2. Preliminary lemmas

In this section, we introduce some fundamental definitions and auxiliary results. By $\Omega$ we always mean a bounded domain of ${\mathbb {R}}^n$ with Lipschitz regular boundary. If not specified, a constant $C$ is a positive constant, possibly changing line by line. By $C^\infty _c(\Omega )$ we mean the set of compactly supported smooth functions over $\Omega$. We begin with $N$-functions and the Musielak–Orlicz space setting.

For an $N$-function we define the general Musielak–Orlicz class $\mathcal {L}_{M}(\Omega ;{\mathbb {R}^n})$ as the set of all measurable functions $\xi (x):\Omega \to {\mathbb {R}^n}$ such that

\[ \int_{\Omega}M(x, \xi(x))\,{\rm d}x < \infty. \]

The Musielak–Orlicz space $L_{M}(\Omega ;{\mathbb {R}^n})$ is the smallest linear hull of $\mathcal {L}_M(\Omega ;{\mathbb {R}^n})$ equipped with the Luxemburg norm

\[ \|\xi\|_{L_{M}(\Omega)}=\inf \Bigg\{\lambda>0:\int_{\Omega}M\Bigg(x,\frac{\xi(x)}{\lambda}\Bigg){\rm d}x\leq 1\Bigg\}. \]

The space $E_M(\Omega ;{\mathbb {R}^n})$ is the closure in $L_M$-norm of the set of bounded functions. Equivalently, $L_M(\Omega ;{\mathbb {R}^n})$ and $E_M(\Omega ;{\mathbb {R}^n})$ are defined as sets of functions $\xi :\Omega \to {\mathbb {R}}^n$ satisfying

\[ \int_{\Omega}M(x,\lambda \xi(x))\, {\rm d}x<\infty \]

for some $\lambda \in {\mathbb {R}}$ and for every $\lambda \in {\mathbb {R}}$, respectively [Reference Chlebicka, Gwiazda, Świerczewska-Gwiazda and Wróblewska-Kamińska15, Lemma 3.1.8]. We also note that $(E_M(\Omega ;{\mathbb {R}^n}))^\ast = L_{M^\ast }(\Omega ;{\mathbb {R}^n})$ and $(E_{M^\ast }(\Omega ;{\mathbb {R}^n}))^\ast = L_M(\Omega ;{\mathbb {R}^n})$ [Reference Chlebicka, Gwiazda, Świerczewska-Gwiazda and Wróblewska-Kamińska15, Theorem 3.5.3] but no other duality relations are expected.

The complementary function to $M$ is

\[ M^*(x,\eta):=\sup_{\xi\in {\mathbb{R}^n}}(\xi \cdot \eta - M(x,\xi)). \]

If $M$ is an $N$-function and $M^\ast$ is the complementary function to $M$, then the following Fenchel–Young inequality is satisfied

\[ |\xi \cdot \eta| \leq M(x,\xi)+ M^\ast(x,\eta) \quad \text{for all}~~\xi,\eta\in {\mathbb{R}^n} ~\text{and a.e.} ~x\in \Omega. \]

Moreover, if $M$ is an $N$-function and $M^\ast$ its complementary, then the generalized Hölder inequality holds, e.g.

\[ \Bigg|\int_{\Omega}\xi\cdot \eta \,{\rm d}x\Bigg|\leq 2 \|\xi\|_{L_{M}}\|\eta\|_{L_{M^\ast}}\quad~\text{for all}~\xi\in L_{M}(\Omega;{\mathbb{R}^n})~\text{and}~ \eta \in L_{M^\ast}(\Omega;{\mathbb{R}^n}). \]

We say that a sequence $\{\xi _n\}_{n=1}^{\infty }\subset L_M(\Omega ;{\mathbb {R}^n})$ converges modularly to $\xi$ in $L_M(\Omega ;{\mathbb {R}^n})$, if there exists $\lambda >0$ such that

\[ \int_{\Omega}M\Bigg(x,\frac{\xi_n-\xi}{\lambda}\Bigg)\mathop{}\!\mathrm{d} x\to 0\text{ as } n\to \infty. \]

For the notion of this convergence, we write $\xi _n \xrightarrow []{M}\xi$.

Then, we shall give some preliminary lemmas related to $N$-functions and Musielak–Orlicz spaces.

Proposition 2.2 [Reference Borowski and Chlebicka9, Theorem 1]

Assume that $\Omega$ is a bounded Lipschitz domain and $M$ is an $N$-function which satisfies the balance condition (B). Then, for every $u\in V^1_0L_M(\Omega )$ there exists a sequence $\{u_{\delta }\}_{\delta }\subset C_c^{\infty }(\Omega )$ such that

\[ u_{\delta}\xrightarrow[]{}u \text{ in } L^1(\Omega) \text{ and } \nabla u_{\delta}\xrightarrow[]{M}\nabla u\text{ in }L_M( \Omega;{\mathbb{R}^n}) \]

Furthermore, there exists a constant $c=c(\Omega )$, such that $||u_{\delta }||_{L^{\infty }(\Omega )}\leq c||u||_{L^{\infty }(\Omega )}$.

Lemma 2.3 de la Vallée Poussin theorem [Reference Chlebicka, Gwiazda, Świerczewska-Gwiazda and Wróblewska-Kamińska15, Lemma 3.4.2]

Suppose $M$ is an $N$-function and let $\{\xi _n\}_{n\in \mathbb {N}}$ be a sequence of measurable functions $\xi _n:\Omega \to \mathbb {R}^n$ satisfying

\[ \sup_{n\in \mathbb{N}}\int_{\Omega}M(x,\xi_n(x))\,{\rm d}x <\infty. \]

Then the sequence $\{\xi _n\}_{n\in \mathbb {N}}$ is uniformly integrable in $L^1(\Omega ).$

Next, we point out that the existence of weak solutions to the following problem follows directly from [Reference Chlebicka, Karppinen and Li18, Theorem 1.1].

Proposition 2.6 Let $\Omega$ be a bounded Lipschitz domain in ${\mathbb {R}^n}$. Suppose that an $N$-function $M$ is regular enough so that $C^{\infty }_{c}(\Omega )$ is dense in $V^1_0L_M(\Omega )$ in the modular topology. Assume further that $g\in L^\infty (\Omega )$, $F\in E_{M^*}(\Omega ;{\mathbb {R}^n})$, function ${\mathcal {A}}$ satisfies assumptions (A1), (A2) and (A3), $\Phi$ is a bounded and continuous function, and $b$ satisfies (b). Then there exists a weak solution to the problem

\[ \left\{\begin{array}{@{}cl} -\operatorname{div} \big({\mathcal{A}}(x,\nabla u)+\Phi(u)\big)+b(x,u)= g+\operatorname{div}{F} & \qquad \mathrm{ in}\qquad \Omega,\\ u(x)=0 & \qquad \mathrm{ on}\qquad \partial\Omega, \end{array}\right. \]

Namely, there exists a function $u\in V^{1}_{0}L_{M}(\Omega )$ satisfying

\[ \int_\Omega {\mathcal{A}}(x,\nabla u) \cdot \nabla \phi + \Phi(u)\cdot \nabla \phi + b(x,u)\phi \, {\rm d}x = \int_{\Omega}g\phi\,{\rm d}x+ \int_\Omega F \cdot \nabla \phi \, {\rm d}x \]

for all $\phi \in V^1_0L_M(\Omega ) \cap L^\infty (\Omega )$.

In fact, for each $g \in L^\infty (\Omega )$, we know that there exists $H:\Omega \to {\mathbb {R}}^n$, such that $g=\operatorname {div} H$ and $H\in E_{M^\ast }(\Omega ;{\mathbb {R}^n})$. The fact one can take $H\in E_{M^\ast }(\Omega ;{\mathbb {R}^n})$ is a consequence of properties of Bogovski operator. This is explained in [Reference Chlebicka, Gwiazda, Świerczewska-Gwiazda and Wróblewska-Kamińska15, Remark 4.1.7] with the use of [Reference Sohr52, Lemma II.2.1.1].

Lemma 2.7 [Reference Chlebicka, Gwiazda, Świerczewska-Gwiazda and Wróblewska-Kamińska15, Theorem 4.1.1]

Suppose $\boldsymbol {A}:\Omega \times {\mathbb {R}^n}\to {\mathbb {R}^n}$ satisfies condition (A1)–(A2) with an $N$-function $M:\Omega \times {\mathbb {R}^n} \to [0,\,\infty ).$ Moreover, assume that there exist

\[ \boldsymbol{\alpha} \in L_{M^*}(\Omega;{\mathbb{R}^n}) \text{ and }\boldsymbol{\xi}\in L_M(\Omega;{\mathbb{R}^n}) \]

such that

\[ \int_{\Omega}\big(\boldsymbol{\alpha}-\boldsymbol{A}(x,\boldsymbol{\eta})\big)\cdot(\boldsymbol{\xi}-\boldsymbol{\eta})\mathop{}\!\mathrm{d} x \geq 0 \text{ for all } \boldsymbol{\eta}\in {\mathbb{R}^n}. \]

Then

\[ \boldsymbol{A}(x,\boldsymbol{\xi})=\boldsymbol{\alpha}\text{ a.e. in } \Omega. \]

Lemma 2.8 [Reference Chlebicka, Gwiazda, Świerczewska-Gwiazda and Wróblewska-Kamińska15, Lemma 8.22]

Suppose $z_s\stackrel {s\to \infty }\rightharpoonup z$ in $L^1(\Omega )$ and $w_s,\,w\in L^{\infty }(\Omega )$. Assume further that there exists a constant $C>0$ such that $\sup _{s\in \mathbb {N}}||w_s||_{\infty }< C$ and $w_s\xrightarrow [s\to \infty ]{a.e.}w$. Then

\[ \lim\limits_{s\to\infty} \int_{\Omega}w_sz_s\,{\rm d}x=\int_{\Omega}wz\,{\rm d}x. \]

Young measures are now a standard tool for non-linear analysis. We will need the following version of the generalized fundamental theorem on Young measures from [Reference Chlebicka, Gwiazda, Świerczewska-Gwiazda and Wróblewska-Kamińska15], where by $\mathcal {M}({\mathbb {R}^n})$ we denote the space of bounded Radon measures. A sequence $\{z_j\}_{j\in \mathbb {N}}$ of measurable function $z_{j}:\Omega \to {\mathbb {R}^n}$ is said to satisfy the tightness condition if

\[ \lim\limits_{R\to \infty}\sup_{j\in \mathbb{N}}|\{x:|z_j(x)|\geq R\}|=0. \]

Lemma 2.10 [Reference Müller45, Corollary 3.3]

$z^j:\Omega \to \mathbb {R}^{n}$ generates the Young measure $\mathit {v}$, $B:\Omega \times {\mathbb {R}^n}\to \mathbb {R}^+$ is a Carathéodory function. Then

\[ \liminf_{j\to \infty}\int_{\Omega}B(x,z^{j}(x))\,{\rm d}x\geq \int_{\Omega}\int_{{\mathbb{R}^n}}B(x,\lambda){\rm d}v_{x}(\lambda)\,{\rm d}x. \]

Lemma 2.13 [Reference Pedregal48, Lemma 6.9]

Let $f_j\in L^1(\Omega )$ for every $j\in \mathbb {N}$, $0\leq f_j(x)$ for a.e. $x\in \Omega$. Moreover, suppose

\[ f_j\xrightarrow[]{b} f \quad \text{and}\quad \limsup_{j\to\infty}\int_{\Omega}f_j\,{\rm d}x \leq \int_{\Omega}f \,{\rm d}x. \]

Then

\[ f_j\rightharpoonup f \quad \text{weakly in}\quad L^1(\Omega)\quad \text{for} \quad j\to\infty. \]

3. Existence of renormalized solutions—main proof

Now, we are in the position to prove our main result. The whole proof is divided into $6$ steps. We begin with the existence of a solution to a problem with truncated data, then we show a priori estimates and the energy control condition for solutions to the problem. After that we focus on the most challenging part—passing to the limit with the level of truncation. In the last step, we show that the function $u$ that we obtained as a weak limit, is in fact a renormalized solution. This is the place where Young measures appear.

Proof of theorem 1.3 Step 1. Problems with truncated data.

The existence of a weak solution to the problem

\[ \left\{\begin{array}{@{}cl} -\operatorname{div} \big({\mathcal{A}}(x,\nabla u)+\Phi_s(u)\big)+b(x,u)= T_s(f)+\operatorname{div} F & \qquad \mathrm{ in}\qquad \Omega,\\ u(x)=0 & \qquad \mathrm{ on}\qquad \partial\Omega, \end{array}\right. \]

for every $s>0$ is a consequence of proposition 2.6 with $g=T_s(f)$, $\Phi _s(u)=\Phi (T_{s}(u))$. Namely, there exists a function $u_s\in V^{1}_{0}L_{M}(\Omega )$ satisfying

(3.1)\begin{equation} \int_\Omega {\mathcal{A}}(x,\nabla u_s) \cdot \nabla \phi + \Phi_s(u_s)\cdot \nabla \phi + b(x,u_s)\phi \, {\rm d}x =\int_{\Omega} T_s(f)\phi\,{\rm d}x + \int_\Omega F \cdot \nabla \phi \, {\rm d}x \end{equation}

for all $\phi \in V^1_0L_M(\Omega ) \cap L^\infty (\Omega )$.

