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This article is cited in 2 scientific papers (total in 2 papers) Entropy solution for an equation with measure-valued potential in a hyperbolic space V. F. Vil'danovaa, F. Kh. Mukminovb a Institute of Mathematics with Computing Centre, Ufa Federal Research Centre of the Russian Academy of Sciences, Ufa, Russia b Bashkir State Pedagogical University n. a. M. Akmulla, Ufa, Russia
Russian version:
Matematicheskii Sbornik, 2023, Volume 214, Number 11, DOI: https://doi.org/10.4213/sm9875 
Bibliographic databases:
    § 1. IntroductionIn this paper we show that the equation
$$
\begin{equation*}
-\operatorname{div}_g(a(x,u,\nabla u))+b_0(x,u)+ b_1(x,u)\mu=f, \qquad f\in L_1(\mathbb{H}^n), \quad n\geqslant2,
\end{equation*}
\notag
$$
where $\mathbb{H}^n$ is a hyperbolic space and $\mu$ is a nonnegative Radon measure, has an entropy solution. The concept of an entropy solution of a Dirichlet problem was proposed in [1]. In that paper, in a (not necessarily bounded) domain $\Omega\subset\mathbb{R}^n$, $n\geqslant2 $, the authors considered the elliptic equation with $L_1$-data
$$
\begin{equation*}
-\operatorname{div}(a(x,\nabla u))=f(x,u), \qquad \sup_{|u|<c}|f(x,u)|\in L_{1,\mathrm{loc}}(\Omega), \quad c>0.
\end{equation*}
\notag
$$
The function $a$ satisfies certain conditions of boundedness, monotonicity and coercivity. It was proved in [1] that a unique entropy solution of this Dirichlet problem exists. After that, since the end of the last century entropy solutions became an object of investigation for many experts in Russia and other countries. Note that we know of no papers where the uniqueness of an entropy solution of a Dirichlet problem is established for an elliptic equation with flow $a$ depending directly on the unknown function $u$. In the recent paper [2] its authors considered the problem
$$
\begin{equation*}
-\Delta u+\mu g(u)=\sigma, \qquad u|_{\partial\Omega}=0.
\end{equation*}
\notag
$$
They showed that this problem has a unique very weak solution under certain conditions on the function $g$, Radon measure $\sigma$ and the nonnegative measure $\mu$ from the Morrey class $\mu$. In [3] it was shown that the Dirichlet problem
$$
\begin{equation}
\mathcal{L}u= -\operatorname{div}(a(x,u,\nabla u)+F(u))+\mu u=f, \qquad x\in\Omega, \quad u|_{\partial\Omega}=0,
\end{equation}
\tag{1.1}
$$
where $F(u)$ denotes a vector field $F(u)^i= F^i u_{ x_i }$, $F^i\in L_{1,\mathrm{loc}}(\Omega)$, is well posed. The case $\Omega=\mathbb{R}^n$ was also treated there. The class $\mathcal{L}$ contains the Schrödinger operator $-\Delta+\mu$ with singular potential $\mu$. Such operators were considered in [4]. The interest of Russian authors to operators with singular coefficients increased after the paper [5] appeared (see [3] and the references there). In [6], for the equation with Radon measure $\sigma$, it was shown that a unique renormalized solution of the Dirichlet problem exists in an arbitrary domain $\Omega$.In unbounded domains with infinite measure and $N$-function $M(x,u) = |u|^{ p(x)}$ the existence of an entropy solution and a renormalized solution of the equation
$$
\begin{equation*}
-\operatorname{div}(a(x,u,\nabla u))+ b(x,u,\nabla u) + |u|^{ p(x)-1} = \sigma
\end{equation*}
\notag
$$
was established for the first time in [7] and [8], for measures $\sigma$ of a special form. In [9] the Dirichlet problem
$$
\begin{equation*}
-\operatorname{div}a(x,u,\nabla u)+M'(x,u)+b(x,u,\nabla u)=\sigma,\qquad u|_{\partial \Omega}=0,
\end{equation*}
\notag
$$
was considered in an unbounded domain, where the growth of the functions $a$ and $b$ is determined by the generalized $N$-function $M(x,u)$ and the bounded Radon measure $\sigma$ is insignificantly different from a function in $L_1(\Omega)$. It was assumed that the conjugate function $ \overline{M}(x,u)$ satisfies the $\Delta_2$-condition and $b(x,u,\nabla u)u\geqslant0$. It was proved that the Dirichlet problem has an entropy solution, and it was shown that it is a renormalized solution at the same time. The reader can find a more comprehensive survey of results on entropy and renormalized solutions in [9]. In our paper we assume that the functions $M(x,u)$ and $ \overline{M}(x,u)$ satisfy the $\Delta_2$-condition. It is known that the space $C_0^\infty(\mathbb{R}^n)$ can be completed with respect to the norms $\displaystyle\biggl(\int|\nabla u|^p\,dx\biggr)^{1/p}$ and $\displaystyle\biggl(\int(|u|^p+|\nabla u|^p)\,dx\biggr)^{1/p}$ alike, and in the latter case this yields the more ‘narrow’ space $W_p^1(\mathbb{R}^n)\subset\mathcal{H}_p^1(\mathbb{R}^n)$. Usually authors take the second way (for instance, see [6] and [9]). In our paper we use the space $\mathcal{H}_M^1(\mathbb{R}^n)$ obtained on the first way. Note that, in contrast to [1] and other papers, we do not assume here that the functions $b_i$, $i=0,1$, are monotone in $u$. § 2. Statement of the problem and main resultsWe consider the hyperbolic space $\mathbb{H}^n$, $n\geqslant2$, in the form of the Poincaré model in the unit ball $B_1$ with Riemannian metric
$$
\begin{equation}
g_{ij}(x)=\frac{4}{(1-|x|^2)^2}\delta_{ij}, \qquad x\in B_1, \quad i,j=1,\dots, n.
\end{equation}
\tag{2.1}
$$
Also let $ g^{ij}$ be the entries of the inverse matrix of $(g_{ij})$, and set
$$
\begin{equation*}
\partial B_1\equiv\partial_{\infty}\mathbb{H}^n=\{\infty\}.
\end{equation*}
\notag
$$
The geodesic distance between an arbitrary point $x\in \mathbb{H}^n$ and $0$ is defined by
$$
\begin{equation}
\rho(x)=\int_0^{|x|}\frac{2}{1-s^2}\,ds=\log \frac{1+|x|}{1-|x|}, \qquad x\in B_1\equiv \mathbb{H}^n,
\end{equation}
\tag{2.2}
$$
so that
$$
\begin{equation}
|x|=\operatorname{tanh}\biggl(\frac{\rho(x)}{2}\biggr).
\end{equation}
\tag{2.3}
$$
In $\mathbb{H}^n$ the volume element has the form
$$
\begin{equation}
d\nu=\sqrt{g}\,dx_1\,dx_2\cdots dx_n=\frac{2^n}{(1-|x|^2)^n}\,dx,
\end{equation}
\tag{2.4}
$$
where $dx$ is the Lebesgue measure in $\mathbb{R}^n$. For each $x\in \mathbb{H}^n$ let $T_x \mathbb{H}^n$ denote the tangent space at this point. Clearly, $({\partial}/{\partial x_1}, {\partial}/{\partial x_2},\dots ,{\partial}/{\partial x_n})$ is a basis of $T_x \mathbb{H}^n$. We denote the scalar product of two vectors by round brackets:
$$
\begin{equation}
(\beta,\xi)_g\equiv(\beta,\xi)_{g,x}=\sum_{i,j=1}^n g_{ij}(x)\beta^i\xi^j;
\end{equation}
\tag{2.5}
$$
furthermore, for any $\beta,\xi\in T_x \mathbb{H}^n$, $x\in \mathbb{H}^n$, where $\beta=\sum_{i=1}^n\beta^i{\partial}/{\partial x_i}$ and $\xi=\sum_{i=1}^n\xi^i\partial/\partial x_i$ for some $(\beta^1,\dots ,\beta^n)\in {\mathbb{R}}^n$ and $(\xi^1,\dots ,\xi^n)\in {\mathbb{R}}^n$. The gradient $\nabla_g u=((\nabla_g u)^1, \dots,(\nabla_g u)^n)$ is defined by
$$
\begin{equation}
(\nabla_g u)^i=\sum_{j=1}^ng^{ij}\frac{\partial u}{\partial x_j}, \qquad i=1,\dots ,n.
