Definition 2.1. An admissible pair of anisotropies $(\sigma, \mu )$ is comprised of an admissible surface tension $\sigma \;:\; {\mathbb{R}}^{d} \to{\mathbb{R}}_{\geq 0}$ that satisfies

1. (S1) positive $1$ -homogeneity, i.e., $\sigma (\lambda p)=\lambda \sigma (p)$ for all $p\in{\mathbb{R}}^{d}$ , $\lambda \gt 0$ ,

2. (S2) positive definiteness, i.e., $\sigma (p)=0$ if and only if $p=0$ ,

3. (S3) smoothness, i.e., $\sigma \in C^{2}({\mathbb{R}}^{d}\setminus \{0\})$ ,

4. (S4) uniform convexity, i.e., $\sigma ^{2}\;:\;{\mathbb{R}}^{d}\to{\mathbb{R}}_{\geq 0}$ is strongly convex in the sense that there exists a constant $c\gt 0$ satisfying

   \begin{equation*} \sigma ^{2}((1-t)p+tq)\leq (1-t)\sigma ^{2}(p)+t\sigma ^{2}(q)-c\frac {t(1-t)}{2}|p-q|^{2} \end{equation*}
   for all $p, q \in{\mathbb{R}}^{d}$ , $t \in (0,1)$ ,

and an admissible mobility $\mu \;:\;{\mathbb{R}}^{d} \to{\mathbb{R}}_{\geq 0}$ that satisfies

1. (M1) positive $1$ -homogeneity, i.e., $\mu (\lambda p)=\lambda \mu (p)$ for all $p\in{\mathbb{R}}^{d}$ , $\lambda \gt 0$ ,

2. (M2) positive definiteness, i.e., $\mu (p)=0$ if and only if $p=0$ ,

3. (M3) regularity, i.e., $\mu \in C^{0,1}({\mathbb{R}}^{d})$ .

The polar norm of $\sigma$ is

(2.1) \begin{equation} \sigma{^{\circ }}\;:\;{\mathbb{R}}^{d}\to{\mathbb{R}}_{\geq 0}, \qquad \sigma{^{\circ }}(q)=\sup _{p:\, \sigma (p)\leq 1}p\cdot q. \end{equation}

Proof. (ii) can be shown directly from the definition of $\sigma{^{\circ }}$ and the positive $1$ -homogeneity of $\sigma$ .

(iii) and (iv) follow from the positive $1$ -homogeneity and continuity resp. smoothness of $\sigma$ .

(v) is a result of (iv) and the fundamental theorem of calculus, cf. [Reference Giga30, Section 1.7.2]:

\begin{equation*} \sigma (p)=\int _{0}^{1}\frac {d}{d\lambda }\sigma (\lambda p)\,d\lambda =\int _{0}^{1}D\sigma (\lambda p)\cdot p \,d\lambda =D\sigma (p)\cdot p. \end{equation*}

Giga [Reference Giga30, Section 1.7.2] also provides the proofs for (vi), (vii), and (i) except the uniform convexity of $\sigma{^{\circ }}$ and the duality statement $\sigma ^{\circ \circ }=\sigma$ . A proof for $\sigma ^{\circ \circ }=\sigma$ can be found in [Reference Rockafellar50, Theorem 15.1 and Corollary 15.1.1]. The uniform convexity of $\sigma ^{\circ }$ can be proved in a similar way to [Reference Lindenstrauss and Tzafriri45, Proposition 1.e.2].

Proof. Variants of (i) and (ii) were provided by Dziuk [Reference Dziuk20, Proposition 2.2] and Laux [Reference Laux39, Lemma 3.3 and Section 4.2], respectively. The proof is included here for the reader’s convenience.

For inequality (i), we consider two cases with respect to $p^{\prime }$ : First, if $D\sigma (p^{\prime })\cdot p\lt \frac{\sigma (p)}{2}$ or $p^{\prime }=0$ , we use the estimate $|p-p^{\prime }|^{2}\leq 4$ to obtain

(2.5) \begin{align} \sigma (p)-|p^{\prime }|\psi (|p^{\prime }|)D\sigma (p^{\prime })\cdot p&\geq \sigma (p)\left (1-\frac{1}{2}|p^{\prime }|\psi (|p^{\prime }|)\right )\nonumber \\[5pt] &\geq \frac{1}{2}\sigma (p)\nonumber \\[5pt] &\geq \frac{1}{8}(\!\min _{S^{d-1}}\sigma )|p-p^{\prime }|^{2}. \end{align}

Second, if $p^{\prime }\neq 0$ and $D\sigma (p^{\prime })\cdot p\geq \frac{\sigma (p)}{2}$ , we have $\left |p+\frac{p^{\prime }}{|p^{\prime }|}\right |\geq \frac{\min _{S^{d-1}}\sigma }{\max _{S^{d-1}}|D\sigma |}\gt 0$ since

\begin{align*} 0\leq D\sigma (p^{\prime })\cdot p & \leq D\sigma (p^{\prime })\cdot \left (-\frac{p^{\prime }}{|p^{\prime }|}\right )+ (\!\max _{S^{d-1}}|D\sigma |)\left |p+\frac{p^{\prime }}{|p^{\prime }|}\right | \\[5pt] & \leq - (\!\min _{S^{d-1}}\sigma ) + (\!\max _{S^{d-1}}|D\sigma |)\left |p+\frac{p^{\prime }}{|p^{\prime }|}\right |. \end{align*}

The uniform convexity assumption (S4) can equivalently be stated as follows (see [Reference Giga30, Remark 1.7.5]): There exists a constant $\underline{\sigma }\gt 0$ such that

(2.6) \begin{equation} D^{2}\sigma (p^{*})\geq \frac{\underline{\sigma }}{|p^{*}|} \qquad \text{on } \{p^{*}\}^{\perp } \end{equation}

for all $p^{*}\neq 0$ . We use a second-order Taylor expansion around $\frac{p^{\prime }}{|p^{\prime }|}$ and write the remainder in terms of an intermediate point $p^{*}=(1-t)p+t\frac{p^{\prime }}{|p^{\prime }|}$ with $t \in [0,1]$ . Together with the convexity property (2.6), it follows that

(2.7) \begin{align*} \sigma (p)-D\sigma (p^{\prime })\cdot p &= \sigma (p)- \sigma\!\left(\frac{p^{\prime }}{|p^{\prime }|}\right)-D\sigma\!\left (\frac{p^{\prime }}{|p^{\prime }|}\right )\cdot \left (p-\frac{p^{\prime }}{|p^{\prime }|}\right )\nonumber \\[5pt] &= \frac{1}{2}\left (p-\frac{p^{\prime }}{|p^{\prime }|}\right )\cdot D^{2}\sigma (p^{*})\left (p-\frac{p^{\prime }}{|p^{\prime }|}\right )\nonumber \\[5pt] &\geq \frac{\underline{\sigma }}{2|p^{*}|}{\left |\left (p-\frac{p^{\prime }}{|p^{\prime }|}\right ) - \left (\left (p-\frac{p^{\prime }}{|p^{\prime }|}\right )\cdot \frac{p^{*}}{|p^{*}|}\right )\frac{p^{*}}{|p^{*}|}\right |}^{2} \\&\geq \frac{\underline{\sigma }}{8}\left |p+\frac{p^{\prime }}{|p^{\prime }|}\right |^{2}\left |p-\frac{p^{\prime }}{|p^{\prime }|}\right |^{2}\nonumber \\[5pt] &\geq \frac{\underline{\sigma }}{8}\frac{(\!\min _{S^{d-1}}\sigma )^{2}}{(\!\max _{S^{d-1}}|D\sigma |)^{2}}\left |p-\frac{p^{\prime }}{|p^{\prime }|}\right |^{2}.\end{align*}

Furthermore, the assumption $D\sigma (p^{\prime })\cdot p \geq \frac{\sigma (p)}{2}$ allows us to compute

(2.8) \begin{equation} \left (1-|p^{\prime }|\psi (|p^{\prime }|)\right )D\sigma (p^{\prime })\cdot p \geq \left (1-|p^{\prime }|\right )\frac{\sigma (p)}{2}\geq \frac{\min _{S^{d-1}}\sigma }{2}{\left |\frac{p^{\prime }}{|p^{\prime }|}-p^{\prime }\right |}^{2}. \end{equation}

Finally, a combination of (2.7) and (2.8) together with an application of Young’s inequality yields

(2.9) \begin{align} \sigma (p)-|p^{\prime }|\psi (|p^{\prime }|)D\sigma (p^{\prime })\cdot p &= \sigma (p)-D\sigma (p^{\prime })\cdot p + \left (1-|p^{\prime }|\psi (|p^{\prime }|)\right )D\sigma (p^{\prime })\cdot p\nonumber \\[5pt] &\geq 2c_{\sigma }\left (\left |p-\frac{p^{\prime }}{|p^{\prime }|}\right |^{2}+ \left |\frac{p^{\prime }}{|p^{\prime }|}-p^{\prime }\right |^{2}\right )\nonumber \\[5pt] &\geq c_{\sigma }\left |p-p^{\prime }\right |^{2}, \end{align}

which completes the proof of (i) in the second case.

For inequality (ii), we distinguish two cases again. First, in the easier case $\left |p-p^{\prime }\right |\geq 1$ , we can estimate

(2.10) \begin{equation} \sigma (p)-|p^{\prime }|\psi (|p^{\prime }|)D\sigma (p^{\prime })\cdot p \leq \max _{S^{d-1}}\sigma +\max _{S^{d-1}}|D\sigma | \leq \left (\!\max _{S^{d-1}}\sigma +\max _{S^{d-1}}|D\sigma |\right ) \left |p-p^{\prime }\right |^{2}. \end{equation}

Second, if $|p-p^{\prime }|\lt 1$ , then

\begin{equation*} {\left |p-\frac {p^{\prime }}{|p^{\prime }|} \right |}^{2}= 2 - 2 p \cdot \frac {p^{\prime }}{|p^{\prime }|}\leq 2+\frac {1}{|p^{\prime }|}\left (|p-p^{\prime }|^{2}-1\right )\lt 2 \end{equation*}

and, therefore,

\begin{equation*} {\left |(1-t)p+t\frac {p^{\prime }}{|p^{\prime }|}\right |}^{2} = 1-\left (t-t^{2}\right )\left |p-\frac {p^{\prime }}{|p^{\prime }|}\right |^{2}\geq 1-\frac {1}{4}{\left |p-\frac {p^{\prime }}{|p^{\prime }|}\right |}^{2}\gt \frac {1}{2} \end{equation*}

for all $t \in [0,1]$ . As in the proof of inequality (i) above, we introduce a second-order Taylor expansion around $\frac{p^{\prime }}{|p^{\prime }|}$ with an intermediate point $p^{*}=(1-t)p+t\frac{p^{\prime }}{|p^{\prime }|}$ , where $t \in [0,1]$ . Using this expansion, Lemma 2.3(ii), (vi), and Young’s inequality, we obtain

(2.11) \begin{align} \sigma (p)-|p^{\prime }|\psi (|p^{\prime }|)D\sigma (p^{\prime })\cdot p &= \sigma (p)-\sigma\!\left (\frac{p^{\prime }}{|p^{\prime }|}\right )-D\sigma\!\left (\frac{p^{\prime }}{|p^{\prime }|}\right )\cdot \left(p-\frac{p^{\prime }}{|p^{\prime }|}\right)\nonumber \\[5pt] &\quad +\left (1-|p^{\prime }|\psi (|p^{\prime }|)\right )D\sigma (p^{\prime })\cdot p\nonumber \\[5pt] &\leq \frac{1}{2}\left (p-\frac{p^{\prime }}{|p^{\prime }|}\right )\cdot D^{2}\sigma (p^{*})\left (p-\frac{p^{\prime }}{|p^{\prime }|}\right ) +\left (1-|p^{\prime }|\psi (|p^{\prime }|)\right )\sigma (p)\nonumber \\[5pt] &\leq \frac{\sqrt{2}}{2}(\!\max _{S^{d-1}}|D^{2}\sigma |)\left |p-\frac{p^{\prime }}{|p^{\prime }|}\right |^{2}+2 (\!\max _{S^{d-1}}\sigma )\left (1-|p^{\prime }|\right )\nonumber \\[5pt] &\leq \sqrt{2}(\!\max _{S^{d-1}}|D^{2}\sigma |)|p-p^{\prime }|^{2} \nonumber \\[5pt] &\quad + \left (\sqrt{2}(\!\max _{S^{d-1}}|D^{2}\sigma |)+2 (\!\max _{S^{d-1}}\sigma )\right )\left (1-|p^{\prime }|\right ), \end{align}

from which (ii) follows in the case $|p-p^{\prime }|\lt 1$ .

