\section{Introduction}

Let $\Omega \subset \mathbb R^m$ be an open, bounded domain with Lipschitz boundary and let $F\colon \mathbb R^{m \times n} \to [0, \infty[$ be a~convex function. We consider the functional $\widetilde{\mathcal F} \colon C^1(\Omega, \mathbb R^{m \times n})\to [0, \infty]$ given by
\begin{equation} \label{tildeF}
	\widetilde{\mathcal F}(w) = \int_{\Omega} F(w_x) \,\mathrm d \mathcal L^m
\end{equation}
and $\mathcal F \colon L^2(\Omega, \mathbb R^{m \times n}) \to [0, \infty]$ defined as a lower semicontinuous envelope of $\widetilde{\mathcal F}$, i.\,e.
\begin{equation}\label{L2_env}
	\mathcal F(w) = \inf\left\{\liminf_{\ell \to \infty} \widetilde{\mathcal F}(w^\ell), \ (w^\ell) \subset C^1(\Omega, \mathbb R^{m \times n}),\ w^\ell \to w \text{ in } L^1_{loc}(\Omega, \mathbb R^n)  \right\}.
\end{equation}
We define a generalized Rudin-Osher-Fatemi functional $\mathcal E \colon L^2(\Omega, \mathbb R^n) \to [0, \infty]$ by
   \begin{equation} \label{ROF_def}
   	 \mathcal E_h^\lambda(w) = \lambda \mathcal F(w) + \frac{1}{2} \int_{\Omega} |w - h|^2 \,\mathrm d \mathcal L^m,
   \end{equation}
where $h \in L^2(\Omega, \mathbb R^n)$, $\lambda > 0$. We will drop indices $h, \lambda$ as long as we are not interested in dependence on them. By convexity, this functional is weakly lower semicontinuous. Moreover, it is proper and its sublevel sets are weakly compact. Therefore it admits a minimizer $u$, which is unique since $\mathcal E$ is strictly convex on its domain (i.\,e.\ the subset of $L^2(\Omega, \mathbb R^n)$ where $\mathcal E$ is finite).

In the superlinear case, where $F(\xi) \sim |\xi|^p$, $p > 1$, the domain of $\mathcal F$ (and $\mathcal E$) coincides with $L^2(\Omega, \mathbb R^n)\cap W^{1,p}(\Omega, \mathbb R^n)$, and on this set $\mathcal F$ is given by formula \eqref{tildeF}. Analogous statement is no longer true if $p = 1$, since $W^{1,1}(\Omega, \mathbb R^n)$ is not reflexive. Here we consider general $F$ of \emph{linear growth}, i.\,e.
\begin{equation} \label{lin_growth}
	F(\xi) \leq C_F^+ |\xi| + B_F^+ \quad \text{for } \xi \in \mathbb R^{m \times n}
\end{equation}
with $C_F^+ > 0$, $B_F^+ \in \mathbb R$. In this case, the domain of $\mathcal F$ includes $L^2(\Omega, \mathbb R^n)\cap BV(\Omega, \mathbb R^n)$. Recall that $BV(\Omega, \mathbb R^n)$ is the space of those elements of $L^1(\Omega, \mathbb R^n)$ whose distributional derivative $w_x$ belongs to $M(\Omega, \mathbb R^{m \times n})$, the space of vector-valued Radon measures. The characterization of $\mathcal F$ restricted to $BV(\Omega, \mathbb R^n)$ was obtained by Goffman and Serrin in \cite{GS64} and is described in more detail in Section 2. Given a~$BV$ mapping $w$, one can write decomposition of its derivative $w_x$ into absolutely continuous and singular parts with respect to the Lebesgue measure as $\,\mathrm d w_x = w_x^{ac} \,\mathrm d  \mathcal L^m + \,\mathrm d w_x^s$. Consequently, we can define a measure $F(w_x^s) \in M(\Omega)$ by
\begin{equation} \label{Fwx}
	\,\mathrm d F(w_x) = F(w_x^{ac}) \,\mathrm d \mathcal L^m + F^{\infty}\left(\tfrac{w_x^s}{|w_x^s|}\right) \,\mathrm d |w_x^s|, \quad \text{where } F^{\infty}(\nu) = \lim_{t \to \infty} \tfrac{1}{t}F(t \nu) \text{ for } \nu \in \mathbb S^{n-1}.
\end{equation}
Then, one can prove that for $w \in BV(\Omega, \mathbb R^n)$
\begin{equation}
    \mathcal F(w) = F(w_x)(\Omega).
\end{equation}
If moreover $F$ is \emph{coercive}, i.\,e.
\begin{equation} \label{coercivity}
	F(\xi) \geq C_F^-|\xi| + B_F^-\quad \text{for } \xi \in \mathbb R^{m \times n}
\end{equation}
with $C_F^- > 0$, $B_F^- \in \mathbb R$, then the domain of $\mathcal F$ coincides with $L^2(\Omega, \mathbb R^n)\cap BV(\Omega, \mathbb R^n)$. In the case $F = |\cdot|$, $\mathcal F$ coincides with the standard $BV$ seminorm also known as the total variation, while $\mathcal E$ coincides with the classical Rudin-Osher-Fatemi functional, see \cite{rof, ChambolleLions}.

A natural question arises, what can be said about the singular part of $u_x$. So far, this question has been mostly investigated in the scalar-valued case $n=1$. Let us report briefly on known results in this direction. Under the assumption that $\Omega$ is convex, it has been proved in \cite{LasicaRybka} that
\[F^\infty(u_x^s)(\Omega) \leq F^\infty(h_x^s)(\Omega).\]
In particular, if $h \in W^{1,1}(\Omega)$, then $u \in W^{1,1}(\Omega)$. On the other hand, in the case where $F$ is the $\ell^1$ norm, one can construct an example of non-convex polygon $\Omega \subset \mathbb R^2$ and $h \in C^\infty(\overline{\Omega})$ such that $u$ has a jump discontinuity in $\Omega$ \cite{tetris}. This sort of singularity does not appear if $F$ is the Euclidean norm, in which case $|u_x^j| \leq |h_x^j|$ as measures, i.\;e.
\[|u_x^j|(V) \leq |h_x^j|(V) \quad \text{for Borel } V \subset \Omega\]
for $h \in BV(\Omega) \cap L^\infty(\Omega)$, where $w_x^j$ denotes the jump part of $w_x$ for $w \in BV(\Omega)$ \cite{jalalzaijump}. This estimate has been generalized to a wider class of functions $F$ in \cite{Valkonen2015}. To our knowledge, it remains an open question whether an analogous estimate holds for the whole singular part of $u_x$. Moreover, no similar estimate on the singular or jump part of $u_x$ is known in the vector-valued setting $n > 1$, except in the one-dimensional case $m=1$.

From now on we will restrict ourselves to $m=1$. In the one-dimensional, scalar-valued case $m=n=1$, for $F = |\cdot|$, it has been proved in \cite{bcno, BonforteFigalli} that
\begin{equation} \label{1d_homo_est}
	|u_x| \leq |h_x| \quad \text{as measures.}
\end{equation}
The proof in \cite{bcno} follows by analysis of level sets of $u$, thus depending on the linear order of the range $\mathbb R$. On the other hand, the reasoning in \cite{BonforteFigalli} is based on explicit description of minimizers for piecewise constant $h$. Inequality \eqref{1d_homo_est} was generalized to the vector-valued case $n > 1$ in \cite{giacomellilasica} using integral estimates.

In this paper, our goal is to obtain estimates on the singular part of the minimizer in the case $m=1$, $n>1$ for possibly general $F$. We will need a structural assumption formally similar to the one in \cite{mercier} (note however that in \cite{mercier} $m>1$, $n=1$ and the results are not closely related). We recall the following definition (compare e.\,g.\ \cite{ChambolleNovaga2015}). We say that a function $\varphi \colon \mathbb R^n \to [0, +\infty[$ is an \emph{anisotropy} if it is convex and positively $1$-homogeneous, i.\,e.
	\[\varphi(t p) = t \varphi(p) \quad \text{for } p \in \mathbb R^n,\ t \geq 0.\]
	For any anisotropy $\varphi$, we denote
	\[c_{\varphi}^+ = \max_{0 \neq p \in \mathbb R^n} \frac{\varphi(p)}{|p| }, \quad c_{\varphi}^- = \min_{0 \neq p \in \mathbb R^n} \frac{\varphi(p)}{|p|}. \]
An anisotropy $\varphi$ will be called \emph{coercive} if $c_\varphi^- > 0$.  Note that any even anisotropy is a seminorm (and vice versa) and any coercive, even anisotropy is a norm (and vice versa). We prove the following three local estimates on the variation measure of minimizers of $\mathcal E$ in terms of the variation measure of the datum $h$ under varying assumptions on $F$.
\begin{thm} \label{thm:homo}
	Let $F\equiv\varphi \colon \mathbb R^n \to [0, \infty[$ be an anisotropy. Suppose that $h \in BV(U, \mathbb R^n)$ for an open interval $U \subset I$. Then the minimizer $u$ of $\mathcal E$ satisfies $u \in BV(U, \mathbb R^n)$ and $|u_x| \leq |h_x|$ in the sense of Borel measures on $U$, i.\,e.
	\begin{equation} \label{homo_est}
		|u_x|(E) \leq |h_x|(E) \text{ for any Borel set } E \subset U.
	\end{equation}
\end{thm}

\begin{thm} \label{thm:constants}
	Let $F = f \circ \varphi$, where $f\colon[0, \infty[ \to [0, \infty[$ is convex, increasing and of linear growth, and $\varphi\colon \mathbb R^n \to \mathbb R$ is a coercive anisotropy. Suppose that $h \in BV(U, \mathbb R^n)$ for an open interval $U \subset I$. Then the minimizer $u$ of $\mathcal E$ satisfies $|u_x^{s}| \leq \left( \frac{c_\varphi^+}{c_\varphi^-} \right)^2|h_x^{s}|$ in the sense of Borel measures on $U$, i.\,e.
	\begin{equation} \label{eq:constants}
		|u_x^{s}|(E) \leq \left( \tfrac{c_\varphi^+}{c_\varphi^-} \right)^2 |h_x^{s}|(E) \text{ for any Borel set } E \subset U.
	\end{equation}
\end{thm}

\begin{thm} \label{thm:regular norm}
	Let $F = f \circ \varphi$, where $f\colon [0, \infty[ \to [0, \infty[$ is strictly convex, increasing and of linear growth, $\varphi\colon \mathbb R^n \to \mathbb R$ is a~coercive anisotropy which is strictly convex in the sense of Reshetnyak (see Definition \ref{def: Reshetnyak}). Moreover, assume that $\varphi$ is differentiable in $\mathbb R^n \setminus \{0\}$ and that $f$ is differentiable with $f'(0) = 0$. Suppose that $h \in BV(U, \mathbb R^n)$ for an open interval $U \subset I$. Then the minimizer $u$ of $\mathcal E$ satisfies $|u_x^{s}| \leq |h_x^{s}|$ in the sense of Borel measures on $U$, i.e.
	\begin{equation} \label{eq: regular norm}
		|u_x^{s}|(E) \leq |h_x^{s}|(E) \text{ for any Borel set } E \subset U.
	\end{equation}
\end{thm}

Without the assumption on coercivity of the anisotropy in Theorem~\ref{thm:homo}, it is not immediately clear that the minimizer $u$ is a~BV function. This difficulty does not appear in Theorems \ref{thm:constants} and \ref{thm:regular norm} since any convex, increasing function $f$ on $[0, \infty[$ satisfies $f(s) \geq cs + b$ for some $a > 0$, $b \in \mathbb R$. Hence, if $\varphi$ is a~coercive anisotropy, then $F = f \circ \varphi$ satisfies \eqref{coercivity}.

The proofs of the three theorems are similar in spirit -- a two-level approximation of the functional $\mathcal E$ is used. On the more regular level, the minimizer $u^{\varepsilon, \eta}$ satisfies the Euler-Lagrange equation in strong sense. We test the equation with a suitable function of form $\zeta^2 G(u_x^{\varepsilon, \eta})$, where $\zeta \in C^1_c$, while $G$ is chosen so that $G(w_x)_x \cdot DF(w_x)_x \geq 0$ for $w \in W^{2,2}(\Omega, \mathbb R^n)$ and $G(\xi) \cdot \xi \sim |\xi|$ for large $|\xi|$. It is essential for our argument to take $G$ of form $G(\xi) = g(\xi) \xi$. Due to such choice, the bulk of terms involving $\zeta_x$ vanishes, owing to equality $\xi \cdot D^2\varphi(\xi) = 0$, which follows from homogeneity of $\varphi$. In the homogeneous case $F = \varphi$, performing this procedure with a relatively simple $g$ readily leads to the strongly local estimate \eqref{homo_est}.

In the inhomogeneous case $F = f \circ \varphi$, the argument becomes more involved and, expectedly, yields a bound only on the singular part of $u_x$. One of essential difficulties in the proof of such estimates is posed by very weak continuity properties of the operator $w \mapsto w_x^s$ on $BV(I, \mathbb R^n)$. Notably, maps such as $w \mapsto |w_x^s|(\Omega)$ fail to be semicontinuous w.\,r.\,t.\;weak* as well as strict or area-strict convergence in $BV(I, \mathbb R^n)$. Therefore, we need to work instead with quantities of type
\[ (|w_x| - k)_+ = \int (|w_x^{ac}| - k)_+ \,\mathrm d \mathcal L^1 +  |w_x^s| \]
and obtain estimates for their decay as $k \to \infty$. However, our previously described procedure delivers estimates for more complicated, non-convex functions of $u_x$ which again fail to be lower semicontinuous.
Nonetheless, by exploiting equivalence between $\varphi$ and $|\cdot|$, we can deduce estimates for a convex function of $u_x$. This is why in Theorem \ref{thm:constants} we require coercivity of $\varphi$ and only obtain an estimate with a multiplicative constant.

We note that such a constant is undesirable in the context of image processing, since in applications the minimization procedure tends to be iterated many times (in the spirit of the  \emph{minimizing movements scheme} -- we recall that notion in the next paragraph). In the case that $F$ is strictly convex and differentiable, we obtain a better estimate \eqref{eq: regular norm}. In the proof we use spherical compactification of $\mathbb R^n$ (see Section~\ref{sec: prelim}) and the fact that the derivative $DF$ is a~homeomorphism to improve the convergence of approximate minimizers. This is one of the reasons why we need additional assumptions on $F$ in this case. However, we do not know whether \eqref{eq: regular norm} can fail otherwise.

