates together in view of (3.3), we obtain a first crude estimate:

(3.4)\begin{align} |a^{(\varepsilon)} \!-\! a^{(\varepsilon')}| \le [h]_{0,\alpha} \Bigg( \Bigg[ o(\varepsilon^{2\alpha-1}) \!+\! {\varepsilon'}^{{-}1} o({\varepsilon'}^\alpha) \varepsilon^\alpha\Bigg] [f]_{0, \alpha} \|\psi\|_{L^1} \!+\! o( \varepsilon^{\alpha}) \|f\|_{0} \|\partial_j \psi\|_{L^1} \Bigg). \end{align}

This estimate is not enough to establish the Cauchy property of the sequence $a^{(\varepsilon )}$. Therefore, we proceed as follows. If $\varepsilon '/\varepsilon \ge 1/2$, we set $\rho = \varepsilon '/\varepsilon$, $m=1$. Otherwise there exists $m \in \mathbb {N}$, depending on $\varepsilon '/\varepsilon <1$, such that $1/4 \le \rho = ({\varepsilon '}/{\varepsilon })^{1/m} <1/2$. We now write

\[ |a^{(\varepsilon)} - a^{(\varepsilon')}| \le \sum_{k=1}^m |a^{(\rho^{{k-1}}\varepsilon)} - a^{(\rho^{k}\varepsilon)}|. \]

We apply (3.4) successively to $0 <\rho ^{k}\varepsilon < \rho ^{k-1}\varepsilon$ (as the new $\varepsilon '$ and $\varepsilon$) to obtain

\begin{align*} |a^{(\varepsilon)} - a^{(\varepsilon')}| & \le [h]_{0,\alpha} \Bigg(o(\varepsilon^{2\alpha-1}) [f]_{0, \alpha} \|\psi\|_{L^1} (1+ \rho^{\alpha-1} ) \sum_{k=1}^m \rho^{{(k-1)}(2\alpha -1)} \\ & \quad+ o( \varepsilon^{\alpha}) \|f\|_{0} \|\partial_j \psi\|_{L^1} \sum_{k=1}^m \rho^{(k-1)\alpha} \Bigg) \\ & \le (\rho^{\alpha-1} + 1 ) \frac{1- \rho^{(2\alpha-1)m}}{1- \rho^{2\alpha-1}}o(\varepsilon^{2\alpha-1}) [h]_{0,\alpha} [f]_{0, \alpha} \|\psi\|_{L^1} \\ & \quad+ \frac{1- \rho^{\alpha m}}{1- \rho^{\alpha}}o( \varepsilon^{\alpha}) [h]_{0,\alpha} \|f\|_{0} \|\partial_j \psi\|_{L^1} \\ & \le o(\varepsilon^{2\alpha-1}) [h]_{0,\alpha} [f]_{0, \alpha} \|\psi\|_{L^1} + o( \varepsilon^{\alpha}) [h]_{0,\alpha} \|f\|_{0} \|\partial_j \psi\|_{L^1}, \end{align*}

since by our choice either $1/2 \le \rho <1$ and $m=1$, or $1/4\le \rho <1/2$, independent of $\varepsilon ',\, \varepsilon$. This implies that $a^{(\varepsilon )}$ is Cauchy sequence. Hence, the limit

\[ f \partial_j h [\psi]:= a = \lim_{\varepsilon \to 0} a^{(\varepsilon)}, \]

exists.

Now, we estimate for a fixed $\varepsilon >0$, as $\varepsilon '\to 0$:

\begin{align*} \Bigg|\int_{V} f_\varepsilon \partial_j h_\varepsilon \psi(x) {\rm d}x - f\partial_j h[\psi] \Bigg| & = |a^{(\varepsilon)} -a| \le \|\psi\|_{L^1} [f]_{0,\alpha} [h]_{0,\alpha} o(\varepsilon^{2\alpha -1})\\ & \quad + \|\partial_j \psi\|_{L^1} \|f\|_{0} [h]_{0,\alpha} o(\varepsilon^\alpha), \end{align*}

which establishes (3.2). To obtain (3.1), applying lemma 3.1-(ii) and proposition 3.5 to $\partial _j h_\varepsilon$ we observe that:

\begin{align*} |f\partial_j h [\psi]| & \le |a^{(\varepsilon)}| + |a^{(\varepsilon)} - a| \le |a^{(\varepsilon)}| + [f]_{0,\alpha} [h]_{0,\alpha} \|\psi\|_{L^1} o( \varepsilon^{2\alpha -1} )\\ & \quad + \|f\|_{0} [h]_{0,\alpha} \|\partial_j \psi\|_{L^1} o(\varepsilon^\alpha) \\ & \le \|f\|_0 [h]_{0,\alpha} \|\psi\|_{L^1} o(\varepsilon^{\alpha-1}) + [f]_{0,\alpha} [h]_{0,\alpha} \|\psi\|_{L^1} o( \varepsilon^{2\alpha -1}) \\ & \quad + \|f\|_{0} [h]_{0,\alpha} \|\partial_j \psi\|_{L^1} o( \varepsilon^\alpha) \\ & \le \|f\|_{0,\alpha} [h]_{0,\alpha} \Bigg(\|\psi\|_{L^1} o(\varepsilon^{\alpha-1}) + \|\partial_j \psi\|_{L^1} o(\varepsilon^{\alpha}) \Bigg) \end{align*}

In other words,

(3.5)\begin{equation} \begin{aligned} & \mbox{for any}\,\, \sigma >0, \mbox{there exists}\,\, \delta_0= \delta_0(\sigma)>0\,\, \mbox{such that} \\ & \varepsilon \le \min\{\delta_0, {\rm dist} (\overline V, \partial U)\} \implies |f\partial_j h [\psi]|\\ & \quad \le c \|f\|_{0,\alpha} [h]_{0,\alpha} \Bigg( \|\psi\|_{L^1} \varepsilon^{\alpha-1}+ \|\partial_j \psi\|_{L^1} \varepsilon^\alpha \Bigg) \sigma, \end{aligned} \end{equation}

with $\delta _0 (1) = +\infty$.

We conclude the proof of (3.1): Given, $\sigma >0$, we let

\[ \displaystyle \delta = \left(1+\frac{{\rm diam} (V)}{ {\rm dist} (\overline V, \partial U)}\right) \delta_0. \]

Assume $\|\psi \|_{L^1} \le \delta \|\partial _j \psi \|_{L^1}$ for a given nonzero $\psi$, then letting

\[ \varepsilon = \left(1+\frac{{\rm diam} (V)}{ {\rm dist} (\overline V, \partial U)}\right)^{{-}1} \frac{\|\psi\|_{L^1}}{\|\partial_j \psi\|_{L^1}}, \]

we obtain $\varepsilon < {\rm dist} (\overline V,\, \partial U)$ by the Poincaré inequality $\|\psi \|_{L^1} \le {\rm diam}(V) \|\partial _j \psi \|_{L^1}$ on $V$. On the other hand we also have $\varepsilon \le \delta _0$ and therefore applying (3.5) we obtain as required:

\[ |f\partial_j h [\psi]| \le c \left(1+\frac{ {\rm diam} (V)}{{\rm dist} (\overline V, \partial U)}\right)^{1-\alpha} \|f\|_{0,\alpha} [h]_{0,\alpha} \|\psi\|^\alpha_{L^1}{\|\partial_j \psi\|^{1-\alpha}_{L^1}} \sigma. \]

Proof. The proof of the proposition follows classical ideas relating the decay of mean oscillations to pointwise behaviour of functions, as pioneered by Morrey and Campanato, see for instance [Reference Campanato8]. We prove the statement by induction over the dimension.

First let $n=1$ and $U \subset {\mathbb {R}}$ and be an open interval and let $I\subset U$ be any open subinterval. Note that for any $\phi \in C^\infty _c (I)$, we can construct a sequence $\tilde \phi _k \in C^\infty _c(I)$ such that $\|\tilde \phi _k\|_{0}\le 1+ 2 \|\phi \|_{0}$, and $\tilde \phi _k$ converges strongly in $L^1$ to $\phi - {\unicode{x2A0F}} _I \phi$. Now fix $\varphi \in C^\infty _c(I)$ with ${\unicode{x2A0F}} _I \varphi =1$ and let

\[ \phi_k := \tilde \phi_k - \Bigg({\unicode{x2A0F}}_I \tilde \phi_k \Bigg) \varphi \xrightarrow{L^1} \phi - {\unicode{x2A0F}}_I \phi. \]

By the dominated convergence theorem we have

\[ \lim_{k\to \infty} \int_I f\phi_k = \int_I f (\phi -{\unicode{x2A0F}}_I \phi). \]