Step 2. A priori estimates

We test the function (3.1) by $T_k(u_s)$ to get

\begin{align*} & \int_{\Omega}{\mathcal{A}}\big(x,\nabla T_k(u_s)\big)\cdot \nabla T_k(u_s) \,{\rm d}x+\int_{\Omega} \Phi(T_s(u_s))\cdot \nabla T_k(u_s)\,{\rm d}x +\int_{\Omega}b(x,u_s)T_k(u_s) \, {\rm d}x \\ & \quad = \int_{\Omega}T_s(f)T_k(u_s) + F \nabla T_k(u_s)\,{\rm d}x. \end{align*}

Using condition (A2) we get an estimate

\[ \frac{1}{2}\int_{\Omega} M\big(x,c_1^{\mathcal{A}}\nabla T_k(u_s)\big) \,{\rm d}x \leq \frac{1}{2}\int_{\Omega}A\big(x,\nabla T_k(u_s)\big)\cdot \nabla T_k(u_s) \, {\rm d}x. \]

Since $\Phi$ is continuous, we can apply the chain rule theorem for Sobolev functions. We notice that there exists a function $G:\mathbb {R}\to {\mathbb {R}^n}$ such that $G(0)=0$ and we have $\operatorname {div} G(T_{k}(u_{s}))=\Phi (T_{k}(u_{s}))\cdot \nabla T_{k}(u_{s})$. Therefore, the Gauss–Green theorem yields:

\[ \int_\Omega \Phi(T_{k}(u_s))\cdot\nabla T_k(u_{s})\,{\rm d}x =0. \]

Moreover, the Fenchel–Young inequality and the definition of $N$ functions allow us to infer that

\begin{align*} \int_{\Omega}\frac{4}{c_1^{\mathcal{A}}}F\cdot \frac{c_1^{\mathcal{A}}}{4}\nabla T_k(u_s) {\rm d}x & \leq \int_{\Omega} M\left(x,\frac{c_1^{\mathcal{A}}}{4}\nabla T_k(u_s)\right)\, {\rm d}x +\int_{\Omega} M^\ast \left(x,\frac{4}{c_1^{\mathcal{A}}}F\right)\, {\rm d}x \\ & \leq \frac{1}{4} \int_{\Omega} M\big(x,c_1^{\mathcal{A}} \nabla T_{k}(u_s)\big) {\rm d}x +\int_{\Omega} M^\ast \left(x,\frac{4}{c_1^{\mathcal{A}}}F\right)\, {\rm d}x. \end{align*}

And thus, according to $M(x,\,\xi ) \geq 0$ for almost every $x \in \Omega$ and every $\xi \in {\mathbb {R}}^n$, $b(x,\,u_s)T_{k}(u_s) \geq 0$ as $b$ satisfies condition (b), we have

\begin{align*} & \frac{1}{4} \int_{\Omega} M\big(x,c_1^{\mathcal{A}}\nabla T_{k}(u_s)\big) \,{\rm d}x + \frac{1}{2}\int_{\Omega}{\mathcal{A}}\big(x,\nabla T_{k}(u_s)\big)\cdot \nabla T_{k}(u_s) \,{\rm d}x\\ & \quad \leq \int_{\Omega} M^\ast \left(x,\frac{4}{c_1^{\mathcal{A}}}F\right ) + k\|f\|_{L^{1}(\Omega)}. \end{align*}

Since $F\in E_{M^\ast }(\Omega )$ by assumption, the right-hand side of the latter inequality is finite and we infer that

\[ \int_\Omega {\mathcal{A}}\big(x,\nabla T_k(u_s)\big)\cdot \nabla T_{k}(u_s) \, {\rm d}x \leq C \]

and

\[ \int_\Omega M(x,c^{\mathcal{A}}_1\nabla T_{k}(u_s)) \, {\rm d}x \leq C. \]

Furthermore, by lemmas 2.4 and 2.5, we have

(3.2)\begin{equation} \|\nabla T_{k}(u_s)\|_{L_M} \leq \frac{1}{c_1^{\mathcal{A}}}\left( \int_\Omega M\big(x,c_1^{\mathcal{A}} \nabla T_{k}(u_s)\big) \, {\rm d}x + 1 \right) \leq C \end{equation}

and

(3.3)\begin{equation} \|{\mathcal{A}}({\cdot},\nabla T_{k}(u_s))\|_{L_{M^\ast}} \leq C, \end{equation}

where $C$ is independent of $s$.

Step 3. Energy control.

Now we will show that for every weak solution $u_s$ to (3.1) there exists a $\gamma :[0,\,\infty )\to [0,\,\infty )$ independent of $s$ and $l$ such that $\lim _{t\to 0}\gamma (t)=0$ and for every $l>0$

(3.4)\begin{equation} \int_{\{l<|u_s|< l+1\}}{\mathcal{A}}(x,\nabla u_s)\cdot \nabla u_s \mathop{}\!\mathrm{d} x \leq \gamma\Bigg(\frac{l}{m_1(c_1l)}\Bigg) \end{equation}

for some $c_1=c_1(\Omega )>0$, where $m_1$ is the minorant of $M$ in the definition of an $N$-function.

Considering properties of truncations, we infer

(3.5)\begin{equation} \begin{aligned} \int_{\{l<|u_s|< l+1\}}{\mathcal{A}}(x,\nabla u_s)\cdot \nabla u_s \,{\rm d}x & = \int_{\{l<|u_s|< l+1\}}{\mathcal{A}}\big(x,\nabla T_{l+1}(u_s)\big)\cdot \nabla T_{l+1}(u_s) \,{\rm d}x \\ & = \int_{\Omega}{\mathcal{A}}(x,\nabla u_s)\cdot \nabla \big(T_{l+1}(u_s)-T_l(u_s)\big) \,{\rm d}x. \end{aligned} \end{equation}

Testing (3.1) by $(T_{l+1}(u_s)-T_l(u_s))$ , we have

\begin{align*} & \int_{\{l<|u_s|< l+1\}}{\mathcal{A}}\big(x,\nabla T_{l+1}(u_s)\big)\cdot \nabla T_{l+1}(u_s) \,{\rm d}x + \int_{\{l<|u_s|< l+1\}}\Phi_s (u_s)\cdot \nabla T_{l+1}(u_s) \, {\rm d}x \\ & \qquad + \int_{\{l\leq |u_s|\}}b(x,u_s)\big(T_{l+1}(u_s) -T_{l}(u_s)\big)\,{\rm d}x \\ & \quad=\int_{\Omega}{\mathcal{A}}(x,\nabla u_s)\cdot \nabla \big(T_{l+1}(u_s)-\!T_l(u_s)\big) \,{\rm d}x +\! \int_{\Omega}\Phi_s(u_s)\cdot \nabla \big(T_{l+1}(u_s)-T_l(u_s)\big) \,{\rm d}x\\ & \qquad + \int_{\Omega}b(x,u_s)\big(T_{l+1}(u_s)-T_l(u_s)\big) \,{\rm d}x \\ & \quad= \int_{\Omega}T_s(f)\big(T_{l+1}(u_s)-T_l(u_s)\big) \,{\rm d}x + \int_{\Omega} F\nabla \big(T_{l+1}(u_s)-T_l(u_s)\big) \,{\rm d}x \\ & \quad\leq \int_{\{|u_s|\geq l\}}|f| \,{\rm d}x+ \int_{\{l\leq |u_s|< l+1\}} F\nabla T_{l+1}(u_s)\,{\rm d}x. \end{align*}

Since $\int _{\Omega }\Phi _s(u_s)\cdot \nabla (T_{l+1}(u_s)-T_l(u_s)) \,{\rm d}x=0$ and the term involving function $b$ can be dropped as it is non-negative, we get

\begin{align*} & \int_{\{l<|u_s|< l+1\}}{\mathcal{A}}(x,\nabla T_{l+1}(u_s))\cdot \nabla T_{l+1}(u_S) \,{\rm d}x \leq \int_{\{l\leq |u_s|\}}|f| \,{\rm d}x\\ & \quad + \int_{\{l\leq |u_s|< l+1\}} F\nabla T_{l+1}(u_s)\,{\rm d}x. \end{align*}

Moreover,

\begin{align*} \int_{\{l\leq |u_s|< l+1\}} F\nabla T_{l+1}(u_s)\,{\rm d}x & \leq \frac{1}{4} \int_{\{l\leq |u_s|< l+1\}} M\big(x,c_1^{\mathcal{A}} \nabla T_{l+1}(u_s)\big) {\rm d}x \\ & \quad +\int_{\{l\leq |u_s|< l+1\}} M^\ast \left(x,\frac{4}{c_1^{\mathcal{A}}}F\right)\, {\rm d}x. \end{align*}

Therefore, we have

(3.6)\begin{equation} \begin{aligned} & \frac{1}{4} \int_{\{l<|u_s|< l+1\}} M\big(x,c_1^{\mathcal{A}} \nabla T_{l+1}(u_s)\big) {\rm d}x +\frac{1}{2} \int_{\{l<|u_s|< l+1\}}{\mathcal{A}}(x,\nabla u_s)\cdot \nabla u_s \,{\rm d}x \\ & \quad \leq\int_{\{ l\leq |u_s|\}}|f| \,{\rm d}x+ \int_{\{l\leq|u_s|\}}M^\ast \left(x,\frac{4}{c_1^{\mathcal{A}}}F\right) \,{\rm d}x. \end{aligned} \end{equation}

We want to estimate the right-hand side of (3.6). To do so, we must find some control over the measure of the set $\{|u_s|\geq l\}$, which is the domain of integration. Note that for $m_1$ we obtain

\[ |\{|u_s|\geq l\}|=|\{|T_l(u_s)|=l\}|=|\{|T_l(u_s)|\geq l\}|=\big|\big\{m_1\big(c_1|T_l(u_s)|\big)\geq m_1(c_1l)\big\}\big|. \]

Applying the Chebyshev inequality [Reference Bogachev and Soares Ruas8, Theorem 2.5.3], the modular Poincaré inequality [Reference Chlebicka, Gwiazda, Świerczewska-Gwiazda and Wróblewska-Kamińska15, Theorem 9.3 ] involving $m_1$ and using the fact it is a convex minorant of $M$, we arrive at

(3.7)\begin{equation} \begin{aligned} |\{|u_s|\geq l\}| & \leq \int_{\Omega}\frac{m_1\big(c_1|T_l(u_s)|\big)}{m_1(c_1l)}\mathop{}\!\mathrm{d} x\\ & \leq \frac{c_2}{m_1(c_1l)}\int_{\Omega}m_1\big(\big|c^{{\mathcal{A}}}_{1}\nabla T_l(u_s)\big|\big) \mathop{}\!\mathrm{d} x \\ & \leq \frac{c_2}{m_1(c_1l)}\int_{\Omega}M\big(x,c^{{\mathcal{A}}}_{1}\nabla T_l(u_s)\big) \mathop{}\!\mathrm{d} x \\ & \leq C\frac{l}{m_1(c_1l)}. \end{aligned} \end{equation}

Since $m_1$ is an $N$-function, it is superlinear at infinity. Hence the right-hand side of the inequality above vanishes when $l\to \infty$. As a result, there exists $\gamma : [0,\,\infty )\to [0,\,\infty )$ independent of $s$ and $l$, such that $\lim _{t\to 0}\gamma (t)=0$ and we have

\[ \int_{A} M^\ast \left(x,\frac{4}{c_1^{\mathcal{A}}}F\right) + |f| \, {\rm d}x \leq \frac{1}{2}\gamma(|A|). \]

Thanks to (3.7), we get

(3.8)\begin{equation} \int_{\{l\leq|u_s|\}}M^\ast \left(x,\frac{4}{c_1^{\mathcal{A}}}F\right) +|f| \, {\rm d}x \leq\frac{1}{2} \gamma\Bigg(\frac{l}{m_1(c_1l)}\Bigg). \end{equation}

Combining this with (3.6) we arrive at the claim, which is

\[ \int_{\{l<|u_s|< l+1\}}{\mathcal{A}}(x,\nabla u_s)\cdot \nabla u_s \, {\rm d}x \leq \gamma\Bigg(\frac{l}{m_1(c_1l)}\Bigg). \]

Step 4. Convergence of trucations.