Let us recall that the convex, lower semicontinuous function $\mathcal F$ on the Hilbert space $L^2(\Omega, \mathbb R^n)$ defines a unique gradient flow \cite{maximaux}. In other words, for any $v_0 \in L^2(\Omega, \mathbb R^n)$ there exists a unique function $v \in C([0, \infty[, L^2(\Omega, \mathbb R^n))\cap H^1_{loc}(0,T;L^2(\Omega, \mathbb R^n))$ such that
\begin{equation} \label{grad_flow}
	v_t(t) \in - \partial \mathcal F(v(t)) \quad \text{ for a.\,e.\ } t > 0, \qquad v(0) = v_0.
\end{equation}
The map associating the minimizer of $\mathcal E_h^\lambda$ to a given $h$ coincides with the resolvent operator for the subdifferential $- \partial \mathcal F$. This is the basis of the \emph{minimizing movements scheme}: for a given $N \in \mathbb N$ we iteratively define \[v^N_0 = v_0, \qquad v^N_j = \mathrm{arg\,min}\, E_{v^N_{j-1}}^{1/N} \quad \text{for } j=1,2,\ldots\]
Then, for a.\,e.\ $t > 0$ we have
\begin{equation} \label{grad_flow_conv}
	v^N_j \to v(t) \quad \text{in } L^2(I, \mathbb R^n) \quad \text{if } N \to \infty,\ j/N \to t.
\end{equation}
Let $F=\varphi$ is an anisotropy. Then, by Theorem \ref{thm:homo}, for all $N,j \in \mathbb N$, $v^N_j \in BV(U,\mathbb R^n)$ and
\begin{equation*}
	|(v^N_j)_x |(V) \leq |(v^N_{j-1})_x |(V) \leq \ldots \leq |(v_0)_x |(V)
\end{equation*}
for any open $V \subset U$. Using \eqref{grad_flow_conv} and Theorem \ref{thm-gs-lsc}, we deduce $|v_x(t)|(V) \leq |(v_0)_x|(V)$ for any open $V \subset U$ and a.\,e.\ $t>0$. Thus, we obtain
\begin{cor}\label{thm: grad flows}
	Suppose that $F = \varphi$ is an anisotropy. If $v_0 \in BV(U, \mathbb R^n)$ with open $U \subset I$, then $v(t) \in BV(U, \mathbb R^n)$ and $|v_x(t)| \leq |v_{0,x}|$ in the sense of Borel measures on $U$, i.\,e.
	\begin{equation}\label{gf_thm_1}
	|v_x(t)|(E) \leq |v_{0,x}|(E) \quad \text{for any Borel set } E \subset U.
	\end{equation}
\end{cor}
We are unable to transfer Theorems \ref{thm:constants} and \ref{thm:regular norm} to the gradient flow setting since $|u_x^s|$ does not have good semicontinuity properties. On the other hand, estimates \eqref{eq:constants final} and \eqref{eq: before k limit} do not behave well under iteration.

In the next section we recall some facts that will be used in the sequel. In Section~\ref{sec: approximate} approximate functionals are defined and in Sections \ref{sec:homo}, \ref{sec:coercive}, \ref{sec:regular} we prove Theorems \ref{thm:homo}, \ref{thm:constants} and\ref{thm:regular norm}, respectively.

\section{Preliminaries} \label{sec: prelim}

Throughout the paper, $\varrho \in C^\infty_c(\mathbb R^d)$ denotes a~positive, radially symmetric function such that $\int_{\mathbb R^d} \varrho \,\mathrm d \mathcal L^d = 1$ and $\varrho_\varepsilon(p) = \varepsilon^{-d} \varrho(p/\varepsilon)$. We will say that measures $\mu_j \in M(\Omega, \mathbb R^d)$ converge weakly* to $\mu \in M(\Omega, \mathbb R^d)$ and write $\mu_j \mathrel{\ensurestackMath{\stackon[0.5pt]{\rightharpoonup}{\scriptstyle\ast}}} \mu$, if for any $\zeta \in C_c(\Omega)$ it is true that $\lim_{j \to \infty} \int_\Omega \zeta \,\mathrm d \mu_j = \int_\Omega \zeta \,\mathrm d \mu$.

\subsection*{Functions of measures}

Definition \eqref{Fwx} from the Introduction can be applied to any Radon measure, not necessarily a~derivative of a~$BV$ function. Throughout this subsection, $G$ denotes~a convex function $G:\mathbb R^d \to [0, \infty[$ of linear growth. Given such a~function and a~measure $\mu \in M(\Omega, \mathbb R^d)$, set
\begin{equation} \label{Fmu}
	\,\mathrm d G(\mu) = G(\mu^{ac}) \,\mathrm d \mathcal L^m + G^{\infty}\left(\tfrac{\mu^s}{|\mu^s|}\right) \,\mathrm d |\mu^s|, \quad \text{where } G^{\infty}(\nu) = \lim_{t \to \infty} \tfrac{1}{t}F(t \nu) \text{ for } \nu \in \mathbb S^{n-1}.
\end{equation}
The function $G^\infty$ is often called a~recession function. In the following paragraphs, facts about properties of $G(\mu)$ are collected.

\begin{thm}\cite[p.\ 172]{GS64} \label{thm-gs-lsc}
	Let $\mu_j, \mu \in M(\Omega, \mathbb R^d)$ and assume that $\mu_j \mathrel{\ensurestackMath{\stackon[0.5pt]{\rightharpoonup}{\scriptstyle\ast}}} \mu$ on $\Omega$. Then for any open $U \subset \Omega$,
	\begin{equation*}
		G(\mu)(U) \leq \liminf_{j \to \infty} G(\mu_j)(U).
	\end{equation*}
\end{thm}

\begin{cor} \label{cor-gs-lsc-zeta}
	Let $\mu_j, \mu \in M(\Omega, \mathbb R^n)$ and assume that $\mu_j \mathrel{\ensurestackMath{\stackon[0.5pt]{\rightharpoonup}{\scriptstyle\ast}}} \mu$ on $\Omega$. Then for any non-negative $\zeta \in C_c(\Omega)$,
	\begin{equation*}
		\int_\Omega \zeta \,\mathrm d G(\mu) \leq \liminf_{j \to \infty} \int_\Omega \zeta \,\mathrm d G(\mu_j).
	\end{equation*}
\end{cor}

\begin{proof}
	Suppose that $\liminf_{j \to \infty} \int_\Omega \zeta \,\mathrm d G(\mu_j) < + \infty$ (if not, the inequality clearly holds). Choose a~subsequence $\mu_{j_\ell}$ such that
\begin{equation*}
	\lim_{\ell \to \infty} \int_\Omega \zeta \,\mathrm d G(\mu_{j_\ell}) = \liminf_{j \to \infty} \int_\Omega \zeta \,\mathrm d G(\mu_j).
\end{equation*}
Since $G$ is of at most linear growth, the sequence of measures $\left \{ G(\mu_{j_\ell}) \right \}_\ell$ is bounded and hence there exists a~subsequence (which we do not relabel) weakly* convergent to a~measure $\nu$. Both $\nu$ and $G(\mu)$ are finite on $\Omega$ and therefore there exists only countably many balls $V$ for which $\nu(\partial V) >0$ and $G(\mu)(\partial V) > 0$. Therefore, at each point $x \in \Omega$ there exists a~family of open balls $\left \{ V_x^i \right \}_{i=1}^{N(x)}$ centered at $x$ with $\mathrm{diam}\left ({V_x^i} \right) \to 0$ as $i \to \infty$ for which
\[
	\nu(\partial V) = G(\mu)(\partial V) = 0.
\]

The family $\left \{ \overline{V_x^i}: \, x \in J, \, i = 1, \ldots N(x) \right \}$ of closed balls satisfies assumptions of a~corollary to Besicovitch's covering theorem (see \cite[Section 1.5, Corollary I]{evansgariepy}). For any open $U \subset \Omega$, it is then possible to choose a~countable subfamily $\mathcal{G}$ of disjoint balls such that
\begin{equation} \label{eq: family G}
	\mathsmaller{\bigcup} \mathcal{G} \subset U \text{ and } \left [\nu + G(\mu) \right] \left( U \setminus \mathsmaller{\bigcup} \mathcal{G} = 0 \right).
\end{equation}
For any ball $V$ with $\overline{V} \in \mathcal{G}$, it is true that $G(\mu_{j_\ell})(V) \to \nu(V)$ and using Theorem \ref{thm-gs-lsc}, we may write
\begin{equation*}
	\nu (\overline{V}) = \nu(V) = \lim_{\ell \to \infty} G(\mu_{j_\ell})(V) \geq G(\mu)(V) = G(\mu)(\overline{V}).
\end{equation*}
Summing over balls from the family $\mathcal{G}$ and using \eqref{eq: family G} results in the estimate $\nu(U) \geq G(\mu)(U)$ for any open $U \subset \Omega$, which implies that
\[
\nu(E) \geq G(\mu)(E) \text{ for any Borel } E \subset \Omega.
\]
As a result, in view of the choice of the subsequence $\mu_{j_\ell}$, one gets
\[
\int_\Omega \zeta \,\mathrm d G(\mu) \leq \int_\Omega \zeta \,\mathrm d \nu = \lim_{\ell \to \infty} \int_\Omega \zeta \,\mathrm d G(\mu_{j_\ell}) = \liminf_{j \to \infty} \int_\Omega \zeta \,\mathrm d G(\mu_j).
\]
\end{proof}

\begin{lemma}\cite[p.\ 171]{GS64} \label{lem: jensen GS}
Let $U \subset \mathbb R^m$ be an open set and $b: U \to [0, \infty[$ a~function with $\int_U b \,\mathrm d \mathcal L^m \neq 0$. For $\mu \in M(U, \mathbb R^d)$, it is true that
\[
	G \left( \frac{\int_U b \,\mathrm d \mu}{\int_U b \,\mathrm d \mathcal L^m} \right) \leq \frac{\int_U b \,\mathrm d G\mu}{\int_U b \,\mathrm d \mathcal L^m}.
\]
\end{lemma}

\begin{lemma} \label{lem: G weaklystar}
Suppose that $w \in L^1(\Omega) \cap BV(U)$ for some open set $U \subsetneq \Omega$ and let $w^\varepsilon = w \ast \varrho_\varepsilon$. Then, $G(w_x^\varepsilon) \mathrel{\ensurestackMath{\stackon[0.5pt]{\rightharpoonup}{\scriptstyle\ast}}} G(w_x)$ on $U$.
\end{lemma}

\begin{proof}
	The statement follows from a small modification of \cite[Theorem 2.2]{ambrosiofuscopallara}, which requires the use of Lemma \ref{lem: jensen GS}.
\end{proof}

The following theorem establishes coincidence of the measure $G(w_x)(\Omega)$ with the following relaxation, as mentioned in the Introduction.
\begin{thm}\cite[Theorem 5]{GS64}
Let $G: \mathbb R^{m \times n} \to [0, \infty[$ be a~convex function of linear growth and $w \in BV(\Omega, \mathbb R^n)$. Then,
	\begin{align*}
	G(w_x)(\Omega)
	&= \inf \left \{ \liminf_{\ell \to \infty} \int_\Omega G(w_x^\ell) \,\mathrm d \mathcal L^m: \, w^\ell \in C^1(\Omega), \, w^\ell \to w \text{ in } L^1_{\mathrm{loc}}(\Omega) \right \}.
	\end{align*}
\end{thm}

Lastly, let us quote the following fact from measure theory.
\begin{lemma} \label{lem: mu leq nu}
	Let $\mu, \nu$ be positive Radon measures on an open set $U \subset \mathbb R^m$ which satisfy
\[
	\int_U \zeta^2 \,\mathrm d \mu \leq \int_U \zeta^2 \,\mathrm d \nu
\]
for any $\zeta \in C_c^1(U)$. Then, $\mu(E) \leq \nu(E) \text{ for any Borel set } E \subset U.$
\end{lemma}

\begin{proof}
	Given any open set $V \subset U$, it is possible to find an increasing family $\{K_i\}_i$ of compact sets contained in $V$ whose union equals $V$. For each $i$ choose a~non-negative function $\zeta_i \in C_c^1(V)$ with $||\zeta_i||_\infty \leq 1$ which equals $1$ on $K_i$. Consequently, the assumption on $\mu$ and $\nu$ yields
\begin{equation*}
\mu(K_i) \leq \int_U \zeta_i^2 \,\mathrm d \mu \leq \int_V \zeta_i^2 \,\mathrm d \nu \leq  \nu(V),
\end{equation*}
which by continuity of measure $|u_x^s|$ means that $\mu(V) \leq \nu(V)$ for any open $V \subset U$. Now, if $E$ is an arbitrary Borel set, then
\begin{equation*}
	\mu(E) = \inf{ \left \{\mu(V):\, V \text{ open, } E \subset V \right \}} \leq \inf{ \left\{ \nu(V):\, V \text{ open, } E \subset V \right \}} = \nu(E),
\end{equation*}
which finishes the proof.
\end{proof}

\subsection*{Convex functions}

\begin{defn} \label{def: Reshetnyak}
Let $\psi: \mathbb R^n \to [0, \infty)$ be a~positively $1$-homogeneous function. It is convex if and only if for any two points $p, q \in \mathbb R^n$ it is true that
\begin{equation} \label{ineq: convexity}
	\psi(p + q) \leq \psi(p) + \psi(q).
\end{equation}
Following Reshetnyak's seminal paper \cite{resh}, we will say that a~positively $1$-homogeneous function $\psi$ is \emph{strictly convex in the sense of Reshetnyak} if and only if inequality \eqref{ineq: convexity} turns into equality only if $p = 0$ or $q = tp$ for some $t \geq 0$.
\end{defn}

\begin{lemma} \label{lem: strict convexity}
Let $\psi: \mathbb R^n \to [0, \infty)$ be positively $1$-homogeneous function which vanishes only at the origin and $g:\mathbb R \to \mathbb R$ be strictly convex and increasing. If $\psi$ is strictly convex in the sense of Reshetnyak, then $G := g \circ \psi$ is strictly convex, i.\;e., $G(\lambda p + (1 - \lambda)q) < \lambda G(p) + (1 - \lambda)G(q)$ for any distinct $p, q \in \mathbb R^n$. If $\psi$ is not strictly convex in the sense of Reshetnyak, then $G$ is not strictly convex.
\end{lemma}

\begin{proof}
Choose any distinct $p, q \in \mathbb R^n$ and $\lambda > 0$ and assume strict convexity (in the sense of Reshetnyak) of $\psi$. If $p = 0$, then the statement holds since $g$ is strictly convex and $\psi$ vanishes at the origin (and only there). If $q = tp$ for $t > 0$ and $t \neq 1$, then $\psi(q) = t\psi(p) \neq \psi(p)$ and again it suffices to use strict convexity of $g$. At last, for points for which inequality \eqref{ineq: convexity} is strict, one can write
\begin{align*}
	G\left(\lambda p + (1-\lambda)q \right) &< g \left( \lambda \psi(p) + (1- \lambda) \psi(q) \right) \\
	& \leq \lambda g(\psi(p)) + (1-\lambda)g(\psi(q)).
\end{align*}
The first inequality is strict as $g$ under our assumptions is strictly increasing whereas the second is not as it might happen that $\psi(p) = \psi(q)$.