\end{equation}
\tag{2.6}
$$
We let $\chi^k(\mathbb{H}^n)$ denote the set of vector fields in the class $C^k$ on $\mathbb{H}^n$, $k\geqslant0$. The differential $df$ of a function $f$ has the local coordinates $\partial f/{\partial x^i}$, and ${|df|_g=|\nabla_g f|_g}$. In a local coordinate chart the Lie derivative of $f\in C^1(\mathbb{H}^n)$ along a vector field $X\in \chi^0(\mathbb{H}^n)$ is defined by
$$
\begin{equation*}
(df,X)=(\nabla f, X)_g=\sum_{i=1}^nX^i\frac{\partial f}{\partial x^i}.
\end{equation*}
\notag
$$
The divergence of a vector field $X$ is defined in a local chart by
$$
\begin{equation*}
\operatorname{div}_gX=\sum_{i=1}^n\frac{1}{\sqrt{g}}\,\frac{\partial }{\partial x^i}\bigl(X^i\sqrt{g}\bigr).
\end{equation*}
\notag
$$
It what follows we let $\mathcal{L}(V;W)$ denote the set of bounded linear operators in the Banach space $V$ that range in the Banach space $W$. We denote the action of a functional $l\in V^*$ on a vector $v\in V$ by corner brackets: $\langle l, v\rangle$. Here are some requisite facts from the theory of Musielak-Orlicz spaces (see [10]). Definition 1. Let $M(x,z)\colon \mathbb{H}^n \times\mathbb{R}\to\mathbb{R}_+$ be a function satisfying the following conditions: 1) $M(x,\,\cdot\,)$ is an $N$-function in $z\in\mathbb{R}$, that is, it is (downward) convex, nondecreasing, even and continuous, $M(x,0)=0$ for $\nu$-almost all $x\in \mathbb{H}^n$ and $\inf_{x\in\mathbb{H}^n} M(x,z)>0$ for all $z\neq 0$; furthermore,
$$
\begin{equation}
\lim_{z\to 0}\sup_{x\in\mathbb{H}^n} \frac{{M}(x,z)}{z}=0
\end{equation}
\tag{2.7}
$$
and
$$
\begin{equation}
\lim_{z\to \infty}\inf_{x\in\mathbb{H}^n}\frac{{M}(x,z)}{z}=\infty;
\end{equation}
\tag{2.8}
$$
it is obvious that the function $M(x,z)/z$ is nondecreasing; 2) $M(\cdot,z) $ is a measurable function of $x\in \mathbb{H}^n$ for each $z\in\mathbb{R}$. Then $M(x,z)$ is called a Musielak-Orlicz function. The conjugate function $\overline{M}(x,\,\cdot\,)$ of a Musielak-Orlicz function $M(x,\,\cdot\,)$ is defined for $\nu$-almost all $x\in \mathbb{H}^n$ and all $z\geqslant 0$ by the equality Hence we have Young’s inequality
$$
\begin{equation}
|zy| \leqslant M(x,z)+ \overline{M}(x,y), \qquad z,y \in \mathbb{R}, \quad x \in \mathbb{H}^n.
\end{equation}
\tag{2.9}
$$
We assume that
$$
\begin{equation}
{M}(x,z),\overline{M}(x,z)\in L_{1,\mathrm{loc}}(\mathbb{H}^n), \qquad z>0.
\end{equation}
\tag{2.10}
$$
The function $M(x,z)$ satisfies the $\Delta_2$-condition if there exist a constant $C$ and a function $G(x)\in L_1(\mathbb{H}^n)$ such that
$$
\begin{equation*}
M(x,2z)\leqslant CM(x,z)+G(x) \quad \forall\, z\in \mathbb{R}, \quad x\in \mathbb{H}^n.
\end{equation*}
\notag
$$
We define the Lebesgue space $L_{M}(\mathbb{H}^n)$ as the set of measurable functions $u$ on $\mathbb{H}^n$ with finite Luxemburg norm
$$
\begin{equation*}
\|u\|_{M}= \inf\{k>0\colon\varrho(k^{-1}u)\leqslant 1\}\quad\text{and} \quad \varrho(u)= \int_{\mathbb{H}^n}M(x,u)\,d\nu.
\end{equation*}
\notag
$$
In particular, if $k<\|u\|_{M}$, then $\varrho(k^{-1}u)> 1. $ Since $M$ is convex in the second argument, the inequality $\|u\|_{M}>1$ yields the relation $\|u\|_{M}\leqslant \varrho(u)$. We denote the space of measurable covectors $w(x)\colon x\to T_x^*(\mathbb{H}^n)$ such that $|w(x)|_g\in L_{M}(\mathbb{H}^n)$ by $\mathbf{L}_{M}(\mathbb{H}^n)$. We assume that both $M$ and $\overline{M}$ satisfy the $\Delta_2$-condition and condition (2.10). The space $L_M(\mathbb{H}^n)$ is separable and reflexive, and $(L_M(\mathbb{H}^n))^*=L_{\overline{M}}(\mathbb{H}^n)$ (see [11], Corollary 3.6.7). The convergence $\|u_j-u\|_{M}\to0$ is equivalent to the modular convergence $\varrho(u_j-u)\to0$. Given two conjugate Musielak-Orlicz functions $M$ and $\overline{M}$, if $u \in L_M(\mathbb{H}^n)$ and $v \in L_{\overline{M}}(\mathbb{H}^n)$, then we have Hölder’s inequality
$$
\begin{equation}
\biggl|\int_{\mathbb{H}^n} u(x)v(x)\,d\nu\biggr| \leqslant 2\|u\|_{M}\|v\|_{\overline{M}}.
\end{equation}
\tag{2.11}
$$
We define the space $\mathcal{H}_{M}^1(\mathbb{H}^n)$ as the completion of the function space $\mathcal{D}(\mathbb{H}^n)$ with respect to the norm For brevity we also denote it by $V$. We denote the dual space of $V$ with induced norm by $V^*$. The following estimate is known for the hyperbolic space (see [12], § 8, Theorem 2.28):
$$
\begin{equation*}
\|u\|_{p^*,\mathbb{H}^n}\leqslant C\|\nabla_gu\|_{p,\mathbb{H}^n}\quad\text{and} \quad u\in C^\infty_0(\mathbb{H}^n), \quad p^*= \frac{np}{n-p}.
\end{equation*}
\notag
$$
Hence if a function $u$ has a finite norm $\|\nabla_gu\|_{p,\mathbb{H}^n}$, then it also has a finite norm $\|u\|_{p^*,\mathbb{H}^n}$, so that it decays to zero at infinity in a certain sense. We interpret this as the ‘Dirichlet boundary condition’ in the problem under consideration. For any $\sigma>0$ let
$$
\begin{equation*}
\mathfrak{B}_\sigma=\{x\in\mathbb{H}^n \mid \rho(x)<\sigma\}
\end{equation*}
\notag
$$
be a ball in the hyperbolic space. Then for each $r\in (0,1)$ we have More precisely, the Euclidean ball $B_r$ is a chart for the hyperbolic ball. Throughout what follows the quantities $r$ and $\sigma$ are related by $\sigma=\log((1+r)/(1-r))$. Since the metric $(g_{ij})$ in $B_{r}$ is smooth, the geodesic distances $\rho'$ between pairs of points in $\mathfrak{B}_{\sigma}$ can be estimated in terms of the Euclidean distances between the corresponding points in $B_{r}$: $C^{-1}r'\leqslant \rho'\leqslant Cr'$. The spaces $W_p^1(\mathfrak{B}_\sigma)$ and $W_p^1(B_r)$ are identified in a natural way, and the relevant norms are equivalent. In Lemma 1 we establish the existence of $p\in(1,n)$ such that
$$
\begin{equation}
\|u\|_{W_p^1(\mathfrak{B}_\sigma)}\leqslant h(\sigma)\|u\|_{V}, \qquad \sigma>0,
\end{equation}
\tag{2.12}
$$
where $h(\sigma)$ is an increasing nonnegative function. Consider an operator of the form where $\mu$ is a nonnegative Radon measure. We assume that for $u,v\in C_0^{\infty}(\mathbb{H}^n)$ this operator acts by the formula
$$
\begin{equation*}
\langle\mathcal{B}u,v\rangle=\int_{\mathbb{H}^n } b_0(x,u)v\,d\nu + \int_{\mathbb{H}^n}b_1(x,u)v\,d\mu.