In this part of the proof, we would like to show that there exists a subsequence of $\{u_s\}_{s>0}$ which has a limit $u:\Omega \to \mathbb {R}$ in the sense that

(3.9)\begin{equation} u_s\xrightarrow[s\to \infty]{}u\text{ a.e. in }\Omega \end{equation}

such that $T_k(u)\in V^1_0L_M(\Omega )$ for every $k>0$ and also

(3.10)\begin{equation} |\{|u|>l\}|\xrightarrow[l\to \infty]{}0. \end{equation}

For every $k\in \mathbb {N}$, passing with $s\to \infty$ we have

(3.11)\begin{equation} \begin{aligned} T_k(u_s) & \to T_k(u) \text{strongly in}\ L^1(\Omega),\\ & \nabla T_k(u_s)\rightharpoonup \nabla T_k(u) \text{weakly in}\ L^1(\Omega;\mathbb{R}^n),\\ & \nabla T_k(u_s)\overset{\ast}{\rightharpoonup} \nabla T_k(u) \text{weakly}-* \text{ in } L_M(\Omega;\mathbb{R}^n),\\ & {\mathcal{A}}(x,\nabla T_k(u_s))\overset{\ast}{\rightharpoonup} {\mathcal{A}}_k \text{weakly-}* \text{ in } L_{M^*}(\Omega;\mathbb{R}^n) \end{aligned} \end{equation}

for some ${\mathcal {A}}_k\in L_{M^*}(\Omega ;{\mathbb {R}^n}).$

Fix $k\in \mathbb {N}$. We have already proven the following a priori estimate (3.2), namely

\[ \|\nabla T_k(u_s)\|_{L_{M}} \leq C. \]

Using the Banach–Alaoglu theorem [Reference Brézis11, Corollary 3.30] we infer that the sequence $\{\nabla T_k(u_s)\}_{s>0}$ is weakly-$*$ compact in $L_M(\Omega ;{\mathbb {R}^n})$. The fact that $M$ is an $N$-function together with lemma 2.3 imply that $\{\nabla T_k(u_s)\}_{s>0}$ is uniformly integrable $L^{1}(\Omega ;{\mathbb {R}^n})$. The Dunford–Pettis theorem [Reference Brézis11, Theorem 4.30], i.e.

\begin{align*} & \left\{f_{n}\right\}_{n} ~~ \mbox{is uniformly integrable in}~~ L^{1}(\Omega) ~\Leftrightarrow~ \left\{f_{n}\right\}_{n}\\ & \quad\mbox{ is relatively compact in the weak topology}, \end{align*}

implies that for every $k\in \mathbb {N}$ the sequence $\{\nabla T_k(u_s)\}_{s>0}$ is relatively compact in the weak topology of $L^1(\Omega ;{\mathbb {R}^n})$. As the set $\Omega$ is bounded, the Rellich–Kondrachov theorem [Reference Brézis11, Theorem 9.16] for $W^{1,1}(\Omega )$ yields uniform integrability of the sequence $\{T_k(u_s)\}_{s>0}$ in the space $L^1(\Omega )$. Hence, there exists a function $u$ such that

\begin{align*} T_k(u_s)& \to T_k(u) \text{ strongly in } L^1(\Omega),\\ & \nabla T_k(u_s)\rightharpoonup \nabla T_k(u) \text{ weakly in } L^1(\Omega;\mathbb{R}^n).\\ \end{align*}

Thus, up to a subsequence, we have $u_s\to u$ in measure and almost everywhere, which gives (3.9). Additionally, the Dunford–Pettis theorem together with (3.2) imply that, up to a subsequence, we have

(3.12)\begin{equation} \nabla T_k(u_s)\overset{\ast}{\rightharpoonup} \nabla T_k(u) \text{ weakly-}*\text{ in } L_M(\Omega;\mathbb{R}^n). \end{equation}

Since $u_s\to u$ in measure, using (3.7) we obtain (3.10).

Now we focus on the last convergence in (3.11). For every $k\in \mathbb {N}$ we define

\[ {\mathcal{A}}_{s,k}={\mathcal{A}}(x,\nabla T_k(u_s)). \]

Using the other a priori estimate (3.3) and repeating the arguments from above we infer that, up to a subsequence, there exists ${\mathcal {A}}_k\in L_{M^*}(\Omega ;\mathbb {R}^n)$ such that

(3.13)\begin{equation} {\mathcal{A}}_{s,k}\overset{\ast}{\rightharpoonup}{\mathcal{A}}_k\text{ weakly-}*\text{ in } L_{M^*}(\Omega;\mathbb{R}^n). \end{equation}

Step 5. Identification of the limit of ${\mathcal {A}}(x,\,\nabla T_k(u_s(x)))$.

We want to show that our limit obtained above in (3.13) is precisely of the form

(3.14)\begin{equation} {\mathcal{A}}_k={\mathcal{A}}(x,\nabla T_k(u))\ \text{a.e. in} \ \Omega. \end{equation}

We are going to prove it via monotonicity trick. To use it, we must first show that

(3.15)\begin{equation} \int_{\Omega}({\mathcal{A}}_k-{\mathcal{A}}(x,\eta))\cdot(\nabla T_k(u)-\eta)\,{\rm d}x \geq 0 \quad \text{for every} \quad \eta \in {\mathbb{R}^n}. \end{equation}

We begin with showing that

(3.16)\begin{equation} \limsup_{s\to \infty}{\int_{\Omega}{\mathcal{A}}_{s,k}\cdot \nabla T_k(u_s)}\,{\rm d}x = \int_{\Omega} {\mathcal{A}}_k\cdot \nabla T_k(u) \,{\rm d}x. \end{equation}

Firstly, we consider a function $\Psi _l:\mathbb {R}\to [0,\,1]$ defined as

(3.17)\begin{equation} \Psi_l(r)=\min\{(l+1-|r|)_+,1\} \end{equation}

and using proposition 2.2, we can take an approximate sequence $\{\nabla (T_k(u))_{\delta }\}_{\delta }$ of smooth functions such that

\[ \nabla (T_k(u))_{\delta} \xrightarrow[\delta \to 0]{M}\nabla T_k(u) \text{ modularly in } L_M(\Omega;{\mathbb{R}^n}). \]

Having this in mind, we will show that

(3.18)\begin{equation} \lim_{\delta\to 0}\limsup_{s\to \infty}\int_{\Omega}{\mathcal{A}}_{s,k}\cdot \nabla\big(T_k(u_s)-(T_k(u))_{\delta}\big)\,{\rm d}x =0. \end{equation}

The condition ($A2$) for the operator ${\mathcal {A}}$ implies ${\mathcal {A}}(x,\,0)=0.$ Hence, for $l\geq k$ this observation yields

\begin{align*} & \int_{\Omega}{\mathcal{A}}_{s,k}\cdot \nabla\big(T_k(u_s)-(T_k(u))_{\delta}\big)\Psi_l(u_s)\,{\rm d}x\\ & \quad=\int_{\Omega}{\mathcal{A}}_{s,k}\cdot \nabla\big(T_k(u_s)-(T_k(u))_{\delta}\big)\, {\rm d}x \\ & \qquad +\int_{\{|u_s|>l\}}{\mathcal{A}}(x,0)\cdot \nabla \big(0-(T_k(u))_{\delta}\big)(\Psi_l(u_s)-1)\,{\rm d}x\\ & \quad=\int_{\Omega}{\mathcal{A}}_{s,k}\cdot \nabla \big(T_k(u_s)-(T_k(u))_{\delta}\big) \, {\rm d}x. \end{align*}

Therefore, (3.18) is equivalent to

(3.19)\begin{equation} \lim_{l\to \infty}\lim_{\delta\to 0}\limsup_{s\to \infty}\int_{\Omega}{\mathcal{A}}_{s,k}\cdot \nabla\big(T_k(u_s)-(T_k(u))_{\delta}\big)\Psi_l(u_s)\mathop{}\!\mathrm{d} x =0. \end{equation}

Now we notice that it is enough to have

(3.20)\begin{equation} \lim_{l\to \infty}\lim_{\delta\to 0}\limsup_{s\to \infty}\int_{\Omega}{\mathcal{A}}_{s,l+1}\cdot \nabla\big(T_k(u_s)-(T_k(u))_{\delta}\big)\Psi_l(u_s)\,{\rm d}x =0. \end{equation}

Indeed, having this result, (3.19) will be satisfied if we manage to prove that for $l\geq k$,

\begin{align*} {J}:& =\int_{\Omega}\big({\mathcal{A}}_{s,k}-{\mathcal{A}}_{s,l+1}\big)\cdot \nabla\big(T_k(u_s)-(T_k(u))_{\delta}\big)\Psi_l(u_s)\,{\rm d}x \\ & =\int_{\Omega}\big({\mathcal{A}}_{s,l+1}-{\mathcal{A}}(x,0)\big)\cdot \nabla(T_k(u))_{\delta}\mathbb{1}_{\{k<|u_s|\}}\Psi_l(u_s)\, {\rm d}x \\ & = \int_{\Omega}{\mathcal{A}}_{s,l+1}\cdot \nabla(T_k(u))_{\delta}\mathbb{1}_{\{k<|u_s|\}}\Psi_l(u_s)\,{\rm d}x \end{align*}

tends to zero as $s\to \infty$ and $\delta \to 0$. In order to do so, we just need to prove that

(3.21)\begin{equation} \begin{aligned} \lim_{\delta\to 0}\limsup_{s\to\infty} |J| & \leq \lim_{\delta\to 0}\limsup_{s\to\infty}\int_{\Omega}|{\mathcal{A}}_{s,l+1}| \mathbb{1}_{\{k<|u_s|\}}\Psi_l(u_s) |\nabla(T_k(u))_{\delta}|\,{\rm d}x \\ & \leq \lim_{\delta\to 0} \int_{\Omega}|{\mathcal{A}}_{l+1}| \mathbb{1}_{\{k<|u|\}}\Psi_l(u) |\nabla(T_k(u))_{\delta}|\,{\rm d}x \\ & = \int_{\Omega}|{\mathcal{A}}_{l+1}|\mathbb{1}_{\{k<|u|\}}\Psi_l(u) |\nabla T_k(u)|\,{\rm d}x =0. \end{aligned} \end{equation}

For the limit as $s\to \infty$ we will use lemma 2.8 with

\[ z_s:=|{\mathcal{A}}_{s,l+1}|\cdot|\nabla(T_k(u))_{\delta}| \stackrel{s\to \infty}\rightharpoonup |{\mathcal{A}}_{l+1}|\cdot|\nabla(T_k(u))_{\delta}|=z \ \ \ \ \ \text{ weakly in} \ L^1(\Omega) \]

and $w_s=\Psi _{l}(u_s)\mathbb {1}_{\{k<|u_s|\}}$. The convergence $z_s\rightharpoonup z$ in $L^1(\Omega )$ is a consequence of (3.13). And (3.9) gives us that $w_s\to w=\Psi _{l}(u)\mathbb {1}_{\{k<|u_s|\}}$ a.e. in $\Omega$. The external limit with $\delta \to 0$ arises from modular convergence in (3.21). In addition, since we have

\[ \nabla T_k(u)\mathbb{1}_{\{k<|u\}}=0, \]

the last equality in (3.21) follows.

Now, to obtain (3.20), we test (3.1) by the sequence

\[ \varphi:=\Psi_l(u_s)\big(T_k(u_s)-(T_k(u))_{\delta}\big), \]

where $\Psi _l$ is defined in (3.17). Thus, we have

(3.22)\begin{equation} \begin{aligned} & \int_{\Omega}{\mathcal{A}}(x,\nabla u_s)\cdot \nabla\big(\Psi_l(u_s)(T_k(u_s)-(T_k(u))_{\delta})\big) \,{\rm d}x\\ & \quad +\int_{\Omega}\Phi_s(u_s)\cdot \nabla \big(\Psi_l(u_s)(T_k(u_s)-(T_k(u))_{\delta})\big) \,{\rm d}x \\ & \quad+\int_{\Omega}b(x,u_s)\Psi_l(u_s)\big(T_k(u_s)-(T_k(u))_{\delta}\big) \,{\rm d}x\\ & = \int_{\Omega}T_s(f)\Psi_l(u_s)\big(T_k(u_s)-(T_k(u))_{\delta}\big) \,{\rm d}x\\ & \quad + \int_{\Omega}F\cdot \nabla \big(\Psi_l(u_s)(T_k(u_s)-(T_k(u))_{\delta})\big) \,{\rm d}x. \end{aligned} \end{equation}

Firstly, we consider the first term in the right-hand side of (3.22). Since we know that

\[ u_s\xrightarrow[]{}u \text{ a.e. in } \Omega, \]

we would like to use the Lebesgue-dominated convergence theorem. To do so, we note that

\begin{align*} & \lim_{\delta\to 0}\lim_{s\to \infty}\Bigg|\int_{\Omega}T_s(f)\Psi_l(u_s)\big(T_k(u_s)-(T_k(u))_{\delta}\big) \mathop{}\!\mathrm{d} x\Bigg| \\ & \quad\leq \lim_{\delta\to 0}\lim_{s\to \infty}\int_{\Omega}\big|T_s(f)\big|\Psi_l(u_s)\big|\big(T_k(u_s)-T_k(u)\big)\big| \mathop{}\!\mathrm{d} x \\ & \qquad + \lim_{\delta\to 0}\lim_{s\to \infty}\int_{\Omega}\big|T_s(f)\big|\Psi_l(u_s)\big|\big(T_k(u)-(T_k(u))_{\delta}\big)\big| \mathop{}\!\mathrm{d} x \\ & \quad\leq \lim_{\delta\to 0}\lim_{s\to \infty}\int_{\Omega}2k|f|\mathop{}\!\mathrm{d} x + \lim_{\delta\to 0}\lim_{s\to \infty}\int_{\Omega}|f|\cdot |T_k(u)-(T_k(u))_{\delta}|\,{\rm d}x \\ & \quad= 2k||f||_{L^1(\Omega)}+ \lim_{\delta\to 0}\int_{\Omega}|f|\cdot|T_k(u)-(T_k(u))_{\delta}|\,{\rm d}x. \end{align*}

Using the modular approximation result (see proposition 2.2) we get $|(T_k(u))_{\delta }|\leq ck$, which implies

(3.23)\begin{equation} |T_k(u)-(T_k(u))_{\delta}|\leq (1+c)k. \end{equation}

Thus, we infer that

(3.24)\begin{equation} \lim_{l\to \infty}\lim_{\delta \to 0}\limsup_{s\to \infty} \int_{\Omega}T_s(f)\Psi_l(u_s)\big(T_k(u_s)-(T_k(u))_{\delta}\big)\,{\rm d}x=0. \end{equation}