Suppose now that $\psi$ is not strictly convex (in the sense of Reshetnyak) so that there exist non-zero points $p, q$ such that $p \neq tq$ for any $t > 0$ for which inequality \eqref{ineq: convexity} yields equality. The same is true for $1/2p$ and $1/2q$, which implies that $\psi$ has to be affine on the segment $[p, q]$. As a~result, for $\lambda = (1 + \psi(p)/\psi(q))^{-1}$ it is true that
\begin{equation*}
	\psi \left( p + \frac{\psi(p)}{\psi(q)}q \right) = \lambda^{-1} \psi( \lambda p + (1 - \lambda)q) = \psi(p) + \psi \left( \frac{\psi(p)}{\psi(q)}q \right).
\end{equation*}
Therefore, we have found points $\tilde{p} = p$, $\tilde{q} = (\psi(p)/\psi(q)) q$ such that $\tilde{p} \neq t\tilde{q}$, whose $\psi$-norms coincide and for which inequality \eqref{ineq: convexity} turns into equality. Consequently, for any $\lambda \in (0,1)$
\begin{equation*}
	G\left(\lambda \tilde{p} + (1-\lambda)\tilde{q} \right)  = g \left( \lambda \psi(\tilde{p}) + (1- \lambda) \psi(\tilde{q}) \right) = g(\psi(\tilde{p})) = \lambda G(\tilde{p}) + (1- \lambda) G(\tilde{q}),
\end{equation*}
what finishes the proof.
\end{proof}

\begin{remark} \label{remark: f(t)/t}
	Given a~convex and differentiable function $f: [0, \infty[ \to [0, \infty[$ of at most linear growth (with constant $C_f^+$) it is true that
	\begin{equation*}
		f^\infty := \lim_{t \to \infty} \frac{f(t)}{t} = \lim_{t \to \infty} f'(t) \leq C_f^+.
	\end{equation*}
Moreover, convolution of such $f$ with a~standard mollifier yields a~function which is of at most linear growth with the same constant. This inequality also shows that such $f$ is Lipschitz, so mollifications of $f$ converge uniformly to $f$ (and not only locally). For any convex $\psi: \mathbb R^n \to \mathbb R$, its mollification $\psi_\eta$ with a~radially symmetric mollifier gives $\psi_\eta \geq \psi$.
\end{remark}

Given $f: [0, +\infty[ \to [0, +\infty[$, set
\begin{equation} \label{eq: f eta}
	f_\eta(t) = \left( f(t) \indi_{\{t>\eta\}} + f(\eta)\indi_{\{-\eta \leq t \leq \eta\}} \right) \ast \varrho_{\frac{\eta}{2}}.
\end{equation}
To construct a~suitable sequence approximating a~given anisotropy $\varphi$ we use a~slightly modified mollification procedure which retains homogeneity and which has been described by Schneider in his monograph on convex bodies \cite{schneider}.

\begin{lemma} \label{lem: schneider} \cite[Theorem 3.3.1]{schneider}
	Take $\varrho(p) := \rho(|p|)$ with $\rho: [0, +\infty[ \to [0, +\infty[$ being a~smooth function with support contained in $[1/2,1]$ such that $\int_{\mathbb R^n} \rho(|z|) \,\mathrm d z = 1$. For any anisotropy $\varphi$, the function
	\begin{equation}\label{goodnorm_eq}
		\overline{\varphi}_\eta(p) := \int_{\mathbb R^n} \varphi(p + |p|z) \varrho_\eta(z) \,\mathrm d z
	\end{equation}
	is an anisotropy of class $C^{\infty}(\mathbb R^n \setminus \{0\})$.
\end{lemma}

\begin{prop}\label{goodnorm}
	Let $\varphi$ be an anisotropy on $\mathbb R^n$. There exists a~sequence of anisotropies $\{ \widetilde{\varphi}_\eta \}_\eta$ with the following properties
	\begin{enumerate}[(i)]
		\item $\widetilde{\varphi}_\eta \in C^\infty(\mathbb R^n \setminus \{0\})$ and $D^2 \left( \frac{1}{2} \widetilde{\varphi}_\eta^{\,2} \right) \geq \eta \mathrm{Id}$ in $\mathbb R^n \setminus \{0\}$;
		\item $\widetilde{\varphi}_\eta \geq \varphi$ and $\widetilde{\varphi}_\eta$ converge to $\varphi$ locally uniformly on $\mathbb R^n$.
	\end{enumerate}
	Assume additionally that $\varphi \in C^{1}(\mathbb R^n \setminus \{0\})$. Then,
	\begin{enumerate}[(i)]
		\setcounter{enumi}{2}
		\item $D \widetilde{\varphi}_\eta$ converges to $D \varphi$ uniformly on $\mathbb R^n \setminus \{0\}$.
    \end{enumerate}
    If $f: [0, +\infty[ \to [0, +\infty[$ is strictly convex, increasing and differentaible with $f'(0) = 0$,
    \begin{enumerate}[(i)]
        \setcounter{enumi}{3}
		\item $D(f_\eta \circ \widetilde{\varphi}_\eta)$ converges to $D(f \circ \varphi)$ uniformly on $\mathbb R^n$, where $f_\eta$ is defined in \eqref{eq: f eta}.
	\end{enumerate}
\end{prop}
Anisotropies which satisfy condition (i) are often referred to as anisotropies of class $C^\infty_+$ (see \cite{schneider}, \cite{BellettiniCasellesNovaga2005}).

\begin{proof}
	We define
	\[\widetilde{\varphi}_\eta(p) = \sqrt{\overline{\varphi}_\eta^2(p) + \eta |p|^2}, \]
	where $\overline{\varphi}_\eta$ is given by \eqref{goodnorm_eq}. Condition (i) is clearly satisfied whereas $\widetilde{\varphi}_\eta \geq \varphi$ follows from the fact that $\overline{\varphi}_\eta \geq \varphi$. Indeed, due to homogeneity, it suffices to check it only for $p$ from the unit sphere, for which \eqref{goodnorm_eq} amounts to the ordinary convolution for which this property can be easily shown to be true. Local uniform convergence of $\widetilde{\varphi}_\eta$ to $\varphi$ holds since $\overline{\varphi}_\eta$ converges to $\varphi$ locally uniformly as well.

	The rest of the proof is devoted to the regular case. The key to uniform convergence of derivatives is the fact that derivative of positively $1$-homogeneous function is $0$-homogeneous, whence it is enough to check it over the unit sphere. Passing with the differentiation under the integral sign gives for $|p| = 1$
	\begin{equation*}
		D \overline{\varphi}_\eta (p) = \int_{\mathbb R^n} D\varphi(p + z) \left( \mathrm{Id} + z \cdot p \otimes p \right) \varrho_\eta(|z|) \,\mathrm d z.
	\end{equation*}
	Uniform continuity of $D\varphi$ on compact sets implies that $D \overline{\varphi}_\eta$ converges uniformly to $D\varphi$. Then, simple calculations lead to the conclusion that property (iii) is satisfied.

	The proof of (iv) starts with showing that $f' \circ \widetilde{\varphi}_\eta$ converges to $f'\ \circ \varphi$ uniformly on $\mathbb R^n$. Recall that for any $\delta$ there exists an $M$ such that if only $t \geq M$, then $f^\infty - f'(t) \leq \delta$. Fix any $\delta > 0$, choose $M$ accordingly and take any $p \in \mathbb R^n$. On the compact ball $\{p: \, \varphi(p) \leq M\}$ functions $\widetilde{\varphi}_\eta$ converge uniformly to $\varphi$, i.e, there exists $\eta_0$ such that for all $\eta \leq \eta_0$ and $p$ with $\varphi(p) \leq M$ it is true that
	\[
	|\widetilde{\varphi}_\eta(p) - \varphi(p)| \leq \delta.
	\]
	Whenever $\varphi(p) \geq M$, then by (ii) $\widetilde{\varphi}_\eta(p) \geq M$ and as a~result, after adding and subtracting $f^\infty$, we get
	\begin{equation*}
		| f'(\widetilde{\varphi}_\eta(p)) - f'(\varphi(p)) | \leq 2\delta,
	\end{equation*}
	which shows the desired claim.

	Clearly, $D(f_\eta \circ \widetilde{\varphi}_\eta)(0) = D(f \circ \varphi)(0) = 0$. Take any $p \neq 0$ and observe that
	\begin{equation*}
		|D(f_\eta \circ \widetilde{\varphi}_\eta)(p) - D(f \circ \varphi)(p)| \leq f^\infty |D\widetilde{\varphi}_\eta(p) - D\varphi(p)| + |f'_\eta(\widetilde{\varphi}_\eta(p)) - f'(\varphi(p))||D\varphi(p)|.
	\end{equation*}
	By (iii), it is evident that the first term converges uniformly to zero. Moreover, $|D\varphi|$ is bounded. Adding and subtracting $f'(\widetilde{\varphi}_\eta(p))$ in the second term together with the claim established in the previous paragraph and the fact that $f'_\eta$ converge uniformly to $f'$ allows us to conclude that the second term converges uniformly to zero as well.
\end{proof}

\begin{lemma} \label{lem: smooth phi sqr}
	There exists a convex function $\widetilde{\widetilde{\varphi}}_\eta \in C^2(\mathbb R^n)$, which equals $\widetilde{\varphi}_\eta^2$ outside $B_\eta(0)$.
\end{lemma}

\begin{proof}
	To construct such a~function, it is sufficient to mollify $\widetilde{\varphi}_\eta^2$ with $\varrho_{\varepsilon}$ for a~sufficiently small $\varepsilon$ and glue it to $\widetilde{\varphi}_\eta^2$ along a~small annulus in $B_\eta(0)$. Convexity is achieved thanks to the fact that $D^2\widetilde{\varphi}_\eta^2 \geq \eta \mathrm{Id}$ on the boundary of the ball $B_\eta(0)$ and the appropriate choice of $\varepsilon$, as proved in \cite[Section 2]{ghomi}.
\end{proof}

\subsection*{Spherical compactification}
\begin{defn}
Let $\overline{\mathbb R^n} := \mathbb R^n \cup \mathbb{S}^{n-1}_\infty$, where $\mathbb{S}^{n-1}_\infty := \{ \infty \, \omega: \, \omega \in \mathbb{S}^{n-1}_1 \}$. We call $\overline{\mathbb R^n}$ the spherical compactification of $\mathbb R^n$. Furthermore, define $\Phi: \overline{\mathbb R^n} \to \overline{\mathbb{B}^n_1}$ with the formula
\begin{equation*}
	\Phi(p) = \begin{cases}
		\frac{p}{1 + |p|} & \text{if } p \in \mathbb R^n, \\
		\omega & \text{if } p = \infty \omega \in \mathbb{S}^{n-1}_\infty.
	\end{cases}
\end{equation*}
The space $\overline{\mathbb R^n}$ can be equipped with the metric
\begin{equation*}
	\overline{d}(p, q) := \lvert \Phi(p) - \Phi(q) \rvert.
\end{equation*}
\end{defn}

The space $\overline{\mathbb R^n}$ with topology induced by the norm $\overline{d}$ is indeed compact since $\Phi$ is a~homeomorphism between this space and the closed unit ball with the Euclidean subspace topology. Clearly, any point $p$ in $\mathbb R^n$ can be uniquely represented as $q = r \omega$ with $\omega \in \mathbb{S}^{n-1}_\infty$. If one looks at a~sequence of points $p_j = r_j \omega_j$, it is evident that it converges to a~point $p = \infty \omega \in \mathbb{S}^{n-1}_\infty$ if and only if $r_j \to +\infty$ and $\omega_j \to \omega$. On the other hand, if a~sequence of points $p_j$ converges in $\overline{d}$ to a~point $p \in \mathbb R^n$, then for all but a~finite number of $j$, $p_j$ lies in $\mathbb R^n$ as well. Moreover, uniform continuity of $\Phi^{-1}$ implies that in such case $|p_j - p| \to 0$. Additionally, since $\Phi$ is $1$-Lipschitz, a~sequence of points $p_j \in \mathbb R^n$ which converges in Euclidean distance to $p \in \mathbb R^n$, is also convergent to $p$ in $\overline{\mathbb R^n}$ with metric $\overline{d}$.

\begin{lemma} \label{lem: injective on spheres}
Let $\varphi: \mathbb R^n \to \mathbb R$ be a~$C^1(\mathbb R^n \setminus \{0\})$ strictly convex (in the sense of Reshetnyak) anisotropy. Then for any $r>0$ the mapping $D\varphi: \{ p\in \mathbb R^n: \, \varphi(p) = r\} \to \mathbb R^n $ is injective.
\end{lemma}

\begin{proof}
The function $\varphi^2$ is $C^1$ on the whole $\mathbb R^n$ and is strictly convex as shown in Lemma \ref{lem: strict convexity}. Therefore, $D\varphi^2$ is strictly monotone and hence injective. For any $p \neq q$ such that $\varphi(p) = \varphi(q) = r$ for some positive $r > 0$ it is then true that
\[
	2 r D\varphi(p) \neq 2 r D\varphi(q),
\]
which means that $D\varphi(p) \neq D \varphi(q)$.
\end{proof}

\begin{lemma} \label{lem: DF homeo}
The mapping $\overline{DF}: \overline{R^n} \to \overline{R^n}$ defined as
\begin{equation*}
	\overline{DF} = \begin{cases}
		DF(p) &  \text{if } p \in \mathbb R^n,  \\
		f^\infty D\varphi(\omega) & \text{if } p = \infty \omega \in \mathbb{S}^{n-1}_\infty
	\end{cases}
\end{equation*}
is a homeomorphism.
\end{lemma}

\begin{proof}
Since $\overline{\mathbb R^n}$ is compact, it suffices to prove that $\overline{DF}$ is continuous and injective. Continuity is easily established by considering three cases. Firstly, take a~point $p_0 \in \mathbb R^n$ and any sequence $p_n$ converging to $p_0$ in metric $\overline{d}$. Then, necessarily, almost all $p_n$ lie in $\mathbb R^n$ as well, so continuity at such a~point follows from continuity of $DF$. Secondly, if one approaches $p_0 = \infty \omega_0 \in \mathbb{S}_\infty^{n-1}$ with a~sequence of points in $R^n$, then writing $p_n = r_n \omega_n$ gives
\[
	DF(p_n) = f'(r_n \varphi(\omega_n))D\varphi(\omega_n) \to f^\infty D\varphi(\omega_0) = \overline{DF}(p_0),
\]
as $\{\varphi(\omega_n)\}_n$ is bounded and $r_n \to +\infty$, and $D\varphi$ is continuous on the unit sphere. Lastly, if one approaches such a~point with a~sequence of points lying on $\mathbb{S}_{n-1}^\infty$, then continuity of $\overline{DF}$ follows directly from continuity of $D\varphi$ on the unit sphere.