Let $\psi _k (t) = \int _{-\infty }^t \phi _k(s) {\rm d}s$ and note that $\psi _k \in C^\infty _c(I)$. We have therefore for all $\phi \in C^\infty _c(I)$ and $\sigma >0$, provided $|I|\le \delta (\sigma )$:

\begin{align*} \Bigg|\int_I (f - {\unicode{x2A0F}}_I f) \phi \Bigg| & = \Bigg| \int_I f (\phi -{\unicode{x2A0F}}_I \phi) \Bigg| = \lim_{k\to \infty} \Bigg| \int_I f\phi_k \Bigg| \!=\! \lim_{k\to \infty} \Bigg| \int_I f\psi'_k\Bigg| = \lim_{k\to \infty} \Bigg|T [\psi_k]\Bigg| \\ & \le\liminf_{k\to \infty} \|\psi_k\|_{L^1}^{\alpha} \|\psi'_k\|^{1-\alpha} _{L^1} \sigma \le \liminf_{k\to \infty} \|\psi'_k\|_{L^1} |I|^\alpha \sigma \\ & =\liminf_{k\to \infty} \|\phi_k - {\unicode{x2A0F}}_I \phi_k\|_{L^1} |I|^\alpha \sigma \!=\! \|\phi - {\unicode{x2A0F}}_I \phi\|_{L^1} |I|^\alpha\sigma\le 2\|\phi\|_{L^1} |I|^\alpha \sigma, \end{align*}

where we used the Poincaré inequality on $I$. Since $(L^1)' = L^\infty$, and $C^\infty _c(I)$ is dense in $L^1(I)$, we conclude that

(3.6)\begin{equation} \forall \sigma>0 \quad |I| \le \delta(\sigma) \implies \|f- {\unicode{x2A0F}}_I f\|_{L^\infty(I)} \le 2 |I|^\alpha \sigma. \end{equation}

Now, let $y\in U$, and let $d= {\rm dist}(y,\, \partial U)$, and for $z\in (y-d/2,\, y+d/2)$ and $r< d/2$ define

\[ h_r(z):= {\unicode{x2A0F}}_{z-r}^{z+r} f(x) {\rm d}x. \]

$h_r$ is continuous in $z$ and converges a.e. to $f$ on $(y-d/2,\, y+d/2)$ as $r\to 0$. On the other hand, for all $r'< r< d/2$:

\begin{align*} & \displaystyle \Bigg| h_r (z) - h_{r'}(z)\Bigg| \le {\unicode{x2A0F}}_{(z-r, z+r) \times (z-r', z+r') } |f(x) - f(x')| {\rm d}x {\rm d}x'\\ & \quad \le 4|(z-r, z+r)|^\alpha = 2^{\alpha+2} r^\alpha, \end{align*}

where we applied bound (3.6) to $I=(z-r,\, z+r)$, $\sigma =1$, $\delta (1)=+\infty$. As a consequence, $h_r$ is a Cauchy sequence in the uniform norm and $f$ is continuous as the uniform limit of the $h_r$ for $r \to 0$. Now, applying once again (3.6) to $I= (x,\,y) \subset U$ we obtain that

\[ \forall \sigma>0 \quad \forall x,y \in U\quad |x-y| \le \delta(\sigma) \implies |f(x) - f(y)| \le 4 |x-y|^\alpha \sigma, \]

which implies $[f]_{0,\alpha ;U|\delta (\sigma )} \le 4 \sigma$, as required.

Now, assume that $n>1$ and that the statement is true for $n-1$. Let $Q\subset U$ be any coordinate rectangular box, i.e.

\[ Q:=\displaystyle \prod_{k=1}^n I_k \subset U, \]

with the open intervals $I_k\subset {\mathbb {R}}$. Let $\widehat {x}:= (x_1,\, \cdots,\, x_{n-1})$ and note that for all $h\in L^1(Q)$,

\[ {\rm for\ \,\,\ a.e.} \,\, \widehat{x} \in Q^{n-1} := \prod_{j=1}^{n-1} I_k , \,\, \hat h(\widehat{x}) := {\unicode{x2A0F}}_{I_n} h(x) {\rm d}x_n \]

is well defined and belongs to $L^1(Q)$. We claim that:

(3.7)\begin{equation} \forall \sigma>0\quad |I_n| \le \delta(\sigma) \implies \mathop{\textrm{ess sup}}\limits_{x\in Q} \Bigg| f(x) - \hat f(\widehat{x}) \Bigg| \le 2 |I_n| ^\alpha \sigma. \end{equation}

Let us first show that the conclusion holds assuming claim (3.7) is true for all coordinate boxes $Q\subset U$. For $a,\,b$ the extremities of $I_n$, choose $0 \le \theta _k \le 1$ in $C^\infty _c (I_n)$ such that $\theta _k \equiv 1$ on $(a+1/k,\, b-1/k)$. Given $\sigma >0$, for $1 \le j \le n-1$, and $\psi \in C^\infty _c (Q^{n-1})$, assuming that $\|\psi \|_{L^1(Q^{n-1})} \le \delta (\sigma ) \|\partial _j \psi \|_{L^1(Q^{n-1})}$, we have by our main assumption on $f$:

\begin{align*} \Bigg| \int_{Q^{n-1}} \hat f \partial_j \psi \Bigg| & = \frac{1}{|I_n|} \Bigg| \int_Q f(x) \partial_j \psi(\widehat x) {\rm d}x \Bigg| = \lim_{k \to \infty} \frac{1}{|I_n|} \Bigg| \int_Q f(x) \theta_k (x_n)\partial_j \psi(\widehat{x}) {\rm d}x \Bigg| \\ & = \lim_{k \to \infty} \frac{1}{|I_n|} \Bigg| \int_Q f \partial_j (\theta_k \psi) \Bigg|\le \lim_{k \to \infty} \frac{1}{|I_n|} \|\theta_k \psi\|^\alpha_{L^1(Q)} \|\partial_j(\theta_k \psi)\|^{1-\alpha}_{L^1(Q)} \sigma \\ & = \lim_{k \to \infty} \frac{1}{|I_n|} \|\theta_k \psi\|^\alpha_{L^1(Q)} \| \theta_k \partial_j\psi \|^{1-\alpha}_{L^1(Q)} \sigma = \frac{1}{|I_n|} \|\psi\|^\alpha_{L^1(Q)} \|\partial_j\psi\|^{1-\alpha}_{L^1(Q)} \sigma \\ & = \|\psi\|^\alpha_{L^1(Q^{n-1})} \|\partial_j \psi\|^{1-\alpha}_{L^1(Q^{n-1})} \sigma, \end{align*}

since $\theta _k \psi \in C^\infty _c(Q)$ and

\begin{align*} \|\theta_k \psi\|_{L^1(Q)} & =\Bigg(\int_a^b \theta_k \Bigg) \|\psi\|_{L^1(Q^{n-1})} \le \delta(\sigma) \Bigg(\int_a^b \theta_k \Bigg) \|\partial_j \psi\|_{L^1(Q^{n-1})}\\ & = \delta(\sigma) \|\partial_j (\theta_k \psi)\|_{L^1(Q)}. \end{align*}

Applying the induction assumption, we deduce that

(3.8)\begin{equation} [\hat f]_{0,\alpha; Q^{n-1}|\delta(\sigma)} \le 2^{n}. \end{equation}

To prove that $f$ is continuous, fix $y\in U$ and consider a box $Q_d = Q^{n-1} \times (y_n-d,\, y_n+d) \subset U$ containing $y$. For all $z\in Q^{n-1} \times (y_n -d/2,\, y_n+ d/2)$, and $r< d/2$ we define

\[ h_r(z):= {\unicode{x2A0F}}_{z_n - r}^{z_n+ r} f(\widehat{z}, s) {\rm d}s \]

Note that, if $r$ is fixed, the vertical averages of $f$ are continuous in $\widehat {z}$ as established in (3.8), and so $h_r$ is continuous in $z$. Applying (3.7) with $\sigma =1$ to $Q=Q^{n-1} \times (z_n -r,\, z_n+ r)$ we have for $r'< r< d/2$:

\begin{align*} & \displaystyle \Bigg| h_r (z) - h_{r'}(z)\Bigg| \le {\unicode{x2A0F}}_{(z-r, z+r) \times (z-r', z+r') } |f(\widehat{z},s) - f(\widehat{z},s')| {\rm d}s {\rm d}s'\\ & \quad \le 4|(z-r, z+r)|^\alpha = 2^{\alpha+2} r^\alpha, \end{align*}

which implies that $h_r$ locally uniformly converge to their limit, which happens to be $f$. Hence $f$ is continuous in $U$.

To obtain a Hölder estimate, let $x,\,y\in U$. First, assume $x_j=y_j$ for some $1 \le j\le n$. We re-arrange the coordinates so that $j=n$ and note that for any sequence of coordinate rectangular boxes $Q_k$ of height $1/k$, containing $x$ and $y$, such that $x_n=y_n$ is the midpoint of $I_{n,k}$, we have by (3.8)

\begin{align*} \forall \sigma>0\quad |x-y| & \le \delta(\sigma) \implies |f(x) - f(y)| = \lim_{k\to \infty} |\hat f_{Q_k}(\widehat{x})- \hat f_{Q_k}(\widehat{y})|\\ & \le 2^{n} |\widehat{x} - \widehat{y}|^\alpha\sigma \le 2^n |x-y|^\alpha \sigma. \end{align*}