\end{equation*}
\notag
$$
We regard the Radon measure $\mu$ on $\mathbb{H}^n$ as a measure on the ball $B_1$. Let $\mu$ be a Radon measure with finite total variation and with support in a bounded domain $\Omega\subset\mathbb{R}^n$. Assume that this measure is extended by zero outside $\Omega$. Recall that $\mu$ is in the Morrey class $\mathbb{M}_s(\Omega)$, $s\geqslant1$, if for any ball with centre $x$ we have
$$
\begin{equation*}
|B_r(x)|_\mu:=\int_{B_r(x)}\,d|\mu|\leqslant cr^{n(1-1/s)}, \qquad r>0, \quad x\in\Omega.
\end{equation*}
\notag
$$
Using another notation $\mu\in \mathbb{M}_{n/(n-\theta)} (\Omega)$ for $\theta\in[0,n]$, $\theta=n(1-1/s)$, if we have We assume that there exists $s>np/(np+p-n)$ such that Consider the equation
$$
\begin{equation}
-\operatorname{div}_g(a(x,u,d u))+\mathcal{B} u=f, \qquad f\in L_1(\mathbb{H}^n).
\end{equation}
\tag{2.14}
$$
We prove here that it has an entropy solution. The vector field $a(x,u,dv)$ in (2.14) satisfies the following conditions for $x\in\mathbb{H}^n$.
Furthermore, let $b_i$ be Carathéodory functions satisfying the inequalities
$$
\begin{equation}
|b_i(x,s)|\leqslant \operatorname{g}(r)\widetilde{G}_i(x), \qquad |s|\leqslant r, \quad\rho(x)\leqslant r, \quad i=0,1,
\end{equation}
\tag{2.18}
$$
where $\widetilde{G}_0\in L_{1,\mathrm{loc}}(\mathbb{H}^n)$ and $\widetilde{G}_1\in L_{1,\mu,\mathrm{loc}}(\mathbb{H}^n)$; Assume that there exists an increasing function $\widetilde{g}(r)$, $r>0$, where ${\lim_{r\to\infty}\widetilde{g}(r)=\infty}$, such that
$$
\begin{equation}
|b_1 (x,s)|>\widetilde{g}(r), \qquad s\geqslant r, \quad x\in \mathbb{H}^n.
\end{equation}
\tag{2.20}
$$
We define a function $T_k(r)$ by
$$
\begin{equation*}
T_k(r)= \begin{cases} k & \text{for }r>k, \\ r & \text{for }|r|\leqslant k, \\ -k & \text{for }r<-k. \end{cases}
\end{equation*}
\notag
$$
Let $\mathring{\mathcal{T}}_{M}^1(\mathbb{H}^n)$ denote the set of measurable functions $u \colon \mathbb{H}^n\to \mathbb{R}$ such that ${T_k(u)\in V}$ for each $k>0$. Definition 2. An entropy solution of the Dirichlet problem for equation (2.14) is a function $u\in \mathring{\mathcal{T}}_{M}^1(\mathbb{H}^n)$ such that for all $k>0$ and $\xi\in C_0^1(\mathbb{H}^n)$ the inequality is well defined and valid, that is, all terms in it are finite. The main result in our paper is as follows. Theorem 1. Assume that conditions (2.13) and (2.15)–(2.20) are met. Then the Dirichlet problem for equation (2.14) has an entropy solution. § 3. Technical lemmas Proof. It follows from the $\Delta_2$-conditions on the functions $M$ and $ \overline{M} $ (see the proof of Corollary 3.6.7 and Definition 2.1.2 in [11]) that there exist $p\in(1,n)$ and $\beta\in(0,1)$ such that for all $\lambda>1$ the inequality holds for all $r>0$ and almost all $x\in\mathbb{H}^n$. It is obvious that
$$
\begin{equation*}
\inf M(x,1)\int_{\mathfrak{B}_R} |du|_g^p\,d\nu\leqslant \int_{\mathfrak{B}_R} M(x,1)\,d\nu +\int_{\mathfrak{B}_R, |du|_g>1} M(x,1) |du|_g^p\,d\nu.
\end{equation*}
\notag
$$
Let $\|u\|_{V}\leqslant1$. Then by (3.2) we have
$$
\begin{equation}
\beta\int_{\mathfrak{B}_R, |du|_g>1} M(x,1)|du|_g^p\,d\nu \leqslant\int_{\mathbb{H}^n} M(x,|du|_g)\,d\nu\leqslant1.
\end{equation}
\tag{3.3}
$$
Taking (2.10) into account we see that the integral $\displaystyle\int_{\mathfrak{B}_R} |du|_g^p\,d\nu$ is bounded in the ball $\|u\|_{V}\leqslant1$. This yields the inequality
$$
\begin{equation}
\||du|_g\|_{p,\mathfrak{B}_R}\leqslant C(R)\| u\|_{V}.
\end{equation}
\tag{3.4}
$$
Hence we have a continuous embedding and (3.1) holds. In fact, if the embedding operator is unbounded on $C^\infty_0(\mathbb{H}^n)$, then there exists a sequence of smooth functions $v^k$ such that
$$
\begin{equation*}
\|v^k\|_{W^1_{p}(\mathfrak{B}_R)}\geqslant k\| v^k\|_{V}.
\end{equation*}
\notag
$$
Multiplying both sides by an appropriate scalar we reduce this inequality to where $\| v^k\|_{W^1_{p}({\mathfrak{B}_R})}=1$. By (3.5) we have By Kondrashov’s theorem the $v^k$ converge strongly in $L_{p}(\mathfrak{B}_R)$. Taking (3.4) into account we see that $v^k\to C\ne0$ in the space $W^1_{p}({\mathfrak{B}_R})$. This contradicts inequality (3.5), which implies that $v^k\to 0$ in $V$.
The proof is complete. Lemma 2. Let $u(x)$ be a measurable function in $\mathbb{H}^n$. Then $\{k\colon \operatorname{meas} \{x\in\mathbb{H}^n$: $|u(x)|= k\}>0\}$ is finite or countable. Proof. We select $k_i$ such that $\operatorname{meas}\{x\in\mathfrak{B}_{r}\colon |u(x)|= k_i\}>1/N$. These sets are disjoint, so that Hence there can be at most $N \operatorname{meas} \mathfrak{B}_{r}$ such sets. Then the set $\{k \colon \operatorname{meas}\{{x\in\mathfrak{B}_{r}}$: $|u(x)|= k\}>0\}$ is finite or countable. Now the required result follows easily. Proof. Using Sobolev’s inequality and (3.1) we obtain
$$
\begin{equation*}
\|T_k(v)\|_{p^*,\mathfrak{B}_{r}}\leqslant C(r)\|T_k(v)\|_{W_p^1(\mathfrak{B}_{r})}\leqslant C(r)h(r)\|T_k(v)\|_V.
\end{equation*}
\notag
$$
For $k_1\in(0,k]$ it is obvious that
$$
\begin{equation*}
\operatorname{meas}\{\mathfrak{B}_{r} \colon |v|\geqslant k_1\} \leqslant \frac{\|T_k(v)\|_{p^*,\mathfrak{B}_{r}}^{p^*}}{k_1^{p^*}}\leqslant C_1(r) \frac{\|T_k(v)\|_V^{p^*}}{k_1^{p^*}}\leqslant C_1(r) \frac{Ck^{p^*/p}}{k_1^{p^*}}.
\end{equation*}
\notag
$$
Hence, setting $k_1=k$ we obtain (3.7).
The proof is complete. Lemma 4. Let $Q\subset \mathbb{H}^n$, let $\{v^m\}_{m\in \mathbb{N}}$ be a bounded sequence in $L_M(Q)$, let ${v\in L_M(Q)}$, and assume that Then For a bounded domain $Q\subset \mathbb{R}^n$ Lemma 4 was proved in [13]. For $Q\subset\mathbb{H}^n$ the proof is similar. Remark 1. In what follows, to avoid cumbersome reasoning, in place of statements of the form “from a sequence $u^m$ we can extract a subsequence converging $\nu$-almost everywhere in $\mathbb{H}^n$ as $m\to \infty$” we will simply write “a sequence $u^m$ contains a subsequence converging $\nu$-almost everywhere in $\mathbb{H}^n$ as $m\to \infty $”. We will also write “converges weakly along a subsequence” and so on, omitting a subscript indicating the subsequence. Lemma 5. Let $v^j$, $j\in \mathbb{N}$, and $v$ be functions in $L_M(Q)$ such that Lemma 5 is a consequence of Lebesgue’s theorem. Lemma 6. Let $x\in \mathbb{H}^n$ be a point such that the function $a(x,\cdot\,{,}\,\cdot\,)$ is continuous for $v\in\mathbb{R}$ and $z\in T^*_x\mathbb{H}^n$ and the monotonicity condition (2.17) is met. Let $v_m\in\mathbb{R}$ and $z_m\in T^*_x\mathbb{H}^n$ be sequences such that $v_m\to v$ and Then $ z_m\to z$. Proof. Let $s_0=|v|+|z|_g+1$ and
$$
\begin{equation*}
\Lambda(x,r,y,z)=\bigl(a(x,r,y)-a(x,r,z),y-z), \qquad y,z\in T_x^{*}\mathbb{H}^n, \quad r\in\mathbb{R}.