To prove injectivity, observe that $F$ being strictly convex implies $DF$ being strictly monotone, which means that $DF: \mathbb R^n \to \mathbb R^n$ is one-to-one. Therefore, $\overline{DF}(p) \neq \overline{DF}(q)$ for any distinct $p, q \in \mathbb R^n$. In view of Lemma \ref{lem: injective on spheres}, it suffices to check that $\overline{DF}(p) \neq \overline{DF}(q)$ for $p \in \mathbb R^n$ and $q \in \mathbb{S}_\infty^{n-1}$. To this end, let us show that for any $p \in \mathbb R^{n}\setminus \{0\}$ it is true that
\begin{equation} \label{dual norm derivative}
	\varphi^* \left(D\varphi(p) \right) = 1, \text{ where } \varphi^*(p) := \sup \left \{ p \cdot q: \, \varphi(q) \leq 1 \right \}.
\end{equation}
Indeed, taking $q = p/\varphi(p)$ gives $\varphi^*\left(D\varphi(p)\right) \geq 1$. On the other hand, the product $q \cdot D\varphi(p)$ equals the derivative of $\varphi$ in direction $q$ and hence by triangle inequality one gets
\begin{equation*}
	\varphi^*\left( D\varphi(p) \right) = \lim_{t \to 0^+} \frac{1}{t} \left( \varphi(p + tq) - \varphi(p) \right) \leq \varphi(q) \leq 1.
\end{equation*}

Assume now that $DF(p) = f^\infty D\varphi(\omega)$ for some $p \in \mathbb R^n \setminus \{0\}$ and $\omega \in \mathbb{S}^{n-1}_1$. Applying $\varphi^*$ to both sides and using its positive homogeneity results in
\begin{equation*}
	f'(\varphi(p)) \varphi^* \left( D \varphi(p) \right) = f^\infty \varphi^*\left( D\varphi(\omega) \right).
\end{equation*}
Given \eqref{dual norm derivative}, one gets $f'(\varphi(p)) = f^\infty$, which contradicts strict convexity of $f$. Therefore, $\overline{DF}$ is injective.
\end{proof}

\section{Approximate functionals} \label{sec: approximate}

Suppose $h \in L^2(I) \cap BV(U, \mathbb R^n)$ for an open interval $U \subset I$ and extend $h$ by zero over a~slightly larger than $I$ interval in order to properly define mollifications of $h$. We set
\begin{equation}
	h^\varepsilon = h \ast \varrho_\varepsilon.
\end{equation}
We suppose that $F_{\eta}$ is given either by
\begin{equation} \label{Feta1def}
	F_\eta = f_\eta \circ \overline{\varphi}_\eta + \frac{\eta}{2}|\cdot|^2
\end{equation}
or
\begin{equation} \label{Feta2def}
	F_\eta = f_\eta \circ \widetilde{\varphi}_\eta + \frac{\eta}{2} \widetilde{\widetilde{\varphi}}_\eta.
\end{equation}
In \eqref{Feta1def}, \eqref{Feta2def} we have used $f_\eta$, $\overline{\varphi}_\eta$, $\widetilde{\varphi}_\eta$ and $\widetilde{\widetilde{\varphi}}_\eta$ defined in \eqref{eq: f eta}, Lemma \ref{lem: schneider}, Proposition \ref{goodnorm} and Lemma \ref{lem: smooth phi sqr} for given anisotropy $\varphi \colon \mathbb R^n \to [0, \infty[$ and convex, increasing $f \colon [0, \infty[ \to [0, \infty[$ of linear growth. Observe that $F_\eta$ is smooth and that $D^2 F_\eta \geq \eta \mathrm{Id}$. Moreover, since $\widetilde{\varphi}_\eta \geq \varphi$ and $f_\eta \geq f$ due to convexity of $f$, it is true that $F_\eta \geq F$.

We will use a two-layer approximation of $\mathcal E$:
\begin{align*}
	\mathcal E_{\varepsilon, \eta} (w) &= \lambda \int_I F_\eta(w_x) \,\mathrm d \mathcal L^1 + \frac{1}{2} \int_I |w - h^{\varepsilon} |^2 \,\mathrm d \mathcal L^1 \text{ for } w \in W^{1,2}(I, \mathbb R^n) \text{ with mimimizer } u^{\varepsilon, \eta}, \\
	\mathcal E_\varepsilon(w) &= \lambda \int_I F(w_x) \,\mathrm d \mathcal L^1  + \frac{1}{2} \int_I |w-h^\varepsilon|^2 \,\mathrm d \mathcal L^1 \text{ for } w \in L^2(I, \mathbb R^n) \text{ with mimimizer } u^{\varepsilon}.
\end{align*}
Observe that $F_\eta$ and $F$ are strictly convex, non-negative, weakly lower semicontinuous in $W^{1,2}$ and $L^2$, respectively. Moreover the sublevels of $\mathcal E_{\varepsilon, \eta}$, $\mathcal E_\varepsilon$ are weakly compact in $W^{1,2}$ and $L^2$, respectively. Consequently, there indeed exist unique minimizers of these functionals.

\begin{lemma} \label{lem: gamma conv eta}
	Functionals $\mathcal E_{\varepsilon, \eta}$ $\Gamma$-converge to $\mathcal E_{\varepsilon}$ w.\,r.\,t.\;weak $W^{1,2}$ convergence.
\end{lemma}
\begin{proof}
	As $F_\eta \geq F$, then $\mathcal E_{\varepsilon, \eta} \geq \mathcal E_\varepsilon$, whence lower semicontinuity follows from a~standard result \cite[Chapter 8.2, Theorem 1]{evans}.
	As for the recovery sequence, choose $w_{\eta} = w$, then one only needs to show that
	\begin{equation} \label{f eta limit}
		\lim_{\eta \to 0}\int_I F_\eta(w_x) \,\mathrm d \mathcal L^1 = \int_I F(w_x) \,\mathrm d \mathcal L^1.
	\end{equation}
	Since $\overline{\varphi}$ (resp.\ $\widetilde{\varphi}_\eta$) is an anisotropy and $f_\eta$ is of at most linear growth, we have $f_\eta \circ \overline{\varphi} (w_x) \leq C( 1 + |w_x|)$ (resp.\ $f_\eta \circ \widetilde{\varphi}_\eta (w_x) \leq C( 1 + |w_x|)$), which is integrable over $I$, implying that dominated convergence theorem can be applied. Additionally, we have
	\begin{equation*}
		\lim_{\eta \to 0} \frac{\eta}{2} \int_I \widetilde{\widetilde{\varphi}}_\eta (w_x) \,\mathrm d \mathcal L^1 \leq \lim_{\eta \to 0} \frac{\eta}{2} \int_I C (1 + |w_x|^2) \,\mathrm d \mathcal L^1 = 0,
	\end{equation*}
	which proves \eqref{f eta limit} and concludes the proof of the lemma.
\end{proof}

\begin{lemma} \label{lem: gamma conv eps 1}
	Functionals $\mathcal E_\varepsilon$ $\Gamma$-converge to $\mathcal E$ w.\,r.\,t\;weak convergence in $L^2$.
\end{lemma}
\begin{proof}
	Since $h^\varepsilon$ converge strongly to $h$ in $L^2(I, \mathbb R^n)$, for each $w \in BV(I, \mathbb R^n)$ there exists a~trivial recovery sequence.

	Then, take $w \in BV(I, \mathbb R^n)$ and a~sequence $w^\varepsilon$ converging to $w$ weakly in $L^2$. Since the norm is lower semicontinuous w.\,r.\,t.\;weak convergence,
	\begin{equation*}
		\liminf_{\varepsilon \to 0} \int_I |w^\varepsilon - h^\varepsilon|^2 \,\mathrm d \mathcal L^1 \geq \int_I |w - h|^2 \,\mathrm d \mathcal L^1.
	\end{equation*}

	Lower semicontinuity of $\mathcal F$ with respect to strong $L^2(I, \mathbb R^n)$ convergence follows from the definition of $\mathcal F$, as $L^2$ convergence is stronger than $L^1_{loc}$ convergence. Since $\mathcal F$ is convex, it is also lower semicontinuous with respect to weak $L^2$ convergence.
\end{proof}

Since $\mathcal E_{\varepsilon, \eta}$ is smooth, the minimizer admits weak second derivatives and satisfies the following Euler-Lagrange equation
 \begin{equation} \label{Euler-Lagrange-eps-eta}
	u^{\varepsilon, \eta} - h^\varepsilon = \lambda \left(DF_\eta(u_x^{\varepsilon, \eta})\right)_x \text{ in } I, \quad u_x^{\varepsilon, \eta} = 0 \text{ on } \partial I.
\end{equation}
Due to the fact that $D^2F_\eta \geq \eta \mathrm{Id}$, we gather that $u^{\varepsilon, \eta} \in W^{2,2}(I, \mathbb R^n)$, which in particular means that $u^{\varepsilon, \eta} \in C^{1}(\bar{I}, \mathbb R^n)$.

\begin{lemma} \label{lem: cpt eta}
As $\eta \to 0$, the sequence of minimizers $u^{\varepsilon, \eta}$ converges strongly in $L^2(I, \mathbb R^n)$ and weakly in $W^{1,2}(I, \mathbb R^n)$ to the minimizer $u^\varepsilon$.
\end{lemma}
\begin{proof}
Firstly, let us prove that the sequence of minimizers $u^{\varepsilon, \eta}$ is bounded in $W^{1,2}$. The fact that $\mathcal E_{\varepsilon, \eta}(u^{\varepsilon, \eta}) \leq \mathcal E_{\varepsilon, \eta}(0)$ together with triangle inequality for $L^2$ norm  show that $u^{\varepsilon, \eta}$ are bounded in $L^2$. For the boundedness of the derivatives, let us multiply the Euler-Lagrange equation by $u_{xx}^{\varepsilon, \eta}$ and integrate the l.\,h.\,s.\;by parts. The r.\,h.\,s\;is non-negative due to convexity of $F_\eta$ and the boundary term vanishes, hence
	\begin{equation*}
		\int_{I} |u_x^{\varepsilon, \eta}|^2 \,\mathrm d \mathcal L^1 \leq \int_I u_x^{\varepsilon, \eta} \cdot h_x^\varepsilon \,\mathrm d \mathcal L^1.
	\end{equation*}
Applying Young's inequality gives the desired bound for $L^2$ norm of the sequence of derivatives of minimizers.

Therefore, from any subsequence $u^{\varepsilon, \eta}$ it is possible to choose a~weakly convergent subsequence $u^{\varepsilon, \eta_j}$. In view of the $\Gamma$-convergence proved in Lemma \ref{lem: gamma conv eta}, the weak limit of $\{u^{\varepsilon, \eta_j}\}_j$ must equal the minimizer $u^\varepsilon$. A standard application of Rellich-Kondrashov theorem allows us to choose a~further subsequence (which we do not relabel) which additionally converges strongly in $L^2$ to $u^\varepsilon$. Therefore, a~properly convergent subsequence can be chosen from arbitrary subsequence of $\{u_x^\varepsilon\}_\varepsilon$, which shows the desired convergence of the whole sequence.
\end{proof}

\begin{lemma} \label{lem: gamma conv eps}
As $\varepsilon \to 0$, the minimizers $u^\varepsilon$ converge weakly in $L^2(I, \mathbb R^n)$ to the minimizer~$u$. If $F$ is coercive (i.e., satisfies \eqref{coercivity}), then the sequence $u^{\varepsilon, \eta}$ converges strongly in $L^2(I, \mathbb R^n)$ and $u_x^\varepsilon \mathrel{\ensurestackMath{\stackon[0.5pt]{\rightharpoonup}{\scriptstyle\ast}}} u_x$.
\end{lemma}
\begin{proof}
    To see that $u^\varepsilon$ are bounded in $L^2$, it suffices to write $\mathcal E_\varepsilon(u^\varepsilon) \leq \mathcal E_\varepsilon(0)$ and use triangle inequality for $L^2$ norm. Consequently, from every sequence $u^{\varepsilon_i}$ it is possible to choose a~further subsequence which, by $\Gamma$-convergence proved in Lemma \ref{lem: gamma conv eps 1}, converges weakly in $L^2$ to $u$.

    If $F$ is coercive, one can use it to bound $\int_I F(u^{\varepsilon_i}_x)$ from below to see that $\{u^{\varepsilon_i}\}_i$ is bounded in $BV$, which implies an existence of a~subsequence strongly convergent in $L^2$ (see \cite[Corollary 3.49]{ambrosiofuscopallara}) such that the derivatives converge weakly* as measures.
\end{proof}

\section{The homogeneous case} \label{sec:homo}

\begin{proof}[Proof of Theorem \ref{thm:homo}]
Let
\begin{equation}
	F_\eta:= f_\eta \circ \overline{\varphi}_\eta + \frac{\eta}{2} |\cdot|^2,
\end{equation}
where $f = |\cdot|$ is the absolute value function. We calculate
\[
DF_{\eta}(p) = f_\eta'(\overline{\varphi}_\eta(p)) D \overline{\varphi}_\eta(p) + \eta p,
\]
\[
\left(DF_\eta(u_x^{\varepsilon, \eta})\right)_x = f_\eta'(\overline{\varphi}_\eta(u_x^{\varepsilon, \eta})) D^2 \overline{\varphi}_\eta(u_x^{\varepsilon, \eta}) u_{xx}^{\varepsilon, \eta} + f_\eta''(\overline{\varphi}_\eta(u_x^{\varepsilon, \eta})) \overline{\varphi}_\eta(u_x^{\varepsilon, \eta})_x D \overline{\varphi}_\eta(u_x^{\varepsilon, \eta}) + \eta\, u_{xx}^{\varepsilon, \eta}.
\]
Testing the Euler-Lagrange equation \eqref{Euler-Lagrange-eps-eta} with $u^{\varepsilon, \eta}_{xx}$, integrating by parts on the l.\,h.\,s.\;and using Cauchy's inequality we obtain a uniform estimate
\begin{equation} \label{homo_eta_unif_bd}
\int_I |u_x^{\varepsilon, \eta}|^2 + 2 \lambda \eta \int_I |u_{xx}^{\varepsilon, \eta}|^2 \leq \int_I |h_x^{\varepsilon}|^2 .
\end{equation}