Secondly, for the second term in the right-hand side of (3.22), we define

\begin{align*} & \int_{\Omega}F\cdot \nabla \big(\Psi_l(u_s)(T_k(u_s)-(T_k(u))_{\delta})\big) \,{\rm d}x \\ & \quad = \int_{\Omega}F\cdot \nabla \Psi_l(u_s)\big(T_k(u_s)-(T_k(u))_{\delta}\big) \,{\rm d}x + \int_{\Omega}F\cdot \Psi_l(u_s) \nabla T_k(u_s)\,{\rm d}x \\ & \qquad -\int_{\Omega}F\cdot\Psi_l(u_s) \nabla (T_k(u))_{\delta} \,{\rm d}x\\ & \quad =:h_1+h_2-h_3. \end{align*}

Thanks to the Fenchel–Young inequality and (3.23), we get

(3.25)\begin{equation} \begin{aligned} & \lim_{l\to \infty}\lim_{\delta\to 0}\limsup_{s\to \infty}|h_1|\\ & = \lim_{l\to \infty}\lim_{\delta\to 0}\limsup_{s\to \infty}\big| \int_{\Omega}F\cdot \nabla \Psi_l(u_s)\big(T_k(u_s)-(T_k(u))_{\delta}\big) \,{\rm d}x \big|\\ & \leq C\lim_{l\to \infty}\lim_{\delta\to 0}\limsup_{s\to \infty}\int_{\{l\leq |u_s|\leq l+1\}}\big|F\cdot \nabla T_{l+1}(u_s) \big|\,{\rm d}x\\ & \leq C\lim_{l\to \infty}\limsup_{s\to \infty}\Bigg( \int_{\{l<|u_s|< l+1\}}\frac{1}{4} M\big(x,c_1^{\mathcal{A}} \nabla T_{l+1}(u_s)\big)\,{\rm d}x \\ & \quad +\int_{\{l<|u_s|< l+1\}} M^\ast \left(x,\frac{4}{c_1^{\mathcal{A}}} F\right)\, {\rm d}x\Bigg)\\ & \leq C\lim_{l\to \infty} \gamma\Bigg(\frac{l}{m_1(c_1l)}\Bigg)=0. \end{aligned} \end{equation}

where in the last line we use (3.4), (3.8) and the fact that $m_1$ is superlinear at infinity as an $N$-function. Note that

\[ \nabla T_k(u_s) \overset{s\to\infty}{\rightharpoonup} \nabla T_k(u) \text{ weakly in } L^1(\Omega;\mathbb{R}^n), \]

combining with the fact

\begin{align*} \int_{\Omega}|F\nabla T_{k}(u_s)|\,{\rm d}x & \leq \int_{\Omega}M^{*}\left(x,\frac{1}{c^{{\mathcal{A}}}_{1}}F\right)\,{\rm d}x + \int_{\Omega}M(x,c_1^{\mathcal{A}} \nabla T_{l+1}(u_s))\,{\rm d}x \\ & \leq C, \end{align*}

where C is independent of $s$, we deduce that

\[ F\nabla T_k(u_s) \overset{s\to\infty}{\rightharpoonup} F\nabla T_k(u) \ \ \ \ \text{weakly in}\ L^1(\Omega). \]

Recall that $|\Psi _l(u_s)|\leq 1$ and $\Psi _l(u_s)\xrightarrow [s\to \infty ]{a.e.} \Psi _l(u)$, it follows from lemma 2.8 that

\[ \lim_{s\to\infty}\int_{\Omega} F \Psi_l(u_s) \nabla T_k(u_s)\,{\rm d}x= \int_{\Omega} F\Psi_l(u) \nabla T_k(u)\,{\rm d}x. \]

Notice that by definition of the function $\Psi _l$ (see (3.17)), we have

(3.26)\begin{equation} \Psi_l(u)\xrightarrow[]{l\to \infty}1 \text{ a.e. in } \Omega. \end{equation}

Thus, collecting all the facts mentioned above and using the Lebesgue-dominated convergence theorem we infer that

\begin{align*} \lim_{l\to \infty}\lim_{s\to \infty} h_{2} & =\lim_{l\to \infty}\lim_{s\to \infty}\int_{\Omega} F\cdot\Psi_l(u_s)\nabla T_k(u_s) \,{\rm d}x\\ & =\lim_{l\to \infty}\int_{\Omega}F\cdot \Psi_l(u) \nabla T_k(u)\,{\rm d}x\\ & =\int_{\Omega} F\cdot\nabla T_k(u)\,{\rm d}x. \end{align*}

Additionally, by the Lebesgue-dominated convergence theorem and the fact $\nabla (T_{k}(u))_{\delta } \rightarrow \nabla T_{k}(u)$ modularly in $L_M(\Omega ;{\mathbb {R}^n})$, we see that

\begin{align*} \lim_{l\to \infty}\lim_{\delta\to 0}\lim_{s\to \infty} h_{3} & =\lim_{l\to \infty} \lim_{\delta \to 0}\lim_{s \to \infty} \int_{\Omega} F\cdot \Psi_l(u_s) \nabla (T_k(u))_{\delta} \,{\rm d}x \\ & =\lim_{l\to \infty}\lim_{\delta\to 0}\int_{\Omega}F\cdot \Psi_l(u) \nabla (T_k(u))_{\delta}\,{\rm d}x.\\ & =\int_{\Omega} F\cdot\nabla T_k(u)\,{\rm d}x. \end{align*}

All in all, we conclude that

(3.27)\begin{equation} \begin{aligned} & \lim_{l\to \infty}\lim_{\delta \to 0}\limsup_{s\to \infty} \int_{\Omega}F\cdot \nabla \big(\Psi_l(u_s)(T_k(u_s)-(T_k(u))_{\delta})\big) \,{\rm d}x \\ & \quad= \lim_{l\to \infty}\lim_{\delta \to 0}\limsup_{s\to \infty} \big(h_1+h_2-h_3\big) =0. \end{aligned} \end{equation}

Now, we will focus on the left-hand side of (3.22). Let us denote

(3.28)\begin{equation} \begin{aligned} & \int_{\Omega}{\mathcal{A}}(x,\nabla u_s)\cdot \nabla\big(\Psi_l(u_s)(T_k(u_s)-(T_k(u))_{\delta})\big) \,{\rm d}x \\ & \quad + \int_{\Omega}\Phi_s(u_s)\cdot \nabla \big(\Psi_l(u_s)(T_k(u_s)-(T_k(u))_{\delta})\big) \,{\rm d}x \\ & \quad + \int_{\Omega}b(x,u_s)\Psi_l(u_s)\big(T_k(u_s)-(T_k(u))_{\delta}\big) \,{\rm d}x\\ & =: I_1+I_2+I_3. \end{aligned} \end{equation}

At first we concentrate on the easier terms. We are going to show that both

\[ \lim_{\delta\to 0}\lim_{s\to \infty}(I_2)=0 \text{ and } \lim_{\delta\to 0}\lim_{s\to \infty}(I_3)=0. \]

Indeed, for $I_3$ we have

\[ I_3=\int_{\Omega}b(x,T_{l+1}(u_s))\Psi_l(u_s)\big(T_k(u_s)-(T_k(u))_{\delta}\big) \mathop{}\!\mathrm{d} x, \]

due to the definition of the function $\Psi _l$ (see (3.17)). Thanks to the assumption $(b)$, we know that $b(\cdot,\,s)\in L^1(\Omega )$ for each $s\in \mathbb {R}$. This fact, together with (3.23) yields

(3.29)\begin{equation} \begin{aligned} \lim_{\delta\to 0}\lim_{s\to \infty}|I_3| & =\lim_{\delta\to 0}\lim_{s\to \infty}\Bigg|\int_{\Omega}b(x,T_{l+1}(u_s))\Psi_l(u_s)\big(T_k(u_s)-(T_k(u))_{\delta}\big) \mathop{}\!\mathrm{d} x\Bigg| \\ & \leq \lim_{\delta\to 0}\lim_{s\to \infty} \int_{\Omega}|b(x,T_{l+1}(u_s))|\Psi_l(u_s)\big|T_k(u_s)-T_k(u)\big| \mathop{}\!\mathrm{d} x \\ & \quad + \lim_{\delta\to 0}\lim_{s\to \infty} \int_{\Omega}|b(x,T_{l+1}(u_s))|\Psi_l(u_s)\big|T_k(u)-(T_k(u))_{\delta}\big| \mathop{}\!\mathrm{d} x\\ & =:I^1_3+I^2_3. \end{aligned} \end{equation}

Now, from the Lebesgue-dominated convergence theorem we have

\[ \lim_{\delta\to 0}\lim_{s\to \infty}(I_3^1)=\lim_{\delta\to 0}\lim_{s\to \infty}(I_3^2)=0. \]

To justify the convergence of $I_2$, note that by definition of $\Psi _l$ and the chain rule we can also rewrite it as

\begin{align*} I_2& =\int_{\Omega}\Phi(T_{s}(u_s))\cdot \nabla \big(T_k(u_s)-(T_k(u))_{\delta}\big)\Psi_l(u_s) \mathop{}\!\mathrm{d} x\\ & \quad+\int_{\Omega}\Phi(T_{s}(u_s))\cdot \nabla \Psi_l(u_s)(T_k(u_s)-(T_k(u))_{\delta}) \mathop{}\!\mathrm{d} x\\ & =:I_2^1+I_2^2. \end{align*}

For $s\geq l+1$, we have

\[ I^1_2=\int_{\Omega}\Phi(T_{l+1}(u_s))\cdot \nabla \big(T_k(u_s)-(T_k(u))_{\delta}\big)\Psi_l(u_s) \mathop{}\!\mathrm{d} x. \]

Since $\Phi$ is continuous and $u_s \to u$ almost everywhere in $\Omega$, we obtain

\[ \Psi_{l}(u_s) \Phi(T_{l+1}(u_s)) \to \Psi_{l}(u)\Phi(T_{l+1}(u)) \quad \text{a.e. in} \quad \Omega. \]

As $\Phi (T_{l+1}(u_s))$ is uniformly bounded with respect to $s$, i.e.

\[ \|\Phi(T_{l+1}(u_s))\|_{L^{\infty}(\Omega;{\mathbb{R}^n})} \leq \sup_{\tau\in [{-}l-1,l+1]}|\Phi(\tau)|< C, \]

where the constant $C>0$ is independent of $s\in \mathbb {N}$ and as the following facts

\begin{align*} & |\Psi_{l}(u_s)|\leq 1 & \text{a.e. in } \Omega, \\ \nabla T_k(u_s)\rightharpoonup \nabla T_k(u) & \text{weakly in}\ L^1(\Omega;\mathbb{R}^n),\\ \nabla (T_k(u))_{\delta} \xrightarrow[\delta \to 0]{M}\nabla T_k(u) & \text{modularly in} \ L_M(\Omega;{\mathbb{R}^n}), \end{align*}

it follows from lemma 2.8 that

\[ \lim_{\delta \to 0}\limsup_{s\to \infty}I_2^1=0. \]

Let us write

\[ I_2^2=\int_{\Omega}\operatorname{div}{\Bigg(\int_0^{T_{l+1}(u_s)}\Phi(r)\Psi_l'(r)\mathop{}\!\mathrm{d} r\Bigg)}\big(T_k(u_s)-(T_k(u))_{\delta}\big)\mathop{}\!\mathrm{d} x, \]

we may use the Gauss–Green theorem and obtain

\[ I_2^2={-}\int_{\Omega}\int_0^{T_{l+1}(u_s)}\Phi(r)\Psi_l'(r)\mathop{}\!\mathrm{d} r \cdot \nabla \big(T_k(u_s)-(T_k(u))_{\delta}\big)\mathop{}\!\mathrm{d} x. \]

Using the same arguments as above, we infer that

\[ \lim_{\delta \to 0}\limsup_{s\to \infty}I_2^2=0. \]

Therefore, we have

\[ \lim_{l\to \infty}\lim_{\delta \to 0}\limsup_{s\to \infty}I_2=0. \]

Finally, we will concentrate on the most challenging and difficult term $I_1$. As before, we rewrite it as follows:

(3.30)\begin{equation} \begin{aligned} I_1 & = \int_{\Omega}{\mathcal{A}}(x,\nabla u_s)\cdot \nabla\Bigg(\Psi_l(u_s)(T_k(u_s)-(T_k(u))_{\delta})\Bigg) \mathop{}\!\mathrm{d} x\\ & = \int_{\Omega}{\mathcal{A}}(x,\nabla u_s)\cdot \nabla\Psi_l(u_s)\big(T_k(u_s)-(T_k(u))_{\delta} \big) \mathop{}\!\mathrm{d} x\\ & \quad + \int_{\Omega}{\mathcal{A}}(x,\nabla u_s)\cdot \nabla\big(T_k(u_s)-(T_k(u))_{\delta}\big)\Psi_l(u_s) \mathop{}\!\mathrm{d} x\\ & =:I_1^{1}+I^{2}_1. \end{aligned} \end{equation}