If, on the other hand, $x_j\neq y_j$ for all $j$, we can choose a coordinate rectangular box $Q\subset U$ which has $x$ and $y$ as its opposite vertices on the largest diameter. We obtain by applying (3.7) and (3.8) to the now continuous $f$, provided $|x-y| \le \delta (\sigma )$:

\begin{align*} |f(x) - f (y) | & \le |f(x) - \hat f (\widehat{x})| + | \hat f (\widehat{x}) - \hat f (\widehat{y}) | + |f(y) - \hat f (\widehat{y}) | \\ & \le 4 |x_n-y_n|^\alpha \sigma + 2^{n} |\widehat{x} -\widehat{y}|^\alpha\sigma \le (4+ 2^n) |x-y|^\alpha \sigma \\ & \le 2^{n+1} |x-y|^\alpha \sigma. \end{align*}

In both cases we can conclude with the desired bound, i.e. $[f]_{0,\alpha ; U |\delta (\sigma )} \le 2^{n+1}\sigma$. Note that the sequence $Q_k$ in the first case and the box $Q$ in the second case exist since $U$ is assumed to be a coordinate box itself.

It remains to prove claim (3.7). Given $\sigma >0$ assume that $|I_n|\le \delta (\sigma )$ as required in (3.7). For any $\phi \in C^\infty _c (Q)$, consider a sequence $\tilde \phi _k \in C^\infty _c(Q)$, $\|\tilde \phi _k\|_0 \le 1 + 2\|\phi \|_0$, converging strongly in $L^1$ to $\phi - \hat \phi (\widehat {x})$. As a consequence,

\[ \lim_{k\to \infty} \int_{I_n} \tilde \phi_k(x) {\rm d}x_n =0 \quad \mbox{in}\,\, L^1(Q^{n-1}). \]

Choose a $\varphi \in C^\infty _c(I_n)$ such that ${\unicode{x2A0F}} _{I_n} \varphi =1$ and define

\[ \phi_k (x) = \tilde \phi_k(x) - \Bigg({\unicode{x2A0F}}_{I_n} \tilde \phi_k (\widehat{x}, s ) {\rm d}s \Bigg)\varphi (x_n). \]

Then $\phi _k \in C^\infty _c(Q)$ converges in $L^1$ to $\phi - \hat \phi (\widehat {x})$ and $\hat \phi _k (\widehat {x}) = {\unicode{x2A0F}} _{I_n} \phi _k(x) {\rm d}x_n =0$. Letting $\psi _k (\widehat {x},\, t) = \int _{-\infty }^t \phi _k(\widehat {x},\, s) {\rm d}s$ we obtain for all $\phi \in C^\infty _c (Q)$, $\psi _k \in C^\infty _c(Q)$. We hence obtain

\begin{align*} \Bigg|\int_Q (f - \hat f (\widehat{x})) \phi \Bigg| & = \Bigg| \int_Q f (\phi - \hat \phi (\widehat{x}))\Bigg| =\lim_{k\to \infty} \Bigg| \int_Q f\phi_k \Bigg| = \lim_{k\to \infty} \Bigg| \int_Q f \partial_n \psi_k \Bigg|\\ & \quad = \lim_{k\to \infty} \Bigg| T_n [\psi_k] \Bigg| \\ & \le \liminf_{k\to \infty} \|\psi_k\|^\alpha _{L^1} \|\partial_n \psi_k\|^{1-\alpha}_{L^1}\sigma\le \liminf_{k\to \infty} |I_n|^\alpha \|\partial_n \psi_k\|^\alpha _{L^1} \|\partial_n \psi_k\|^{1-\alpha}_{L^1} \sigma \\ & =\lim_{k\to \infty} |I_n|^\alpha \|\phi_k - {\unicode{x2A0F}}_{I_n} \phi_k(x) {\rm d}x_n \|_{L^1} \sigma = |I_n|^\alpha \|\phi \\ & \quad - {\unicode{x2A0F}}_{I_n} \phi (x) {\rm d}x_n\|_{L^1(Q)}\sigma \\ & \le 2 |I_n|^\alpha \|\phi\|_{L^1(Q)}\sigma, \end{align*}

where we used the Poincaré inequality

\[ \|\psi_k\|_{L^1(Q)} \le |I_n| \|\partial _n \psi_k \|_{L^1(Q)} \le \delta(\sigma) \|\partial_n \psi_k \|_{L^1(Q)} \]

on $I_n$. Once again, using the fact that $L^\infty$ is the dual of $L^1$, and the density of $C^\infty _c(Q)$ in $L^1(Q)$, we conclude with (3.7).