Proof. For $\eta \in C^{1}(\Omega )^{d}$ such that $\sigma{^{\circ }}(\eta )\leq 1$ , we have that

\begin{equation*}\int _\Omega \sigma (B) \, dx \geq \int _{\Omega }\sigma (B) \sigma {^{\circ }}(\eta ) \, dx \geq \int _{\Omega }B\cdot \eta \, dx\end{equation*}

by Lemma 2.3. To see the other bound, whenever $B \neq 0$ note that $\sigma (B) = B\cdot D\sigma (B)$ and that $\sigma{^{\circ }}(D\sigma (B)) = 1$ by Lemma 2.3. As $\{\nu \;:\; \sigma ^{\circ }(\nu )\leq 1\}$ is a convex set, if $\eta$ is given by a smooth mollification of $D\sigma (B)$ , the inequality $\sigma ^{\circ }(\eta )\leq 1$ still holds. Approximating $D\sigma (B)$ by such $\eta$ shows that $\int _{\Omega }\sigma (B)dx\leq \sup _{\eta }\int _{\Omega }B\cdot \eta \, dx$ , concluding the proof.

Proof. The identity for the first variation (2.12) was stated by Almgren, Taylor, and Wang [Reference Almgren, Taylor and Wang1, Section 2.2] and proved by Bellettini and Paolini in the more general context of Finsler geometry [Reference Bellettini and Paolini7, Theorem 5.1]. The proof of the steepest descent statement follows the idea of [Reference Bellettini and Paolini7, Proposition 5.1] but incorporates the (arbitrary) mobility $\mu$ :

Let $B\in L^{2}(\partial A)^{d}$ , then by Lemma 2.3(ii) and the Cauchy–Schwarz inequality it follows that

\begin{align*} (dE, B)&={c_{0}} \int _{\partial A}H_{\sigma }\nu \cdot B d{\mathcal{H}^{d-1}}\\[5pt] &=-{c_{0}} \int _{\partial A}|H_{\sigma }|\nu \cdot \left (-{\mathrm{sgn}}(H_{\sigma })B\right ) d{\mathcal{H}^{d-1}}\\[5pt] &\geq -{c_{0}} \int _{\partial A}|H_{\sigma }|\sigma (\nu )\,\sigma ^{\circ }(\!-\!{\mathrm{sgn}}(H_{\sigma })B)\, d{\mathcal{H}^{d-1}}\\[5pt] &\geq -{c_{0}} \left (\int _{\partial A}|H_{\sigma }|^{2}\mu (\nu )\,d{\mathcal{H}^{d-1}}\right )^{\frac{1}{2}} \left (\int _{\partial A}\sigma ^{\circ }(\!-\!{\mathrm{sgn}}(H_{\sigma })B)^{2}\frac{\sigma (\nu )^{2}}{\mu (\nu )} d{\mathcal{H}^{d-1}}\right )^{\frac{1}{2}}\\[5pt] &=-\sqrt{{c_{0}}} \left (\int _{\partial A}|H_{\sigma }|^{2}\mu (\nu )\,d{\mathcal{H}^{d-1}}\right )^{\frac{1}{2}} \|B\|_{*}, \end{align*}

and Lemma 2.3(v), (vi) yield that the first inequality is an equality if $B=-\lambda{\mathrm{sgn}}(H_{\sigma })D\sigma (\nu )$ for some nonnegative function $\lambda \;:\; \partial A \to{\mathbb{R}}_{\geq 0}$ . Therefore, if $B$ is a constant positive multiple of $-H_{\sigma }\frac{\mu (\nu )}{\sigma (\nu )}D\sigma (\nu )$ , then both inequalities are equalities. The statement follows by the positive $1$ -homogeneity of $\|\cdot \|_{*}$ .

Remark 2.7. Let $A \subset{\mathbb{R}}^{d}$ be a bounded open set with $C^{2}$ -boundary. For every vector field $B\in C^{1}({\mathbb{R}}^{d})^{d}$ , we have

(2.14) \begin{equation} \int _{\partial A}\left (\sigma (\nu )I_{d}-\nu \otimes D\sigma (\nu )\right ):\;\nabla B\, d{\mathcal{H}^{d-1}} = \int _{\partial A} H_{\sigma }\,B\cdot \nu \, d{\mathcal{H}^{d-1}}. \end{equation}

To see this, we extend the normal $\nu$ to a vector field $\nu \in C^{1}({\mathbb{R}}^{d})^{d}$ , e.g., as a truncation of the normalized gradient $\frac{\nabla \mathrm{sdist}}{|\nabla \mathrm{sdist}|}$ of the signed distance function. There exists an open neighborhood $U$ of $\partial A$ where $\nu \neq 0$ , and in this neighborhood we can decompose the vector field $B$ as

\begin{align*} B&= \left (B-\frac{1}{\sigma (\nu )}(B\cdot \nu )D\sigma (\nu )\right ) + \frac{1}{\sigma (\nu )}(B\cdot \nu )D\sigma (\nu )\\[5pt] &\;=\!:\; B_{1}+B_{2}. \end{align*}

The restriction of $B_{1}$ to $\partial A$ is tangential since, by Lemma 2.3(v), we have $B_{1}\cdot \nu = 0$ . By the divergence theorem on the closed surface $\partial A$ , we obtain

(2.15) \begin{align} 0&= \int _{\partial A}\mathrm{div}_{\partial A}\left (\sigma (\nu )B_{1}\right )d{\mathcal{H}^{d-1}}\nonumber \\[5pt] &= \int _{\partial A}\left (\sigma (\nu )\mathrm{div} B_{1}+ D\sigma (\nu )\cdot \nabla \nu B_{1}-\sigma (\nu )\nu \cdot \nabla B_{1}\nu \right )d{\mathcal{H}^{d-1}}\nonumber \\[5pt] &= \int _{\partial A}\left (\sigma (\nu )\mathrm{div} B_{1}- \nu \cdot \nabla B_{1}D\sigma (\nu )+ \left (D\sigma (\nu )-\sigma (\nu )\nu \right )\cdot \nabla \nu B_{1}+\vphantom{\nu \cdot \nabla B_{1}\left (D\sigma (\nu )-\sigma (\nu )\nu \right )} \right .\nonumber \\[5pt] &\hspace{8.5pt}\left .\vphantom{\sigma (\nu )\mathrm{div} B_{1}- \nu \cdot \nabla B_{1}D\sigma (\nu )+ \left (D\sigma (\nu )-\sigma (\nu )\nu \right )\cdot \nabla \nu B_{1}}+\nu \cdot \nabla B_{1}\left (D\sigma (\nu )-\sigma (\nu )\nu \right ) \right )d{\mathcal{H}^{d-1}}\nonumber \\[5pt] &= \int _{\partial A}\left (\left (\sigma (\nu )I_{d}-\nu \otimes D\sigma (\nu )\right ):\nabla B_{1}+ \left (D\sigma (\nu )-\sigma (\nu )\nu \right )\cdot \nabla \nu B_{1}+\vphantom{\nu \cdot \nabla B_{1}\left (D\sigma (\nu )-\sigma (\nu )\nu \right )} \right .\nonumber \\[5pt] &\hspace{8.5pt}\left .\vphantom{\sigma (\nu )\mathrm{div} B_{1}- \nu \cdot \nabla B_{1}D\sigma (\nu )+ \left (D\sigma (\nu )-\sigma (\nu )\nu \right )\cdot \nabla \nu B_{1}}-B_{1} \cdot \nabla \nu \left (D\sigma (\nu )-\sigma (\nu )\nu \right ) \right )d{\mathcal{H}^{d-1}}\nonumber \\[5pt] &= \int _{\partial A}\left (\sigma (\nu )I_{d}-\nu \otimes D\sigma (\nu )\right ):\nabla B_{1}\,d{\mathcal{H}^{d-1}}. \end{align}

In this computation, the third equality follows by adding zero and using that $\nu \cdot \nabla \nu \,B_{1} =B_{1}\cdot \nabla \left (\frac{1}{2}|\nu |^{2}\right )=0$ . For the fourth equality observe that $(\nabla B_{1}){^{T}} \nu + B_{1}){^{T}} \nu + (\nabla \nu ){^{T}} B_{1}=\nabla (B_{1}\cdot \nu )=0$ since $B_{1}$ is tangential. holds true because the second fundamental form

\begin{equation*} \mathrm {I\!I}_{x}\;:\;T_{x}\partial A \times T_{x}\partial A \to {\mathbb {R}}, \qquad \mathrm {I\!I}_{x}(v,w)\;:\!=\;w\cdot (v\cdot \nabla )\nu (x) \end{equation*}

is symmetric for all $x \in \partial A$ (see [Reference Giga30, Section 1.3]).

On the other hand, we have

\begin{align*} \nabla B_{2}&=-\frac{1}{\sigma ^{2}(\nu )}(B\cdot \nu )D\sigma (\nu )\otimes (\nabla \nu ){^{T}} D\sigma (\nu )+\frac{1}{\sigma (\nu )}D\sigma (\nu )\otimes (\nabla \nu ){^{T}} B \\[5pt] &\quad +\frac{1}{\sigma (\nu )}D\sigma (\nu )\otimes (\nabla B){^{T}} \nu +\frac{1}{\sigma (\nu )}(B\cdot \nu )D^{2}\sigma (\nu )\nabla \nu, \end{align*}

which yields

(2.16) \begin{equation} \left (\sigma (\nu )I_{d}-\nu \otimes D\sigma (\nu )\right ):\nabla B_{2}= \mathrm{div}(D\sigma (\nu ))B\cdot \nu = H_{\sigma }B\cdot \nu . \end{equation}

From (2.15) and (2.16) we obtain (2.14).

Using the integration by parts (2.14), we arrive at a distributional formulation for anisotropic mean curvature which resembles the isotropic version due to Luckhaus and Sturzenhecker [Reference Luckhaus and Sturzenhecker47]. Following [Reference Laux39], our definition for $BV$ solutions to anisotropic mean curvature flow also includes an optimal energy dissipation inequality, which alludes to the gradient flow structure of the problem.

Definition 2.8 (Distributional solutions to anisotropic mean curvature flow). Let $\chi _{0} \in BV(\Omega ;\;\{0,1\})$ . A distributional (or $BV$ ) solution to $V=-\mu (\nu ) H_{\sigma }$ with initial condition $\chi _{0}$ is a function $\chi \in BV\left (\Omega \times (0,T);\left \{0,1\right \}\right )\cap C^{0,\frac{1}{2}}\left ([0,T];\;\;L^{1}(\Omega )\right )\cap L^{\infty }\left (0,T;\;BV(\Omega )\right )$ such that

1. (i) there exists a $\big |\nabla \chi \big |$ -measurable function $V\;:\; \Omega \times (0,T) \to{\mathbb{R}}$ such that

   (2.17) \begin{equation} {\int _{0}^{T}\int _{\Omega }} V^{2}{\big |\nabla \chi \big |} \lt \infty, \end{equation}
   which is the normal velocity in the sense that for all $\zeta \in C^{1}(\Omega \times [0,T])$ and ${T^{\prime }}{T^{\prime }} \in (0,T]$ :
   (2.18) \begin{align} \int _{\Omega } \zeta (x,{T^{\prime }})\chi (x,{T^{\prime }})dx&-\int _{\Omega } \zeta (x,0)\chi _{0}(x)dx \nonumber \\[5pt] &=\int _{0}^{{T^{\prime }}}\int _{\Omega } \partial _{t}\zeta (x,t)\chi (x,t)dx dt + \int _{0}^{{T^{\prime }}}\int _{\Omega } \zeta (x,t)V(x,t){\big |\nabla \chi \big |}, \end{align}

2. (ii) the relation $V=-\mu (\nu ) H_{\sigma }$ is satisfied in a distributional sense, i.e., for all $B \in C^{1}(\Omega \times [0,T])^{d}$ , we have

   (2.19) \begin{equation} -{\int _{0}^{T}\int _{\Omega }} \frac{1}{\mu (\nu )}VB\cdot \nu{\big |\nabla \chi \big |} ={\int _{0}^{T}\int _{\Omega }} \nabla B \;:\; \left (\sigma (\nu )I_{d}-\nu \otimes D\sigma (\nu ) \right ){\big |\nabla \chi \big |}, \end{equation}
   and

3. (iii) the function $\chi$ satisfies the optimal energy dissipation inequality

   (2.20) \begin{equation} \int _{\Omega }\sigma (\nu )|\nabla \chi ({T^{\prime }})|+\int _{0}^{{T^{\prime }}}\int _{\Omega }\frac{1}{\mu (\nu )}V^{2}{\big |\nabla \chi \big |} \leq \int _{\Omega }\sigma (\nu )|\nabla \chi _0| \end{equation}
   for every ${T^{\prime }} \in [0,T]$ .

Proof.