To estimate $I^{1}_1$, we will use (3.23) and (3.4) to get

\begin{align*} & \lim_{l\to \infty}\lim_{\delta \to 0}\limsup_{s\to \infty}|I^{1}_1| \\ & \quad \leq \lim_{l\to \infty}\Bigg(\lim_{\delta \to 0}\limsup_{s\to \infty} \int_{\{l<|u_s|< l+1\}}\big|{\mathcal{A}}(x,\nabla u_s)\cdot \nabla u_s\big|\big| \big(T_k(u_s)-(T_k(u))_{\delta} \big)\big| \mathop{}\!\mathrm{d} x\Bigg) \\ & \quad\leq C\lim_{l\to \infty}\Bigg(\lim_{\delta \to 0}\limsup_{s\to \infty} \int_{\{l<|u_s|< l+1\}}\big|{\mathcal{A}}(x,\nabla u_s)\cdot \nabla u_s\big|\mathop{}\!\mathrm{d} x\Bigg)\\ & \quad = C\lim_{l\to \infty}\Bigg(\limsup_{s\to \infty} \int_{\{l<|u_s|< l+1\}}{\mathcal{A}}(x,\nabla u_s)\cdot \nabla u_s\mathop{}\!\mathrm{d} x\Bigg)\\ & \quad\leq C\lim_{l\to \infty} \gamma\Bigg(\frac{l}{m_1(c_1l)}\Bigg)=0. \end{align*}

Now we notice that (3.30) yields

(3.31)\begin{equation} \begin{aligned} \lim_{l\to \infty} & \lim_{\delta \to 0}\limsup_{s\to \infty}\big(I^{2}_1\big)\\ & =\lim_{l\to \infty}\lim_{\delta \to 0}\limsup_{s\to \infty}\int_{\Omega}{\mathcal{A}}(x,\nabla u_s)\cdot \nabla\big(T_k(u_s)-(T_k(u))_{\delta}\big)\Psi_l(u_s) \mathop{}\!\mathrm{d} x =0. \end{aligned} \end{equation}

Hence, by virtue of all the above limits, (3.31) is actually equivalent to (3.20). Thus, we arrive at (3.18). According to (3.18) and (3.13), we obtain

(3.32)\begin{equation} \begin{aligned} \lim_{\delta \to 0}{\limsup_{s\to \infty}}\int_{\Omega}{\mathcal{A}}_{s,k}\cdot \nabla T_k(u_s)\,{\rm d}x & = \lim_{\delta \to 0}{\limsup_{s\to \infty}}\int_{\Omega}{\mathcal{A}}_{s,k}\cdot \nabla \big(T_k(u)\big)_{\delta}\,{\rm d}x\\ & =\lim_{\delta \to 0}\int_{\Omega}{\mathcal{A}}_{k}\cdot \nabla \big(T_k(u)\big)_{\delta}\,{\rm d}x\\ & =\int_{\Omega}{\mathcal{A}}_{k}\cdot \nabla T_k(u)\,{\rm d}x. \end{aligned} \end{equation}

Eventually, we get (3.16).

We are about to complete the proof of identification of the limit of $\{{\mathcal {A}}_{s,k}\}_{s>0}$. Using the monotonicity of ${\mathcal {A}}$ (condition (A3)) we infer that for every $\eta \in L^{\infty }(\Omega ;{\mathbb {R}^n})$ we have

(3.33)\begin{equation} \int_{\Omega}{\mathcal{A}}_{s,k} \cdot \eta \mathop{}\!\mathrm{d} x + \int_{\Omega}{\mathcal{A}}(x,\eta)\cdot(\nabla T_k(u_s)-\eta)\mathop{}\!\mathrm{d} x \leq \int_{\Omega}{\mathcal{A}}_{s,k}\cdot \nabla T_k(u_s)\mathop{}\!\mathrm{d} x. \end{equation}

Because of (3.12), (3.13) and (3.16), we may take the upper limit with $s\to \infty$ of both sides of (3.33) to get

\[ \int_{\Omega}{\mathcal{A}}_{k}\cdot \eta \mathop{}\!\mathrm{d} x + \int_{\Omega}{\mathcal{A}}(x,\eta)\cdot(\nabla T_k(u)-\eta)\mathop{}\!\mathrm{d} x \leq \int_{\Omega}{\mathcal{A}}_{k}\cdot \nabla T_k(u)\mathop{}\!\mathrm{d} x, \]

which, by rearranging terms is obviously equivalent to (3.15). Hence, we may apply the famous monotonicity trick (lemma 2.7), which ends the proof of this step.

Step 6. Renormalized solutions

Now our goal is to show the existence of renormalized solutions, which will end the proof of theorem 1.3. In fact, we are going to show that the function $u$, obtained as a limit in step $4$ is precisely the renormalized solution. By definition, we must check whether the three conditions $(R1),\, (R2)$ and $(R3)$ are satisfied.

Condition $(R1)$.

We just notice that thanks to the convergence in (3.11), condition $(R1)$ is satisfied.

Condition $(R2)$.

As we know that $T_k(u)\in V_0^1L_M(\Omega )$, proposition 2.2 yields existence of a sequence $\{u_r\}_{r>0}\subset C_c^{\infty }(\Omega )$ for which we have

(3.34)\begin{equation} \begin{aligned} u_r\xrightarrow{}u & \text{ a.e. in } \Omega,\\ \nabla T_k(u_r)\stackrel{*}\rightharpoonup\nabla T_k(u) & \text{weakly-$*$ in}\ L_M(\Omega; \mathbb{R}^n),\\ \nabla h(u_r)\rightharpoonup\nabla h(u) & \text{weakly in}\ L_M(\Omega; \mathbb{R}^n), \end{aligned} \end{equation}

for an arbitrary function $h\in C_c^1(\Omega )$. Now, for such a fixed $h$ and $\phi \in W^{1,\infty }_0(\Omega )$ we test (3.1) by $\Psi _l(u_s)h(u_r)\phi$, where $\Psi _l$ defined in (3.17). Therefore, we get

(3.35)\begin{equation} \begin{aligned} & \int_{\Omega}{\mathcal{A}}(x,\nabla u_s)\cdot \nabla\big(\Psi_l(u_s)h(u_r)\phi\big) \mathop{}\!\mathrm{d} x + \int_{\Omega}\Phi_s(u_s)\cdot \nabla \big(\Psi_l(u_s)h(u_r)\phi\big) \mathop{}\!\mathrm{d} x \\ & \quad + \int_{\Omega}b(x,u_s)\Psi_l(u_s)h(u_r)\phi \mathop{}\!\mathrm{d} x\\ & = \int_{\Omega}T_s(f)\Psi_l(u_s)h(u_r)\phi \mathop{}\!\mathrm{d} x+\int_{\Omega}F\cdot \nabla \big(\Psi_l(u_s)h(u_r)\phi\big) \mathop{}\!\mathrm{d} x. \end{aligned} \end{equation}

Let us denote the terms on the left-hand side of the equation above as $L^1_{s,r,l},\,L^2_{s,r,l}$ and $L^3_{s,r,l}$ respectively. Also, write

\[ L^1_{s,r,l}+L^2_{s,r,l}+ L^3_{s,r,l}=R^1_{s,r,l}+ R^2_{s,r,l}, \]

where $R^1_{s,r,l}$, $R^2_{s,r,l}$ stand for the right-hand side of (3.35) respectively.

At first, we see that by the Lebesgue-dominated convergence theorem, we get

\[ \lim_{l\to \infty}\lim_{r\to \infty}\lim_{s\to \infty}R^1_{s,r,l}=\int_{\Omega}fh(u)\phi \,{\rm d}x. \]

For the second term in the right-hand side of (3.35), we can write

\begin{align*} R^2_{s,r,l}& =\int_{\Omega}F\cdot \nabla \big(\Psi_l(u_s)h(u_r)\phi\big)\,{\rm d}x\\ & = \int_{\Omega}F\cdot \nabla \big(\Psi_l(u_s)\big)h(u_r)\phi \,{\rm d}x + \int_{\Omega}F\cdot \nabla \big(h(u_r)\phi\big)\Psi_l(u_s)\,{\rm d}x. \end{align*}

By the similar arguments to (3.25), we have

\begin{align*} & \lim_{l\to \infty}\lim_{r\to \infty}\limsup_{s\to \infty}\big| \int_{\Omega}F\cdot \nabla \big(\Psi_l(u_s)\big)h(u_r)\phi \,{\rm d}x\big| \\ & \quad \leq \|h\|_{L^{\infty}(\Omega)}\|\phi\|_{L^{\infty}(\Omega)}\lim_{l\to \infty}\limsup_{s\to \infty}\int_{\{l\leq u_s\leq l+1\}}|F|\cdot |\nabla T_{l+1}(u_s)|\,{\rm d}x \\ & \quad\leq C\lim_{l\to \infty} \gamma\Bigg(\frac{l}{m_1(c_1l)}\Bigg)=0. \end{align*}

Moreover, since $F\in E_{M^{*}}(\Omega ;{\mathbb {R}^n})\subseteq L^{1}(\Omega ;{\mathbb {R}^n})$, $\Psi _l(u_s)$ converges to $\Psi _l(u)$ a.e. in $\Omega$ with $|\Psi _l(u_s)|\leq 1$, the product sequence $F \Psi _l(u_s)$ also converges strongly to $F \Psi _l(u)$ in $L^1(\Omega ;{\mathbb {R}^n})$ as $s\to \infty$. Combining with the following facts

\[ \nabla (h(u_r)\phi) \in L^{\infty}(\Omega; \mathbb{R}^n),\quad \nabla \big(h(u_r)\phi\big) \stackrel{\ast}\rightharpoonup\nabla\big( h(u)\phi\big) \ \ \ \text{weakly-* in}\ L_M(\Omega; \mathbb{R}^n), \]

we deduced that

\begin{align*} \lim_{r\to \infty}\lim_{s\to \infty} \int_{\Omega}F\Psi_l(u_s) \cdot \nabla \big(h(u_r)\phi\big)\,{\rm d}x& = \lim_{r\to\infty}\int_{\Omega}F\Psi_l(u)\cdot \nabla \big(h(u_r)\phi\big)\,{\rm d}x\\ & =\int_{\Omega}F\Psi_{l}(u)\cdot \nabla \big(h(u)\phi\big)\,{\rm d}x. \end{align*}

Notice that there exists $m>0$ such that $\text {supp}{(h)}\subset [-m,\,m]$. Choosing such $m$, we may interchange $T_{l+1}$ with $T_m$. Then $\Psi _l(u)=\Psi _l(T_m(u))=1$ for $l>m$. Thus, we have

\begin{align*} \lim_{l\to \infty}\lim_{r\to \infty}\lim_{s\to \infty}R^2_{s,r,l} & =\lim_{l\to \infty} \int_{\Omega}F\Psi_{l}(u)\nabla \big(h(u)\phi\big)\,{\rm d}x\\ & =\int_{\Omega}F\nabla \big(h(u)\phi\big)\,{\rm d}x. \end{align*}

Next, we concentrate on the left-hand side of (3.35).

Firstly, we look at the term $L^3_{s,r,l}$. Thanks to assumption (b), we obtained that $b(x,\,T_{l+1}(u_s)) \to b(x,\,T_{l+1}(u))$ almost everywhere in $\Omega$, and $\{b(\cdot,\,T_{l+1}(u_s)\}$ is uniformly integrable. As $\Omega$ has a finite measure, by Vitali convergence theorem, we have

\[ b\big({\cdot}, T_{l+1}(u_s)\big) \to b\big({\cdot},T_{l+1}(u)\big) \text{ in } L^1(\Omega). \]

This combined with the facts that $\Psi _l(u_s)$ converges to $\Psi _l(u)$ a.e. in $\Omega$ and $|\Psi _l(u_s)|\leq 1$, imply that the product sequence $b(x,\,u_s)\Psi _l(u_s)=b(x,\,T_{l+1}(u_s))\Psi _l(u_s)$ also converges strongly to $b(x,\,u)\Psi _l(u)=b(x,\,T_{l+1}(u))\Psi _{l}(u)$ in $L^1(\Omega )$. Since the term $h(u_r)\phi$ is bounded, we obtain

\[ \lim_{s\to\infty}\int_{\Omega}b(x,u_s)\Psi_l(u_s)h(u_r)\phi \mathop{}\!\mathrm{d} x=\int_{\Omega}b(x,u)\Psi_l(u)h(u_r)\phi \mathop{}\!\mathrm{d} x. \]

Moreover, $h(u_r)\xrightarrow []{}h(u)$ a.e. in $\Omega$, which leads us to

\[ \lim_{r\to \infty}\int_{\Omega}b(x,u)\Psi_l(u)h(u_r)\phi \mathop{}\!\mathrm{d} x=\int_{\Omega}b(x,u)\Psi_l(u)h(u)\phi \mathop{}\!\mathrm{d} x. \]

For $l\geq m$, where $m$ is such that $\text {supp} {(h)}\subset [-m,\,m]$, we infer that

\[ \lim_{l\to \infty}\lim_{r\to \infty}\lim_{s\to \infty}L^3_{s,r,l}= \int_{\Omega}b(x,u)h(u)\phi \mathop{}\!\mathrm{d} x. \]

Secondly, for the term $L^2_{s,r,l}$, choosing $s\geq l+1$, we can rewrite it as follows

(3.36)\begin{equation} \begin{aligned} L^2_{s,r,l} & =\int_{\Omega}\Phi(T_{l+1}(u_s))\cdot \nabla \big(h(u_r)\phi\big)\Psi_l(u_s)\mathop{}\!\mathrm{d} x\\ & \quad +\int_{\Omega}\Phi(T_{l+1}(u_s))\cdot \Psi'_l(u_s) \nabla T_{l+1}(u_s) (h(u_r)\phi)\mathop{}\!\mathrm{d} x\\ & =: L^{2,1}_{s,r,l}+L^{2,2}_{s,r,l}{.} \end{aligned} \end{equation}