We define
\begin{equation}
	\mathfrak G_k (p) = (|p|-k)_+, \quad   \mathfrak G_{k, \delta} (p) = \sqrt{(|p|-k)_+^2 + \delta^2}.
\end{equation}
Therefore,
\[
D \mathfrak G_{k, \delta} (p) = \frac{(|p|-k)_+}{\sqrt{(|p|-k)_+^2 + \delta^2}} \frac{p}{|p|}
\]
and, given $\zeta \in C_c^1(U)$
\begin{multline*}
	\left(D \mathfrak G_{k, \delta} (u_x^{\varepsilon, \eta}) \zeta^2\right)_x  = \frac{(|u_x^{\varepsilon, \eta}|-k)_+}{\sqrt{(|u_x^{\varepsilon, \eta}|-k)_+^2 + \delta^2}} \frac{1}{|u_x^{\varepsilon, \eta}|}\left(I - \frac{u_x^{\varepsilon, \eta}}{|u_x^{\varepsilon, \eta}|} \otimes \frac{u_x^{\varepsilon, \eta}}{|u_x^{\varepsilon, \eta}|}\right)  u_{xx}^{\varepsilon, \eta} \zeta^2 \\ + \frac{\delta^2 \indi_{|u_x^{\varepsilon, \eta}|>k }}{\sqrt{(|u_x^{\varepsilon, \eta}|-k)_+^2 + \delta^2}^3} \frac{u_x^{\varepsilon, \eta}}{|u_x^{\varepsilon, \eta}|} \otimes \frac{u_x^{\varepsilon, \eta}}{|u_x^{\varepsilon, \eta}|}\, u_{xx}^{\varepsilon, \eta} \zeta^2 + \frac{(|u_x^{\varepsilon, \eta}|-k)_+}{\sqrt{(|u_x^{\varepsilon, \eta}|-k)_+^2 + \delta^2}} \frac{u_x^{\varepsilon, \eta}}{|u_x^{\varepsilon, \eta}|} \, 2 \zeta \zeta_x.
\end{multline*}
We observe that
\begin{equation}\label{homo_est1}
	\int_I \left(D \mathfrak G_{k, \delta} (u_x^{\varepsilon, \eta}) \zeta^2\right)_x \cdot u_{xx}^{\varepsilon, \eta} \geq \int_I 2 \zeta \zeta_x \frac{(|u_x^{\varepsilon, \eta}|-k)_+}{\sqrt{(|u_x^{\varepsilon, \eta}|-k)_+^2 + \delta^2}} \frac{u_x^{\varepsilon, \eta}}{|u_x^{\varepsilon, \eta}|} \cdot u_{xx}^{\varepsilon, \eta} \geq - 2 \max |\zeta_x| \left(\int_I  |u_{xx}^{\varepsilon, \eta}|^2 \right)^{\frac{1}{2}}.
\end{equation}

Assuming that $k > \frac{2}{c_\varphi^-} \eta$, we have $|p| \leq k$ or $\overline{\varphi}_\eta(p) \geq 2 \eta$ for all $p \in \mathbb R^n$. In the latter case $f'(\overline{\varphi}_\eta(p)) = 1$ and $f''(\overline{\varphi}_\eta(p)) = 0$. Taking into account \eqref{homo_est1}, \eqref{homo_eta_unif_bd} and the observation that
\[D^2 \overline{\varphi}_\eta(p) \cdot p = 0 \quad \text{for } p \neq 0,\]
we get
\begin{equation} \label{homo_est2}
	\int_I \left(D \mathfrak G_{k, \delta} (u_x^{\varepsilon, \eta}) \zeta^2\right)_x \cdot \left(DF_\eta(u_x^{\varepsilon, \eta})\right)_x \geq - 2 \eta \max |\zeta_x| \left(\int_I  |u_{xx}^{\varepsilon, \eta}|^2 \right)^{\frac{1}{2}} \geq - \frac{\sqrt{2\eta}}{\sqrt{\lambda}} \max |\zeta_x|\left(\int_I |h_x^\varepsilon|^2 \right)^{\frac{1}{2}}.
\end{equation}
On the other hand, by convexity of $\mathfrak G_{k, \delta}$,
\begin{multline} \label{homo_rhs_est}
\int_I \left(D \mathfrak G_{k, \delta} (u_x^{\varepsilon, \eta}) \zeta^2\right)_x \cdot (u^{\varepsilon, \eta} - h^\varepsilon)  = \int_I \zeta^2 D \mathfrak G_{k, \delta} (u_x^{\varepsilon, \eta}) \cdot (h^\varepsilon_x - u^{\varepsilon, \eta}_x) \\ \leq \int_I \zeta^2 \mathfrak G_{k, \delta} (h_x^\varepsilon) - \int_I \zeta^2 \mathfrak G_{k, \delta} (u_x^{\varepsilon, \eta}).
\end{multline}
Collecting \eqref{homo_est2} and \eqref{homo_rhs_est} and passing to the limit $\delta \to 0^+$ using dominated convergence, we infer that
\begin{equation} \label{homo_est3}
	 \int_I \zeta^2 \mathfrak G_k (u_x^{\varepsilon, \eta}) \leq \int_I \zeta^2 \mathfrak G_k (h_x^\varepsilon) + \frac{\sqrt{2\eta}}{\sqrt{\lambda}} \max |\zeta_x|\left(\int_I |h_x^\varepsilon|^2 \right)^{\frac{1}{2}}.
\end{equation}
As shown in Lemma \ref{lem: cpt eta}, $u_x^{\varepsilon, \eta}$ converge weakly in $L^2$ to $u_x^{\varepsilon}$ as $\eta \to 0^+$. Therefore, by lower semicontinuity of $w_x \mapsto \int \mathfrak G_k(w_x) \,\mathrm d \mathcal L^1$, inequality \eqref{homo_est3} yields
\begin{equation} \label{homo_est4}
	\int_I \zeta^2 \mathfrak G_k (u_x^\varepsilon) \leq \int_I \zeta^2 \mathfrak G_k (h_x^\varepsilon).
\end{equation}

What is more, Lemma \ref{lem: mu leq nu} implies that $\mathfrak{G}_k(u_x^\varepsilon)(U) \leq \mathfrak G_k(h_x^\varepsilon(U))$. For any $w \in BV(I, \mathbb R^n)$ \[ |w_x| \leq \mathfrak{G}_k(w_x) + k \leq |w_x| + k\]
as measures, hence, using the fact that the sequence $\mathfrak G_k(h_x^\varepsilon)$ is weakly* convergent by Lemma \ref{lem: G weaklystar} (and therefore bounded), we deduce that $u^\varepsilon$ is uniformly bounded in $BV(U, \mathbb R^n)$. Since by Lemma \ref{lem: gamma conv eps} we already know that $u^\varepsilon$ converge weakly in $L^2$ to $u$, it can be inferred that $u \in BV(U, \mathbb R^n)$ and $u_x^\varepsilon \mathrel{\ensurestackMath{\stackon[0.5pt]{\rightharpoonup}{\scriptstyle\ast}}} u_x$. As a~result, Lemmata \ref{cor-gs-lsc-zeta} and \ref{lem: G weaklystar} mean that the following inequality holds
\[
    \int_U \zeta^2 \,\mathrm d \mathfrak{G}_k(u_x) \leq \int_U \zeta^2 \mathfrak G_k(h_x).
\]
Employing Lemma \ref{lem: mu leq nu} again goes to show that
\begin{equation*}
	\mathfrak{G}_k (u_x) \leq \mathfrak{G}_k (h_x)
\end{equation*}
as measures or, in other words,
\begin{equation*}
	\mathfrak |u_x^s|(V) + \int_V (|u_x^{ac}|-k)_+ \,\mathrm d \mathcal L^1  \leq |h_x^s|(V) + \int_V (|h_x^{ac}|-k)_+ \,\mathrm d \mathcal L^1
\end{equation*}
for any $V$ Borel. Passing to the limit $k\to 0^+$ we obtain the desired assertion.
\end{proof}

\section{The coercive case} \label{sec:coercive}

Let us now multiply both sides of the Euler-Lagrange equation \eqref{Euler-Lagrange-eps-eta} by $(\zeta^2 G_k(u_x^{\varepsilon, \eta}))_x$, where $\zeta \in C_c^1(U)$ and
\begin{equation*}
	G_k(p) = g_k \circ \tilde{\varphi}_\eta(p)p \text{ and } g_k(\sigma) = \frac{(\sigma - k)_+}{\sigma^2} f'_\eta(\sigma)
\end{equation*}
Observe that $G_k(p)$ is bounded since $f_\eta$ is of linear growth and $\widetilde{\varphi}_\eta$ is a~coercive anisotropy. After integrating and differentiating by parts on the l.\,h.\,s. we get
\begin{equation} \label{eq: after mult}
	\frac{1}{\lambda} \int_I \zeta^2 G_k(u_x^{\varepsilon, \eta}) \cdot  (u_x^{\varepsilon, \eta} - h_x^{\varepsilon}) \,\mathrm d \mathcal L^1 = - \int_I (\zeta^2 G_k(u_x^{\varepsilon, \eta}))_x \cdot D^2 F_\eta (u_x^{\varepsilon, \eta}) \cdot u_{xx}^{\varepsilon, \eta} \,\mathrm d \mathcal L^1.
\end{equation}

\begin{prop} \label{prop: general lhs eps}
	The l.\,h.\,s.\;of \eqref{eq: after mult} satisfies
\begin{align*}
	\liminf_{\varepsilon \to 0} \liminf_{\eta \to 0} & \int_I \zeta^2 G_k(u_x^{\varepsilon, \eta}) \cdot (u_x^{\varepsilon, \eta} - h_x^{\varepsilon}) \,\mathrm d \mathcal L^1 \geq \\
	& \frac{r(k)}{ (c_\varphi^+)^{3}} \int_I \zeta^2 \,\mathrm d (c_\varphi^- |u_x| - k)_+(c_\varphi^+ |u_x| - k)_+ |u_x|^{-1} - \frac{f^\infty}{ c_\varphi^- c_\varphi^+} \int_I \zeta^2 \,\mathrm d (c_\varphi^+ |h_x| - k)_+,
\end{align*}
where $\lim_{k \to \infty} r(k) = f^\infty$.
\end{prop}

\begin{proof}
	We will firstly show that for a~fixed $\varepsilon$, it is true that
\begin{align} \label{eq: lhs general eta}
\begin{split}
	\liminf_{\eta \to 0} & \int_I \zeta^2 G_k(u_x^{\varepsilon, \eta}) \cdot (u_x^{\varepsilon, \eta} - h_x^{\varepsilon}) \,\mathrm d \mathcal L^1 \geq \\
	& \frac{r(k)}{ (c_\varphi^+)^{3}} \int_I \zeta^2 (c_\varphi^- |u_x^{\varepsilon}| - k)_+(c_\varphi^+ |u_x^{\varepsilon}| - k)_+ |u_x^{\varepsilon}|^{-1} \,\mathrm d \mathcal L^1 - \frac{f^\infty}{ c_\varphi^- c_\varphi^+} \int_I \zeta^2 (c_\varphi^+ |h_x^\varepsilon| - k)_+ \,\mathrm d \mathcal L^1.
\end{split}
\end{align}
Since $\widetilde{\varphi}_\eta(p) \geq \varphi(p)$ and due to local uniform convergence of $\widetilde{\varphi}_\eta$ to $\varphi$, one can define a~decreasing function $\eta(\delta)$ such that for all $\eta \leq \eta (\delta)$
\begin{equation} \label{eq: norms}
	c_\varphi^- |p| \leq \widetilde{\varphi}_\eta(p) \leq c^+_\varphi + \delta.
\end{equation}
Then, it is possible to get a~function $\delta(\eta)$, which will converge to zero as $\eta$ goes to zero. Let us also note a~consequence of convexity of the function $p \mapsto (C|p| - k)_+$, namely, the inequality
	\begin{equation} \label{ineq: lhs general}
		C \indi_{C|p| > k} \cdot \frac{p}{|p|} (p - q) \geq (C|p| - k)_+ - (C|q| - k)_+.
	\end{equation}

Using the fact that $(\widetilde{\varphi}_\eta(p) - k)_+ \, \indi_{(c^+_\varphi + \delta)|p| > k} = (\widetilde{\varphi}_\eta(p) - k)_+$ and \eqref{ineq: lhs general}, we obtain the following estimate
\begin{align*}
	G_k(u_x^{\varepsilon, \eta}) &\cdot (u_x^{\varepsilon, \eta} - h_x^\varepsilon) = \frac{(\widetilde{\varphi}_\eta(u_x^{\varepsilon, \eta}) - k)_+}{\widetilde{\varphi}_\eta^2(u_x^{\varepsilon, \eta})} f'_\eta(\widetilde{\varphi}_\eta(u_x^{\varepsilon, \eta})) u_x^{\varepsilon, \eta} \cdot (u_x^{\varepsilon, \eta} - h_x^\varepsilon) \\
		&\overset{\eqref{ineq: lhs general}}{\geq} \frac{(\widetilde{\varphi}_\eta(u_x^{\varepsilon, \eta}) - k)_+}{\widetilde{\varphi}_\eta^2(u_x^{\varepsilon, \eta})} f'_\eta(\widetilde{\varphi}_\eta(u_x^{\varepsilon, \eta})) \frac{|u_x^{\varepsilon, \eta}|}{c_\varphi^+ + \delta}\left [ \left ( (c_\varphi^+ + \delta)|u_x^{\varepsilon, \eta}| - k \right)_+ - \left( (c_\varphi^+ + \delta) |h_x^\varepsilon| - k \right)_+ \right ].
\end{align*}
Handling these two terms separately and using \eqref{eq: norms} repeatedly yields
\begin{align} \label{eq: G k ineq}
	G_k&(u_x^{\varepsilon, \eta}) \cdot (u_x^{\varepsilon, \eta} - h_x^\varepsilon) \geq \nonumber \\
		&\frac{f'_\eta(k)}{(c_\varphi^+ + \delta)^3} \underbrace{\left( c_\varphi^-|u_x^{\varepsilon, \eta}| - k \right)_+ \left( c_\varphi^+|u_x^{\varepsilon, \eta}| - k \right)_+ \frac{1}{|u_x^{\varepsilon, \eta}|}}_{A_k(|u_x^{\varepsilon, \eta}|)} - \frac{f'_\eta(\widetilde{\varphi}_\eta(u_x^{\varepsilon, \eta}))}{(c_\varphi^- - \delta)(c_\varphi^+ + \delta)}\left( (c_\varphi^+ + \delta)|h_x^\varepsilon| - k \right)_+ \nonumber \\
		&\geq \frac{f(k) - f(0) -1}{k} (c_\varphi^+ + \delta)^{-3} A_k(|u_x^{\varepsilon, \eta}|) - \frac{f^\infty}{(c_\varphi^- - \delta)c_\varphi^+} \left( (c_\varphi^+ + \delta)|h_x^\varepsilon| - k \right)_+.
\end{align}

The last inequality essentially follows from convexity of $f$. Indeed, due to uniform convergence of $f_\eta$ to~$f$, $||f - f_\eta||_\infty \leq 1$ for sufficiently small $\eta$. Together with convexity of $f_\eta$ and the fact that $f_\eta \geq f$, it means that
\[
	\frac{f(t) - f(0) - 1}{t} \leq \frac{1}{t}(f_\eta(t) - f_\eta(0)) \leq \, f'_\eta(t) \, \leq \frac{1}{t}( f_\eta(2t) - f_\eta(t)) \leq \frac{f(2t) + 1}{t} - \frac{f(t)}{t}.
\]
Taking limit with $t \to \infty$ yields $\lim_{t \to \infty} f'_\eta(t) \leq f^\infty$, which was used in the estimate in the second term, $f'_\eta(\widetilde{\varphi}_\eta(u_x)) \leq \lim_{t \to \infty} f'_\eta(t)$. On the other hand, plugging $t = k$ on the l.\,h.\,s.\;of this inequality explains the estimate in the first term. Let $r(k) := (f(k) - f(0) - 1)/k$ and observe that $r(k) > 0 $ for sufficiently large $k$ and that $\lim_{k \to \infty} r(k) = f^\infty$.