\end{equation*}
\notag
$$
It follows from the hypotheses that The sequence $v_{m}$ tends to $v$ and $|v|<s_0$, so we can assume that $|v_{m}|< s_0$. The following inequality is easy to establish (see [14], formula (2.1)):
$$
\begin{equation}
\Lambda(x,r,y,z)\geqslant|y-z|_g\Lambda\biggl(x,r,z+\frac{y-z}{|y-z|_g},z\biggr), \qquad |y-z|_g\geqslant2.
\end{equation}
\tag{3.9}
$$
Since at the point $x$ the function $a$ is continuous with respect to the other variables, the function $\Lambda(x,r,y,z)$ is continuous in $r,y$ and $z$. The function $\Lambda(x,r,z+e,z)$ is continuous and positive and attains a positive minimum $c(x)>0$ on the compact set $|e|_g=1,|r|\leqslant s_0, |z|_g\leqslant s_0$. Then from (3.9) we obtain
$$
\begin{equation}
\Lambda(x,r,y,z)\geqslant|y-z|_gc(x), \qquad |y-z|_g\geqslant2, \quad|r|\leqslant s_0, \quad |z|_g\leqslant s_0.
\end{equation}
\tag{3.10}
$$
We claim that the sequence $z_{m}$ is bounded. Otherwise a subsequence $z_{m_k}\to \infty$ can be extracted and $\Lambda(x,v_{m_k},z_{m_k},z)\geqslant|z_{m_k}-z|_gc(x)\to\infty$ by (3.10), which contradicts (3.8).
Thus, the sequence $z_{m}$ is bounded. We can extract from it a convergent subsequence $z_{m_k}\to z_0$. Taking the limit we see that This is possible only for $z_0= z$. Then the whole sequence has the same limit. The proof is complete. Lemma 7. Assume that conditions (2.15)–(2.17) are met in $\mathbb{H}^n$ and let $w^j\in V$ be a sequence such that Proof. Set
$$
\begin{equation*}
q^j(x)=\Lambda(x,w^j,dw^j,dw), \qquad x\in \mathbb{H}^n, \quad j\in \mathbb{N}.
\end{equation*}
\notag
$$
By (2.17) the functions $\{q^j(x)\}_{j\in \mathbb{N}}$ are nonnegative, and by (3.14) $q^j\,{\to}\, 0$ in $L_1(\mathfrak{B}_R)$ as $j\to\infty$. Hence along some subsequence. Using a diagonal procedure we can extract a subsequence such that
From (3.12) we obtain Let $Q$ be the full-measure subset of points at which we have (3.16) and (3.13), inequalities (2.15)–(2.17) hold and Using Lemma 6 for $v_j= w^j$, $z_j=d w^j$ and $z=dw$ we obtain the convergence in the whole of $Q$.It follows from (3.13), since $a(x,r,y)$ is continuous in $r$ and $y$, that
$$
\begin{equation*}
a(x,w^j,dw^j)\to a(x,w,dw) \quad \text{$\nu$-a.e. in } \mathbb{H}^n, \qquad j\to\infty.
\end{equation*}
\notag
$$
Furthermore, from (2.15) and (3.11) we obtain
$$
\begin{equation*}
\overline{M}(x,|a(x,w^j,dw)|_g)\leqslant \operatorname{g}(k)( G(x))+M(x,|dw|_g))\in L_1(\mathbb{H}^n), \qquad j\in \mathbb{N}.
\end{equation*}
\notag
$$
Lemma 5 yields the convergence
$$
\begin{equation}
a(x,w^j,dw)\to a(x,w,dw) \quad\text{strongly in } \mathbf L_{\overline{M}}(\mathbb{H}^n), \qquad j\to\infty.
\end{equation}
\tag{3.19}
$$
By (3.17) we have the weak convergence
$$
\begin{equation}
a(x,w^j,dw^j)\rightharpoonup a(x,w,dw) \quad\text{weakly in } \mathbf L_{\overline{M}}(\mathbb{H}^n), \qquad j\to\infty.
\end{equation}
\tag{3.20}
$$
We introduce the notation $\lambda^j=(a(x,w^j,dw^j),dw^j)$ and $\lambda= (a(x,w,d w),dw)$. Then it is easy to see that
$$
\begin{equation*}
q^j=\lambda^j-(a(x,w^j,dw),dw^j)-(a(x,w^j,dw^j),dw)+(a(x,w^j,dw),dw).
\end{equation*}
\notag
$$
Using (3.12), (3.14) and (3.19) we can show that
$$
\begin{equation}
\int_{\mathfrak{B}_R}\lambda^jd\nu\to \int_{\mathfrak{B}_R}\lambda \,d\nu, \qquad j\to\infty.
\end{equation}
\tag{3.21}
$$
It follows from (3.13) and (3.18) that
$$
\begin{equation}
\lambda^j\to \lambda \quad \text{$\nu$-a.e. in } \mathbb{H}^n, \qquad j\to\infty.
\end{equation}
\tag{3.22}
$$
Using (2.16) and the $\Delta_2$-condition we obtain
$$
\begin{equation*}
\begin{aligned} \, \lambda^j+\lambda &=(a(x,w^j,dw^j),dw^j)+ (a(x,w,d w),dw) \\ &\geqslant c_0 M(x,|d w^j|_g)+c_0 M(x,|dw|_g)-2G(x) \\ &\geqslant c_0 M\biggl(x,\frac{1}{2}| {d(w^j-w)}|_g\biggr)-2G(x) \\ &\geqslant \frac{c_0}{C} M(x,|d(w^j-w)|_g)-2G(x). \end{aligned}
\end{equation*}
\notag
$$
From Fatou’s lemma, (3.15) and (3.22) we deduce the inequality
$$
\begin{equation*}
\int_{\mathfrak{B}_R}2\lambda \,d\nu\leqslant \lim_{j\to\infty}\inf \int_{\mathfrak{B}_R} \biggl(\lambda^j+\lambda-\frac{c_0}{C} M(x,|d(w^j-w)|_g)\biggr)d\nu.
\end{equation*}
\notag
$$
Taking (3.21) into account we conclude that
$$
\begin{equation*}
0\leqslant -\lim_{j\to\infty}\sup\int_{\mathfrak{B}_R} M(x,|d(w^j-w^j)|_g)\,d\nu.
\end{equation*}
\notag
$$
Hence
$$
\begin{equation*}
0\leqslant \lim_{j\to\infty}\inf\int_{\mathfrak{B}_R} M(x,|d(w^j-w)|_g)\,d\nu \leqslant\lim_{j\to\infty}\sup\int_{\mathfrak{B}_R} M(x,|d(w^j-w)|_g)\,d\nu\leqslant 0,
\end{equation*}
\notag
$$
so that
$$
\begin{equation*}
\int_{\mathfrak{B}_R} M(x,|d(w^j-w)|_g)\,d\nu\to 0, \qquad j\to\infty,
\end{equation*}
\notag
$$
which yields the convergence (3.15).
Lemma 7 is proved. Lemma 8. Let $\{v^j\}_{j\in \mathbb{N}}$ be a bounded sequence in $L_{\infty}(\mathbb{H}^n)$, let $v\in L_{\infty}(\mathbb{H}^n)$, and assume that Then In addition, if $h^j$, $j\in \mathbb{N}$, and $h$ are functions in $L_M(\mathbb{H}^n)$ such that thenLemma 8 is a consequence of Lebesgue’s theorem. The next result is usually called Beppo Levi’s theorem. Lemma 9. Let ($S,\sum,\mu$) be a space with positive measure and $\{f_n\}$ be a nondecreasing sequence of nonnegative measurable, but not necessarily integrable functions. Then The proof can be found in [15], Ch. III, § 6, Corollary 17. § 4. Weak solution of approximation problem4.1. An abstract result on the existence of a weak solutionIn fact, here we formulate a known result in a form suited for our purposes. Let $W$ be a reflexive separable Banach space and $ K\colon W\to W^* $ be a weakly continuous operator. Let $A(u)=\widetilde{A}(u,u)+K(u)$, where the operator $\widetilde{A}$ maps $W\times W$ to $W^* $ and is monotone in the second argument:
$$
\begin{equation}
\langle \widetilde{A}(u,v),v-w\rangle\geqslant\langle \widetilde{A}(u,w),v-w\rangle, \qquad u,v,w\in W.