As $\Phi (T_{l+1}(u_s))\Psi _{l}(u_s)$ is uniformly bounded, the $a.e.$ convergence of $\{u_s\}_{s>0}$ and the Vitali theorem provide that $\Phi (T_{l+1}(u_s))\Psi _{l}(u_s)\to \Phi (T_{l+1}(u))\Psi _{l}(u)$ in $L^1(\Omega ;{\mathbb {R}^n})$, thus

\[ \lim_{s\to \infty}\int_{\Omega}\Phi(T_{l+1}(u_s))\cdot \nabla \big(h(u_r)\phi\big)\Psi_l(u_s)\mathop{}\!\mathrm{d} x=\int_{\Omega}\Phi(T_{l+1}(u))\cdot \nabla \big(h(u_r)\phi\big)\Psi_l(u)\mathop{}\!\mathrm{d} x. \]

Since $\nabla (h(u_r)\phi )\stackrel {\ast }\rightharpoonup \nabla (h(u)\phi )$ in $L_M(\Omega ;{\mathbb {R}^n})$ and $\Phi$ is Lipschitz continuous, we find that

(3.37)\begin{equation} \lim_{r\to \infty}\int_{\Omega}\Phi(T_{l+1}(u))\cdot \nabla \big(h(u_r)\phi\big)\Psi_l(u)\mathop{}\!\mathrm{d} x = \int_{\Omega}\Phi(T_{l+1}(u))\cdot \nabla \big(h(u)\phi\big)\Psi_l(u)\mathop{}\!\mathrm{d} x. \end{equation}

For $l\geq m$, where $m$ is such that $\text {supp} {(h)}\subset [-m,\,m]$. Rewriting (3.37), we arrive at

\[ \lim_{l\to \infty}\lim_{r\to \infty}\lim_{s\to \infty} L^{2,1}_{s,r,l}=\int_{\Omega}\Phi(u)\cdot \nabla (h(u)\phi)\mathop{}\!\mathrm{d} x. \]

Now we concentrate on the second term from (3.36). Again, similarly as before we may rewrite it as follows

\[ L^{2,2}_{s,r,l}=\int_{\Omega}\operatorname{div}\Bigg({\int_0^{T_{l+1}(u_s)}\Phi(t)\Psi'_l(t) \mathop{}\!\mathrm{d} t}\Bigg)h(u_r)\phi \mathop{}\!\mathrm{d} x. \]

and using the Gauss–Green theorem, we obtain

\[ L^{2,2}_{s,r,l}={-}\int_{\Omega}\int_0^{T_{l+1}(u_s)}\Phi(t)\Psi'_l(t)\mathop{}\!\mathrm{d} t \cdot \nabla (h(u_r)\phi)\mathop{}\!\mathrm{d} x. \]

For the limit with $s\to \infty$, we observe that

\begin{align*} \big|L^{2,2}_{s,r,l}\big|& =\Bigg|\int_{\Omega}\int_0^{T_{l+1}(u_s)}\Phi(t)\Psi'_l(t)\mathop{}\!\mathrm{d} t \cdot \nabla (h(u_r)\phi)\mathop{}\!\mathrm{d} x\Bigg| \\ & \leq \int_{\Omega}\Bigg|\int_0^{T_{l+1}(u_s)}\Phi(t)\Psi'_l(t)\mathop{}\!\mathrm{d} t \cdot \nabla (h(u_r)\phi)\Bigg|\mathop{}\!\mathrm{d} x \\ & \leq \int_{\Omega}\Bigg|\int_0^{T_{l+1}(u_s)}\Phi(t)\Psi'_l(t)\mathop{}\!\mathrm{d} t\Bigg| \cdot \big|\nabla (h(u_r)\phi)\big|\mathop{}\!\mathrm{d} x \\ & \leq \int_{\Omega}\sqrt{n}\Bigg(\sup_{i\in \{1,\ldots,n\}}\Bigg|\int_0^{T_{l+1}(u_s)}\Phi_i(t)\Psi'_l(t)\mathop{}\!\mathrm{d} t\Bigg|\Bigg)\cdot \big|\nabla (h(u_r)\phi)\big|\mathop{}\!\mathrm{d} x\\ & \leq \int_{\Omega}2\sqrt{n}(l+1)\Bigg(\sup_{i\in \{1,\ldots,n\}}\sup_{y\in [{-}l-1,l+1]}|\Phi_i(y)|\Bigg)\cdot \big|\nabla (h(u_r)\phi)\big|\mathop{}\!\mathrm{d} x. \end{align*}

Since the term $\nabla (h(u_r)\phi )$ is bounded, $\Phi =(\Phi _1,\,\Phi _2,\,\dots \Phi _n)$ is Lipschitz and $u_s \to u$ almost everywhere in $\Omega$, by the Lebesgue-dominated convergence theorem, we infer that

\begin{align*} \lim_{s\to \infty}L^{2,2}_{s,r,l}& ={-}\int_{\Omega}\int_0^{T_{l+1}(u)}\Phi(t)\Psi'_l(t)\mathop{}\!\mathrm{d} t \cdot \nabla (h(u_r)\phi)\mathop{}\!\mathrm{d} x \\ & = \int_{\Omega}\Phi(T_{l+1}(u))\cdot \nabla T_{l+1}(u)\Psi'_l(u)(h(u_r)\phi)\mathop{}\!\mathrm{d} x. \end{align*}

Using the Lebesgue-dominated convergence theorem again, we obtain

(3.38)\begin{equation} \begin{aligned} \lim_{r\to \infty}\int_{\Omega} & \Phi(T_{l+1}(u))\cdot \nabla T_{l+1}(u)\Psi'_l(u)(h(u_r)\phi)\mathop{}\!\mathrm{d} x\\ & =\int_{\Omega}\Phi(T_{l+1}(u))\cdot \nabla T_{l+1}(u)\Psi'_l(u)(h(u)\phi)\mathop{}\!\mathrm{d} x. \end{aligned} \end{equation}

For $m>0$ such that $\text {supp}{(h)}\subset [-m,\,m]$, $T_{l+1}$ can be replaced by $T_m$ in (3.38) and $\Psi '_l(u)=\Psi '_l(T_m(u))=0$ for $l>m$. Rewriting (3.38), we arrive at

\[ \lim_{l\to\infty}\lim_{r\to \infty}\lim_{s\to \infty}L^{2,2}_{s,r,l}=0. \]

Finally, we focus on the most important term, which is $L^1_{s,r,l}$. Let us write

\begin{align*} L^1_{s,r,l}& =\int_{\Omega}{\mathcal{A}}(x,\nabla u_s)\cdot \nabla\Psi_l(u_s)h(u_r)\phi \mathop{}\!\mathrm{d} x\\ & \quad + \int_{\Omega}{\mathcal{A}}(x,\nabla u_s)\cdot \nabla\big(h(u_r)\phi \big)\Psi_l(u_s) \mathop{}\!\mathrm{d} x\\ & =:L^{1,1}_{s,r,l}+L^{1,2}_{s,r,l}. \end{align*}

Convergence of the first term is quite straightforward, namely

\begin{align*} \lim_{l\to \infty}& \lim_{r\to \infty}\limsup_{s\to \infty}\big|L^{1.1}_{s,r,l}\big|\\ & \leq ||h||_{L^{\infty}(\Omega)}||\phi||_{L^{\infty}(\Omega)}\lim_{l\to \infty}\lim_{r\to \infty}\Bigg(\sup_{s>0}\int_{\{l<|u_s|< l+1\}}{\mathcal{A}}_{s,l+1}(x)\cdot \nabla T_{l+1}(u_s)\mathop{}\!\mathrm{d} x \Bigg)\\ & \leq C \lim_{l\to \infty}\gamma\Bigg(\frac{l}{m(c_1l)}\Bigg)\\ & =0, \end{align*}

where in the last inequality, we used the energy control condition, stated in (3.4).

For $L^{1.2}_{s,r,l}$, we need to recall some facts we already know. Firstly, by (3.3) and the de la Vallée Poussin theorem (lemma 2.3) we get the uniform integrability of the sequence $\{{\mathcal {A}}_{s,l+1}\}_{s>0}$. But due to the weak-$*$ convergence in (3.11), the Dunford–Pettis theorem yields (up to a subsequence)

\[ {\mathcal{A}}_{s,l+1}\rightharpoonup{\mathcal{A}}(x,\nabla T_{l+1}(u)) \text{ weakly in } L^1(\Omega;{\mathbb{R}^n}). \]

Furthermore, since we also know the following

\begin{align*} |\Psi_l& (u_s)|\leq 1,\\ \nabla (h(u_r)\phi)& \in L^{\infty}(\Omega; \mathbb{R}^n),\\ \Psi_l(u_s)\xrightarrow[s\to \infty]{}& \Psi_l(u) \text{ a.e. in } \Omega, \end{align*}

we infer from lemma 2.8 that

\begin{align*} & \lim_{r\to \infty}\limsup_{s\to \infty}\int_{\Omega}{\mathcal{A}}_{s,l+1}\Psi_l(u_s)\cdot \nabla (h(u_r)\phi ) \mathop{}\!\mathrm{d} x\\ & \quad =\lim_{r\to \infty} \int_{\Omega}{\mathcal{A}}(x,\nabla T_{l+1}(u))\Psi_l(u)\cdot \nabla \big(h(u_r)\phi\big)\mathop{}\!\mathrm{d} x\\ & \quad = \int_{\Omega}{\mathcal{A}}(x,\nabla T_{l+1}(u))\Psi_l(u)\cdot \nabla \big(h(u)\phi\big)\mathop{}\!\mathrm{d} x. \end{align*}

Choosing $l>m$, we obtain

\begin{align*} \lim_{l\to \infty}& \lim_{r\to \infty}\limsup_{s\to \infty}L^{1,2}_{s,r,l}\\ & =\lim_{l\to \infty}\int_{\Omega}{\mathcal{A}}(x,\nabla T_{l+1} (u))\cdot \nabla\big(h(u)\phi)\big)\Psi_l(u) \mathop{}\!\mathrm{d} x\\ & =\int_{\Omega}{\mathcal{A}}(x,\nabla u)\cdot \nabla \big(h(u)\phi\big)\mathop{}\!\mathrm{d} x. \end{align*}

Using the facts that $C^{\infty }_{c}(\Omega )\subset W^{1,\infty }_{0}(\Omega )$ and the gradients of functions in $V^1_{0}L_M(\Omega )$ can be approximated by smooth functions in the weak-$*$ topology of $L_M(\Omega ;{\mathbb {R}^n})$, we finally arrive at

\begin{align*} & \int_\Omega {\mathcal{A}}(x,\nabla u)\cdot\nabla(h(u)\phi) +\Phi(u)\cdot\nabla(h(u)\phi)+ b(x,u)h(u)\phi \,{\rm d}x\\ & \quad =\int_\Omega fh(u)\phi+ F\cdot\nabla(h(u)\phi)\,{\rm d}x \end{align*}

for every $h\in C^1_c({\mathbb {R}})$ and all $\phi \in V_0^1L_M(\Omega )\cap L^\infty (\Omega )$. Thus, condition $(R2)$ is satisfied.

Condition $(R3)$.

As in the definition of renormalized solutions, we have to show that

\[ \int_{\{l<|u|< l+1\}}{\mathcal{A}}(x,\nabla u)\cdot \nabla u \mathop{}\!\mathrm{d} x = \int_{\{l<|u|< l+1\}}{\mathcal{A}}(x,\nabla T_{l+1}(u))\cdot \nabla T_{l+1}(u)\,{\rm d}x \xrightarrow{l \to \infty}0. \]

The first thing we are going to prove is that

(3.39)\begin{equation} {\mathcal{A}}_{s,l+1}\nabla T_{l+1}(u_s)\rightharpoonup {\mathcal{A}}(x,\nabla T_{l+1}(u))\cdot \nabla T_{l+1}(u) \text{ weakly in } L^1(\Omega) \end{equation}

as $s \to \infty$. Here is the place where the first time we apply Young measures. We begin with showing the uniform integrability of the sequence

\[ \Bigg\{\Bigg({\mathcal{A}}_{s,l+1}-{\mathcal{A}}(x,\nabla T_{l+1}(u))\Bigg)\cdot \Bigg(\nabla T_{l+1}(u_s)-\nabla T_{l+1}(u)\Bigg)\Bigg \}_{s}. \]

At first, for every $s$ we can make the following estimate

\begin{align*} & \int_{\Omega}\Bigg({\mathcal{A}}_{s,l+1}-{\mathcal{A}}(x,\nabla T_{l+1}(u))\Bigg)\cdot \Bigg(\nabla T_{l+1}(u_s)-\nabla T_{l+1}(u)\Bigg) \mathop{}\!\mathrm{d} x\\ & \quad \leq |J_1| +|J_2| + |J_3| + |J_4|, \end{align*}

where

\begin{align*} & J_1:=\int_{\Omega}{\mathcal{A}}_{s,l+1}\cdot \nabla T_{l+1}(u_s)\mathop{}\!\mathrm{d} x,\\ & J_2:=\int_{\Omega}{\mathcal{A}}_{s,l+1}\cdot \nabla T_{l+1}(u)\mathop{}\!\mathrm{d} x,\\ & J_3=\int_{\Omega}{\mathcal{A}}(x,\nabla T_{l+1}(u))\cdot \nabla T_{l+1}(u_s)\mathop{}\!\mathrm{d} x,\\ & J_4:=\int_{\Omega}{\mathcal{A}}(x,\nabla T_{l+1}(u))\cdot \nabla T_{l+1}(u)\mathop{}\!\mathrm{d} x. \end{align*}