Since $h_x^\varepsilon$ is integrable, it follows from dominated convergence theorem that
\begin{equation} \label{eq: lim for h}
	\lim_{\eta \to 0} \frac{f^\infty}{(c_\varphi^- - \delta(\eta))c_\varphi^+} \int_I \left( (c_\varphi^+ + \delta(\eta))|h_x^{\varepsilon} | - k \right)_+ \,\mathrm d \mathcal L^1 = \frac{f^\infty}{c_\varphi^- c_\varphi^+} \int_I (c_\varphi^+|h_x^\varepsilon| - k)_+ \,\mathrm d \mathcal L^1.
\end{equation}
In the term involving $u_x^{\varepsilon, \eta}$ it will not be possible to take the limit but it will suffice to estimate (recall that Lemma \ref{lem: cpt eta} established that $u^{\varepsilon,\eta}$ converges weakly in $W^{1,2}$ to $u^\varepsilon$)
\begin{align} \label{eq: liminf for u}
\begin{split}
	\liminf_{\eta \to 0} r(k)(c_\varphi^+ + \delta)^{-3} \int_I A_k( |u_x^{\varepsilon, \eta}|) \,\mathrm d \mathcal L^1 &\geq r(k) (c_\varphi^+)^{-3} \, \liminf_{\eta \to 0} \int_I A_k(|u_x^{\varepsilon, \eta}|) \,\mathrm d \mathcal L^1 \\
	&\geq r(k) (c_\varphi^+)^{-3} \int_I A_k(|u_x^{\varepsilon}|) \,\mathrm d \mathcal L^1.
\end{split}
\end{align}
The last inequality follows from convexity of $p \mapsto A_k(|p|)$ (it is easy to check that $A_k(s)$ is convex and increasing). Indeed, since $u_x^{\varepsilon, \eta}$ converge weakly in $L^2$ to $u_x^\varepsilon$, lower semicontinuity follows from the standard result \cite[Chapter 8.2, Theorem 1]{evans}.

At this point, one needs to multiply the inequality \eqref{eq: G k ineq} by $\zeta^2$ and integrate it over $I$. Then, take $\liminf$ with $\eta \to 0$ and use \eqref{eq: lim for h} and \eqref{eq: liminf for u} to arrive at the desired result \eqref{eq: lhs general eta}.

We may now turn to the passage with $\varepsilon \to 0$. Plugging $G(p) = (c_\varphi^+|p| - k)_+$ in Lemma \ref{lem: G weaklystar} means that
	\[
		\lim_{\varepsilon \to 0} \int_I \zeta^2(c_\varphi^+|h_x^\varepsilon| - k)_+ \,\mathrm d \mathcal L^1 = \int_I \zeta^2 \,\mathrm d (c_\varphi^+|h_x| - k)_+.
	\]
As shown in Lemma \ref{lem: gamma conv eps}, $u_x^\varepsilon \mathrel{\ensurestackMath{\stackon[0.5pt]{\rightharpoonup}{\scriptstyle\ast}}} u_x$, which after applying Corollary \ref{cor-gs-lsc-zeta} results in the slightly modified lower semicontinuity, i.e.,
\[
\liminf_{\varepsilon \to 0} \int_I \zeta^2 \,\mathrm d A_k(|u_x^{\varepsilon}|) \geq \int_I \zeta^2 \,\mathrm d A_k(|u_x|),
\]
which finishes the proof.
\end{proof}

\begin{prop} \label{prop-rhs-eps-eta}
Set $\mathcal{R}(\varepsilon, k) := \limsup_{\eta \to 0} - \int_I (\zeta^2 G_k(u_x^{\varepsilon, \eta}))_x \cdot D^2 F_\eta (u_x^{\varepsilon, \eta}) \cdot u_{xx}^{\varepsilon, \eta} \,\mathrm d \mathcal L^1$.
Then, the r.\,h.\,s. of equation \eqref{eq: after mult} satisfies
\[
	\lim_{k \to \infty} \limsup_{\varepsilon \to 0} \mathcal{R}(\varepsilon, k) = 0.
\]
\end{prop}

\begin{proof}
Throughout this proof $u_x^{\varepsilon, \eta}$ is denoted without superscripts. We compute the necessary derivatives and deal with the terms coming from $f_\eta \circ \widetilde{\varphi}_\eta$ and $\eta/2 \widetilde{\widetilde{\varphi}}_\eta$ separately.
\begin{comment}
\begin{align*}
	D^2 F_\eta &= f^{''}_\eta \circ \widetilde{\varphi}_\eta D \widetilde{\varphi}_\eta \otimes D \widetilde{\varphi}_\eta + f'_\eta \circ \widetilde{\varphi}_\eta D^2 \widetilde{\varphi}_\eta + \theta(\eta) \left ( D\widetilde{\varphi}_\eta \otimes D \widetilde{\varphi}_\eta + \widetilde{\varphi}_\eta D^2\widetilde{\varphi}_\eta \right ) \\
	(\zeta^2 G_k(u_x))_x &= 2 \zeta \zeta' g_k(\widetilde{\varphi}_\eta(u_x)) u_x + \zeta^2 g_k'(\widetilde{\varphi}_\eta(u_x))(\widetilde{\varphi}_\eta(u_x))_x u_x + \zeta^2 g_k(\widetilde{\varphi}_\eta(u_x)) u_{xx}
\end{align*}
\begin{align*}
	(\zeta^2 G_k(u_x))_x &\cdot D^2 F_\eta(u_x) \cdot u_{xx} =\\
	&\quad 2\zeta \zeta'g_k(\widetilde{\varphi}_\eta(u_x)) f''_\eta(\widetilde{\varphi}_\eta(u_x)) \widetilde{\varphi}_\eta(u_x) (\widetilde{\varphi}_\eta(u_x))_x \\
	&+ \zeta^2 g_k'(\widetilde{\varphi}_\eta(u_x))f''_\eta(\widetilde{\varphi}_\eta(u_x)) \widetilde{\varphi}_\eta(u_x) (\widetilde{\varphi}_\eta(u_x))_x^2 \\
	&+ \zeta^2 g_k(\widetilde{\varphi}_\eta(u_x)) f''_\eta(\widetilde{\varphi}_\eta(u_x)) (\widetilde{\varphi}_\eta(u_x))_x^2 \\
	&+ \zeta^2 g_k(\widetilde{\varphi}_\eta(u_x)) f'_\eta(\widetilde{\varphi}_\eta(u_x)) u_{xx} \cdot D^2 \widetilde{\varphi}_\eta \cdot u_{xx} \\
	&+ \theta(\eta) \Bigl (
	2\zeta\zeta_x g_k(\widetilde{\varphi}_\eta(u_x))u_x \cdot D\widetilde{\varphi}_\eta \otimes D\widetilde{\varphi}_\eta \cdot u_{xx} \\
	&+ \zeta^2g_k'(\widetilde{\varphi}_\eta(u_x))(\widetilde{\varphi}_\eta(u_x))_x u_x \cdot D\widetilde{\varphi}_\eta \otimes D\widetilde{\varphi}_\eta \cdot u_{xx}  \\
	&+ \zeta^2 g_k(\widetilde{\varphi}_\eta(u_x))u_{xx} \cdot D\widetilde{\varphi}_\eta \otimes D\widetilde{\varphi}_\eta \cdot u_{xx} \\
	&+ \zeta^2g_k(\widetilde{\varphi}_\eta(u_x)) \widetilde{\varphi}_\eta(u_x) u_{xx} \cdot D^2\widetilde{\varphi}_\eta \cdot u_{xx}
	\Bigr) \\
\end{align*}
\end{comment}
\begin{align*}
		(\zeta^2 G_k(u_x))_x \cdot D^2(f_\eta \circ \widetilde{\varphi}_\eta )(u_x)\cdot u_{xx} &=
	2\zeta \zeta_x g_k(\widetilde{\varphi}_\eta(u_x)) f''_\eta(\widetilde{\varphi}_\eta(u_x)) \widetilde{\varphi}_\eta(u_x) (\widetilde{\varphi}_\eta(u_x))_x \\
	&+ \zeta^2 f''_\eta(\widetilde{\varphi}_\eta(u_x)) (\widetilde{\varphi}_\eta(u_x))_x^2  \left( g_k'(\widetilde{\varphi}_\eta(u_x)) \widetilde{\varphi}_\eta(u_x)  + g_k(\widetilde{\varphi}_\eta(u_x)) \right)\\
	&+ \zeta^2 g_k(\widetilde{\varphi}_\eta(u_x)) f'_\eta(\widetilde{\varphi}_\eta(u_x)) u_{xx} \cdot D^2 \widetilde{\varphi}_\eta(u_x) \cdot u_{xx} \\
	&\geq -\zeta_x^2 f''_\eta(\widetilde{\varphi}_\eta(u_x)) \underbrace{\frac{g_k^2(\widetilde{\varphi}_\eta(u_x)) \widetilde{\varphi}_\eta^2(u_x)}{g_k(\widetilde{\varphi}_\eta(u_x)) + g'_k(\widetilde{\varphi}_\eta(u_x))\widetilde{\varphi}_\eta(u_x)}}_{\mathcal{K}(\widetilde{\varphi}_\eta(u_x))}.
\end{align*}
The rest vanishes, because $\widetilde{\varphi}_\eta$ is homogeneous and hence $D^2 \widetilde{\varphi}_\eta(u_x) \cdot u_{x} = 0$. The third term is non-negative, whereas for the first and the second and third we use $2ab \geq -a^2 - b^2$ inequality, which yields the above estimate. To this end, one needs to make sure that $g_k(\sigma) + g'_k(\sigma)\sigma$ is non-negative for $\sigma > k$. Clearly, after substituting $\sigma$ for $\widetilde{\varphi}_\eta(u_x)$, we have
\begin{equation*}
\mathcal{K}(\sigma) = \frac{(\sigma - k)_+^2  | f'_{\eta} (\sigma) |^2}{k f'_{\eta}(\sigma) + \sigma(\sigma - k) f''_{\eta}(\sigma)}.
\end{equation*}
Forgetting about the non-negative term $k f'_\eta$ in the denominator and using linear growth of $f_\eta$ and the fact that $\sigma > k$ allows us to conclude that
\begin{equation} \label{ke-small}
\zeta^2_x f''_\eta (\sigma) \mathcal{K}(\sigma) \leq \frac{||\zeta_x||_\infty^2}{\sigma}(\sigma - k)_+ |f'_\eta(\sigma)|^2 \leq \frac{C ||\zeta_x||_\infty^2}{k} (\sigma - k)_+.
\end{equation}

Let us now turn to the neglected terms of the derivative which are multiplied by $\eta/2$. Using an analogous method, we will show that they converge to zero when $\eta$ converges to zero. Recall that $\widetilde{\widetilde{\varphi}}_\eta = \widetilde{\varphi}_\eta^2$ outside a~small ball, where $(\zeta^2 G_k(u_x))_x$ vanishes anyway.

\begin{align*}
	\frac{2}{\eta}(\zeta^2 G_k(u_x))_x \cdot D^2 \left(\frac{1}{2}\eta \widetilde{\varphi}_\eta^2 \right) \cdot u_{xx} &=
	2\zeta\zeta_x g_k(\widetilde{\varphi}_\eta(u_x)) \widetilde{\varphi}_\eta(u_x) (\widetilde{\varphi}_\eta(u_x))_x \\
	&+ \zeta^2 \Bigl (\widetilde{\varphi}_\eta(u_x))_x^2 (g_k'(\widetilde{\varphi}_\eta(u_x))\widetilde{\varphi}_\eta(u_x) + g_k(\widetilde{\varphi}_\eta(u_x) \Bigr) \\
	&+ \zeta^2g_k(\widetilde{\varphi}_\eta(u_x)) \widetilde{\varphi}_\eta(u_x) u_{xx} \cdot D^2\widetilde{\varphi}_\eta \cdot u_{xx} \\
	& \geq -\zeta_x^2 \mathcal{K}(\widetilde{\varphi}_\eta(u_x)).
\end{align*}
Forgetting about the term with $f''_\eta$ in the denominator of $\mathcal{K}$ and keeping in mind, as before, that $f_\eta$ is of linear growth, one gets
\begin{equation} \label{ke-big}
\zeta^2_x \mathcal{K}(\sigma) \leq \frac{C ||\zeta_x||_\infty^2}{k} (\sigma - k)_+^2.
\end{equation}

Using estimates \eqref{ke-big} and \eqref{ke-small} and careful multiplying by $-1$ lead to the conclusion that
\begin{equation*}
- \int_I (\zeta^2 G_k(u_x))_x \cdot D^2 F_\eta (u_x) \cdot u_{xx} \,\mathrm d \mathcal L^1 \leq \frac{C||\zeta_x||_\infty^2}{k} \left(  \int_I (\widetilde{\varphi}_\eta(u_x) - k)_+ \,\mathrm d \mathcal L^1 + \eta  \int_I (\widetilde{\varphi}_\eta(u_x) - k)_+^2 \,\mathrm d \mathcal L^1 \right).
\end{equation*}