1. (i) Let $u, v \in L^{2}(\Omega )$ and $\mu \in [0,1]$ . We will use the equivalent formulation (ii) in Remark 3.1, so that we have to show that

   \begin{equation*} E_{\varepsilon }[(1-\mu )u+\mu v] \leq (1-\mu ) E_{\varepsilon }[u]+ \mu E_{\varepsilon }[v]+\frac {\lambda }{4\varepsilon }\mu (1-\mu )\|v-u\|_{L^{2}}^{2} \end{equation*}
   for all $u,v \in L^{2}(\Omega )$ and $\mu \in (0,1)$ . We can assume without loss of generality that $u, v \in \mathrm{dom}(E_{\varepsilon })$ . Exploiting the convexity of $f$ as well as the $\lambda$ -convexity of $W$ (W4), we obtain
   (3.12) \begin{align} E_{\varepsilon }[(1-\mu )u+\mu v] &= \frac{1}{2}\int _{\Omega }\left (\varepsilon f(\!-(1-\mu )\nabla u - \mu \nabla v)+\frac{1}{\varepsilon }W((1-\mu )u+\mu v)\right )dx \nonumber \\[5pt] &\leq \frac{1}{2}\int _{\Omega }\left (\varepsilon (1-\mu ) f(\!-\!\nabla u) +\varepsilon \mu f(\!-\!\nabla v) \right )dx \nonumber \\[5pt] &\quad + \frac{1}{2}\int _{\Omega }\left (\frac{1-\mu }{\varepsilon }W(u)+\frac{\mu }{\varepsilon }W(v)+\frac{\lambda }{2\varepsilon }\mu (1-\mu )|v-u|^{2}\right )dx \nonumber \\[5pt] &\leq (1-\mu )E_{\varepsilon }[u]+\mu E_{\varepsilon }[v] + \frac{\lambda }{4\varepsilon }\mu (1-\mu )\left \|v-u\right \|_{L^{2}}^{2}. \end{align}

2. (ii) This follows from (i) and quantifying the strong convexity of the metric term: Given $u, v \in L^{2}(\Omega )$ and $\mu \in (0,1)$ , we compute

   (3.13) \begin{align} \frac{1}{2h}\left \|((1-\mu )u\vphantom{+\mu v)-{u_{h}^{n-1}}}\right .& \left .\vphantom{((1-\mu )u}\hspace{-3pt}{+} \mu v)-{u_{h}^{n-1}}\right \|_{{u_{h}^{n-1}}}^{2}\nonumber \\[5pt] &= \frac{1-\mu }{2h}\|u-{u_{h}^{n-1}}\|_{{u_{h}^{n-1}}}^{2}+\frac{\mu }{2h}\|v-{u_{h}^{n-1}}\|_{{u_{h}^{n-1}}}^{2}\nonumber \\[5pt] &\quad - \frac{\mu (1-\mu )}{2h}\int _{\Omega }\varepsilon g(\!-\!\nabla{u_{h}^{n-1}})|v-u|^{2}dx \\ &\leq \frac{1-\mu }{2h}\|u-{u_{h}^{n-1}}\|_{{u_{h}^{n-1}}}^{2}+\frac{\mu }{2h}\|v-{u_{h}^{n-1}}\|_{{u_{h}^{n-1}}}^{2}\nonumber \\[5pt] &\quad - \frac{\varepsilon c_{g}}{2h}\mu (1-\mu )\left \|v-u\right \|_{L^{2}}^{2}. \end{align}
   Combining (3.12) with (3.13) yields strong convexity for $u \mapsto E_{\varepsilon }[u]+\frac{1}{2h}\left \|u-{u_{h}^{n-1}}\right \|_{{u_{h}^{n-1}}}^{2}$ if $\frac{\lambda }{4\varepsilon }\lt \frac{\varepsilon c_{g}}{2h},$ or, equivalently, if $h\lt \frac{2\varepsilon ^{2}c_{g}}{\lambda }$ .

3. (iii) This will follow from the lower semi-continuity of the individual terms. To see that the anisotropic Dirichlet energy is lower semi-continuous, we can assume without restriction that $v_{k} \in H^{1}(\Omega )$ for all $k \in{\mathbb{N}}$ , so that it remains to prove that

   \begin{equation*} \frac {\varepsilon }{2}\int _{\Omega }f(\!-\!\nabla v)dx \leq \liminf _{k \to \infty }\frac {\varepsilon }{2}\int _{\Omega }f(\!-\!\nabla v_{k})dx. \end{equation*}
   Without loss of generality, $v_k$ converge to $v$ weakly in $H^1(\Omega ),$ and lower semi-continuous follows from convexity of $f$ . The lower semicontinuity of the potential term $\int _{\Omega } W(u) \, dx$ and the metric term $u\mapsto \frac{1}{2h}\left \|u-{u_{h}^{n-1}} \right \|_{{u_{h}^{n-1}}}^{2}$ follow from Fatou’s lemma.

4. (iv) The functional to be minimized is proper. Let $\left \{v_{k}\right \}_{k \in{\mathbb{N}}}$ be a minimizing sequence, i.e.,

   \begin{equation*} \lim _{k \to \infty } \left (E_{\varepsilon }[v_{k}]+\frac {1}{2h}\left \|v_{k}-{u_{h}^{n-1}}\right \|_{{u_{h}^{n-1}}}^{2}\right ) =\inf _{u \in L^{2}(\Omega )}\left \{E_{\varepsilon }[u]+\frac {1}{2h}\left \|u-{u_{h}^{n-1}}\right \|_{{u_{h}^{n-1}}}^{2}\right \}. \end{equation*}
   Every minimizing sequence is bounded in $H^{1}(\Omega )$ since
   (3.14) \begin{align} \left \|v_{k}\right \|_{H^{1}}^{2}&\leq 2\left \|v_{k}-{u_{h}^{n-1}}\right \|_{L^{2}}^{2} + 2 \left \|{u_{h}^{n-1}}\right \|_{L^{2}}^{2}+ \left \|\nabla v_{k}\right \|_{L^{2}}^{2}\nonumber \\[5pt] &\leq \frac{2}{\varepsilon c_{g}}\left \|v_{k}-{u_{h}^{n-1}}\right \|_{{u_{h}^{n-1}}}^{2} + 2 \left \|{u_{h}^{n-1}}\right \|_{L^{2}}^{2}+ \frac{1}{(\!\min _{S^{d-1}}\sigma )^{2}}\int _{\Omega }f(\!-\!\nabla v_{k})dx\nonumber \\[5pt] &\leq 2 \left \|{u_{h}^{n-1}}\right \|_{L^{2}}^{2}+\left (\frac{4h}{\varepsilon c_{g}}+\frac{2}{\varepsilon (\!\min _{S^{d-1}}\sigma )^{2}} \right )\left (E_{\varepsilon }[v_{k}]+\frac{1}{2h}\left \|v_{k}-{u_{h}^{n-1}}\right \|_{{u_{h}^{n-1}}}^{2}\right ). \end{align}
   By Rellich’s theorem, there exists a subsequence converging in $L^{2}(\Omega )$ to a limit function $v \in L^{2}(\Omega )$ . The lower semicontinuity (iii) then shows that
   \begin{equation*} v \in \underset {u \in L^{2}(\Omega )}{\arg \min }\left \{E_{\varepsilon }[u]+ \frac {1}{2h}\left \|u-{u_{h}^{n-1}}\right \|_{{u_{h}^{n-1}}}^{2} \right \}. \end{equation*}
   The minimizer $v$ is unique due to the strong convexity (ii). For the $L^{\infty }$ -bound, we define a truncated version of $u_{h}^{n}$ by
   \begin{equation*} v(x)\;:\!=\;\begin {cases} -\alpha ^{n-1}+\tfrac 12 &\text {if } {u_{h}^{n}}(x)\lt -\alpha ^{n-1}+\tfrac 12,\\[5pt] {u_{h}^{n}}(x)&\text {if } |{u_{h}^{n}}(x)-\tfrac 12|\leq \alpha ^{n-1},\\[5pt] \alpha ^{n-1}+\tfrac 12 &\text {if } {u_{h}^{n}}(x)\gt \alpha ^{n-1}+\tfrac 12, \end {cases} \end{equation*}
   where $\alpha ^{n-1} \;:\!=\; \max \{\|{u_{h}^{n-1}}-\tfrac 12\|_{L^{\infty }},\tfrac 12\}.$ Letting $A\;:\!=\; \{x\in \Omega \,\big |\, v(x)\neq{u_{h}^{n}}(x) \}$ , it follows from ${u_{h}^{n}} \in \mathrm{dom}(E_{\varepsilon })\subseteq H^{1}(\Omega )$ and a general fact about Sobolev spaces that $v \in H^{1}(\Omega )$ and
   (3.15) \begin{equation} \nabla v(x) =\begin{cases} \nabla{u_{h}^{n}}(x) &\text{for almost every } x \in \Omega \setminus A,\\[5pt] 0&\text{for almost every } x\in A.\end{cases} \end{equation}
   Then, using (3.15), the positive definiteness of $\sigma$ , the monotonicity assumption (W5), and the pointwise inequality $|v-{u_{h}^{n-1}}|\leq |{u_{h}^{n}}-{u_{h}^{n-1}}|$ , one obtains
   \begin{align*} E_{\varepsilon }[v]+\frac{1}{2h}\left \|v-{u_{h}^{n-1}}\right \|_{{u_{h}^{n-1}}}^{2}&\leq \frac{\varepsilon }{2}\int _{\Omega }f(\!-\!\nabla{u_{h}^{n}})dx+\frac{1}{2\varepsilon }\int _{\Omega }W({u_{h}^{n}})dx\\[5pt] &\quad +\frac{\varepsilon }{2h}\int _{\Omega }g(\!-\!\nabla{u_{h}^{n-1}})|{u_{h}^{n}}-{u_{h}^{n-1}}|^{2}dx\\[5pt] &=E_{\varepsilon }[{u_{h}^{n}}]+\frac{1}{2h}\left \|{u_{h}^{n}}-{u_{h}^{n-1}}\right \|_{{u_{h}^{n-1}}}^{2}. \end{align*}
   By the uniqueness of minimizers, this implies that $v={u_{h}^{n}}$ and, therefore,
   \begin{equation*} \|{u_{h}^{n}}-\tfrac 12\|_{L^{\infty }}=\|v-\tfrac 12\|_{L^{\infty }}\leq \alpha ^{n-1} = \max \left \{\|{u_{h}^{n-1}}-\tfrac 12\|_{L^{\infty }},\tfrac 12 \right \}. \end{equation*}

Proof. It suffices to prove the corresponding inequality involving the time steps $u_{h}^{n}$ , $n \in{\mathbb{N}}$ : Let $h\lt \frac{\varepsilon ^{2}c_{g}}{2\lambda }$ and $n \in{\mathbb{N}}$ . Then for all $v \in L^{2}(\Omega )$ , we have

(3.18) \begin{equation} E_{\varepsilon }[v]\geq E_{\varepsilon }[{u_{h}^{n}}]-\int _{\Omega } \varepsilon g(\!-\!\nabla{u_{h}^{n-1}})\frac{{u_{h}^{n}}-{u_{h}^{n-1}}}{h}\left (v-{u_{h}^{n}}\right )dx - \frac{\lambda }{4\varepsilon } \int _{\Omega } \left |v-{u_{h}^{n}}\right |^{2}dx. \end{equation}

Clearly, one can write the interpolations $u_{h}$ , $\overline{u}_{h}$ , and $\underline{u}_{h}$ on the left-hand side of (3.17) in terms of the time steps, choosing $n=n(t) \in{\mathbb{N}}$ such that $t \in [(n(t)-1)h, n(t)h)$ . We also observe that the piecewise affine interpolation $u_{h} \in C\left ([0,T];\;L^{2}(\Omega )\right )$ has a weak time derivative $\partial _{t}u_{h}$ given by

\begin{equation*} \partial _{t}u_{h}(t)=\frac {\overline {u}_{h}(t)-\underline {u}_{h}(t)}{h} = \frac {{u_{h}^{n(t)}}-{u_{h}^{n(t)-1}}}{h} \end{equation*}

for almost every $t \in (0,T)$ . Therefore, (3.17) follows from (3.18) by reformulating the interpolation in terms of the time steps and using $v=w(t)$ for fixed $t \in (0,T)$ .