\end{equation}
\tag{4.1}
$$
We assume that We also assume that there exist $p>1$ and $C_K>1$ such that Now we state a result we use in what follows. Theorem 2. Assume that conditions (I) and (4.1)–(4.3) are met and ${K}$ is a weakly continuous operator. Also assume that for each sequence $u_j\rightharpoonup u$ converging weakly in $W$ the condition is satisfied. Then $A$ is a surjective operator. Proof. That the equation $A(u)=f$ is solvable for $f\in W^*$ follows from the fact that $A(u)= \widetilde{A}(u,u)+{K}(u)$ is a pseudomonotone coercive operator (see [16], Ch. 2, Theorem 2.7).
Verifying coercivity. From conditions (II) we obtain
$$
\begin{equation*}
\begin{aligned} \, \langle {A}(u),u\rangle &=\langle \widetilde{A}(u,u)+{K}(u),u\rangle\geqslant \langle {K}(u),u\rangle+ \alpha\|u\|_{W}^p-C(\|u\|_{W}+1) \\ &\geqslant \frac{\alpha}{2} \|u\|_{W}^p-C(\|u\|_{W}+1). \end{aligned}
\end{equation*}
\notag
$$
Hence $A$ is coercive.
Verifying pseudomonotonicity. Let $u_j\rightharpoonup u$ weakly in $W$, and assume that We must prove that
$$
\begin{equation}
\liminf\langle A(u_j),u_j-v\rangle\geqslant\langle A(u),u-v\rangle
\end{equation}
\tag{4.6}
$$
for all $v\in W$.
It follows from (4.4) and (4.5) that
$$
\begin{equation}
\limsup \langle \widetilde{A}(u_j,u_j),u_j-u\rangle\leqslant0.
\end{equation}
\tag{4.7}
$$
By (4.1) we have
$$
\begin{equation}
\langle \widetilde{A}(u_j,v),v-w\rangle\geqslant\langle\widetilde{A}(u_j,w),v-w\rangle \quad \forall\, v,w\in W.
\end{equation}
\tag{4.8}
$$
By condition (I) the sequence $\widetilde{A}(u_j,v)$ converges strongly in $W^*$, and therefore
$$
\begin{equation}
\langle \widetilde{A}(u_j,v),u_j-u\rangle\to0 \quad \forall\, v\in W.
\end{equation}
\tag{4.9}
$$
Hence by (4.8)
$$
\begin{equation*}
\langle \widetilde{A}(u_j,u_j),u_j-u\rangle\geqslant\langle \widetilde{A}(u_j,u),u_j-u\rangle\to0.
\end{equation*}
\notag
$$
From this, taking (4.7) into account we obtain We substitute $v=u_j $ and $ w=(1-\theta)u+\theta z$, $\theta\in(0,1)$, into (4.8). Then we have
$$
\begin{equation*}
\begin{aligned} \, &\theta\langle \widetilde{A}(u_j,u_j),u-z\rangle \\ &\qquad\geqslant-\langle \widetilde{A}(u_j,u_j),u_j-u\rangle +\langle \widetilde{A}(u_j,w),u_j-u\rangle+\theta\langle \widetilde{A}(u_j,w),u-z\rangle. \end{aligned}
\end{equation*}
\notag
$$
Using (4.10) and (4.9) we derive the inequality
$$
\begin{equation*}
\theta\liminf\langle \widetilde{A}(u_j,u_j),u-z\rangle\geqslant\theta\langle \widetilde{A}(u,w),u-z\rangle.
\end{equation*}
\notag
$$
Dividing by $\theta$ and adding the result to (4.10) yields
$$
\begin{equation*}
\liminf\langle \widetilde{A}(u_j,u_j),u_j-z\rangle\geqslant\langle \widetilde{A}(u,(1-\theta)u+\theta z),u-z\rangle.
\end{equation*}
\notag
$$
Letting $\theta$ tend to zero and using the fact that $\widetilde{A}$ is semicontinuous we obtain
$$
\begin{equation*}
\liminf\langle \widetilde{A}(u_j,u_j),u_j-z\rangle\geqslant\langle \widetilde{A}(u,u),u-z\rangle.
\end{equation*}
\notag
$$
Using the fact that the operator ${K}$ is weakly continuous and (4.4) we can write
$$
\begin{equation*}
\langle \widetilde{K}(u),u-z\rangle=\lim\langle \widetilde{K}(u_j),u-z\rangle=\lim\langle \widetilde{K}(u_j),u_j-z\rangle.
\end{equation*}
\notag
$$
From this we deduce (4.6).
Theorem 2 is proved. 4.2. Examples of operators $K$ and vector fields $a(x,u,du)$. An approximation problemNow let $W$ coincide with the space $V$. In view of (2.15) the vector field $a^m(x,r,y)=a(x,T_m(r),y)$ defines an operator $ \widetilde{A}\colon V\times V\to V^* $. In the current context it acts by the formula
$$
\begin{equation*}
\langle \widetilde{A}(u,v),w\rangle=\int_{\mathbb{H}^n}(a^m(x,u,dv), d w) \,d\nu, \qquad u,v,w\in V.
\end{equation*}
\notag
$$
From (2.16) it follows that
$$
\begin{equation*}
\langle \widetilde{A}(u,u),u\rangle\geqslant c_0\varrho(|du|_g)-\int_{\mathbb{H}^n}G(x)\,d\nu.
\end{equation*}
\notag
$$
If $\|u\|_{V}>1$, then by (3.2), for $k=\|u\|_{V}-\varepsilon$ we have
$$
\begin{equation}
\begin{aligned} \, \notag \varrho(|du|_g) &=\int_{\mathbb{H}^n} M(x, |du|_g)\,d\nu =\int_{\mathbb{H}^n} M\biggl(x, k\frac{|du|_g}{k}\biggr)\,d\nu \\ &\geqslant\int_{\mathbb{H}^n} \beta k^p M\biggl(x,\frac{|du|_g}{k} \biggr)\,d\nu >\beta k^p = \widetilde{\beta}\|u\|_{V}^p. \end{aligned}
\end{equation}
\tag{4.11}
$$
Hence the operator $\widetilde{A}$ satisfies (4.2). The operator $K$ satisfies condition (4.3), provided that
$$
\begin{equation}
\langle 2{K}(u),u\rangle+c_0\varrho(|du|_g)\geqslant0, \qquad u\in V,
\end{equation}
\tag{4.12}
$$
for $\|u\|_{V}\geqslant C_K$. Now we turn to operators $K$ defined by formulae of the form
$$
\begin{equation}
\langle K(u),v\rangle=\int_{\Omega}b(x,u)v(x)\,d\mu.
\end{equation}
\tag{4.13}
$$
In the case when the measure $\mu$ is nonnegative and $b(x,u)u(x)\geqslant0$, inequality (4.3) clearly holds. Also note that the compact operator $K$ satisfies (4.4). Let $\mu$ be a Radon measure with finite total variation and with support in a bounded domain $\Omega\subset\mathbb{R}^n$. Recall that $\mu\in \mathbb{M}_{n/(n-\theta)} (\Omega)$ for $\theta\in[0,n]$ if The delta function $\delta$ belongs to the class $\mathbb{M}_1(\Omega)$. Functions in $L_s(\Omega)$ define measures in the class $\mathbb{M}_s(\Omega)$ by Hölder’s inequality. If
$$
\begin{equation*}
f\in L_q(\Omega\cap\{x^1=\dots =x^k=0\})\quad\text{and} \quad x'=(0,\dots ,0,x^{k+1},\dots ,x^{n}),
\end{equation*}
\notag
$$
then
$$
\begin{equation*}
\int_{B_r(x_0)}|f(x)|\,dx'\leqslant\|f\|_q\biggl(\int_{B_r(x_0)\cap\{x^1=\dots =x^k=0\}}\,dx'\biggr)^{1-1/q} \leqslant cr^{(n-k)(1-1/q)},
\end{equation*}
\notag
$$
and this function also defines a measure in the Morrey class with support on a plane of dimension $n-k$. For $q<{\theta p}/(n-p)$ and a nonnegative measure $\mu\in\mathbb{M}_{n/(n-\theta)} (\Omega)$ it is known that there is a compact embedding This is a special case of a general result (see [2], Proposition 2.5). In the case of the Lebesgue measure we have a compact embedding for $q_0<n p/(n-p)$. Assume that the Nemytskii operator $u\to b(x,u(x))$ is continuous from $L_{q,\mu}(\Omega)$ to $L_{q',\mu}(\Omega)$, $ {1}/{q}+{1}/{q'}=1 $. It is sufficient for this that $b$ be a Carathéodory function and
$$
\begin{equation}
|b(x,r)|^{q'}\leqslant C(|r|^q+G_\mu(x)), \qquad r\in \mathbb{R}, \quad G_\mu\in L_{1,\mu}(\Omega).