Then, using the Fenchel–Young inequality and our a priori estimates (3.2) and (3.3), we infer that

\[ |J_1|\!=\!\big|\int_{\Omega}{\mathcal{A}}_{s,l+1}\cdot \nabla T_{l+1}(u_s)\mathop{}\!\mathrm{d} x\big|\!\leq \! 2\|{\mathcal{A}}(x,T_{l+1}(u_s))\|_{L_{M^*}(\Omega)}\|\nabla T_{l+1}(u_s)\|_{L_{M}(\Omega)} \!\leq\! C, \]

where $C$ is independent of $s$. Notice that we may obtain the same estimate for each $J_i$, where $i=1,\,2,\,3,\,4$. Thus, since it does not depend on $s$, we get the uniform boundedness of the sequence

\[ \Bigg\{\Bigg({\mathcal{A}}_{s,l+1}-{\mathcal{A}}(x,\nabla T_{l+1}(u))\Bigg)\cdot \Bigg(\nabla T_{l+1}(u_s)-\nabla T_{l+1}(u)\Bigg)\Bigg \}_{s} \]

in $L^1(\Omega )$. Therefore, we may use Chacon's biting lemma (lemma 2.12) and lemma 2.9, up to a subsequence, get the following biting convergence

(3.40)\begin{equation} \begin{aligned} 0 & \leq \Bigg({\mathcal{A}}_{s,l+1}-{\mathcal{A}}(x,\nabla T_{l+1}(u))\Bigg)\cdot \Bigg(\nabla T_{l+1}(u_s)-\nabla T_{l+1}(u)\Bigg)\\ & \xrightarrow[s\to \infty]{b} \ \ \int_{\mathbb{R}^n}\Bigg({\mathcal{A}}(x,\lambda)-{\mathcal{A}}(x,\nabla T_{l+1}(u))\Bigg) \cdot \Bigg(\lambda - \nabla T_{l+1}(u)\Bigg)\mathop{}\!\mathrm{d} \nu_x(\lambda). \end{aligned} \end{equation}

Here, $\nu _x$ is a Young measure, which is generated by the sequence $\big \{\nabla T_{l+1}(u_s)\big \}$. Since $\nabla T_{l+1}(u_s)\rightharpoonup \nabla T_{l+1}(u)$ in $L^1(\Omega ;{\mathbb {R}^n})$ (obtained in (3.11)), the equality

\[ \int_{\mathbb{R}^N}\lambda \mathop{}\!\mathrm{d} \nu_x(\lambda)=\nabla T_{l+1}(u) \]

holds for a.e. $x\in \Omega$. It follows that

\[ \int_{\mathbb{R}^N}{\mathcal{A}}(x,\nabla T_{l+1}(u))\cdot \big(\lambda - \nabla T_{l+1}(u)\big)\mathop{}\!\mathrm{d} \nu_x(\lambda)=0. \]

Thus, our limit simply becomes

(3.41)\begin{equation} \begin{aligned} \int_{\mathbb{R}^n}\Bigg({\mathcal{A}}(x,\lambda) & -{\mathcal{A}}(x,\nabla T_{l+1}(u))\Bigg) \cdot \Bigg(\lambda - \nabla T_{l+1}(u)\Bigg)\mathop{}\!\mathrm{d} \nu_x(\lambda)\\ & =\int_{\mathbb{R}^n}{\mathcal{A}}(x,\lambda)\cdot \lambda \mathop{}\!\mathrm{d} \nu_x(\lambda)-\int_{\mathbb{R}^N}{\mathcal{A}}(x,\lambda)\cdot \nabla T_{l+1}(u)\mathop{}\!\mathrm{d} \nu_x(\lambda). \end{aligned} \end{equation}

Now, the result in (3.4) yields uniform boundedness of the sequence $\big \{{\mathcal {A}}_{s,l+1}\cdot \nabla T_{l+1}(u_s)\big \}_{s}$ and this allows us to use Chacon's biting lemma (lemma 2.12) and lemma 2.9 to get

\[ {\mathcal{A}}_{s,l+1}\cdot \nabla T_{l+1}(u_s)\xrightarrow[s\to \infty]{b}\int_{\mathbb{R}^N}{\mathcal{A}}(x,\lambda)\cdot \lambda \mathop{}\!\mathrm{d} \nu_x(\lambda). \]

Now we look on assumption $(A2)$. In particular, it immediately implies ${\mathcal {A}}_{s,l+1}\cdot \nabla T_{l+1}(u_s)\geq 0$. Thus, by lemma 2.10, we obtain

(3.42)\begin{equation} \limsup_{s\to \infty}\big({\mathcal{A}}(x,\nabla T_{l+1}(u_s))\cdot \nabla T_{l+1}(u_s)\big)\geq \int_{{\mathbb{R}^n}}{\mathcal{A}}(x,\lambda)\cdot \lambda \mathop{}\!\mathrm{d} \nu_x(\lambda). \end{equation}

Since we already considered this limit in (3.32), taking

\[ {\mathcal{A}}_k={\mathcal{A}}(x,\nabla T_{l+1}(u))=\int_{{\mathbb{R}^n}}{\mathcal{A}}(x,\lambda)\mathop{}\!\mathrm{d} \nu_x( \lambda), \]

we may rewrite (3.42) as

\[ \nabla T_{l+1}(u)\int_{{\mathbb{R}^n}}{\mathcal{A}}(x,\lambda)\mathop{}\!\mathrm{d} \nu_x(\lambda)\geq \int_{{\mathbb{R}^n}}{\mathcal{A}}(x,\lambda)\cdot \lambda \,\nu_x(\lambda). \]

Comparing this inequality with (3.41), we see that the limit obtained in (3.40) is less or equal to zero. Thus, we infer that

\[ \Bigg({\mathcal{A}}_{s,l+1}-{\mathcal{A}}(x,\nabla T_{l+1}(u))\Bigg)\cdot \Bigg(\nabla T_{l+1}(u_s)-\nabla T_{l+1}(u)\Bigg) \xrightarrow[s\to \infty]{b} 0. \]

Now we would like to show that there is actually a stronger convergence, namely

(3.43)\begin{equation} {\mathcal{A}}_{s,l+1}\cdot \nabla T_{l+1}(u_s)\xrightarrow[s\to \infty]{b} {\mathcal{A}}(x,\nabla T_{l+1}(u))\cdot \nabla T_{l+1}(u). \end{equation}

If so, then lemma 2.13 combined with (3.32) and also the weak-$*$ convergence of ${\mathcal {A}}$, described in (3.11), will give us (3.39).

Thus, let us prove (3.43). Observe that since ${\mathcal {A}}(x,\,\nabla T_{l+1}(u))\in L_{M^*}(\Omega ; {\mathbb {R}^n})$, there exists a family of ascending sets $\big \{E_j^{l+1}\big \}$, such that

\[ \lim_{j\to \infty}\big|E_j^{l+1}\big|=0 \]

and also

\[ {\mathcal{A}}(x,\nabla T_{l+1}(u))\in L^{\infty}(\Omega\setminus E_j^{l+1}). \]

As stated in (3.11), we have $\nabla T_{l+1}(u_s)\stackrel {\ast }\rightharpoonup \nabla T_{l+1}(u)$ weakly-$*$ in $L_M(\Omega ;{\mathbb {R}^n})$ as $s\to \infty$. Therefore, we infer that

\[ {\mathcal{A}}(x,\nabla T_{l+1}(u))\cdot \big(\nabla T_{l+1}(u_s)-\nabla T_{l+1}(u)\big) \xrightarrow[s\to \infty]{b} 0. \]

Moreover, very similar arguments yield

\[ {\mathcal{A}}_{s,l+1}\cdot \nabla T_{l+1}(u) \xrightarrow[s\to \infty]{b} {\mathcal{A}}(x,\nabla T_{l+1}(u))\cdot \nabla T_{l+1}(u). \]

Collecting the two above convergences, we arrive at (3.43).

Now is the time to make use of (3.39) and (3.4). First, observe that by the properties of truncations, for every $l\in \mathbb {N}$ we get

\[ \nabla u_s=0 \text{ a.e. in } \{x \in \Omega : |u_s|\in \{l,l+1\}\}. \]

Therefore, by (3.4) we obtain

(3.44)\begin{equation} \lim_{l\to \infty}\limsup_{s\to \infty}\int_{\{l-1<|u|< l+2\}}{\mathcal{A}}(x,\nabla u_s)\cdot \nabla u_s \mathop{}\!\mathrm{d} x=0. \end{equation}

Let us define a function $G_l:\mathbb {R}\to \mathbb {R}$ by the formula:

\[ G_l(r)=\left\{ \begin{array}{@{}ll} 1 & \ \ \ \mbox{if } l\leq |r|\leq l+1 \\ 0 & \ \ \ \mbox{if } |r|< l-1 \ \text{or} \ |r|>l+2\\ \text{affine} & \ \ \ \text{otherwise} \end{array} \right. \]

Then we may write

(3.45)\begin{equation} \int_{\{l<|u|< l+1\}}{\mathcal{A}}(x,\nabla u)\cdot \nabla u \,{\rm d}x \leq \int_{\Omega}G_l(u){\mathcal{A}}(x,\nabla T_{l+2}(u))\cdot \nabla T_{l+2}(u)\mathop{}\!\mathrm{d} x. \end{equation}

As we mentioned before, thanks to $(A2)$ we have ${\mathcal {A}}(x,\,\xi )\cdot \xi \geq 0$, hence taking a limit in (3.45), we get

(3.46)\begin{equation} \begin{aligned} 0 & \leq \lim_{l\to \infty}\int_{\{l<|u|< l+1\}}{\mathcal{A}}(x,\nabla u)\cdot \nabla u \mathop{}\!\mathrm{d} x\\ & \leq \lim_{l\to \infty}\int_{\Omega}G_l(u){\mathcal{A}}(x,\nabla T_{l+2}(u))\cdot \nabla T_{l+2}(u)\mathop{}\!\mathrm{d} x. \end{aligned} \end{equation}

But since we have (3.39) and $G_l$ is a bounded continuous function, we obtain

(3.47)\begin{equation} \begin{aligned} \lim_{l\to \infty}\int_{\Omega} & G_l(u){\mathcal{A}}(x,\nabla T_{l+2}(u))\cdot \nabla T_{l+2}(u)\mathop{}\!\mathrm{d} x\\ & =\lim_{l\to \infty}\lim_{s\to \infty}\int_{\Omega}G_l(u){\mathcal{A}}(x,\nabla T_{l+2}(u_s))\cdot \nabla T_{l+2}(u_s)\mathop{}\!\mathrm{d} x\\ & \leq \lim_{l\to \infty}\limsup_{s\to \infty}\int_{\{l-1<|u|< l+2\}}{\mathcal{A}}(x,\nabla u_s)\cdot \nabla u_s \mathop{}\!\mathrm{d} x \\ & =0. \end{aligned} \end{equation}

The last inequality follows directly from (3.44). By (3.47) and (3.46), we obtain

\[ \lim_{l\to \infty}\int_{\{l<|u|< l+1\}}{\mathcal{A}}(x,\nabla u)\cdot \nabla u \mathop{}\!\mathrm{d} x =0, \]

which gives condition $(R3)$. Thus, $u$ is a renormalized solution and the proof is complete.

4. Uniqueness of renormalized solutions

Now we are ready to prove the uniqueness of renormalized solutions for problem (1.1) under the condition that $s\to b(\cdot,\,s)$ is strictly increasing. We would like to point out that our approach is much influenced by [Reference Chlebicka, Giannetti and Zatorska-Goldstein14, Reference Di Nardo, Feo and Guibé26, Reference Wittbold and Zimmermann53].