To prove (3.18), let $v \in L^{2}(\Omega )$ and $\delta \in (0,1)$ . Again, we use the equivalent characterization of $\frac{\lambda }{2\varepsilon }$ -convexity in Remark 3.1(ii). For computational ease, first note, that

\begin{equation*} |\delta v + (1-\delta )u^{n}_h - u^{n-1}_h|^2 = \delta ^2 |v - u^{n}_h|^2 + 2\delta (v-u^{n}_h)(u^n_h - u^{n-1}_h) + |u^n_h-u^{n-1}_h|^2 . \end{equation*}

Subtracting $(1-\delta )|u^n-u^{n-1}|^2$ from both sides of the above equality, we may directly compute

(3.19) \begin{align} E_{\varepsilon }[v]&\geq \frac{1}{\delta }\left (E_{\varepsilon }[\delta v + (1-\delta ){u_{h}^{n}}]-(1-\delta )E_{\varepsilon }[{u_{h}^{n}}]\right )-(1-\delta )\frac{\lambda }{4\varepsilon }\int _{\Omega }|v-{u_{h}^{n}}|^{2}dx \nonumber \\[5pt] &=\frac{1}{\delta }\left (E_{\varepsilon }[\delta v + (1-\delta ){u_{h}^{n}}]+\frac{1}{2h}\left \|\delta v +(1-\delta ){u_{h}^{n}}-{u_{h}^{n-1}}\right \|_{{u_{h}^{n-1}}}^{2} \vphantom{-(1-\delta )E_{\varepsilon }[{u_{h}^{n}}]-(1-\delta )\frac{1}{2h}\left \|{u_{h}^{n}}-{u_{h}^{n-1}}\right \|_{{u_{h}^{n-1}}}^{2}}\right .\nonumber \\[5pt] &\hspace{8.5pt}\left .\vphantom{E_{\varepsilon }[\delta v + (1-\delta ){u_{h}^{n}}]+\frac{1}{2h}\left \|\delta v +(1-\delta ){u_{h}^{n}}-{u_{h}^{n-1}}\right \|_{{u_{h}^{n-1}}}^{2}}-(1-\delta )E_{\varepsilon }[{u_{h}^{n}}]-(1-\delta )\frac{1}{2h}\left \|{u_{h}^{n}}-{u_{h}^{n-1}}\right \|_{{u_{h}^{n-1}}}^{2}\right )\nonumber \\[5pt] &\quad -\frac{1}{2h}\int _{\Omega }\varepsilon g(\!-\!\nabla{u_{h}^{n-1}}) \left (2({u_{h}^{n}}-{u_{h}^{n-1}})(v-{u_{h}^{n}})+({u_{h}^{n}}-{u_{h}^{n-1}})^{2}+\delta ({u_{h}^{n}}-v)^{2}\right )dx\nonumber \\[5pt] &\quad -(1-\delta )\frac{\lambda }{4\varepsilon }\int _{\Omega }|v-{u_{h}^{n}}|^{2}dx \nonumber \\[5pt] &\geq \frac{1}{\delta }\left (E_{\varepsilon }[{u_{h}^{n}}]+\frac{1}{2h}\left \|{u_{h}^{n}}-{u_{h}^{n-1}}\right \|_{{u_{h}^{n-1}}}^{2} \vphantom{-(1-\delta )E_{\varepsilon }[{u_{h}^{n}}]-(1-\delta )\frac{1}{2h}\left \|{u_{h}^{n}}-{u_{h}^{n-1}}\right \|^{2}}\right .\nonumber \\[5pt] &\hspace{8.5pt}\left .\vphantom{E_{\varepsilon }[\delta v + (1-\delta ){u_{h}^{n}}]+\frac{1}{2h}\left \|\delta v +(1-\delta ){u_{h}^{n}}-{u_{h}^{n-1}}\right \|_{{u_{h}^{n-1}}}^{2}}-(1-\delta )E_{\varepsilon }[{u_{h}^{n}}]-(1-\delta )\frac{1}{2h}\left \|{u_{h}^{n}}-{u_{h}^{n-1}}\right \|_{{u_{h}^{n-1}}}^{2}\right )\nonumber \\[5pt] &\quad -\frac{1}{2h}\int _{\Omega }\varepsilon g(\!-\!\nabla{u_{h}^{n-1}}) \left (2({u_{h}^{n}}-{u_{h}^{n-1}})(v-{u_{h}^{n}})+({u_{h}^{n}}-{u_{h}^{n-1}})^{2}+\delta ({u_{h}^{n}}-v)^{2}\right )dx\nonumber \\[5pt] &\quad -(1-\delta )\frac{\lambda }{4\varepsilon }\int _{\Omega }|v-{u_{h}^{n}}|^{2}dx \nonumber \\[5pt] &=E_{\varepsilon }[{u_{h}^{n}}]- \int _{\Omega }\varepsilon g(\!-\!\nabla{u_{h}^{n-1}}) \frac{{u_{h}^{n}}-{u_{h}^{n-1}}}{h}(v-{u_{h}^{n}})dx-\frac{\delta }{2h}\int _{\Omega }\varepsilon g(\!-\!\nabla{u_{h}^{n-1}}) ({u_{h}^{n}}-v)^{2}dx\nonumber \\[5pt] &\quad -(1-\delta )\frac{\lambda }{4\varepsilon }\int _{\Omega }|v-{u_{h}^{n}}|^{2}dx, \end{align}

where the third step is an application of (3.11), with $\delta v + (1-\delta ){u_{h}^{n}}$ acting as a contender for the minimization problem. Taking the limit $\delta \searrow 0$ finally yields (3.18).

Proof.

1. (i) The first step is to prove a $H^{1}$ -bound for the minimizers $u_{h}^{n}$ , $n \in{\mathbb{N}}_{0}$ :

   (3.20) \begin{align} \left \|{u_{h}^{n}}\right \|_{H^{1}(\Omega )}^{2}&=\|{u_{h}^{n}}\|_{L^{2}(\Omega )}^{2}+\left \|\nabla{u_{h}^{n}}\right \|_{L^{2}(\Omega )}^{2}\nonumber \\[5pt] &\leq \left \|{u_{h}^{n}}\right \|_{L^{\infty }(\Omega )}^{2}+ \frac{1}{\min _{S^{d-1}}\sigma ^{2}}\int _{\Omega }f(\!-\!\nabla{u_{h}^{n}})dx\nonumber \\[5pt] &\leq \left \|{u_{h}^{n}}\right \|_{L^{\infty }(\Omega )}^{2}+ \frac{2}{\varepsilon \min _{S^{d-1}}\sigma ^{2}}E_{\varepsilon }[{u_{h}^{n}}]\nonumber \\[5pt] &\leq \max \left \{\left \|{u_{h}^{0}}\right \|_{L^{\infty }(\Omega )}^{2},1\right \}+ \frac{2}{\varepsilon \min _{S^{d-1}}\sigma ^{2}}E_{\varepsilon }[{u_{h}^{0}}], \end{align}
   where the last inequality uses Lemma 3.6(iv) and (3.11) repeatedly.

   For any $t \in [0,T]$ , the function $u_{h}(t)$ is a convex combination of two minimizers $u_{h}^{n-1}$ , $u_{h}^{n-1}$ , $u_{h}^{n}$ of (3.11). Applying the triangle inequality yields

   (3.21) \begin{align} \sup _{t\in [0,T]}\limsup _{h \searrow 0}\left \|u_{h}(t)\right \|_{H^{1}(\Omega )}&\leq \left (\!\max \left \{\limsup _{h\searrow 0}\left \|{u_{h}^{n}}\right \|_{L^{\infty }(\Omega )}^{2},1\right \}\vphantom{+\frac{2}{\varepsilon \min _{S^{d-1}}\sigma ^{2}}\limsup _{h\searrow 0}E_{\varepsilon }[{u_{h}^{0}}]}\right .\nonumber \\[5pt] &\hspace{8.5pt}{\left . \vphantom{\max \left \{\limsup _{h\searrow 0}\left \|{u_{h}^{n}}\right \|_{L^{\infty }(\Omega )}^{2},1\right \}} +\frac{2}{\varepsilon \min _{S^{d-1}}\sigma ^{2}}\limsup _{h\searrow 0}E_{\varepsilon }[{u_{h}^{0}}]\right )}^{\frac{1}{2}}\nonumber \\[5pt] &\lt \infty . \end{align}
   It follows from Rellich’s compact embedding theorem that any sequence $\{u_{h_{k}}(t)\}_{k \in{\mathbb{N}}}{\mathbb{N}}$ with $t \in [0,T]$ and $\lim _{k\to \infty }h_{k}=0$ has a subsequence converging in $L^{2}(\Omega )$ . To prove equicontinuity in time, which is the second requirement for the Arzelà–Ascoli theorem, we will – as is standard with minimizing movement approximations – show the stronger statement that $u_{h} \in C\left ([0,T];\;L^{2}(\Omega )\right )$ are $\frac{1}{2}$ -Hölder continuous and the Hölder constants are uniformly bounded as $h \searrow 0$ . Indeed, for $n \in{\mathbb{N}}$ , we use the definition of the metric $(\cdot,\cdot )_{{u_{h}^{n-1}}}$ and the minimization property (3.11) to show that
   (3.22) \begin{align} \left \|{u_{h}^{n}}-{u_{h}^{n-1}} \right \|_{L^{2}(\Omega )}^{2} &\leq \frac{1}{\varepsilon c_{g}}\left \|{u_{h}^{n}}-{u_{h}^{n-1}}\right \|_{{u_{h}^{n-1}}}^{2} \nonumber \\[5pt] &= \frac{2h}{\varepsilon c_{g}}\left (E_{\varepsilon }[{u_{h}^{n}}]+\frac{1}{2h}\left \|{u_{h}^{n}}-{u_{h}^{n-1}}\right \|_{{u_{h}^{n-1}}}^{2} - E_{\varepsilon }[{u_{h}^{n}}] \right ) \nonumber \\[5pt] &\leq \frac{2h}{\varepsilon c_{g}}\left (E_{\varepsilon }[{u_{h}^{n-1}}]-E_{\varepsilon }[{u_{h}^{n}}] \right ). \end{align}
   Inequality (3.22) allows us to bound the $L^{2}$ -norms of the weak time derivates $\partial _{t}u_{h}$ as
   (3.23) \begin{align} \|\partial _{t}u_{h}\|_{L^{2}\left (0,T;\;L^{2}(\Omega )\right )}^{2}&\leq \sum _{n=1}^{\left \lceil \frac{T}{h} \right \rceil } \int _{(n-1)h}^{nh}\|\partial _{t}u_{h}(t) \|_{L^{2}(\Omega )}^{2}dt \nonumber \\[5pt] &= \sum _{n=1}^{\left \lceil \frac{T}{h} \right \rceil } \int _{(n-1)h}^{nh} \frac{\|{u_{h}^{n}}-{u_{h}^{n-1}}\|_{L^{2}(\Omega )}^{2}}{h^{2}}dt \\ &\leq \sum _{n=1}^{\left \lceil \frac{T}{h} \right \rceil } \frac{2}{\varepsilon c_{g}}\left (E_{\varepsilon }[{u_{h}^{n-1}}]-E_{\varepsilon }[{u_{h}^{n}}] \right ) \nonumber \\[5pt] &\leq \frac{2}{\varepsilon c_{g}}E_{\varepsilon }[{u_{h}^{0}}]. \end{align}
   The fundamental theorem of calculus for vector-valued functions together with the Cauchy–Schwarz inequality now yields
   (3.24) \begin{align} \|u_{h}(t)-u_{h}(s)\|_{L^{2}(\Omega )}&\leq \int _{(s,t)}\|\partial _{t}u_{h}(r)\|_{L^{2}(\Omega )}dr \nonumber \\[5pt] &\leq \|\partial _{t}u_{h}\|_{L^{2}\left (s,t;L^{2}(\Omega )\right )}\sqrt{|t-s|}\nonumber \\[5pt] &\leq \sqrt{\frac{2E_{\varepsilon }[{u_{h}^{0}}]}{\varepsilon c_{g}}}\sqrt{|t-s|} \end{align}
   for all $s,t \in [0,T]$ . This is the one-dimensional case of Morrey’s inequality.

   By (3.24) and the assumption on the initial data, the Hölder constants are bounded as $h \searrow 0$ . Families of functions with bounded Hölder constants are equicontinuous. These equicontinuity and precompactness statements allow us to apply the Arzelà–Ascoli theorem, which states that there exists a subsequence $h \searrow 0$ and a function $u \in C\left ([0,T];\;L^{2}(\Omega )\right )$ such that $u_{h} \to u$ in $C\left ([0,T];\;L^{2}(\Omega )\right )$ . Furthermore, the Hölder continuity carries over to the limit function, i.e.,

   (3.25) \begin{equation} \|u(t)-u(s)\|_{L^{2}(\Omega )}\leq \sqrt{\frac{2\limsup _{h\searrow 0}E_{\varepsilon }[{u_{h}^{0}}]}{\varepsilon 0}E_{\varepsilon }[{u_{h}^{0}}]}{\varepsilon c_{g}}\sqrt{|t-s|} \qquad \text{for all }s,t \in [0,T]. \end{equation}
   One can extract a further subsequence $\{u_{h_{k}} \}_{k\in{\mathbb{N}}}$ such that $h_{k}\searrow 0$ and $u_{h_{k}}(x,t)\to u(x,t)$ almost everywhere in $\Omega \times (0,T)$ . This pointwise convergence together with the triangle inequality yields
   \begin{align*} \|u-\tfrac 12\|_{L^{\infty }(\Omega \times (0,T))}&\leq \limsup _{k\to \infty }\|u_{h_{k}}-\tfrac 12\|_{L^{\infty }(\Omega \times (0,T))}\\[5pt] &\leq \limsup _{h\searrow 0}\|u_{h}-\tfrac 12\|_{L^{\infty }(\Omega \times (0,T))}\\[5pt] &\leq \limsup _{h\searrow 0}\sup _{n\in{\mathbb{N}}}\|{u_{h}^{n}}-\tfrac 12\|_{L^{\infty }(\Omega )}\\[5pt] &\leq \max \left \{\limsup _{h\searrow 0}\|{u_{h}^{0}}-\tfrac 12\|_{L^{\infty }(\Omega )},\tfrac 12\right \}. \end{align*}
   In a last step, we argue briefly that the piecewise constant interpolations $\overline{u}_{h}$ and $\underline{u}_{h}$ converge uniformly.