Eventually, recall that if $\varepsilon$ is fixed, then $\left \{ \int_I (\widetilde{\varphi}_\eta(u_x^{\varepsilon, \eta}) - k)_+^2 \,\mathrm d \mathcal L^1 \right \}$ is bounded w.\,r.\,t.\;$\eta$ (which was essentially proved in Lemma \ref{lem: cpt eta}), whence the second term vanishes when $\eta$ converges to zero. Lastly, the fact that
\[
\mathcal E_{\varepsilon, \eta}(u^{\varepsilon, \eta}) \leq \int_I |h^\varepsilon|^2 \,\mathrm d \mathcal L^1 \leq 2\int_I |h|^2 \,\mathrm d \mathcal L^1
\]
and convexity of $f$ imply that $\left \{\int_I (\widetilde{\varphi}_\eta(u_x^{\varepsilon, \eta})  - k)_+ \,\mathrm d \mathcal L^1 \right\}$ is bounded (w.\,r.\,t.\;both $\eta$ and $\varepsilon$).
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:constants}]
	Propositions \ref{prop: general lhs eps} and \ref{prop-rhs-eps-eta} imply that the minimizer $u$ of $\mathcal E$ satisfies
	\begin{equation} \label{eq:constants final}
		\frac{r(k)}{ (c_\varphi^+)^{3}} \int_I \zeta^2 \,\mathrm d (c_\varphi^- |u_x| - k)_+(c_\varphi^+ |u_x| - k)_+ |u_x|^{-1} \leq  \frac{f^\infty}{ c_\varphi^- c_\varphi^+} \int_I \zeta^2 \,\mathrm d (c_\varphi^+ |h_x| - k)_+ + \lambda \limsup_{\varepsilon \to 0} \mathcal{R}(\varepsilon, k),
	\end{equation}
where $\lim_{k \to \infty} r(k) = f^\infty$. Let us now pass with $k \to \infty$. Since
\[
	\int_I \zeta^2 \,\mathrm d (c_\varphi^+|h_x| - k)_+ = \int_I \zeta^2 \,\mathrm d (c_\varphi^+|h_x^{ac}| - k)_+ \,\mathrm d \mathcal L^1 + c_\varphi^+ \int_I \zeta^2 \,\mathrm d |h_x^s|,
\]
we can use dominated convergence theorem (as $\zeta^2 |h_x^{ac}|$ is integrable) to pass with the limit in the first integral and see that it vanishes. Therefore, the r.\,h.\,s.\;of \eqref{eq:constants final} converges to $f^\infty(c_\varphi^-)^{-1} \int_I \zeta^2 \,\mathrm d |h_x^s|$. In an analogous manner, limit passage on the l.\,h.\,s\;can be performed, which results in
\begin{equation} \label{eq:constants final2}
	\int_I \zeta^2 \,\mathrm d |u_x^s| \leq \left ( \frac{c_\varphi^+}{c_\varphi^-} \right)^2 \int_I \zeta^2 \,\mathrm d |h_x^s|.
\end{equation}
To finish the proof, it suffices to apply Lemma \ref{lem: mu leq nu} for $\mu = |u_x^s|$ and $\nu = (c_\varphi^+)^2(c_\varphi^-)^{-2}|h_x^s|$.
\end{proof}

\section{The regular case} \label{sec:regular}

Firstly, let us note that under assumptions of Theorem~\ref{thm:regular norm}, $F$ is differentiable in $\mathbb R^n$. Additionally, any differentiable convex function on $\mathbb R^n$ is in fact continuously differentiable (\cite[Corollary 25.5.1]{rockafellar}). Since $\mathbb R^n \setminus \{0\}$ can be easily covered with a~finite number of open convex sets, this fact remains true if one replaces $\mathbb R^n$ with $\mathbb R^n \setminus \{0\}$. In view of Lemma \ref{lem: strict convexity}, $F$ is strictly convex and the assumption of strict convexity in the sense of Reshetnyak for $\varphi$ is necessary for $F$ to be strictly convex.

We define $\mathcal F_M \colon M(I, \mathbb R^n) \to [0, \infty[$ by
\[\mathcal F_M(\mu) = F(\mu)(\mathcal I).\]
Clearly, we have $\mathcal F(w) = \mathcal F_M(w_x)$ for $w \in BV(I, \mathbb R^n)$.

\begin{thm} \cite[Theorem 2.4]{anzellotti2} \label{thm:anzellotti}
	Functional $\mathcal F_M$ is differentiable at $\alpha \in M(I, \mathbb R^n)$ in direction $\beta \in M(I, \mathbb R^n)$ iff $|\beta^s| \ll |\alpha^s|$ and that the derivative is of the form
\begin{equation*}
	\int_I DF(\alpha^{ac}) \cdot \beta^{ac} \,\mathrm d \mathcal L^1 + \int_I DF^{\infty} \left( \frac{\alpha^s}{|\alpha^s|} \right) \cdot \,\mathrm d \beta^s,
\end{equation*}
where $DF^\infty(p) = \lim_{t \to \infty} \tfrac{1}{t}DF(tp)$.
\end{thm}

\begin{lemma} \label{lem: vanish at endpoints}
	The following statements hold:
	\begin{enumerate}[(i)]
		\item $DF_\eta(u_x^{\varepsilon, \eta}) \in W^{1,2}_0(I, \mathbb R^n)$ and $\lambda \left( DF_\eta(u_x^{\varepsilon, \eta}) \right)_x = u^{\varepsilon, \eta} - h^\varepsilon$ a.e. in $I$;
		\item $DF(u_x^{\varepsilon}) \in W^{1,2}_0(I, \mathbb R^n)$ and $\lambda \left( DF(u_x^{\varepsilon}) \right)_x = u^\varepsilon - h^\varepsilon$ a.e. in $I$;
		\item $DF(u_x^{ac}) \in W^{1,2}_0(I, \mathbb R^n)$ and $\lambda \left( DF (u_x^{ac}) \right)_x = u - h$ a.e. in $I$.
	\end{enumerate}
\end{lemma}

\begin{proof}
We will show only proof of (iii) as it is the most interesting one. Statements (i) and (ii) are proved in the same way only without invoking the additional Theorem \ref{thm:anzellotti}.

	Firstly, observe that since $DF$ is bounded, the function $DF(u_x^{ac}) \in L^2(I, \mathbb R^n)$. Secondly, applying Theorem \ref{thm:anzellotti} allows us to write the derivative of $\mathcal F$ at point $u$ and in direction $\xi \in C^\infty_c(I)$ as
\begin{equation*}
	\int_I \xi_x DF(u_x^{ac})  \,\mathrm d \mathcal L^1,
\end{equation*}
since the integral with respect to the singular part vanishes. As $u$ is the minimizer of $\mathcal E$, it is true that
\begin{equation*}
	- \lambda  \int_I \xi_x DF(u_x^{ac}) \,\mathrm d \mathcal L^1 = \int_I \xi (u - h) \,\mathrm d \mathcal L^1,
\end{equation*}
which means that the distributional derivative of $DF(u_x^{ac})$ equals $(u - h)/\lambda \in L^2(I, \mathbb R^n)$. Consequently, $DF(u_x^{ac}) \in W^{1,2}(I, \mathbb R^n)$, which implies also that there exists a~continuous representative of this function.

	To show that any continuous representative equals zero at the endpoints of the interval, let $\xi^\ell(x) = (-\ell x + 1 + al) \cdot \indi_{[a, a + 1/\ell]}$ be a~piece-wise linear function equal $1$ for $x = a$ and $0$ on $[a, a + 1/\ell]$. Once again the Euler-Lagrange equation given by Theorem \ref{thm:anzellotti} yields
\begin{equation*}
	-\lambda \int_I \xi^\ell_x DF(u_x^{ac})  \,\mathrm d \mathcal L^1 = \int_I \xi^\ell (u - h)  \,\mathrm d \mathcal L^1.
\end{equation*}
As $\xi^\ell_x = -\ell \indi_{[a, a + 1/\ell]}$, continuity of $DF(u_x^{ac})$ implies that the l.\,h.\,s. converges to $DF(u_x^{ac}(a))$ as $\ell \to \infty$. Morover, dominated convergence theorem enables us to pass with the limit under the integral sign in order to get zero on the r.\,h.\,s, which implies that $DF(u_x^{ac}(a)) = 0$. To get $DF(u_x^{ac}(b)) = 0$, one needs to repeat this argument with $\xi^\ell = (\ell x + 1 -\ell b) \cdot \indi_{[b-1/\ell, b]}$.
\end{proof}

Observe that for fixed representatives of $u^{\varepsilon, \eta}$ and $DF_\eta(u_x^{\varepsilon, \eta})$ it is true that almost everywhere on $I$, $DF_\eta(u_x^{\varepsilon, \eta})$ is an actual composition of $DF$ and $u_x^{\varepsilon, \eta}$, i.e., $DF_\eta(u_x^{\varepsilon, \eta})(x_0) = DF_\eta(u_x^{\varepsilon, \eta}(x_0))$. The same holds for analogous pairs of mappings from Lemma \ref{lem: vanish at endpoints}.

\subsection*{Limit passage with $\eta$}

Thanks to the additional regularity of $F$, it is possible to perform a~more subtle limit passage with $\eta \to 0$ on the l.\,h.\,s\;than in the previous case.

\begin{lemma} \label{lem: a.e. conv eta}
There exists a sequence $\eta_j$ with $\eta_j \to 0$ as $j \to \infty$ such that $u_x^{\varepsilon, \eta_j}$ converges to $u_x^\varepsilon$ a.\,e.\;on $I$.
\end{lemma}

\begin{proof}
To begin with, we will show that the sequence $DF_{\eta}( u_x^{\varepsilon, \eta})$ converges to $DF(u_x^\varepsilon)$ in $C( \overline{I}, \mathbb R^n)$. Due to the choice of $h^\varepsilon$ and properties established in Lemma \ref{lem: cpt eta}, functions $ u^{\varepsilon, \eta} - h^\varepsilon$ converge in $L^2$ to $u^{\varepsilon}-h^{\varepsilon}$. In light of Lemma \ref{lem: vanish at endpoints}, it is then evident that $(DF_{\eta}(u_x^{\varepsilon, \eta_j}))_x$ converge to $(DF(u_x^{\varepsilon}))_x$ in $L^2$ as well. For every $x_0 \in \overline{I}$, due to the boundary condition established in the same lemma, it is true that
\begin{equation*}
	DF_{\eta} (u_x^{\varepsilon, \eta})(x_0) = \int_a^{x_0} \left(DF_{\eta} (u_x^{\varepsilon, \eta}) \right)_x \,\mathrm d \mathcal L^1 \quad \text{and} \quad DF (u_x^{\varepsilon})(x_0) = \int_a^{x_0} \left(DF(u_x^{\varepsilon}) \right)_x \,\mathrm d \mathcal L^1.
\end{equation*}
The $L^2$ convergence of the integrands allows to conclude that
\begin{equation} \label{eq: a.e. conv eta}
	\sup_{x_0 \in \overline{I}} \left \lvert DF_{\eta}( u_x^{\varepsilon, \eta}) (x_0) - DF(u_x^{\varepsilon})(x_0) \right \rvert \to 0.
\end{equation}

Secondly, let us observe that
\begin{align} \label{eq: l1 convergence}
\begin{split}
	\int_I \left| DF_{\eta} (u_x^{\varepsilon, \eta}) - DF(u_x^{\varepsilon, \eta}) \right| \,\mathrm d \mathcal L^1 &\leq \int_I \left| D(f_{\eta} \circ \widetilde{\varphi}_\eta)(u_x^{\varepsilon, \eta}) - D(f \circ \varphi)(u_x^{\varepsilon, \eta}) \right | \,\mathrm d \mathcal L^1 +\\
	&+ \frac{\eta}{2} \int_I |D \widetilde{\widetilde{\varphi}}_{\eta} (u_x^{\varepsilon, \eta})| \,\mathrm d \mathcal L^1.
\end{split}
\end{align}
In view of property (iv) from Proposition \ref{goodnorm}, it is evident that the first integral converges to zero. The second one (up to a negligible constant coming from the fact that $\widetilde{\widetilde{\varphi}}$ coincides with $\widetilde{\varphi}_{\eta}$ outside a small ball) does not exceed
\[
	\eta \int_I \widetilde{\varphi}_\eta(u_x^{\varepsilon, \eta}) |D\widetilde{\varphi}_\eta(u_x^{\varepsilon, \eta}) | \,\mathrm d \mathcal L^1.
\]
As $|D\widetilde{\varphi}_\eta|$ is a~bounded function and $\int_I \widetilde{\varphi}_\eta(u_x^{\varepsilon, \eta}) \,\mathrm d \mathcal L^1$ is bounded w.r.t $\eta$ (see proof of Proposition \ref{prop-rhs-eps-eta}), one can conclude that \eqref{eq: l1 convergence} converges to zero. This implies existence of a~subsequence $\eta_j$ with $\eta_j \to 0$ as $j \to \infty$ for which
\begin{equation*}
	\lim_{j \to \infty} \left| DF_{\eta_{j}} (u_x^{\varepsilon, \eta_j}) - DF(u_x^{\varepsilon, \eta_j}) \right| = 0 \text{ a.e. on } I.
\end{equation*}

In \eqref{eq: a.e. conv eta} it has been already checked that the sequence $DF_{\eta_j} (u_x^{\varepsilon, \eta_j})$ converges to $DF (u_x^\varepsilon)$ in $C( \overline{I}, \mathbb R^n)$. As a consequence,
\begin{equation*}
	\lim_{j \to \infty} \left| DF (u_x^{\varepsilon, \eta_j}) - DF(u_x^{\varepsilon}) \right| = 0 \text{ a.e. on } I.
\end{equation*}
As $F$ is strictly convex and $C^1$, $DF$ is strictly monotone and continuous. Strict monotonicity of $DF$ implies that the map is injective and hence, by Brouwer's invariance of domain theorem, there exists its continuous inverse $DF^{-1}$. Therefore, at any point in which $DF(u_x^{\varepsilon, \eta_j})(x_0) \to DF(u_x^{\varepsilon})(x_0)$ and the mappings $DF(u_x^{\varepsilon, \eta_j})$ and $DF(u_x^{\varepsilon})$ are actual compositions,
we have
\begin{equation*}
	u_x^{\varepsilon, \eta_j}(x_0) = DF^{-1} \left( DF(u_x^{\varepsilon, \eta_j}(x_0)) \right) \to DF^{-1} \left( DF(u_x^{\varepsilon} (x_0)) \right) = u_x^{\varepsilon}(x_0).
\end{equation*}
Since the set of such points is of full measure, we have established the desired a.e. convergence of $u_x^{\varepsilon, \eta_j}$ to $u_x^\varepsilon$.