\end{equation}
\tag{4.15}
$$
For a nonnegative measure $\mu\in\mathbb{M}_s (\Omega)$ and $1\leqslant q<\theta p/(n-p)$, $\theta=n(1- 1/s)$, consider the bounded operator $K\colon W_p^1(\Omega)\to L_{q',\mu} (\Omega)$ acting by formula (4.13). It is obvious that
$$
\begin{equation}
|\langle K(u),v\rangle|\leqslant C(\|u\|_{W_p^1(\Omega)}+C_1)\|v\|_{W_p^1(\Omega)}.
\end{equation}
\tag{4.16}
$$
Thus, as a map from $W_p^1(\Omega)$ to $(W_p^1(\Omega))^*$, $K$ is a compact operator. Next consider the hyperbolic space $\mathbb{H}^n$. Recall that the geodesic distances $\rho'$ between pairs of points in $\mathfrak{B}_{\sigma}$ are estimated in terms of the Euclidean distances between the corresponding points in $B_{r}$: $C^{-1}r'\leqslant \rho'\leqslant Cr'$. Hence the Morrey class $\mathbb{M}_s(\mathfrak{B}_{\sigma})$ of measures on $\mathbb{H}^n$ can also be viewed as a Morrey class $\mathbb{M}_s({B}_{r})$ of measures on the ball $B_{r}$ in Euclidean space. Then, under the assumption (4.15), for $\mu\in\mathbb{M}_s(\mathfrak{B}_{\sigma})$ it is easy to see that the operator
$$
\begin{equation*}
K\colon W_p^1(\mathfrak{B}_{\sigma})\to L_{q',\mu} (\mathfrak{B}_{\sigma}), \qquad q<\frac{n\theta}{n-p}, \quad \theta=n\biggl(1-\frac 1s\biggr),
\end{equation*}
\notag
$$
that acts by the formula
$$
\begin{equation}
\langle K(u),v\rangle=\int_{\mathfrak{B}_{\sigma}}b(x,u(x))v(x)\,d\mu
\end{equation}
\tag{4.17}
$$
is compact. By (3.1), $K$ is compact as a map from $V$ to $V^*$. Set
$$
\begin{equation*}
\begin{gathered} \, f^m(x)=T_m(f(x))\chi_{m}(x), \\ \chi_{m}(x)= \begin{cases} 1 & \text{for } x\in \mathfrak{B}_m, \\ 0 & \text{for } x\notin \mathfrak{B}_m, \end{cases} \end{gathered}
\end{equation*}
\notag
$$
and Clearly, for $ r\in \mathbb{R}$
$$
\begin{equation}
|b_i^m(x,r)|\leqslant |b_i(x,r)|\quad\text{and} \quad |b_i^m(x,r)|\leqslant m\chi_{m}(x), \qquad x\in \mathbb{H}^n.
\end{equation}
\tag{4.18}
$$
In addition, using (2.19) we see that
$$
\begin{equation}
b_i^m(x,r)r\geqslant 0, \qquad x\in \mathbb{H}^n, \quad r\in \mathbb{R}.
\end{equation}
\tag{4.19}
$$
Using inequality (3.1) it is easy to show that $f^m\in V^*$,
$$
\begin{equation}
f^m\to f \quad \text{in } L_1(\mathbb{H}^n), \qquad m\to\infty,
\end{equation}
\tag{4.20}
$$
and, furthermore,
$$
\begin{equation}
|f^m(x)|\leqslant |f(x)|\quad\text{and} \quad |f^m(x)|\leqslant m\chi_{m}(x), \qquad x\in \mathbb{H}^n, \quad m\in\mathbb{N}.
\end{equation}
\tag{4.21}
$$
The operator $\mathcal{B}_m \colon \ V\to V^*$ acts by the formula
$$
\begin{equation*}
\langle\mathcal{B}_m u,v\rangle=\int_{\mathbb{H}^n}b_0^m(x,u)v\,d\nu +\int_{\mathbb{H}^n}b_1^m(x,u)v\,d\mu=\langle K_0 u,v\rangle+\langle K_1 u,v\rangle.
\end{equation*}
\notag
$$
Using (4.18) and the above arguments we can easily prove that $\mathcal{B}_m $ is compact. Consider the equation
$$
\begin{equation}
-\operatorname{div}_g a^m(x,u,du)+\mathcal{B}_mu=f^m(x), \qquad x\in \mathbb{H}^n, \quad m\in\mathbb{N},
\end{equation}
\tag{4.22}
$$
where $a^m(x,r,y)=a(x,T_m(r),y)$. A weak solution of the Dirichlet problem for equation (4.22) is a function $u^m\in V$ satisfying the integral identity
$$
\begin{equation}
\int_{\mathbb{H}^n}(a(x,T_m(u^m),du^m),dv) \,d\nu + \langle\mathcal{B}_m u^m,v\rangle=\langle f^m,v\rangle
\end{equation}
\tag{4.23}
$$
for each function $v\in \mathcal{D}(\mathbb{H}^n)$. It can easily be proved that relation (4.23) also holds for all $v\in V$. By Theorem 2, for each $m\in\mathbb{N}$ the Dirichlet problem for equation (4.22) has a weak solution $u^m\in V$. § 5. Existence of a solution Proof of Theorem 1. In (4.23) we set $v=T_{k,h} (u^m)=T_k(u^m-T_h(u^m))$. Taking (4.19) into account we obtain
$$
\begin{equation}
\begin{aligned} \, &\int_{\{\mathbb{H}^n \colon h\leqslant|u^m|<k+h\}}(a^m(x,u^m,du^m),du^m)\, d\nu \nonumber \\ &\qquad\qquad +k \int_{\{\mathbb{H}^n \colon |u^m|\geqslant k+h\}}|b_0^m(x,u^m)|\, d\nu+k \int_{\{\mathbb{H}^n \colon |u^m|\geqslant k+h\}}|b_1^m(x,u^m)|\,d\mu \nonumber \\ &\qquad\qquad +\int_{\{\mathbb{H}^n \colon h\leqslant|u^m|< k+h\}}b_0^m(x,u^m)u^m\biggl(1-\frac{h}{|u_m|}\biggr) \,d\nu \nonumber \\ \nonumber &\qquad\qquad +\int_{\{\mathbb{H}^n \colon h\leqslant|u^m|< k+h\}} b_1^m(x,u^m)u^m\biggl(1-\frac{h}{|u_m|}\biggr)\, d\mu \\ &\qquad\leqslant k\int_{\{\mathbb{H}^n \colon |u^m|\geqslant h\}}|f^m|\,d\nu. \end{aligned}
\end{equation}
\tag{5.1}
$$
Taking (4.19) into account and using (4.21) and (2.16), from (5.1) we deduce
$$
\begin{equation}
\begin{aligned} \, &\int_{\{\mathbb{H}^n \colon h\leqslant|u^m|<k+h\}}((a^m(x,u^m,du^m),du^m)+G(x))\,d\nu \nonumber \\ &\qquad\qquad +k\int_{\{\mathbb{H}^n \colon |u^m|\geqslant k+h\}}|b_0^m(x,u^m)| \,d\nu+ k\int_{\{\mathbb{H}^n \colon |u^m|\geqslant k+h\}}|b_1^m(x,u^m)|\,d\mu \nonumber \\ &\qquad \leqslant\int_{\{\mathbb{H}^n \colon |u^m|\geqslant h\}}(k|f|+|G|)\,d\nu, \qquad m\in\mathbb{N}. \end{aligned}
\end{equation}
\tag{5.2}
$$
Setting $h=0$ in (5.1) and using inequality (2.16) we obtain
$$
\begin{equation}
\begin{aligned} \, &\int_{\{\mathbb{H}^n \colon |u^m|<k\}}c_0 M(x,|d u^m|_g) \,d\nu \nonumber \\ &\qquad\qquad+\int_{\{\mathbb{H}^n \colon |u^m|<k\}}b_0^m(x,u^m)u^m \,d\nu +k\int_{\{\mathbb{H}^n \colon |u^m|\geqslant k\}}|b_0^m(x,u^m)| \,d\nu \nonumber \\ &\qquad\qquad+\int_{\{\mathbb{H}^n \colon |u^m|<k\}}b_1^m(x,u^m)u^m\,d\mu +k\int_{\{\mathbb{H}^n \colon |u^m|\geqslant k\}}|b_1^m(x,u^m)|\,d\mu \nonumber \\ &\qquad\leqslant (k+1)C_1, \qquad m\in\mathbb{N}. \end{aligned}
\end{equation}
\tag{5.3}
$$
It follows from (5.3) that
$$
\begin{equation}
\begin{aligned} \, \nonumber &\int_{\{\mathbb{H}^n \colon |u^m|<k\}}M(x,|du^m|_g)\,d\nu \\ &\qquad=\int_{\mathbb{H}^n}M(x,|dT_k (u^m)|_g)\,d\nu\leqslant C_1(k+1), \qquad k>1, \quad m\in\mathbb{N}. \end{aligned}
\end{equation}
\tag{5.4}
$$
Hence $T_k (u^m)\in V$ for each $k>1$, and it follows from (4.11) that
$$
\begin{equation}
\| T_k(u^m)\|_{V}^p\leqslant C_2k, \qquad m\in\mathbb{N}.