   Given $t \in [0,T]$ , we find $n \in{\mathbb{N}}$ such that $t \in [(n-1)h,nh)$ . Then

   \begin{equation*} \|\overline {u}_{h}(t)-u_{h}(t)\|_{L^{2}(\Omega )}=\|u_{h}(nh)-u_{h}(t)\|_{L^{2}(\Omega )}\leq \sqrt {\frac {2E_{\varepsilon }[{u_{h}^{0}}]}{\varepsilon c_{g}}}\sqrt {nh-t}\leq \sqrt {\frac {2E_{\varepsilon }[{u_{h}^{0}}]}{\varepsilon c_{g}}}\sqrt {h}, \end{equation*}
   and a similar argument shows that
   \begin{equation*} \|\underline {u}_{h}(t)-u_{h}(t)\|_{L^{2}(\Omega )}\leq \sqrt {\frac {2E_{\varepsilon }[{u_{h}^{0}}]}{\varepsilon c_{g}}}\sqrt {h}. \end{equation*}
   Combining this bound with the uniform convergence $u_{h}\to u$ in $C\left ([0,T];\;L^{2}(\Omega )\right )$ , we obtain
   \begin{align*} \limsup _{h\searrow 0}\sup _{t\in [0,T]}\max &\left \{\|\overline{u}_{h}(t)-u(t)\|_{L^{2}(\Omega )},\|\underline{u}_{h}(t)-u(t)\|_{L^{2}(\Omega )} \right \}\\[5pt] &\leq \limsup _{h\searrow 0}\sup _{t \in [0,T]}\|u_{h}(t)-u(t)\|_{L^{2}(\Omega )}+ \lim _{h\searrow 0}\sqrt{\frac{2E_{\varepsilon }[{u_{h}^{0}}]}{\varepsilon c_{g}}}\sqrt{h} \\[5pt] &=0. \end{align*}

2. (ii) It follows from (i) that $u_{h}\to u$ as $h \searrow 0$ in $L^{2}\left (0,T;\;L^{2}(\Omega )\right )\cong L^{2}(\Omega \times (0,T))$ . Furthermore, by (3.23), the time derivatives $\{\partial _{t}u_{h} \}$ are bounded in $L^{2}(\Omega \times (0,T))$ as $h \searrow 0$ . By, e.g., [Reference Ambrosio, Fusco and Pallara2, Prop. 2.5(b)], one concludes that $u$ has a weak time derivative $\partial _{t}u \in L^{2}(\Omega \times (0,T))$ , and that $\partial _{t}u_{h}\rightharpoonup \partial _{t}u$ in $L^{2}(\Omega \times (0,T))$ as $h \searrow 0$ .

3. (iii) We will prove the desired convergence result for the piecewise constant interpolation $\overline{u}_{h}$ first. Integrating (3.20) from $0$ to $T$ shows that the gradients $\{\nabla \overline{u}_{h} \}$ are bounded in $L^{2}(\Omega \times (0,T))^{d}$ as $h \searrow 0$ . Just like in (ii), it then follows from the strong convergence $\overline{u}_{h}\to u$ in $L^{2}(\Omega \times (0,T))$ that the limit function $u$ has a weak gradient $\nabla u \in L^{2}(\Omega \times (0,T))^{d}$ , and that $\nabla \overline{u}_{h}{\rightharpoonup } \nabla u$ as $h \searrow 0$ in $L^{2}(\Omega \times (0,T))^{d}$ .

   The next step is to upgrade this weak convergence statement for the gradients to strong convergence in $L^{2}$ . The key ingredient for the argument will be the energy convergence

   (3.26) \begin{equation} E_{\varepsilon }[\overline{u}_{h}(\!\cdot\!)]{\rightharpoonup } E_{\varepsilon }[u(\!\cdot\!)] \qquad \text{in } L^{1}(0,T). \end{equation}
   Let us first argue how (3.26) implies strong convergence of the gradients. Since the Cahn–Hilliard energy $E_{\varepsilon }$ is made up of a Dirichlet energy and a nonconvex term involving $W$ , we will use the convergence result (3.26) for the Cahn–Hilliard energies and a liminf inequality for the nonconvex part to derive a limsup inequality for the Dirichlet energy, which will allow us to prove the strong convergence statement.

   Similarly to the proof of Lemma 3.6(iii), an application of Fatou’s lemma yields

   (3.27) \begin{equation} \liminf _{h\searrow 0}{\int _{0}^{T}\int _{\Omega }} W(\overline{u}_{h})dxdt \geq{\int _{0}^{T}\int _{\Omega }} W(u)dxdt. \end{equation}
   One can now test (3.26) with a constant test function and combine this convergence statement with (3.27), which yields
   (3.28) \begin{align} \limsup _{h\searrow 0}{\int _{0}^{T}\int _{\Omega }} f(\!-\!\nabla \overline{u}_{h})dx dt&=\frac{2}{\varepsilon }\lim _{h\searrow 0}\int _{0}^{T}E_{\varepsilon }[\overline{u}_{h}(t)]dt-\frac{1}{\varepsilon ^{2}}\liminf _{h\searrow 0}{\int _{0}^{T}\int _{\Omega }} W(\overline{u}_{h})dxdt \nonumber \\[5pt] &\leq \frac{2}{\varepsilon }\int _{0}^{T}E_{\varepsilon }[u(t)]dt-\frac{1}{\varepsilon ^{2}}{\int _{0}^{T}\int _{\Omega }} W(u)dxdt \nonumber \\[5pt] &={\int _{0}^{T}\int _{\Omega }} f(\!-\!\nabla u)dx dt. \end{align}
   The function $f$ is strongly convex, i.e., there exists a constant $c\gt 0$ such that $f(p^{\prime })\geq f(p)+Df(p)\cdot (p^{\prime }-p)+\frac{c}{2}|p^{\prime }-p|^{2}$ for all $p, p^{\prime } \in{\mathbb{R}}^{d}$ . Thus, if (3.28) holds true, then
   \begin{align*} \limsup _{h\searrow 0}\frac{c}{2}{\int _{0}^{T}\int _{\Omega }}& \left |\nabla \overline{u}_{h} - \nabla u\right |^{2}dx dt \\[5pt] &\leq \limsup _{h \searrow 0}\left ({\int _{0}^{T}\int _{\Omega }} f(\!-\!\nabla \overline{u}_{h})dx dt-{\int _{0}^{T}\int _{\Omega }} f(\!-\!\nabla u)dxdt \vphantom{-{\int _{0}^{T}\int _{\Omega }} Df(\!-\!\nabla u)\cdot (\nabla u - \nabla u_{h}) dxdt}\right .\\[5pt] &\hspace{8.5pt}\left . \vphantom{{\int _{0}^{T}\int _{\Omega }} f(\!-\!\nabla u_{h})dx dt- f(\!-\!\nabla u)dxdt} -{\int _{0}^{T}\int _{\Omega }} Df(\!-\!\nabla u)\cdot (\nabla u - \nabla \overline{u}_{h})dx dt\right ) \\ &\leq \limsup _{h \searrow 0}{\int _{0}^{T}\int _{\Omega }} f(\!-\!\nabla \overline{u}_{h})dx dt -{\int _{0}^{T}\int _{\Omega }} f(\!-\!\nabla u)dxdt\\[5pt] &\quad -\liminf _{h \searrow 0}{\int _{0}^{T}\int _{\Omega }} Df(\!-\!\nabla u)\cdot (\nabla u - \nabla \overline{u}_{h}) dxdt\\[5pt] &\leq 0, \end{align*}
   where the last step also uses the fact that $\nabla u_{h}{\rightharpoonup } \nabla u$ in $L^{2}(\Omega \times (0,T))^{d}$ . This computation implies that $\nabla \overline{u}_{h}\to \nabla u$ strongly in $L^{2}(\Omega \times (0,T))$ .

   In order to prove (3.26), it suffices to consider nonnegative test functions $\zeta \in L^{\infty }\left (0,T;\;{\mathbb{R}}_{\geq 0}\right )$ . We will prove a liminf inequality and a limsup inequality separately.

   On the one hand, by Fatou’s lemma, the lower semicontinuity of $E_{\varepsilon }$ , and (i), we obtain

   (3.29) \begin{equation} \liminf _{h\searrow 0}\int _{0}^{T}\zeta (t)E_{\varepsilon }[\overline{u}_{h}(t)]dt\geq \int _{0}^{T}\zeta (t)\liminf _{h\searrow 0}E_{\varepsilon }[\overline{u}_{h}(t)]dt\geq \int _{0}^{T}\zeta (t)E_{\varepsilon }[u(t)]dt. \end{equation}
   On the other hand, choosing $w(x,t)\;:\!=\;u(x,t)$ in (3.17) and integrating against $\zeta$ yields
   \begin{align*} \int _{0}^{T}\zeta (t)E_{\varepsilon }[u(t)]dt& \geq \int _{0}^{T}\zeta (t)E_{\varepsilon }[\overline{u}_{h}(t)]dt\\[5pt] & \quad -\int _{0}^{T}\zeta (t)\int _{\Omega }\left (\varepsilon g(\!-\!\nabla \underline{u}_{h})\partial _{t}u_{h}(u-\overline{u}_{h}) +\frac{\lambda }{4\varepsilon }|u-\overline{u}_{h}|^{2}\right )dx dt. \end{align*}
   In the limit $h\searrow 0$ , the second integral on the right-hand side vanishes since $\overline{u}_{h}\to u$ in $L^{2}(\Omega \times (0,T))$ and $g$ is bounded in $L^\infty$ by Lemma 3.2. Thus,
   (3.30) \begin{equation} \int _{0}^{T}\zeta (t)E_{\varepsilon }[u(t)]dt\geq \limsup _{h\searrow 0}\int _{0}^{T}\zeta (t)E_{\varepsilon }[\overline{u}_{h}(t)]dt. \end{equation}
   Combining (3.29) and (3.30) yields (3.26).

   We have shown that $\nabla \overline{u}_{h}\to \nabla u$ in $L^{2}(\Omega \times (0,T))$ as $h \searrow 0$ . As for $\nabla \underline{u}_{h}$ , one computes

   \begin{align*} \|\nabla \underline{u}_{h}-&\nabla u\|_{L^{2}(\Omega \times (0,T))}^{2}\\[5pt] &=\int _{0}^{h}\int _{\Omega }|\nabla \underline{u}_{h}-\nabla u|^{2}dx dt + \int _{h}^{T}\int _{\Omega }|\nabla \underline{u}_{h}-\nabla u|^{2}dx dt \\[5pt] &=\int _{0}^{h}\int _{\Omega }|\nabla{u_{h}^{0}}-\nabla u|^{2}dx dt + \int _{h}^{T}\int _{\Omega }|\nabla \overline{u}_{h}(x,t-h)-\nabla u(x,t)|^{2}dx dt\\[5pt] &\leq 2h\int _{\Omega }|\nabla{u_{h}^{0}}|^{2}dx + 2 \int _{0}^{h}\int _{\Omega }|\nabla u|^{2}dxdt\\[5pt] &\quad +2\int _{0}^{T-h}\int _{\Omega }|\nabla \overline{u}_{h}(x,t)-\nabla u(x,t)|^{2}dx dt\\[5pt] &\quad + 2\int _{0}^{T-h}\int _{\Omega }|\nabla u(x,t)-\nabla u(x, t+h)|^{2}dx dt\\[5pt] &\longrightarrow 0 \qquad \text{as } h \searrow 0, \end{align*}
   where the last integral vanishes as $h \searrow 0$ due to the continuity of translation.

   Lastly, $\nabla u_{h}(x,t)$ is a convex combination of $\nabla \overline{u}_{h}(x,t)$ and $\nabla \underline{u}_{h}(x,t)$ for almost all $(x,t)\in \Omega \times (0,T)$ . Hence, it follows from $\nabla \overline{u}_{h}\to \nabla u$ and $\nabla \underline{u}_{h}\to \nabla u$ that, as $h\searrow 0$ , we have $\nabla u_{h}\to \nabla u$ in $L^{2}(\Omega \times (0,T))$ as well.

Proof. First, by the weak convergence (3.26), it follows that

(3.32) \begin{equation} \lim _{h\searrow 0}\int _{0}^{T}\zeta (t)E_{\varepsilon }[\overline{u}_{h}(t)]dt = \int _{0}^{T}\zeta (t)E_{\varepsilon }[u(t)]dt \end{equation}

for any function $\zeta \in L^{\infty }(0,T)$ such that $\zeta \geq 0$ .

Second, for the term involving the time derivative, we know from Lemma 3.8(ii) that $\partial _{t}u_{h}{\rightharpoonup } \partial _{t}u$ in $L^{2}(\Omega \times (0,T))$ . Thus, to derive the convergence of the integrals, it suffices to show that $g(\!-\!\nabla \underline{u}_{h})(w-\overline{u}_{h}) \to g(\!-\!\nabla u)(w-u)$ (strongly) in $L^{2}(\Omega \times (0,T))$ .