\end{proof}

\begin{prop} \label{prop-lhs-eps-eta}
	There exists a sequence $\eta_j$ with $\eta_j \to 0$ as $j \to \infty$ for which the l.\,h.\,s. of equation \eqref{eq: after mult} satisfies
	\begin{equation*}
		\lim_{j \to \infty} \int_I \zeta^2 G_k(u_x^{\varepsilon, \eta_j}) \cdot  (u_x^{\varepsilon, \eta_j} - h_x^{\varepsilon}) \,\mathrm d \mathcal L^1
		= \int_I \zeta^2 \frac{(\varphi(u_x^\varepsilon) - k)_+}{\varphi^2(u_x^{\varepsilon})} f'(\varphi(u_x^\varepsilon)) (u_x^{\varepsilon} - h_x^{\varepsilon}) \cdot u_x^\varepsilon \,\mathrm d \mathcal L^1.
	\end{equation*}
\end{prop}

\begin{proof}
Let us choose the sequence from Lemma \ref{lem: a.e. conv eta}, for which we know that the integrand is a.\,e convergent. Then, it suffices to prove its uniform integrability and apply the Vitali convergence theorem. Bearing in mind the formula for $G_k(\cdot)$, the absolute value of the integrand can be estimated from above by $C |u_{x}^{\varepsilon, \eta} - h_x^{\varepsilon}|$ for some constant $C$. As a~result, it suffices to check that
\begin{equation*}
	\int_{|u_x^{\varepsilon, \eta} - h_x^{\varepsilon}| \geq t} |u_x^{\varepsilon, \eta} - h_x^{\varepsilon}| \,\mathrm d \mathcal L^1 \xrightarrow{t \to \infty} 0,
\end{equation*}
which follows from H\"{o}lder's and Chebyshev's inequalities and again the fact that the sequence $\{u_x^{\varepsilon, \eta}\}_\eta$ is bounded in $L^2$.
\end{proof}

\begin{prop} \label{prop: inequality with eps}
	Recall that $u^\varepsilon$ is a~minimizer of the functional $\mathcal E_\varepsilon$. The following estimate holds
	\begin{equation*}
		f'(k) \int_I \zeta^2 |u_x^\varepsilon|^2 \frac{(\varphi(u_x^\varepsilon) - k)_+}{\varphi^2(u_x^\varepsilon)} \,\mathrm d \mathcal L^1 \leq f^\infty \int_I \zeta^2 |h_x^\varepsilon| |u_x^\varepsilon| \frac{(\varphi(u_x^\varepsilon) - k)_+}{\varphi^2(u_x^\varepsilon)} \,\mathrm d \mathcal L^1 + \lambda \mathcal{R}(\varepsilon, k),
	\end{equation*}
where $\mathcal{R}(\varepsilon, k)$ was defined in Proposition \ref{prop-rhs-eps-eta}.
\end{prop}

\begin{proof}
The proof boils down to observing what has already been proved about the Euler--Lagrange equation \eqref{Euler-Lagrange-eps-eta} of $\mathcal E_{\varepsilon, \eta}$. Combining Propositions \ref{prop-rhs-eps-eta} and \ref{prop-lhs-eps-eta} gives
\begin{equation*}
	\int_I \zeta^2 \frac{(\varphi(u_x^\varepsilon) - k)_+}{\varphi^2(u_x^\varepsilon)} f'(\varphi(u_x^\varepsilon)) |u_x^{\varepsilon}|^2\,\mathrm d \mathcal L^1
	\leq
	 \int_I \zeta^2 \frac{(\varphi(u_x^\varepsilon) - k)_+}{\varphi^2(u_x^\varepsilon)} f'(\varphi(u_x^\varepsilon)) h_x^{\varepsilon} \cdot u_x^\varepsilon \,\mathrm d \mathcal L^1 + \mathcal{R}(\varepsilon, k).
\end{equation*}
Then, on the l.\,h.\,s.\;one needs to use the fact that $f'$ is non-decreasing and the integration takes place on the set where $\varphi(u_x^\varepsilon) > k$, whereas on the r.\,h.\,s.\;the fact that $f'(t) \leq f^\infty$ for all~$t$.
\end{proof}

\subsection*{Limit passage with $\varepsilon$}

\begin{lemma} \label{lem: u eps a.e.}
There exist functions $\overline{u}_x^{\varepsilon_i}, \overline{u}_x^{ac} \in C(\overline{I}, \overline{R^n})$ s.t.\;$\overline{u}_x^{\varepsilon_i} \to \overline{u}_x^{ac}$ in $C(\overline{I}, \overline{R^n})$ and for a.\,e.\;$x_0 \in \overline{I}$, $\overline{u}_x^{\varepsilon_i}(x_0) = u_x^{\varepsilon_i}(x_0)$ and for a.\,e.\;$x_0 \in \overline{I}$, $\overline{u}_x^{ac}(x_0) = u_x^{ac}(x_0)$.
\end{lemma}
\begin{proof}
	Lemma \ref{lem: vanish at endpoints} implies (by a~reasoning similar to the beginning of the proof of Lemma \ref{lem: a.e. conv eta}) that $DF(u_x^\varepsilon)$ converge to $DF(u_x^{ac})$ in $C(\overline{I}, \mathbb R^n)$. This implies convergence in $C(\overline{I}, \overline{\mathbb R^n})$. The map $\overline{DF}$ is a~homeomorphism, as established in Lemma \ref{lem: DF homeo}, which prompts the following definition
\begin{equation*}
	\overline{u}_x^{\varepsilon_i} := \overline{DF}^{-1} \circ DF(u_x^{\varepsilon_i}) \text{ and } \overline{u}_x^{ac} := \overline{DF}^{-1} \circ DF (u_x^{ac}).
\end{equation*}
Requested properties of these functions now follow from continuity of $\overline{DF}^{-1}$ and the fact that $u_x^{\varepsilon_i}$ and $u_x^{ac}$ are almost everywhere finite.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:regular norm}]
Let us recall the inequality proved in Proposition \ref{prop: inequality with eps}
\begin{equation*}
		L_{\varepsilon, k} :=  f'(k) \int_I \zeta^2 |u_x^\varepsilon|^2 \frac{(\varphi(u_x^\varepsilon) - k)_+}{\varphi^2(u_x^\varepsilon)} \,\mathrm d \mathcal L^1 \leq f^\infty \int_I \zeta^2 |h_x^\varepsilon| |u_x^\varepsilon| \frac{(\varphi(u_x^\varepsilon) - k)_+}{\varphi^2(u_x^\varepsilon)} \,\mathrm d \mathcal L^1 + \lambda \mathcal{R}(\varepsilon, k) := R_{\varepsilon, k},
\end{equation*}
where $\zeta \in C_c^1(I)$ and $\lim_{k \to \infty} \limsup_{\varepsilon \to 0} \mathcal{R}(\varepsilon, k) = 0$ as stated in Proposition \ref{prop-rhs-eps-eta}. We will firstly pass with $\varepsilon$ to zero. Define a~continuous function $b_k: \overline{\mathbb R^n} \to \mathbb R$
\begin{equation*}
	b_k(p) = \begin{cases}
		\frac{|p| \left( \varphi(p) - k \right)_+}{\varphi^2(p)} & \text{ if } p \in \mathbb R^n \vspace{1mm} \\
		\frac{1}{\varphi(\omega)} & \text{ if } p = \infty \omega \in \mathbb{S}^{n-1}_\infty.
	\end{cases}
\end{equation*}
Clearly, $u_x^\varepsilon = \overline{u}_x^{\varepsilon}$ $|u_x^{\varepsilon}|$-a.e. due to absolute continuity of this measure w.r.t $\mathcal L^1$. Therefore,
\begin{equation*}
	L_{\varepsilon, k} = f'(k) \int_I \zeta^2 |u_x^{\varepsilon}| \left( b_k(\overline{u}_x^\varepsilon) - b_k(\overline{u}_x^{ac}) \right) \,\mathrm d \mathcal L^1 + f'(k) \int_I \zeta^2 b_k(\overline{u}_x^{ac}) \,\mathrm d |u_x^{\varepsilon}|.
\end{equation*}
In view of Lemma \ref{lem: u eps a.e.} (ii), it is true that $b_k(\overline{u}_x^{\varepsilon_i})$ converges to $\overline{u}_x^{ac}$ in $C(\overline{I}, \mathbb R)$, which implies that the first term converges to zero (recall that $\{u_x^\varepsilon\}$ is bounded in $L^1$). The second term is lower semicontinuous w.\,r.\,t\;weak* convergence of $u_x^\varepsilon \to u_x$ (see Corollary \ref{cor-gs-lsc-zeta}) and whence
\begin{equation*}
	\lim_{i \to \infty} L_{\varepsilon_i, k} \geq f'(k) \int_I \zeta^2 b_k(\overline{u}_x^{ac}) \,\mathrm d |u_x|.
\end{equation*}
As $h^\varepsilon$ has been chosen so that $|h_x^\varepsilon| \mathrel{\ensurestackMath{\stackon[0.5pt]{\rightharpoonup}{\scriptstyle\ast}}} |h_x|$, one shows in a~similar manner that
\begin{equation*}
	\limsup_{\varepsilon \to 0} R_{\varepsilon, k} \leq f^\infty \int_I \zeta^2 b_k(\overline{u}_x^{ac}) \,\mathrm d |h_x| + \limsup_{\varepsilon \to 0} \mathcal{R}(\varepsilon, k).
\end{equation*}
As a result,
\begin{align} \label{eq: before k limit}
	f'(k) \int_I \zeta^2 b_k(\overline{u}_x^{ac}) \,\mathrm d |u_x^{s}| 		&\leq f^\infty \int_I \zeta^2 b_k(\overline{u}_x^{ac}) \,\mathrm d |h_x| + \limsup_{\varepsilon \to 0} \mathcal{R}(\varepsilon, k) \leq \\
	& \leq f^\infty \int_I \zeta^2 b_k(\overline{u}_x^{ac}) \,\mathrm d |h_x^{ac}| + f^\infty \int_I \zeta^2 b_k(\overline{u}_x^{ac}) \,\mathrm d |h_x^{s}| + \limsup_{\varepsilon \to 0} \mathcal{R}(\varepsilon, k). \nonumber
\end{align}

We will now pass with $k$ to infinity. At any point $x \in I$ for which $\overline{u}_x^{ac}(x) \in \mathbb{S}^{n-1}_\infty$, set $\omega(x) \in \mathbb{S}_{1}^{n-1}$ to be such that $\overline{u}_x^{ac}(x) = \infty \, \omega(x)$. Clearly, for any $x \in I$
\[
	b_k(\overline{u}_x^{ac}(x)) \xrightarrow{k \to \infty} b(x) := \frac{1}{\varphi(\omega(x))} \indi_{ \left\{x: \, \overline{u}_x^{ac}(x) \in \mathbb{S}_\infty^{n-1} \right\}}(x).
\]
Therefore, the fact that $|\overline{u}_x^{ac}|$ is a.e.\;finite together with dominated convergence theorem allow us to conclude that \eqref{eq: before k limit} after passing with $k$ to infinity and dividing by $f^\infty$ becomes
\begin{equation}\label{eq: after k limit}
	\int_I \zeta^2 b \,\mathrm d |u_x^s| \leq \int_I \zeta^2 b \,\mathrm d |h_x^s|.
\end{equation}
Function $b$ is measurable w.r.t.\;any Radon measure as a~pointwise limit of continuous functions, whence both $ b \,\mathrm d |u_x^{s}|$ and $ b \,\mathrm d |h_x^s|$ are~Radon measures. Lemma \ref{lem: mu leq nu} for $\mu =  b \,\mathrm d |u_x^s|$ and $\nu =  b \,\mathrm d |h_x^s|$ implies that
\begin{equation*}
	\int_E b \,\mathrm d |u_x^s| \leq \int_E b \,\mathrm d |h_x^s| \text{ for any Borel } E \subset U.
\end{equation*}

It is shown in Lemma \ref{lem: ac sphere} below that $b(x) > 0$ for $|u_x^s|$-a.e. $x$, which implies absolute continuity of $|u_x^s|$ w.r.t $|h_x^s|$. Writing $|u_x^s|/|h_x^s|$ for the Radon--Nikodym derivative of $|u_x^s|$ w.\,r.\,t.\;$|h_x^s|$, we can see that the measure
\[
	b\,\left( 1 - \frac{|u_x^s|}{|h_x^s|} \right) \,\mathrm d |h_x^s|
\]
is non-negative on Borel sets. Multiplying it by $\varphi(\omega(x)) \, \indi_{ \left \{x: \, \overline{u}_x^{ac}(x) \in \mathbb{S}_\infty^{n-1} \right\}}(x)$, which is clearly non-negative, yields a measure
\[
	\indi_{ \left \{x: \, \overline{u}_x^{ac}(x) \in \mathbb{S}_\infty^{n-1} \right\}} \, \left( 1 - \frac{|u_x^s|}{|h_x^s|} \right) \,\mathrm d |h_x^s|
\]
whose non-negativity implies that for all Borel sets $E \subset U$ one has $|u_x^s|(E) \leq |h_x^s|(E)$, which concludes the proof.

\end{proof}

\begin{lemma} \label{lem: ac sphere}
	For $|u_x^s|$-a.e. $x$ it is true that $\overline{u}_x^{ac}(x) \in \mathbb{S}_\infty^{n-1}$.
\end{lemma}

\begin{proof}
Choose $x_0$ such that $|\overline{u}_x^{ac}(x_0)| < \infty$. Since $\overline{u}_x^{ac}$ is continuous, there exists an open interval $J \subset I$ such that for any $y \in J$ it is true that $\overline{u}_x^{ac}(y) \in \mathbb R^n$. Then for any $\xi \in C_c(J)$,
\begin{equation*}
	\left | \int_J u_x^{\varepsilon} \xi \,\mathrm d \mathcal L^1 - \int_J u_x^{ac} \xi \,\mathrm d \mathcal L^1 \right | = \left | \int_J \overline{u}_x^{\varepsilon} \xi \,\mathrm d \mathcal L^1 - \int_J \overline{u}_x^{ac} \xi \,\mathrm d \mathcal L^1 \right | \leq \sup_{y \in J} |\overline{u}_x^\varepsilon(y) - \overline{u}_x^{ac}(y) | \int_J \xi \,\mathrm d \mathcal L^1 \xrightarrow{\varepsilon \to 0} 0,
\end{equation*}
due to the facts that $\overline{u}_x^{\varepsilon}$ coincides with $u_x^\varepsilon$ a.e. and $\overline{u}_x^{ac}$ coincides with $u_x^{ac}$ a.e.  and that $\overline{u}_x^{\varepsilon}$ converge in $C(\overline{I}, \overline{\mathbb R^n})$ to $\overline{u}_x^{ac}$. As a result, $u_x^\varepsilon \mathrel{\ensurestackMath{\stackon[0.5pt]{\rightharpoonup}{\scriptstyle\ast}}} u_x^{ac}$ on $J$. On the other hand, Lemma \ref{lem: gamma conv eps} says that $u_x^\varepsilon \mathrel{\ensurestackMath{\stackon[0.5pt]{\rightharpoonup}{\scriptstyle\ast}}} u_x$ on $J$, which implies that $u_x^s \mres J = 0$ and, consequently, that
\[
	|u_x^s| \mres J = 0,
\]
which in turn means that $x_0 \notin \mathrm{supp}\,|u_x^s|$ (the support of the measure $|u_x^s|$). We have therefore established that if $x_0 \in \mathrm{supp}\,|u_x^s|$, then $\overline{u}_x^{ac} \in \mathbb{S}_\infty^{n-1}$. It remains to recall that the set $\mathrm{supp}\,|u_x^s|$ is of full $|u_x^s|$-measure, which concludes the proof. \\
\end{proof}

\noindent
{\bf \Large Acknowledgements} \\

The second author was partially supported by the grant nr.\ 2020/36/C/ST1/00492 of the National Science Center, Poland. Part of this work was created during second author's JSPS Postdoctoral Fellowship at the University of Tokyo.
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