\end{equation}
\tag{5.5}
$$
Since $V$ is reflexive, we can extract a subsequence
$$
\begin{equation}
T_k (u^m)\rightharpoonup v_k\in V, \qquad m\to \infty,
\end{equation}
\tag{5.6}
$$
converging weakly in $V$. In view of (5.5) we can use Lemma 3, which yields the estimate
$$
\begin{equation}
\operatorname{meas}\{x\in\mathfrak{B}_{R}\colon |u^m(x)|\geqslant k\}\leqslant \frac{C(R)}{k^{p^*(1-p^{-1})}}, \qquad m>k>k_0.
\end{equation}
\tag{5.7}
$$
Then, taking a sufficiently large $R$ first and then selecting $k$, we obtain
$$
\begin{equation}
\begin{aligned} \, \nonumber &\int_{\{\mathbb{H}^n \colon |u^m|\geqslant k\}}(|f|+|G|)\,d\nu\leqslant\int_{\{\mathfrak{B}_R \colon |u^m|\geqslant k\}}(|f|+|G|)\,d\nu \\ &\qquad\qquad +\int_{\{\mathbb{H}^n \colon \rho(x)\geqslant R\}}(|f|+|G|)\,d\nu\leqslant\varepsilon(k), \qquad m>k, \end{aligned}
\end{equation}
\tag{5.8}
$$
where $\varepsilon(k)\to0$ as $k\to\infty$.
By (2.16) the first integral in (5.2) is nonnegative, so that for $k=1$ we can write the inequality
$$
\begin{equation}
\begin{gathered} \, \int_{\{\mathbb{H}^n \colon |u^m|\geqslant h\}}|b_0^m(x,u^m)|\,d\nu+\int_{\{\mathbb{H}^n \colon |u^m|\geqslant h\}}|b_1^m(x,u^m)|\,d\mu <C, \qquad m\in\mathbb{N}. \end{gathered}
\end{equation}
\tag{5.9}
$$
Using (2.20), for $m> \widetilde{g}(h)$ we obtain
$$
\begin{equation*}
\int_{\{\mathbb{H}^n \colon |u^m|\geqslant h\}}\widetilde{g}(h)\chi_{m}(x)\,d\mu <C.
\end{equation*}
\notag
$$
Hence for each $R>0$ we have
$$
\begin{equation}
\operatorname{meas}_\mu\{x\in\mathfrak{B}_{R}\colon |u^m(x)|\geqslant h\}\leqslant \frac{C}{\widetilde{g}(h)}, \qquad m>\widetilde{g}(h)+R.
\end{equation}
\tag{5.10}
$$
Now we establish convergence with respect to a subsequence, namely,
$$
\begin{equation}
u^m\to u \quad \nu\text{-a.e. and } \mu\text{-a.e. in } \mathbb{H}^n, \qquad m\to \infty.
\end{equation}
\tag{5.11}
$$
The sequence $T_s (u^m)$ is bounded in $V$, and by (3.1) it is bounded in $W_p^1(\mathfrak{B}_{R})$. By Kondrashov’s theorem we can extract a convergent subsequence such that $T_s (u^m) \to \widetilde{v}_{s}$ in $L_{p}(\mathfrak{B}_{R})$ as $m\to\infty$. Hence $T_s(u^m)\to \widetilde{v}_{s}$ $\nu$-almost everywhere in $\mathfrak{B}_{R}$. By (5.6) we have $v_{s}=\widetilde{v}_{s}$ $\nu$-almost everywhere in $\mathfrak{B}_{R}$. Now using the diagonal procedure with respect to $R\in \mathbb{N}$ we find a subsequence such that $T_s(u^m)\to {v}_{s}$ $\nu$-almost everywhere in $\mathbb{H}^n$. Let $\Omega$ denote the set of points in $\mathbb{H}^n$ at which the sequence $u^m(x)$ has a finite limit. We denote this limit by $ u(x)$. Then for $x\in\Omega$ we have If $\lim |T_s(u^m(x))|<s$ at some points $x$, then $\lim T_s(u^m(x))=v_s(x)=\lim u^m(x)$, that is, $x\in\Omega$. Thus, for $\nu$-almost all $x\notin\Omega$ we have for each $s>0$. In particular,
$$
\begin{equation*}
\lim T_{s+h}(u^m(x))=(s+h) \operatorname{sign}(v_{s+h}(x)).
\end{equation*}
\notag
$$
Then $|u^m(x)|>s$ for large $m$, and therefore $\lim |u^m(x)|=\infty$. By (5.7) the set of such points in $\mathfrak{B}_{R}$ has measure zero. Hence we conclude that the difference $\mathbb{H}^n\setminus \Omega$ has measure zero, which proves the convergence (5.11) regarding the measure $\nu$. Then $v_s(x)=T_s(u)$ for $\nu$-almost all $x\in \mathbb{H}^n$.
Also note that $T_s (u^m) \to v_{s}$ in $L_{q,\mu}(\mathfrak{B}_{R})$, as follows from (4.14) and (3.1). Then $T_s(u^m)\to v_{s}$ $\mu$-almost everywhere in $ \mathfrak{B}_{R}$ (along a subsequence). Next, using a diagonal procedure with respect to $R\in \mathbb{N}$ we establish the following convergence along a subsequence: $\mu$-almost everywhere in $\mathbb{H}^n$, so that we also prove (5.11).Relation (5.6) can be written as
$$
\begin{equation}
dT_k (u^m)\rightharpoonup dT_k(u) \quad \text{in } L_{M}(\mathbb{H}^n), \qquad m\to \infty.
\end{equation}
\tag{5.13}
$$
We show that for all $s>0$ and $R>0$
$$
\begin{equation}
b_0^m(x,T_s(u^m)) \to b_0(x, T_s(u)) \quad\text{in } L_{1}(\mathfrak{B}_R), \qquad m\to \infty,
\end{equation}
\tag{5.14}
$$
and
$$
\begin{equation}
b_1^m(x,T_s(u^m)) \to b_1(x, T_s(u)) \quad\text{in } L_{1,\mu}(\mathfrak{B}_R), \qquad m\to \infty.
\end{equation}
\tag{5.15}
$$
It follows from (5.11) that
$$
\begin{equation}
b_0(x,T_s(u^m)) \to b_0(x, T_s(u)) \quad \nu\text{-a.e. in } \mathbb{H}^n, \qquad m\to \infty.
\end{equation}
\tag{5.16}
$$
By (2.18) we have $|b_0^m(x,T_s(u^m))|\leqslant \operatorname{g}(s)\widetilde{G_0}(x)\in L_{1}(\mathfrak{B}_R)$, so that (5.14) is a consequence of Lebesgue’s theorem. In a similar way we prove (5.15).
At this step we establish the strong convergence
$$
\begin{equation}
dT_k(u^m)\to dT_k(u) \quad \text{in } L_{M,\mathrm{loc}}(\mathbb{H}^n), \qquad m\to \infty.
\end{equation}
\tag{5.17}
$$
From (5.4) and (2.15), for each $k>1$ we obtain the estimate
$$
\begin{equation}
\|a(x,T_k(u^m),d T_k(u^m))\|_{ \overline{M},\mathbb{H}^n}\leqslant C_5(k), \qquad m\in\mathbb{N}.