By Lemma 3.8(iii), we know that $\nabla \overline{u}_{h},\nabla \underline{u}_{h}\to \nabla u$ in $L^{2}(\Omega \times (0,T))$ . Since $g$ is a continuous function, every subsequence of $g(\!-\!\nabla \underline{u}_{h})$ as $h\searrow 0$ has a further subsequence that converges pointwise almost everywhere to $g(\!-\!\nabla u)$ . By the uniform boundedness of $g$ and the dominated convergence theorem, it follows that

\begin{equation*} g(\!-\!\nabla \underline {u}_{h}(t))(w(t)-\overline {u}_{h}(t)) \to g(\!-\!\nabla u(t))(w(t)-u(t)) \qquad \text {in } L^{2}(\Omega ) \text { for almost all } t\in (0,T). \end{equation*}

It remains to show the $L^{2}$ -convergence on the product space $\Omega \times (0,T)$ . Using the generalized dominated convergence theorem with

\begin{align*} \left \|g(\!-\!\nabla \underline{u}_{h}(t))(w(t)-\overline{u}_{h}(t)) \right \|_{L^{2}(\Omega )}&\leq (\!\sup _{{\mathbb{R}}^{d}}g) \left (\int _{\Omega }(w(t)-\overline{u}_{h}(t))^{2}dx\right )^{\frac{1}{2}} \\[5pt] &\longrightarrow (\!\sup _{{\mathbb{R}}^{d}}g) \left (\int _{\Omega }(w(t)-u(t))^{2}dx\right )^{\frac{1}{2}} \qquad \text{in } L^{2}(0,T), \end{align*}

it follows that

\begin{equation*} \int _{0}^{T}\left \|g(\!-\!\nabla \underline {u}_{h}(t))(w(t)-\overline {u}_{h}(t))-g(\!-\!\nabla u(t))(w(t)-u(t)) \right \|_{L^{2}(\Omega )}^{2}dt\longrightarrow 0, \end{equation*}

i.e., by Fubini’s theorem, $g(\!-\!\nabla \underline{u}_{h})(w-\overline{u}_{h}) \to g(\!-\!\nabla u)(w-u)$ in $L^{2}(\Omega \times (0,T))$ . In total, we obtain

(3.33) \begin{align} \lim _{h\searrow 0}\int _{0}^{T}\zeta (t)&\int _{\Omega }\varepsilon g(\!-\!\nabla \underline{u}_{h}(t))\partial _{t}u_{h}(t)(w(t)-\overline{u}_{h}(t))dx dt\nonumber \\[5pt] & =\int _{0}^{T}\zeta (t)\int _{\Omega }\varepsilon g(\!-\!\nabla u(t))\partial _{t}u(t)(w(t)-u(t))dx dt. \end{align}

Third, it follows from Lemma 3.8(i) that

(3.34) \begin{equation} \lim _{h\searrow 0}\frac{\lambda }{4\varepsilon }\int _{0}^{T}\zeta (t)\int _{\Omega }|w(t)-\overline{u}_{h}(t)|^{2}dxdt = \frac{\lambda }{4\varepsilon }\int _{0}^{T}\zeta (t)\int _{\Omega }|w(t)-u(t)|^{2}dxdt. \end{equation}

Integrating (3.17) against $\zeta$ , taking the limit $h \searrow 0$ , and plugging in (3.32)–(3.34) proves the desired inequality.

Proof of Theorem 3.5. Choosing ${u_{h}^{0}}=u_{0}$ , one can construct the functions $u_{h}$ , $\overline{u}_{h}$ , and $\underline{u}_{h}$ by Lemma 3.6 and (3.16). Invoking Lemma 3.8 yields a subsequence $h \searrow 0$ as well as a function $u \in H^{1}(\Omega \times (0,T))\cap L^{\infty }(\Omega \times (0,T))$ such that $\|u-\tfrac 12\|_{L^{\infty }(\Omega \times (0,T))}\leq \max \left \{\|u_{0}-\tfrac 12\|_{L^{\infty }(\Omega )},\tfrac 12 \right \}$ , and such that the three convergence results in Lemma 3.8(i)–(iii) hold true. It remains to be shown that $u$ is a weak solution to the anisotropic Allen–Cahn equation with initial data $u_{0}$ in the sense of Definition 3.3.

As for the initial condition, it follows from the uniform convergence in Lemma 3.8(i) and the choice ${u_{h}^{0}}=u_{0}$ that

\begin{equation*} u(0)=\lim _{h\searrow 0}u_{h}(0)=\lim _{h\searrow 0}{u_{h}^{0}}=u_{0}, \end{equation*}

where all limits are in $L^{2}(\Omega )$ .

For the optimal energy dissipation relation (3.4) let ${T^{\prime }} \in (0,T]$ be a fixed time horizon. We define an extension of $u\;:\;\Omega \times [0,{T^{\prime }}]\to{\mathbb{R}}$ to $\Omega \times{\mathbb{R}}$ by

\begin{equation*} \tilde {u}(x,t)=\begin {cases} u(x,t)&\text {if } 0\leq t \leq {T^{\prime }},\\[5pt] u(x,0)&\text {if } t \lt 0,\\[5pt] u(x,{T^{\prime }})&\text {if } t\gt {T^{\prime }}. \end {cases} \end{equation*}

Then $\tilde{u} \in H_{\mathrm{loc}}^{1}(\Omega \times{\mathbb{R}})$ , with the weak time derivative being given by $\partial _{t}\tilde{u}(x,t)=\chi _{\Omega \times (0,{T^{\prime }})}\partial _{t}u(x,t)$ . Taking $a\gt 0$ , $\zeta \;:\!=\;\frac{1}{a}\chi _{(0,{T^{\prime }})}$ , and $w(x,t)\;:\!=\;\tilde{u}(x, t+a)$ in (3.31) yields

(3.35) \begin{align} \int _{0}^{{T^{\prime }}}\int _{\Omega }\varepsilon g(\!-\!\nabla u(t))\partial _{t}u(t)& \frac{\tilde{u}(t+a)-\tilde{u}(t)}{a}dx dt + a\int _{0}^{{T^{\prime }}}\int _{\Omega }\frac{\lambda }{4\varepsilon }\left (\frac{\tilde{u}(t+a)-\tilde{u}(t)}{a}\right )^{2}dxdt\nonumber \\[5pt] &\geq -\int _{0}^{{T^{\prime }}}\frac{E_{\varepsilon }[\tilde{u}(t+a)]-E_{\varepsilon }[\tilde{u}(t)]}{a}dt\nonumber \\[5pt] &=\frac{1}{a}\int _{0}^{a}E_{\varepsilon }[\tilde{u}(t)]dt - \frac{1}{a}\int _{{T^{\prime }}}^{{T^{\prime }}+a}E_{\varepsilon }[\tilde{u}(t)]dt\nonumber \\[5pt] &=\frac{1}{a}\int _{0}^{a}E_{\varepsilon }[u(t)]dt - E_{\varepsilon }[u({T^{\prime }})], \end{align}

where we have replaced $u$ by $\tilde{u}$ on $\Omega \times (0,{T^{\prime }})$ several times and used the definition of $\tilde{u}$ in the last line.

Let us now take $a \searrow 0$ in the inequality above. As a consequence of the lower semicontinuity of $E_{\varepsilon }$ and the $L^{2}$ -continuity of the map $t\mapsto u(t)$ , we have

\begin{equation*}\liminf _{a \searrow 0} \frac {1}{a}\int _{0}^{a}E_{\varepsilon }[u(t)]dt \geq E_{\varepsilon }[u(0)]=E_{\varepsilon }[u_{0}]. \end{equation*}

Furthermore, it follows from a general fact about Sobolev functions that

\begin{equation*}\frac {\tilde {u}(\!\cdot +\, a)-\tilde {u}(\!\cdot\!)}{a}{\rightharpoonup } \partial _{t}\tilde {u}(=\partial _{t}u)\qquad \text {in } L^{2}(\Omega \times (0,T))\text { as }a \searrow 0\end{equation*}

(see [Reference Gilbarg and Trudinger34, Lemma 7.23 and proof of Lemma 7.24]). In particular, the integral $\int _{0}^{{T^{\prime }}}\int _{\Omega }\frac{\lambda }{2\varepsilon }\left (\frac{\tilde{u}(t+a)-\tilde{u}(t)}{a}\right )^{2}dxdt$ is bounded as $a \searrow 0$ .

Therefore, the limiting inequality of (3.35) as $a \searrow 0$ reads

(3.36) \begin{equation} \int _{0}^{{T^{\prime }}}\int _{\Omega }\varepsilon g(\!-\!\nabla u(t))(\partial _{t}u(t))^{2}dx dt \geq E_{\varepsilon }[u_{0}]-E_{\varepsilon }[u({T^{\prime }})], \end{equation}

which is one inequality in the optimal energy dissipation relation (3.4).

For the converse inequality, we proceed analogously by taking $w(x,t)\;:\!=\;\tilde{u}(x,t-a)$ and $a \searrow 0$ . Using first the fact that $\frac{\tilde{u}(\cdot -a)-\tilde{u}(\!\cdot\!)}{a}{\rightharpoonup } -\partial _{t}u$ in $L^{2}(\Omega \times (0,{T^{\prime }}))$ , and second the inequality $\liminf _{a \searrow 0}\frac{1}{a}\int _{{T^{\prime }} - a}^{{T^{\prime }}}E_{\varepsilon }[u(t)]dt\geq E_{\varepsilon }[u({T^{\prime }})]$ , we obtain

(3.37) \begin{equation} \int _{0}^{{T^{\prime }}}\int _{\Omega }\varepsilon g(\!-\!\nabla u(t))(\partial _{t}u(t))^{2}dx dt \leq E_{\varepsilon }[u_{0}]-E_{\varepsilon }[u({T^{\prime }})]. \end{equation}

The combination of (3.36) and (3.37) is precisely the optimal energy dissipation identity (3.4).

For the distributional formulation of the PDE (3.3) let $\varphi \in C^{1}(\Omega \times [0,T])$ and $s \gt 0$ . By plugging in $\zeta \equiv 1$ and $w=u+s\varphi$ into (3.31), we obtain

\begin{align*} \int _{0}^{T}&\left (E_{\varepsilon }[u(t)+s\varphi (t)]- E_{\varepsilon }[u(t)] + \int _{\Omega }\varepsilon g(\!-\!\nabla u(t))\partial _{t}u(t)s\varphi (t)dx \vphantom{ + \frac{\lambda }{4\varepsilon }\int _{\Omega }s^{2}\varphi (t)^{2}dx}\right .\\[5pt] &\qquad \left .\vphantom{ E_{\varepsilon }[u(t)+s\varphi (t)]- E_{\varepsilon }[u(t)] + \int _{\Omega }\varepsilon g(\!-\!\nabla u(t))\partial _{t}u(t)s\varphi (t)\, dx} + \frac{\lambda }{4\varepsilon }\int _{\Omega }s^{2}\varphi (t)^{2}dx\right )dt\geq 0. \end{align*}

Dividing by $s$ and taking $s \searrow 0$ leads to

\begin{align*} \int _{0}^{T}\int _{\Omega }\varepsilon g(\!-\!\nabla u)&\partial _{t}u \varphi \, dx dt \geq -\liminf _{s\searrow 0}\int _{0}^{T}\frac{E_{\varepsilon }[u(t)+s\varphi (t)]- E_{\varepsilon }[u(t)]}{s}dt\\[5pt] &=-\frac{1}{2}\liminf _{s \searrow 0} \left (\varepsilon \int _{0}^{T}\int _{\Omega }\frac{f(\!-\!\nabla u - s\nabla \varphi )-f(\!-\!\nabla u)}{s}dx dt\vphantom{ + \frac{1}{\varepsilon }\int _{0}^{T}\int _{\Omega }\frac{W(u+s\varphi )-W(u)}{s}dx dt}\right . \\[5pt] &\hspace{8.5pt}\left .\vphantom{\varepsilon \int _{0}^{T}\int _{\Omega }\frac{f(\!-\!\nabla u - s\nabla \varphi )-f(\!-\!\nabla u)}{s}dx dt} + \frac{1}{\varepsilon }\int _{0}^{T}\int _{\Omega }\frac{W(u+s\varphi )-W(u)}{s}dx dt\right )\\[5pt] &=\frac{\varepsilon }{2}\int _{0}^{T}\int _{\Omega }Df(\!-\!\nabla u)\cdot \nabla \varphi \, dx dt-\frac{1}{2\varepsilon }\int _{0}^{T}\int _{\Omega }W^{\prime }(u)\varphi \, dxdt, \end{align*}

where the limit and the integrals can be interchanged in the last step because $u$ is essentially bounded.

We divide by $\varepsilon$ to obtain one inequality in (3.3). The converse inequality follows by taking $s \lt 0$ , $s \nearrow 0$ .