Proof of proposition 1.4. We define the auxiliary functions

(4.1)\begin{equation} H_\delta(r)=\left\{\begin{array}{@{}lll} 0, & r < 0,\\ \dfrac{r}{\delta}, & 0 \leq r\leq \delta,\\ 1, & r>\delta, \end{array}\right. h_l(r)=\left\{ \begin{array}{@{}cc} 1, & |r|\leq l-1,\\ l-|r|, & l-1\leq |r| \leq l,\\ 0, & r>l, \end{array}\right. \end{equation}

for $\delta >0$ and $l>1$. By denoting

\[ Z_{l,\delta}=\{0< T_{l}(u_1)-T_{l}(u_2)<\delta\} \]

and testing equation (1.1) with $\phi _1=h_{l}(u_1)H_{\delta }(T_{l}(u_1)-T_{l}(u_2))$ and $\phi _2=h_{l}(u_2)H_{\delta }(T_{l}(u_1)-T_{l}(u_2))$ respectively, subtracting the resulting equations, we get

(4.2)\begin{equation} I^1_{l,\delta}+I^2_{l,\delta}+I^3_{l,\delta}+I^4_{l,\delta}+I^5_{l,\delta}=I^6_{l,\delta}+I^7_{l,\delta}+I^8_{l,\delta} \end{equation}

with

\begin{align*} I^1_{l,\delta}& =\int_{\Omega}\Bigg(b(x,u_1)h_{l}(u_1)-b(x,u_2)h_l(u_2)\Bigg)H_{\delta}\big(T_{l}(u_1)-T_{l}(u_2)\big)\,{\rm d}x,\\ I^2_{l,\delta}& =\int_{\Omega}\Bigg(h'_l(u_1){\mathcal{A}}(x,\nabla u_1)\cdot \nabla u_1-h'_l(u_2){\mathcal{A}}(x,\nabla u_2)\cdot \nabla u_2\Bigg)\\ & \quad \cdot H_{\delta}(T_{l}(u_1)-T_{l}(u_2))\,{\rm d}x,\\ I^3_{l,\delta}& =\frac{1}{\delta}\int_{Z_{l,\delta}}\Bigg(h_l(u_1){\mathcal{A}}(x,\nabla u_1)-h_l(u_2){\mathcal{A}}(x,\nabla u_2)\Bigg)\cdot \nabla \big(T_{l}(u_1)-T_{l}(u_2)\big)\,{\rm d}x,\\ I^4_{l,\delta}& =\int_{\Omega}\Bigg(h'_l(u_1)\Phi(u_1)\cdot \nabla u_1-h'_l(u_2)\Phi(u_2)\cdot \nabla u_2\Bigg)\cdot H_{\delta}(T_{l}(u_1)-T_{l}(u_2))\,{\rm d}x,\\ I^5_{l,\delta}& =\frac{1}{\delta}\int_{Z_{l,\delta}}\Bigg(h_l(u_1)\Phi(u_1)-h_l(u_2)\Phi(u_2)\Bigg)\cdot \nabla \big(T_{l}(u_1)-T_{l}(u_2)\big)\,{\rm d}x,\\ I^6_{l,\delta}& =\int_{\Omega}f(h_l(u_1)-h_l(u_2))\cdot H_{\delta}\big(T_{l}(u_1)-T_{l}(u_2)\big) \,{\rm d}x,\\ I^7_{l,\delta}& =\int_{\Omega}F\Bigg(h'_l(u_1)\cdot \nabla u_1-h'_l(u_2)\cdot \nabla u_2\Bigg)\cdot H_{\delta}(T_{l}(u_1)-T_{l}(u_2))\,{\rm d}x,\\ I^8_{l,\delta}& =\frac{1}{\delta}\int_{Z_{l,\delta}}F(h_l(u_1)-h_l(u_2))\cdot \nabla \big(T_{l}(u_1)-T_{l}(u_2)\big)\,{\rm d}x.\\ \end{align*}

Next, we are going to estimate $I^{i}_{l,\delta }(1\leq i \leq 8)$ one by one.

Estimate of $I^1_{l,\delta }$. Note that we have $H_{\delta }(T_{l}(u_1)-T_{l}(u_2)) \to \operatorname {sign}^+_{0}{(T_l(u_1)-T_{l}(u_2))}$ as $\delta \to 0$. Thus, the Lebesgue-dominated convergence theorem yields

\begin{align*} I^1_l& =\lim_{\delta\to 0}I^1_{l,\delta}\\ & =\int_{\Omega}\Bigg(b(x,u_1)h_{l}(u_1)-b(x,u_2)h_l(u_2)\Bigg)\operatorname{sign}^+_{0}{(T_l(u_1)-T_{l}(u_2))}\,{\rm d}x. \end{align*}

Estimate of $I^2_{l,\delta }$. According to condition $(R3)$, we know that the integrand in $I^2_{l,\delta }$ is bounded in $L^1$ and by the same arguments as above we get

\begin{align*} I^2_l& =\lim_{\delta\to 0}I^2_{l,\delta}=\int_{\Omega}\Bigg(h'_l(u_1){\mathcal{A}}(x,\nabla u_1)\cdot \nabla u_1-h'_l(u_2){\mathcal{A}}(x,\nabla u_2)\cdot \nabla u_2\Bigg)\\ & \quad \cdot\operatorname{sign}^+_{0}{(T_l(u_1)-T_{l}(u_2))}\,{\rm d}x. \end{align*}

Estimate of $I^3_{l,\delta }$. Observing that

\[ I^3_{l,\delta}=I^{3,1}_{l,\delta}+I^{3,2}_{l,\delta}, \]

where

\begin{align*} I^{3,1}_{l,\delta}& =\frac{1}{\delta}\int_{Z_{l,\delta}}(h_l(u_1)-h_l(u_2)){\mathcal{A}}(x,\nabla T_{l}(u_1)) \nabla \big(T_{l}(u_1)-T_{l}(u_2)\big)\,{\rm d}x,\\ I^{3,2}_{l,\delta}& =\frac{1}{\delta}\int_{Z_{l,\delta}}h_l(u_2)({\mathcal{A}}(x,\nabla T_l(u_1))-{\mathcal{A}}(x,\nabla T_{l}(u_2))) \nabla (T_{l}(u_1)-T_{l}(u_2))\,{\rm d}x. \end{align*}

The monotonicity of ${\mathcal {A}}$ implies that $I^{3,2}_{l,\delta }\geq 0$. Since $\|h'_l\|_{\infty }=1$ due to the generalized Hölder inequality, we have

\begin{align*} |I^{3,1}_{l,\delta}|& \leq \int_{\Omega}|{\mathcal{A}}(x,\nabla T_{l}(u_1)) \nabla (T_{l}(u_1)-T_{l}(u_2))\chi_{Z_{l,\delta}}|\,{\rm d}x,\\ & \leq 2\|{\mathcal{A}}(x,\nabla T_{l}(u_1))\|_{L_{M^\ast}(\Omega)}\|\nabla (T_{l}(u_1)-T_{l}(u_2))\chi_{Z_{l,\delta}}\|_{L_{M}(\Omega)}. \end{align*}

Notice that the constant $C$ in (3.2) and (3.3) is independent of $s$, we obtain

\[ \lim_{\delta\to 0}I^{3,1}_{l,\delta}=0. \]

Estimate of $I^4_{l,\delta }$. We note that

\begin{align*} I^4_{l,\delta}& =\int_{\Omega}\Bigg(h'(T_{l}(u_1))\Phi(u_1)\cdot \nabla u_1-h'(T_{l}(u_2))\Phi(u_2)\cdot \nabla u_2\Bigg)\\ & \quad \cdot H_{\delta}(T_{l}(u_1)-T_{l}(u_2))\,{\rm d}x\\ & = \int_{\Omega}\operatorname{div}\Bigg(\int^{T_{l}(u_1)}_{T_{l}(u_2)}h'_{l}(r)\Phi(r)dr\Bigg)\cdot H_{\delta}(T_{l}(u_1)-T_{l}(u_2))\,{\rm d}x\\ & ={-}\int_{\Omega}\Bigg(\int^{T_{l}(u_1)}_{T_{l}(u_2)}h'_{l}(r)\Phi(r)dr\Bigg)\cdot \nabla H_{\delta}(T_{l}(u_1)-T_{l}(u_2))\,{\rm d}x\\ & ={-}\frac{1}{\delta}\int_{\{0< T_{l}(u_1)-T_{l}(u_2)<\delta\}}\Bigg(\int^{T_{l}(u_1)}_{T_{l}(u_2)}h'_{l}(r)\Phi(r)dr\Bigg)\cdot \nabla (T_{l}(u_1)-T_{l}(u_2))\,{\rm d}x.\\ \end{align*}

Then

\[ |I^4_{l,\delta}|\leq \max_{s\in[{-}l,l]}|\Phi(s)|\int_{\{0< T_{l}(u_1)-T_{l}(u_2)<\delta\}} |\nabla (T_{l}(u_1)-T_{l}(u_2))|\,{\rm d}x \xrightarrow{\delta\to 0}0, \]

since on the right-hand side we have integrals of integrable functions over shrinking sets. Therefore, $\lim _{\delta \to 0}I^4_{l,\delta }=0$.

Estimate of $I^5_{l,\delta }$. We can split $I^5_{l,\delta }$ as

\[ I^5_{l,\delta}=I^{5,1}_{l,\delta}+I^{5,2}_{l,\delta}, \]

where

\begin{align*} I^{5,1}_{l,\delta}& = \frac{1}{\delta}\int_{Z_{l,\delta}} h_l(u_1)\Bigg(\Phi(T_{l}(u_1))-\Phi(T_{l}(u_2))\Bigg)\cdot \nabla \big(T_{l}(u_1)-T_{l}(u_2)\big)\,{\rm d}x,\\ I^{5,2}_{l,\delta} & = \frac{1}{\delta}\int_{Z_{l,\delta}} (h_l(u_1)-h_{l}(u_2))\Phi(u_2)\cdot \nabla \big(T_{l}(u_1)-T_{l}(u_2)\big)\,{\rm d}x. \end{align*}

Since $\Phi$ is Lipschitz, we have

\begin{align*} | I^{5,1}_{l,\delta}|& \leq L_{\Phi}\int_{\{0< T_{l}(u_1)-T_{l}(u_2)<\delta\}} |\nabla (T_{l}(u_1)-T_{l}(u_2))|\,{\rm d}x\\ | I^{5,2}_{l,\delta}|& \leq \max_{s\in[{-}l,l]}|\Phi(s)|\int_{\{0< T_{l}(u_1)-T_{l}(u_2)<\delta\}} |\nabla (T_{l}(u_1)-T_{l}(u_2))|\,{\rm d}x. \end{align*}

Argue as above, we have $\lim _{\delta \to 0}I^5_{l,\delta }=0$.

Estimate of $I^6_{l,\delta }$. It can be deduced from the Lebesgue-dominated convergence theorem that

\[ I^6_{l}=\lim_{\delta\to 0}I^6_{l,\delta} =\int_{\Omega}f(h_l(u_1)-h_l(u_2))\cdot \operatorname{sign}^+_{0}{(T_{l}(u_1)-T_{l}(u_2))} \,{\rm d}x. \]

Estimate of $I^7_{l,\delta }$. Similarly, we have

\begin{align*} I^7_{l}& =\lim_{\delta\to 0}I^7_{l,\delta}\\ & =\int_{\Omega}F\Bigg(h'_l(u_1)\cdot \nabla u_1-h'_l(u_2)\cdot \nabla u_2\Bigg)\cdot\operatorname{sign}^+_0{(T_{l}(u_1)-T_{l}(u_2))}\,{\rm d}x. \end{align*}

Estimate of $I^8_{l,\delta }$. According to the fact that $\|h'_l\|_{\infty }=1$ and Fenchel–Young inequality, we have

\begin{align*} |I^8_{l,\delta}|& =\Bigg|\frac{1}{\delta}\int_{Z_{l,\delta}}F(h_l(u_1)-h_l(u_2))\cdot \nabla \big(T_{l}(u_1)-T_{l}(u_2)\big)\,{\rm d}x\Bigg|\\ & \leq\int_{Z_{l,\delta}}\Bigg|F\cdot \nabla (T_{l}(u_1)-T_{l}(u_2))\Bigg| \,{\rm d}x\\ & \leq \int_{Z_{l,\delta}} 2M^\ast\left(x,\frac{4}{c^{\mathcal{A}}_1}F\right)+\frac{1}{4}M(x, c^{\mathcal{A}}_1\nabla T_{l}(u_1))+\frac{1}{4}M(x, c^{\mathcal{A}}_1\nabla T_{l}(u_2))\,{\rm d}x \xrightarrow{\delta \to 0}0. \end{align*}

Combining the above estimates, we find that

\[ I^1_{l}+I^2_{l}+I^3_{l}=I^6_{l}+I^7_{l}. \]

Since $I^3_l \geq 0$, we have

\[ I^1_{l}+I^2_{l}\leq I^6_{l}+I^7_{l}. \]

Thanks to condition $(R3)$, the definition of $h_{l}$ and by the similar arguments as (3.25), we immediately have $I^2_l \to 0$, $I^6_l \to 0$, $I^7_l\to 0$ as $l \to \infty$. Next, we focus on proving that

\[ I^1= \lim_{l\to \infty} I^1_l =\int_{\Omega}(b(x,u_1)-b(x,u_2))^+\,{\rm d}x. \]

For this, we notice that $\lim \limits _{l\to \infty }\operatorname {sign}^+_0{(T_{l}(u_1)-T_{l}(u_2))}=\operatorname {sign}^+_0{(u_1-u_2)}$ almost everywhere in $\Omega$ and weakly-$*$ in $L^{\infty }(\Omega )$. Therefore, we can pass to the limit and obtain

\[ I^1=\int_{\Omega}(b(x,u_1)-b(x,u_2))\operatorname{sign}^+_{0}{(u_1-u_2)}\,{\rm d}x=\int_{\Omega}(b(x,u_1)-b(x,u_2))^+\,{\rm d}x. \]

Thus, we have

\[ \int_{\Omega}(b(x,u_1)-b(x,u_2))^+\,{\rm d}x \leq 0, \]

so that $(b(x,\,u_1)-b(x,\,u_2))^+=0$ almost everywhere due to sign condition of $b$. Since $b$ is strictly increasing with respect to the second variable, we see that $u_1\leq u_2$. Considering $\phi _1=h_{l}(u_1)H_{\delta }(T_{l}(u_2)-T_{l}(u_1))$ and $\phi _2=h_{l}(u_2)H_{\delta }(T_{l}(u_2)-T_{l}(u_1))$ yields the opposite inequality and thus $u_1=u_2$..