Proof. Applying proposition 3.8 to the components $f:= \partial _i u^m$, $h:= \vec n^m$, for $m\in \{1,\,2,\,3\}$, we deduce that for a distribution $A_{ij} \in \mathcal {D'}(V)$,

\[ B^{\varepsilon}_{ij} :={-} \partial_i u_\varepsilon \cdot \partial_j \vec n_\varepsilon \to A_{ij} :={-} \partial_i u \cdot \partial_j \vec n \quad \mbox{in}\,\, \mathcal{D'}(V), \]

and

\[ |B^{\varepsilon}_{ij}[\psi] - A_{ij} [\psi]| \le C \Bigg( o(\varepsilon^{2\alpha-1}) \|\psi\|_{L^1} + o(\varepsilon^\alpha) \|\partial_j \psi\|_{L^1} \Bigg). \]

On the other hand, we have for all $\psi \in W^{1,1}_0(V)$:

\begin{align*} |A^{\varepsilon}_{ij} [\psi] - A_{ij} [\psi]| & \le |B^{\varepsilon}_{ij}[\psi] - A_{ij} [\psi]| + \Bigg|\int_B \partial_i u_\varepsilon(x) \cdot (\partial_j {\vec n}_\varepsilon (x) - \partial_j N^{\varepsilon} (x))\psi(x) {\rm d}x \Bigg| \\ & \le C \Bigg( o(\varepsilon^{2\alpha-1}) \|\psi\|_{L^1} + o(\varepsilon^\alpha) \|\partial_j \psi\|_{L^1} \Bigg) \\ & \quad+ \Bigg|\int_V \partial_{ij} u_\varepsilon \cdot ({\vec n}_\varepsilon - N^{\varepsilon})\psi {\rm d}x \Bigg| + \Bigg|\int_V \partial_i u_\varepsilon \cdot ({\vec n}_\varepsilon - N^{\varepsilon}) \partial_j \psi \Bigg| \\ & \le C \Bigg( o(\varepsilon^{2\alpha-1}) \|\psi\|_{L^1} + o(\varepsilon^\alpha) \|\partial_j \psi\|_{L^1} \Bigg), \end{align*}

where we used proposition 3.8, (4.1) and (4.2). The stated fine $\alpha$-interpolation property follows as in proposition 3.8.

Proof. We recall that the Christoffel symbols associated with a metric $g$ are given by

\[ \Gamma^i_{jk} (g) = \frac 12 g^{im} (\partial_k g_{jm} + \partial_j g_{km} - \partial_m g_{jk}), \]

where the Einstein summation convention is used. Hence, in view of (4.5), we have

\[ \Gamma^{i,\varepsilon}_{kj} := \Gamma^i_{jk} (\mathscr{G}^{\varepsilon}) \to 0 \]

uniformly, if $\alpha \ge 1/2$, with the estimate

(4.7)\begin{equation} \|\Gamma^{i,\varepsilon}_{kj}\|_{0;V}\le C o(\varepsilon^{2\alpha-1}). \end{equation}

Writing the Codazzi–Mainardi equations [Reference Han and Hong27, Equation (2.1.6)] for the immersion $u_\varepsilon$ we have

\begin{align*} \partial_2 A^{\varepsilon}_{11} -\partial_1 A^{\varepsilon}_{12}& = A^{\varepsilon}_{11} \Gamma_{12}^{1, \varepsilon} +A^{\varepsilon}_{12} (\Gamma_{12}^{2,\varepsilon}-\Gamma_{11}^{1,\varepsilon}) -A^{\varepsilon}_{22} \Gamma_{11}^{2,\varepsilon},\\ \partial_2 A^{\varepsilon}_{12} -\partial_1 A^{\varepsilon}_{22}& = A^{\varepsilon}_{11}\Gamma_{22}^{1,\varepsilon} +A^{\varepsilon}_{12} (\Gamma_{22}^{2,\varepsilon}-\Gamma_{12}^{1,\varepsilon})-A^{\varepsilon}_{22} \Gamma_{12}^{2,\varepsilon}. \end{align*}

Hence, by (4.4) and (4.7),

\[ \|\partial_2 A^{\varepsilon}_{i1} - \partial_1 A^{\varepsilon}_{i2}\|_{0;V} \le C o(\varepsilon^{\alpha-1}) o(\varepsilon^{2\alpha-1}), \]

which completes the proof.

Proof. In view of the formula for the $(0,\,4)$-Riemann curvature tensor [Reference Han and Hong27, Equation (2.1.2)]

\[ R_{iljk} = g_{lm} (\partial_k \Gamma^m_{ij} -\partial_j \Gamma^m_{ik} + \Gamma^m_{ks} \Gamma^s_{ij} - \Gamma^m_{js} \Gamma^s_{ik}), \]

and by the Gauss equation [Reference Han and Hong27, Equation (2.1.7)], (4.5) and (4.7) we have on $V$:

\begin{align*} \det A^\varepsilon & = R_{1212}(\mathscr{G}^\varepsilon) = \mathscr{G}^\varepsilon_{1m}(\partial_1 \Gamma^{m,\varepsilon}_{22} - \partial_2 \Gamma^{m,\varepsilon}_{21} + \Gamma^{m,\varepsilon}_{1s} \Gamma^{s,\varepsilon}_{22} - \Gamma^{m,\varepsilon}_{2s} \Gamma^{s,\varepsilon}_{21})\\ & =\partial_1 (\mathscr{G}^\varepsilon_{1m} \Gamma^{m,\varepsilon}_{22} ) - \partial_2 ( \mathscr{G}^\varepsilon_{1m} \Gamma^{m,\varepsilon}_{21}) + o(\varepsilon^{2\alpha-1}) \\ & = 2 \partial_{12} \mathscr{G}^\varepsilon_{12} - \partial_{11} \mathscr{G}^\varepsilon_{22} - \partial_{22} \mathscr{G}^\varepsilon_{11} + o(\varepsilon^{2\alpha-1}) \\ & ={-} {\rm curl}^T {\rm curl}\,\, \mathscr{G}^\varepsilon + o(\varepsilon^{2\alpha-1}). \end{align*}

The conclusion follows by an integration by parts involving the first term and applying (4.5).

Proof. Let $A^{\varepsilon }$ be as defined in (4.3) and let $F^\varepsilon$ be the solution to the Neumann problem

\[ \left\{ \begin{array}{@{}ll} \Delta F^\varepsilon= {\rm div}\, A^{\varepsilon} & \mbox{in}\quad V \\ \partial_\nu F^\varepsilon =A^\varepsilon \cdot \nu & \mbox{on} \quad \partial V \end{array} \right. \]

Since ${\rm div} (\nabla F^\varepsilon - A^{\varepsilon }) =0$, there exists a vector field $E^\varepsilon$ such that $A^\varepsilon = \nabla F^\varepsilon + \nabla ^\perp E^\varepsilon$. $E^\varepsilon$ solves

\[ \left\{\begin{array}{@{}ll} \Delta E^\varepsilon= {\rm curl} (A^\varepsilon - \nabla F^\varepsilon) = {\rm curl}\, A^\varepsilon & \mbox{in}\quad V \\ E^\varepsilon = const. & \mbox{on} \quad \partial V \end{array} \right. \]

Since $A^\varepsilon$ is symmetric, we have $\partial _2 (F_1^\varepsilon + E_2^\varepsilon ) = \partial _1 (F_2^\varepsilon - E_1^\varepsilon )$, and thus we derive the existence of a scalar function $v^{(\varepsilon )} \in C^\infty (V)$ satisfying

\[ \partial_1 v^{(\varepsilon)} = F_1^\varepsilon+ E_2^\varepsilon, \quad \partial_2 v^{(\varepsilon)} = F_2^\varepsilon - E_1^\varepsilon, \]

i.e. $\nabla v^{(\varepsilon )} = F^\varepsilon + (E_1^\varepsilon,\, -E_2^\varepsilon )$. As a consequence

\[ A^\varepsilon = \nabla ^2 v^{(\varepsilon)} - \nabla (E^\varepsilon)^\perp{+} \nabla^\perp E^\varepsilon. \]

Standard elliptic estimates imply that for any $p<\infty$

(4.8)\begin{equation} \|\nabla E^\varepsilon\|_{W^{1,p}(V)} \le c(p,V) \|{\rm curl}\, A^\varepsilon\|_{0;V}. \end{equation}

As a consequence, fixing $p>2$, and in view of (4.6), $\|A^\varepsilon - \nabla ^2 v^{(\varepsilon )}\|_{0;V} \le C o(\varepsilon ^{3\alpha -2}) \to 0$ and therefore $\nabla ^2 v^{(\varepsilon )}$ converges to $A$ in the sense of distributions.

Now, for all $\varphi \in C^\infty _c (V)$, there exists a vector field $\Psi$ in $V$, vanishing on $\partial V$, such that ${\rm div} \, \Psi = \varphi - {\unicode{x2A0F}} _V \varphi$, for which $\|\Psi \|_{W^{1,2}(V)} \le c(V) \|\varphi \|_{L^2(V)}$ (see e.g. [Reference Arnold, Scott and Vogelius1]). By adjusting the $\nabla v^{(\varepsilon )}$ so that it is of average 0 over $V$, and in view of proposition 4.1, we obtain that for all $\varepsilon,\, \varepsilon ' \le \delta$:

\begin{align*} \displaystyle \Bigg| \int_V (\nabla v^{(\varepsilon)} - \nabla v^{(\varepsilon')}) \varphi \Bigg| & = \displaystyle \Bigg| \int_V (\nabla v^{(\varepsilon)} - \nabla v^{(\varepsilon')}) (\varphi - {\unicode{x2A0F}}_V \varphi) \Bigg|\\ & = \Bigg| \int_V (\nabla v^{(\varepsilon)} - \nabla v^{(\varepsilon')}) {\rm div} \, \Psi \Bigg| \\ & = \Bigg| \int_V (\nabla^2 v^{(\varepsilon)} - \nabla^2 v^{(\varepsilon')}) \Psi \Bigg| \\ & \le \Bigg| \int_V (A^\varepsilon -A^{\varepsilon'}) \Psi \Bigg| + \Bigg| \int_V (\nabla^2 v^{(\varepsilon)} - A^{\varepsilon}) \Psi \Bigg|\\ & + \Bigg| \int_V (\nabla^2 v^{(\varepsilon')} -A^{\varepsilon'}) \Psi \Bigg| \\ & \le C \|\Psi\|_{W^{1,1}} o(\delta^{3\alpha-2}) \le C \|\varphi\|_{L^2(V)} o(\delta^{3\alpha-2}) . \end{align*}

By duality, we conclude that $\nabla v^{(\varepsilon )}$ converges strongly in $L^2(V)$ to a vector field $F$ and that $\nabla F = A$. It is now straightforward to see that, if necessary by adjusting the $v^{(\varepsilon )}$ by constants, $v^{(\varepsilon )}$ converges strongly in $W^{1,2}$ to some $v\in W^{1,2}(V)$ and that $A= \nabla ^2 v$.

To complete the proof, it is only necessary to show that (a) ${\mathcal {D}et}\, D^2 v =0$ and (b) $v\in c^{1,\alpha }(V)$.

In order to show (a), we first observe that since $\nabla v^{(\varepsilon )} \to \nabla v$ in $L^2(V)$, for all $\psi \in C^\infty _c(V)$ we obtain:

\begin{align*} \displaystyle -\frac 12{\rm curl}^T {\rm curl} (\nabla v \otimes \nabla v) [\psi] & = \lim_{\varepsilon\to 0} -\frac 12 {\rm curl}^T {\rm curl} (\nabla v^{(\varepsilon)} \otimes \nabla v^{(\varepsilon)}) [\psi]\\ & = \lim_{\varepsilon \to 0} {\rm det}\nabla^2 v^{(\varepsilon)} [\psi] \\ & = \lim_{\varepsilon \to 0} \int_V {\rm det} (A^\varepsilon + \nabla (E^\varepsilon)^\perp{-} \nabla^\perp E^\varepsilon ) \psi \\ & =\lim_{\varepsilon \to 0} \int _V (\det A^\varepsilon) \psi\\ & \quad + \lim_{\varepsilon \to 0} \sum_{\mbox{some indices } i,j,k,l} \int_{V} A_{ij}^\varepsilon (\partial_k E_l^\varepsilon)\psi \\ & \quad+ \lim_{\varepsilon \to 0} \int_V {\rm det} (\nabla (E^\varepsilon)^\perp{-} \nabla^\perp E^\varepsilon) \psi . \end{align*}

The third term obviously converges to $0$, and so does the first term by proposition 4.5. For the second term, we have, using propositions 4.1, (4.6) and (4.8)

\begin{align*} \Bigg| \int_V A_{ij}^\varepsilon (\partial_k E_l^\varepsilon) \psi \Bigg| & \le \Bigg| (A_{ij}^\varepsilon - A_{ij}) [(\partial_k E_l^\varepsilon) \psi] \Bigg| + \Bigg| A [(\partial_k E_l^\varepsilon) \psi] \Bigg| \\ & \le C \|(\partial_k E_l^\varepsilon) \psi\|_{W^{1,1}}o(\varepsilon ^{2\alpha-1}) + C \|(\partial_k E_l^\varepsilon) \psi\|_{W^{1,1}} \\ & \le C \|\psi\|_1 o(\varepsilon^{2\alpha-1}) o(\varepsilon^{3\alpha-2}) + C \|\psi \|_1 o(\varepsilon^{3\alpha-2}) \xrightarrow{\varepsilon\to 0}0. \end{align*}

This completes the proof of (a).

It remains to prove (b) $v\in c^{1,\alpha }(V)$. Since $\nabla ^2 v =A$ is symmetric in $i,\,j$, we have $\partial _{ji} v = \partial _{ij}v = A_{ij}$, as distributions. Proposition 4.1 implies that, for each $i$, $\nabla (\partial _i v)$ satisfies the fine $\alpha$-interpolation inequality on $W^{1,1}_0(V)$. For any $x\in V$, we apply proposition 3.11 to a coordinate rectangular box containing $x$ and compactly included in $V$ to conclude that $\partial _i v \in c^{0,\alpha } (V)$ for $i=1,\,2$, which yields $v\in c^{1,\alpha }(V)$.