Proof of the claim. Given a function $w$ as in the claim, we define the difference quotient

\begin{equation*}D_{l}^{s}w(x)\;:\!=\;\frac {w(x+se_{l})-w(x)}{s} \end{equation*}

for $s\neq 0$ and $l=1,\ldots, d$ . Then $D_{l}^{s}w \in L^{2}\left (0,T;\;H^{1}(\Omega )\right )$ . The weak formulation of the PDE and the substitution $y=x+se_{l}$ yield

(3.38) \begin{align} {\int _{\Omega }} h D_{l}^{s}w\, dx &={\int _{\Omega }} Df(\!-\!\nabla u)\cdot \nabla D_{l}^{s}w \,dx \nonumber \\[5pt] &={\int _{\Omega }} Df(\!-\!\nabla u(x,t))\cdot \frac{\nabla w(x+se_{l})-\nabla w(x)}{s}\,dx \nonumber \\[5pt] &={\int _{\Omega }} \frac{Df(\!-\!\nabla u(x-se_{l}))-Df(\!-\!\nabla u(x))}{s} \cdot \nabla w(x)\,dx . \end{align}

In order to deal with the right-hand side, we will use the following identity for $i=1,2, \ldots, d$ and for almost every $x\in \Omega$ :

(3.39) \begin{align} \partial _{\xi _{i}}f(\!-\!\nabla u(x-se_{l}))\,{-}\,&\partial _{\xi _{i}}f(\!-\!\nabla u(x))\nonumber \\[5pt] &=-\sum _{j=1}^{d}\int _{0}^{1}\partial _{\xi _{i}\xi _{j}}^{2}f\left (-r\nabla u(x-se_{l})-(1-r)\nabla u(x)\right )dr\nonumber \\[5pt] &\quad \quad \quad \quad \cdot \left (\partial _{j}u(x-se_{l})-\partial _{j}u(x)\right )\nonumber \\[5pt] &=\sum _{j=1}^{d}A_{ij}^{s}(x)\left (\partial _{j}u(x)-\partial _{j}u(x-se_{l}) \right ), \end{align}

where the shorthand notation in the last line is to be read as

\begin{equation*}A_{ij}^{s}(x)\;:\!=\;\int _{0}^{1}\partial _{\xi _{i}\xi _{j}}^{2}f\left (-r\nabla u(x-se_{l})-(1-r)\nabla u(x)\right )dr. \end{equation*}

If the line segment between $\nabla u(x-se_{l})$ and $\nabla u(x)$ does not contain the origin, then $f$ is twice continuously differentiable in a neighborhood of the line segment, so that (3.39) follows immediately from the fundamental theorem of calculus. On the other hand, if $0 \in{\mathrm{conv}}\left (\left \{\nabla u(x-se_{l}),\nabla u(x,t)\right \}\right )$ , then (3.39) can be deduced by decomposing the line segment into two parts and applying the fundamental theorem of calculus on each part.

By the strong convexity of $f$ , there exists a constant $c\gt 0$ such that

\begin{equation*} \xi \cdot D^{2}f(p)\xi \geq c|\xi |^{2} \end{equation*}

for all $\xi \in{\mathbb{R}}^{d}$ , $p \in{\mathbb{R}}^{d}\setminus \{0\}$ . Using the definition of $A_{ij}^{s}$ , we obtain

(3.40) \begin{align} \nabla D^{-s}_l u \cdot A^s(x) \nabla D^{-s}_l u\geq c|\nabla D^{-s}_l u|^{2} \end{align}

for almost every $x\in \Omega$ .

One can now choose $w = D_{l}^{-s}u$ in (3.38) and use (3.39) with (3.40) as is standard (see, e.g., [Reference Gilbarg and Trudinger34, Lemma 7.23]) to conclude

\begin{equation*} \int _{\Omega }\left |\nabla D_{l}^{-s}u\right |^{2}dx \leq C{\int _{\Omega }} h^{2}\, dx \lt \infty . \end{equation*}

As $\nabla D_{l}^{-s}u \rightharpoonup \partial _{l}\nabla u$ as $s\to 0$ , this concludes the claim.

Proof.

1. (i) It suffices to show that

   \begin{equation*}\frac {1}{2}{\int _{0}^{T}\int _{\Omega }} \left (\varepsilon f(\!-\!\nabla u_{\varepsilon })+\frac {1}{\varepsilon }W(u_{\varepsilon }) \right ) \zeta dx dt \to {c_{0}} {\int _{0}^{T}\int _{\Omega }} \sigma (\nu )\zeta {\big |\nabla \chi \big |} \end{equation*}
   for all $\zeta \in C^{1}(\Omega \times [0,T])$ with $0\leq \zeta \leq 1$ . By a density and linearity argument, the same convergence then holds true for all $\zeta \in C(\Omega \times [0,T])$ .

   If $\zeta \in C^{1}(\Omega \times [0,T])$ and $0\leq \zeta \leq 1$ , we can use Young’s inequality and the chain rule for Sobolev functions to estimate

   \begin{align*} \frac{1}{2}{\int _{0}^{T}\int _{\Omega }} \left (\varepsilon f(\!-\!\nabla u_{\varepsilon })+\frac{1}{\varepsilon }W(u_{\varepsilon }) \right ) \zeta dx dt &\geq{\int _{0}^{T}\int _{\Omega }} \sqrt{f(\!-\!\nabla u_{\varepsilon })W(u_{\varepsilon }) } \,\zeta dx dt\\[5pt] &={\int _{0}^{T}\int _{\Omega }} \sigma (-\phi ^{\prime }(u_{\varepsilon })\nabla u_{\varepsilon }) \zeta dx dt\\[5pt] &={\int _{0}^{T}\int _{\Omega }} \sigma (\!-\!\nabla (\phi \circ u_{\varepsilon }))\zeta dx dt. \end{align*}
   By Lemma 2.5, we have that
   (4.18) \begin{equation} {\int _{0}^{T}\int _{\Omega }} \sigma (B)\,dxdt=\sup _{\eta \in C^{1}(\Omega \times [0,T])\atop \sigma{^{\circ }}(\eta )\leq 1}{\int _{0}^{T}\int _{\Omega }} B\cdot \eta \,dx dt \end{equation}
   for all $B\in L^{1}(\Omega \times (0,T))^{d}$ .

   We apply (4.18) and the $L^{1}$ -convergence $\phi \circ u_{\varepsilon }\to{c_{0}} \chi$ as $\varepsilon \searrow 0$ to the above estimate, which, recalling Lemma 2.3, yields

   (4.19) \begin{align} \liminf _{\varepsilon \searrow 0} \frac{1}{2}{\int _{0}^{T}\int _{\Omega }} &\left (\varepsilon f(\!-\!\nabla u_{\varepsilon })+\frac{1}{\varepsilon }W(u_{\varepsilon }) \right ) \zeta dx dt\nonumber \\[5pt] &\geq \liminf _{\varepsilon \searrow 0}{\int _{0}^{T}\int _{\Omega }} \sigma (\!-\!\nabla (\phi \circ u_{\varepsilon }))\zeta dx dt \nonumber \\[5pt] &= \liminf _{\varepsilon \searrow 0} \sup _{\eta }{\int _{0}^{T}\int _{\Omega }} \eta \cdot (\!-\!\nabla (\phi \circ u_{\varepsilon }))\zeta dx dt \nonumber \\[5pt] &= \liminf _{\varepsilon \searrow 0} \sup _{\eta }{\int _{0}^{T}\int _{\Omega }}(\phi \circ u_{\varepsilon })\mathrm{div}(\zeta \eta ) dx dt \nonumber \\[5pt] &\geq{c_{0}} \sup _{\eta }{\int _{0}^{T}\int _{\Omega }}\chi \mathrm{div}(\zeta \eta ) dx dt \nonumber \\[5pt] &= -{c_{0}} \sup _{\eta }{\int _{0}^{T}\int _{\Omega }} \zeta \eta \cdot \frac{\nabla \chi }{{\big |\nabla \chi \big |}}{\big |\nabla \chi \big |} \nonumber \\[5pt] &={c_{0}} \sup _{\eta }{\int _{0}^{T}\int _{\Omega }} \zeta \eta \cdot \nu{\big |\nabla \chi \big |}\nonumber \\[5pt] &={c_{0}}{\int _{0}^{T}\int _{\Omega }} \sigma (\nu )\zeta{\big |\nabla \chi \big |}, \end{align}
   where the supremum is taken over all $\eta \in C^{1}(\Omega \times [0,T])^{d}$ such that $\sigma{^{\circ }}(\eta )\leq 1$ . To prove the estimate from above, we observe that (4.19) also applies to the function $1-\zeta$ instead of $\zeta$ , and use the energy convergence assumption (4.3). Indeed,
   (4.20) \begin{align} \limsup _{\varepsilon \searrow 0} \frac{1}{2}{\int _{0}^{T}\int _{\Omega }} &\left (\varepsilon f(\!-\!\nabla u_{\varepsilon })+\frac{1}{\varepsilon }W(u_{\varepsilon }) \right ) \zeta dx dt\nonumber \\[5pt] &=\limsup _{\varepsilon \searrow 0} \left (\frac{1}{2}{\int _{0}^{T}\int _{\Omega }} \left (\varepsilon f(\!-\!\nabla u_{\varepsilon })+\frac{1}{\varepsilon }W(u_{\varepsilon }) \right ) dx dt \right .\nonumber \\[5pt] &\hspace{8.5pt}\left . - \frac{1}{2}{\int _{0}^{T}\int _{\Omega }} \left ({\varepsilon } f(\!-\!\nabla u_{\varepsilon })+\frac{1}{\varepsilon }W(u_{\varepsilon }) \right ) (1-\zeta ) dx dt \right )\nonumber \\[5pt] &=\lim _{\varepsilon \searrow 0} \int _{0}^{T}E_{\varepsilon }[u_{\varepsilon }(t)]dt\nonumber \\[5pt] &\quad{- \liminf _{\varepsilon \searrow 0}\frac{1}{2}}{\int _{0}^{T}\int _{\Omega }} \left (\varepsilon f(\!-\!\nabla u_{\varepsilon })+\frac{1}{\varepsilon }W(u_{\varepsilon }) \right ) (1-\zeta ) dx dt\nonumber \\[5pt] &\leq \int _{0}^{T}E[u(t)]dt -{c_{0}}{\int _{0}^{T}\int _{\Omega }} \sigma (\nu ) (1-\zeta ){\big |\nabla \chi \big |} \nonumber \\[5pt] &={c_{0}}{\int _{0}^{T}\int _{\Omega }} \sigma (\nu ) \zeta{\big |\nabla \chi \big |} . \end{align}
   The combination of the two inequalities (4.19) and (4.20) yields the first claim.

2. (ii) In the same manner as in (i), it suffices to show that

   \begin{equation*}{\int _{0}^{T}\int _{\Omega }} \sigma (\!-\!\nabla (\phi \circ u_{\varepsilon })) \zeta dx dt \to {c_{0}} {\int _{0}^{T}\int _{\Omega }} \sigma (\nu )\zeta {\big |\nabla \chi \big |} \end{equation*}
   whenever $\zeta \in C^{1}(\Omega \times [0,T])$ and $0 \leq \zeta \leq 1$ .

   The $\liminf$ inequality follows from the chain of inequalities in (4.19).

   For the $\limsup$ inequality, we can use Young’s inequality and (i) to compute

   \begin{align*} \limsup _{\varepsilon \searrow 0}{\int _{0}^{T}\int _{\Omega }} \sigma (\!-\!\nabla (\phi \circ u_{\varepsilon })) \zeta dx dt &\leq \lim _{\varepsilon \searrow 0}\frac{1}{2}{\int _{0}^{T}\int _{\Omega }} \left (\varepsilon f(\!-\!\nabla u_{\varepsilon })+\frac{1}{\varepsilon }W(u_{\varepsilon }) \right ) \zeta dx dt\\[5pt] &={c_{0}}{\int _{0}^{T}\int _{\Omega }} \sigma (\nu ) \zeta{\big |\nabla \chi \big |}. \end{align*}

3. (iii) By taking $\zeta \equiv 1$ as test functions for the weak-^* limits in (i) and (ii), we see that

   (4.21) \begin{align} \lim _{\varepsilon \searrow 0}\frac{1}{2}{\int _{0}^{T}\int _{\Omega }} \left (\varepsilon f(\!-\!\nabla u_{\varepsilon })+\frac{1}{\varepsilon }W(u_{\varepsilon }) \right ) &dx dt = \lim _{\varepsilon \searrow 0}{\int _{0}^{T}\int _{\Omega }} \sigma (\!-\!\nabla (\phi \circ u_{\varepsilon })) dx dt \nonumber \\[5pt] & ={c_{0}}{\int _{0}^{T}\int _{\Omega }} \sigma (\nu ){\big |\nabla \chi \big |}. \end{align}
   This observation allows us to compute
   \begin{align*}{\int _{0}^{T}\int _{\Omega }}& \left (\sqrt{\varepsilon f(\!-\!\nabla u_{\varepsilon })}-\sqrt{\frac{1}{\varepsilon }W(u_{\varepsilon })} \right )^{2} dx dt\\[5pt] &={\int _{0}^{T}\int _{\Omega }}\left (\varepsilon f(\!-\!\nabla u_{\varepsilon })+\frac{1}{\varepsilon }W(u_{\varepsilon }) \right ) dx dt - 2{\int _{0}^{T}\int _{\Omega }} \sqrt{f(\!-\!\nabla u_{\varepsilon })W(u_{\varepsilon })}dx dt\\[5pt] &={\int _{0}^{T}\int _{\Omega }}\left (\varepsilon f(\!-\!\nabla u_{\varepsilon })+\frac{1}{\varepsilon }W(u_{\varepsilon }) \right ) dx dt - 2{\int _{0}^{T}\int _{\Omega }} \sigma (\!-\!\nabla (\phi \circ u_{\varepsilon }))dx dt\\[5pt] &\longrightarrow 0 \qquad \text{as } \varepsilon \searrow 0. \end{align*}

4. (iv) Young’s inequality yields

   \begin{align*} \left |\varepsilon f(\!-\!\nabla u_{\varepsilon })-\frac{1}{\varepsilon }W(u_{\varepsilon })\right |&\leq \delta \left (\sqrt{\varepsilon f(\!-\!\nabla u_{\varepsilon })}+\sqrt{\frac{1}{\varepsilon }W(u_{\varepsilon })} \right )^{2} \\[5pt] &\quad + \frac{1}{4\delta }\left (\sqrt{\varepsilon f(\!-\!\nabla u_{\varepsilon })}-\sqrt{\frac{1}{\varepsilon }W(u_{\varepsilon })} \right )^{2} \end{align*}
   for all $\delta \gt 0$ . Taking the limit $\varepsilon \searrow 0$ and using (iii) as well as (4.21), we obtain
   \begin{align*} \limsup _{\varepsilon \searrow 0}{\int _{0}^{T}\int _{\Omega }}&\left |\varepsilon f(\!-\!\nabla u_{\varepsilon })-\frac{1}{\varepsilon }W(u_{\varepsilon })\right | dx dt \\[5pt] &\leq \delta \limsup _{\varepsilon \searrow 0}{\int _{0}^{T}\int _{\Omega }} \left (\sqrt{\varepsilon f(\!-\!\nabla u_{\varepsilon })}+\sqrt{\frac{1}{\varepsilon }W(u_{\varepsilon })} \right )^{2} dx dt\\[5pt] &= \delta \limsup _{\varepsilon \searrow 0}{\int _{0}^{T}\int _{\Omega }} \left (\varepsilon f(\!-\!\nabla u_{\varepsilon })+\frac{1}{\varepsilon }W(u_{\varepsilon }) + 2 \sigma (\!-\!\nabla (\phi \circ u_{\varepsilon })) \right ) dx dt\\[5pt] &= 4 \delta{c_{0}}{\int _{0}^{T}\int _{\Omega }} \sigma (\nu ){\big |\nabla \chi \big |}. \end{align*}
   Since $\delta \gt 0$ is arbitrary, it follows that
   \begin{equation*} \lim _{\varepsilon \searrow 0} {\int _{0}^{T}\int _{\Omega }}\left |\varepsilon f(\!-\!\nabla u_{\varepsilon })-\frac {1}{\varepsilon }W(u_{\varepsilon })\right | dx dt=0. \end{equation*}

5. (v) The two convergence results follow from adding and subtracting (i) and (iv), respectively.

Definition 4.5.

1. (i) Let $u=\chi \in BV(\Omega )$ take values in $\{0,1\}$ almost everywhere. As in the definition of the anisotropic surface energy, let $\nu \;:\!=\;-\frac{\nabla \chi }{|\nabla \chi |}$ be the measure-theoretic outer unit normal. The relative entropy of $u$ with respect to a vector field $\xi \in C(\Omega )^{d}$ is

   (4.28) \begin{equation} {\mathscr{E}\left [u\big |\xi \right ]}\;:\!=\;{c_{0}}\int _{\Omega }\left (\sigma (\nu )-|\xi |\psi (|\xi |)D\sigma (\xi )\cdot \nu \right ){\big |\nabla \chi \big |}, \end{equation}
   where $\psi \in C^{\infty }([0,\infty )$ is a cutoff function such that the three properties in (2.2) hold true.

2. (ii) Let $\varepsilon \gt 0$ and $u_{\varepsilon } \in H^{1}(\Omega )\cap L^{\infty }(\Omega )$ . The $\varepsilon$ -relative entropy of $u_{\varepsilon }$ with respect to a vector field $\xi \in C(\Omega )^{d}$ is

   (4.29) \begin{equation} {\mathscr{E}_{\varepsilon }\left [u_{\varepsilon }\big |\xi \right ]}\;:\!=\;\int _{\Omega }\left (\sigma (\!-\!\nabla u_{\varepsilon })+|\xi |\psi (|\xi |)D\sigma (\xi )\cdot \nabla u_{\varepsilon } \right )\sqrt{W(u_{\varepsilon })} dx. \end{equation}

Remark 4.6.

1. (i) For $u = \chi$ as in Definition 4.5(i), we can define the set of finite perimeter $A\;:\!=\;\left \{x \in \Omega \, \big |\,u(x)=1 \right \}$ (up to a null set). By De Giorgi’s structure theorem (see [Reference Ambrosio, Fusco and Pallara2, Theorem 3.59]), one can replace the total variation measure $|\nabla \chi |$ with the $(d-1)$ -dimensional Hausdorff measure and write

   $@@$
   with $\partial ^{*}$ denoting the reduced boundary of a set of finite perimeter.

2. (ii) One can also show that $|\nabla \chi (t)|\otimes{\mathcal{L}^{1}} (0,T) = |\nabla \chi |$ as Radon measures on $\Omega \times (0,T)$ . Thus, integrating the relative entropy over time yields

   \begin{align*} \int _{0}^{T}{\mathscr{E}\left [u(t)\big |\xi (t)\right ]}dt & ={c_{0}}{\int _{0}^{T}\int _{\Omega }} \left (\sigma (\nu )-|\xi |\psi (|\xi |)D\sigma (\xi )\cdot \nu \right )|\nabla \chi (t)|dt\\[5pt] &={c_{0}}{\int _{0}^{T}\int _{\Omega }} \left (\sigma (\nu )-|\xi |\psi (|\xi |)D\sigma (\xi )\cdot \nu \right ){\big |\nabla \chi \big |}. \end{align*}

Proof. A direct computation yields

(4.31) \begin{align} \int _{0}^{T}{\mathscr{E}_{\varepsilon }\left [u_{\varepsilon }(t)\big |\xi (t)\right ]}dt&= \int _{0}^{T}\int _{\Omega }\left (\sigma (\!-\!\nabla u_{\varepsilon })+|\xi |\psi (|\xi |)D\sigma (\xi )\cdot \nabla u_{\varepsilon } \right )\sqrt{W(u_{\varepsilon })}dxdt\nonumber \\[5pt] &=\int _{0}^{T}\int _{\Omega }\sigma\!\left (\!-\!\nabla \left (\phi \circ u_{\varepsilon }\right )\right )dxdt\nonumber \\[5pt] &\quad +\int _{0}^{T}\int _{\Omega }|\xi |\psi (|\xi |)D\sigma (\xi )\cdot \nabla (\phi \circ u_{\varepsilon })dxdt\nonumber \\[5pt] &\to{c_{0}} \int _{0}^{T}\sigma (\nu ){\big |\nabla \chi \big |} +{c_{0}}\int _{0}^{T}\int _{\Omega }|\xi |\psi (|\xi |)D\sigma (\xi )\cdot \nabla \chi \nonumber \\[5pt] &=\int _{0}^{T}{\mathscr{E}\left [u(t)\big |\xi (t)\right ]}dt, \end{align}

where the first term converges due to Theorem 4.3(ii). The convergence of the second term follows from the observation that $\nabla (\phi \circ u_{\varepsilon }){\overset{{*}}{\rightharpoonup }}{c_{0}} \nabla \chi$ in ${\mathcal{M}}(\Omega \times [0,T])$ , which is a consequence of the bound in (4.7) and the $L^{1}$ -convergence $\phi \circ u_{\varepsilon }\to{c_{0}} \chi$ .

Proof. The outer unit normal $\nu$ is $\big |\nabla \chi \big |$ -measurable, and the estimate

\begin{equation*} \int _{0}^{T}\int _{\Omega }|\nu |^{2}{\big |\nabla \chi \big |} = \int _{0}^{T}\int _{\Omega }{\big |\nabla \chi \big |} \leq \frac {1}{c_{0}\min _{S^{d-1}}\sigma }\int _{0}^{T}E[u(t)]dt\lt \infty \end{equation*}

shows that $\nu \in L^{2}({\big |\nabla \chi \big |})$ . Since $\big |\nabla \chi \big |$ is a Radon measure on $\Omega \times [0,T]$ , there exists an approximating sequence $\{\xi _{n}\}_{n\in \mathbb{N}} \subset C^{\infty }(\Omega \times [0,T])$ such that $\xi _{n}\to \nu$ in $L^{2}({\big |\nabla \chi \big |})$ as $n \to \infty$ .

Clearly, $\nu$ takes values in the closed convex set $\overline{B}_{1}=\left \{p \in{\mathbb{R}}^{d}\,\Big |\,|p|\leq 1 \right \}$ almost everywhere with respect to $\big |\nabla \chi \big |$ , so we can choose $\xi _{n}$ in a way that ensures that $|\xi _{n}|\leq 1$ on $\Omega \times [0,T]$ for all $n \in{\mathbb{N}}$ .

By Lemma 2.4(ii) and the Cauchy–Schwarz inequality, we obtain

\begin{align*} \int _{0}^{T}{\mathscr{E}\left [u(t)\big |\xi _{n}(t)\right ]}dt &\leq{c_{0}}\, C_{\sigma }\int _{0}^{T}\int _{\Omega }\left (|\nu -\xi _{n}|^{2}+(|\nu |-|\xi _{n}|)\right ){\big |\nabla \chi \big |}\\[5pt] &\leq{c_{0}}\, C_{\sigma }\int _{0}^{T}\int _{\Omega }|\nu -\xi _{n}|^{2}{\big |\nabla \chi \big |}\\[5pt] &\quad +{c_{0}}\, C_{\sigma }\left (\int _{0}^{T}\int _{\Omega }{\big |\nabla \chi \big |}\right )^{\frac{1}{2}}\left (\int _{0}^{T}\int _{\Omega }\left |\nu -\xi _{n}\right |^{2}{\big |\nabla \chi \big |}\right )^{\frac{1}{2}}\\[5pt] &\leq{c_{0}}\, C_{\sigma }\left \|\nu -\xi _{n}\right \|_{L^{2}(|\nabla \chi |)}^{2} \\[5pt] &\quad +\sqrt{\frac{{c_{0}}}{\min _{S^{d-1}}\sigma }} C_{\sigma }\left (\int _{0}^{T}E[u(t)]dt\right )^{\frac{1}{2}}\left \|\nu -\xi _{n}\right \|_{L^{2}(|\nabla \chi |)}\\[5pt] &\longrightarrow 0 \end{align*}

as $n \to \infty$ .

Proof. We apply Lemma 2.4(i), the estimate $|\nu _{\varepsilon }-\xi |\leq 2$ , and the Cauchy–Schwarz inequality in order to control the occurring integrals by the anisotropic Cahn–Hilliard energy and the $\varepsilon$ -relative entropy. Precisely,

\begin{align*}{\int _{0}^{T}\int _{\Omega }} &\varepsilon f(\!-\!\nabla u_{\varepsilon })|\nu _{\varepsilon }-\xi |^{2}dxdt\\[5pt] & ={\int _{0}^{T}\int _{\Omega }} \sqrt{W(u_{\varepsilon })}\sigma (\!-\!\nabla u_{\varepsilon })|\nu _{\varepsilon }-\xi |^{2}dxdt\\[5pt] &\quad +{\int _{0}^{T}\int _{\Omega }} \left (\sqrt{\varepsilon f(\!-\!\nabla u_{\varepsilon })}-\sqrt{\frac{1}{\varepsilon }W(u_{\varepsilon })}\right )\sqrt{\varepsilon }\sigma (\!-\!\nabla u_{\varepsilon })|\nu _{\varepsilon }-\xi |^{2}dxdt\\[5pt] &\leq \frac{\max _{S^{d-1}}\sigma }{c_{\sigma }}{\int _{0}^{T}\int _{\Omega }} \sqrt{W(u_{\varepsilon })}|\nabla u_{\varepsilon }|\left (\sigma (\nu _{\varepsilon })-|\xi |\psi (|\xi |)D\sigma (\xi )\cdot \nu _{\varepsilon } \right )dxdt\\[5pt] &\quad +4\left ({\int _{0}^{T}\int _{\Omega }} \varepsilon f(\!-\!\nabla u_{\varepsilon })\,dxdt\right )^{\frac{1}{2}}\left ({\int _{0}^{T}\int _{\Omega }} \left (\sqrt{\varepsilon f(\!-\!\nabla u_{\varepsilon })}-\sqrt{\frac{1}{\varepsilon }W(u_{\varepsilon })}\right )^{2}dxdt\right )^{\frac{1}{2}}\\[5pt] &\leq \frac{\max _{S^{d-1}}\sigma }{c_{\sigma }}\int _{0}^{T}{\mathscr{E}_{\varepsilon }\left [u_{\varepsilon }(t)\big |\xi (t)\right ]}dt\\[5pt] &\quad +4\sqrt{2}\left (\int _{0}^{T}E_{\varepsilon }[u_{\varepsilon }(t)]dt\right )^{\frac{1}{2}}\left ({\int _{0}^{T}\int _{\Omega }} \left (\sqrt{\varepsilon f(\!-\!\nabla u_{\varepsilon })}-\sqrt{\frac{1}{\varepsilon }W(u_{\varepsilon })}\right )^{2}dxdt\right )^{\frac{1}{2}}. \end{align*}

One can now take the limit superior on both sides and note that, by assumption, we have $\limsup _{\varepsilon \searrow 0}\int _{0}^{T}E_{\varepsilon }[u_{\varepsilon }(t)]dt\lt \infty$ . Thus, applying Lemma 4.7 in the first term and Theorem 4.3(iii) in the second term yields

\begin{equation*} \limsup _{\varepsilon \searrow 0}{\int _{0}^{T}\int _{\Omega }}\varepsilon f(\!-\!\nabla u_{\varepsilon })|\nu _{\varepsilon }-\xi |^{2}dxdt\leq \frac {\max _{S^{d-1}}\sigma }{c_{\sigma }}\int _{0}^{T}{\mathscr {E}\left [u(t)\big |\xi (t)\right ]}dt. \end{equation*}