\end{equation}
\tag{5.18}
$$
Let $k>0$, $h>k+1$ and Then by (5.12) we have
$$
\begin{equation}
z^m\to 0 \quad \text{$\nu$-a.e. in } \mathbb{H}^n, \qquad m\to \infty,
\end{equation}
\tag{5.19}
$$
and also We pointed out above that
$$
\begin{equation}
\forall\, R>0\colon \ |z^m| \to 0\quad\text{in } L_{q,\mu}(\mathfrak{B}_{R}),\qquad m\to \infty.
\end{equation}
\tag{5.21}
$$
Using (5.19) and (5.20), from Lemma 8 we derive the convergence
$$
\begin{equation}
|z^m| \stackrel{*}\rightharpoonup 0 \quad\text{in } L_{\infty}(\mathbb{H}^n), \qquad m\to \infty.
\end{equation}
\tag{5.22}
$$
By Lemma 5, for $F\in L_M(\mathbb{H}^n)$ we have
$$
\begin{equation}
Fz^m\to 0 \quad\text{in } L_M(\mathbb{H}^n), \qquad m\to \infty.
\end{equation}
\tag{5.23}
$$
Set $\eta_R(r)=\min(1,\max(0,R+1-r))$. Substituting $z^m\eta_R(\rho(x))\eta_{h-1}(|u^m|)$ into (4.23) as a test function we obtain
$$
\begin{equation}
\begin{aligned} \, I_1^{mh} &=\int_{\mathbb{H}^n}\bigl(a(x,T_h(u^m),dT_h(u^m)),d(\eta_R(\rho(x)) z^m\eta_{h-1}(|u^m|))\bigr)\,d\nu \nonumber \\ &=-\int_{\mathbb{H}^n} z^m\eta_R(\rho(x))\eta_{h-1}(|u^m|)b_0^m(x,u^m)\,d\nu \nonumber \\ &\qquad - \int_{\mathbb{H}^n}b_1^m(x,u^m)z^m\eta_R(\rho(x))\eta_{h-1}(|u^m|)\,d\mu \nonumber \\ &\qquad +\int_{\mathbb{H}^n} f^mz^m\eta_R(\rho(x))\eta_{h-1}(|u^m|)\,d\nu \nonumber \\ &=I_2^{mh}+I_3^{mh}+I_4^{mh},\qquad m\in\mathbb{N}. \end{aligned}
\end{equation}
\tag{5.24}
$$
Estimates for the integrals $I_2^{mh}-I_4^{mh}$. By the inequality
$$
\begin{equation*}
\eta_R(\rho(x))\eta_{h-1}(|u^m|)|b_0^m(x,u^m)|\leqslant \operatorname{g}(h) \widetilde{G}_0(x),
\end{equation*}
\notag
$$
which follows from (2.18), and by the convergence (5.19), from Lebesgue’s theorem we obtain
$$
\begin{equation}
|I_2^{mh}|\leqslant\int_{\mathfrak{B}_{R+1}}|z^m|\operatorname{g}(h) \widetilde{G}_0(x)\,d\nu=\varepsilon_{1}(m,h).
\end{equation}
\tag{5.25}
$$
Here and in what follows In a quite similar way, Using (4.21), (5.19) and (5.20) we obtain
$$
\begin{equation}
|I_4^{mh}|\leqslant\int_{\mathbb{H}^n} |fz^m|\,d\nu=\varepsilon_{3}(m).
\end{equation}
\tag{5.28}
$$
It is obvious that $z^mu^m\geqslant0 $ for $|u^m|\geqslant k$, so for $|u^m|\geqslant h-1> k$ we have $z^m|u^m|=|z^m|u^m$. Using this and (5.20) we obtain the estimate
$$
\begin{equation*}
\begin{aligned} \, I_{12}^{mh} &=-\int_{\{\mathbb{H}^n\colon h-1\leqslant |u^m|<h\}}\bigl(a(x,T_h(u^m),dT_h(u^m)),d|u^m| \bigr) \eta_R(\rho(x)) z^m\,d\nu \\ &=-\int_{\{\mathbb{H}^n\colon h-1\leqslant |u^m|<h\}}\bigl((a(x,u^m,du^m),du^m )+G(x)\bigr) \eta_R(\rho(x)) |z^m|\,d\nu \\ &\qquad + \int_{\{\mathbb{H}^n\colon h-1\leqslant |u^m|<h\}}G(x) \eta_R(\rho(x)) |z^m|\,d\nu. \end{aligned}
\end{equation*}
\notag
$$
In view of (5.2) and (5.8) we see that
$$
\begin{equation}
|I_{12}|^{mh}\leqslant\varepsilon(h), \qquad m\geqslant \widetilde{g}(h)+R,
\end{equation}
\tag{5.29}
$$
where $\varepsilon(h)\to 0$ as $h\to\infty$.
Now, using (5.18) and the inequality $|d\eta_R(\rho(x))|_g\leqslant1$ we obtain
$$
\begin{equation}
\begin{gathered} \, I^{mh}_{13} =\int_{\{\mathbb{H}^n\colon |u^m|<h\}}\bigl(a(x,T_h(u^m),d T_h(u^m)),d\eta_R(\rho(x))\bigr)\eta_{h-1}(|u^m|)z^m\,d\nu, \nonumber \\ |I^{mh}_{13}|\leqslant C_7(h)\|z^m\|_{M,\mathfrak{B}_{R+1}}=\varepsilon_5(m,h). \end{gathered}
\end{equation}
\tag{5.30}
$$
It is easy to show that $I^{mh}_1=I^{mh}_{11}+I^{mh}_{12}+I^{mh}_{13}$, where
$$
\begin{equation*}
I^{mh}_{11}=\int_{\mathbb{H}^n}\bigl(a(x,T_h(u^m),dT_h(u^m)),dz^m\bigr) \eta_R(\rho(x))\eta_{h-1}(|u^m|)\,d\nu.
\end{equation*}
\notag
$$
From (5.25)–(5.30) we can conclude that
$$
\begin{equation}
|I^{mh}_{11}|\leqslant\varepsilon_6(m,h)+\varepsilon(h), \qquad m> \widetilde{g}(h)+R.
\end{equation}
\tag{5.31}
$$
After elementary transformations we arrive at the equalities
$$
\begin{equation*}
\begin{aligned} \, I^{mh}_{11} &=\int_{\mathbb{H}^n}\bigl(a(x,T_h(u^m),dT_h(u^m)),dT_k(u^m)\bigr) \eta_R(\rho(x))\eta_{h-1}(|u^m|)\,d\nu \\ &\qquad -\int_{\mathbb{H}^n}\bigl(a(x,T_h(u^m),dT_h(u^m)),dT_k(u)\bigr) \eta_R(\rho(x))\eta_{h-1}(|u^m|)\,d\nu \\ &=\int_{\mathbb{H}^n}\bigl(a(x,T_k(u^m),dT_k(u^m)),dT_k(u^m)\bigr) \eta_R(\rho(x))\,d\nu \\ &\qquad -\int_{\mathbb{H}^n}\bigl(a(x,T_h(u^m),dT_h(u^m)),dT_k(u) \eta_R(\rho(x))\eta_{h-1}(|u^m|)\bigr)\,d\nu \\ &=\int_{\mathbb{H}^n}\bigl(a(x,T_k(u^m),d T_k(u^m)),dz^m\bigr) \eta_R(\rho(x))\,d\nu \\ &\qquad +\int_{\mathbb{H}^n}\bigl(a(x,T_k(u^m),dT_k(u^m)),dT_k(u)\bigr) \eta_R(\rho(x))\,d\nu \\ &\qquad -\int_{\mathbb{H}^n}\bigl(a(x,T_h(u^m),d T_h(u^m)),dT_k(u)\bigr) \eta_R(\rho(x))\eta_{h-1}(|u^m|)\,d\nu. \end{aligned}
\end{equation*}
\notag
$$
The integrands in the last two integrals coincide for $|u^m|< k$, so it is obvious that
$$
\begin{equation}
\begin{aligned} \, \nonumber I^{mh}_{11} &=\int_{\mathbb{H}^n}\bigl(a(x,T_k(u^m),dT_k(u^m)),dz^m\bigr)\eta_R(\rho(x))\,d\nu \\ &\qquad +\int_{\{\mathbb{H}^n\colon |u^m|\geqslant k\}}\bigl(a(x,T_k(u^m),d T_k(u^m)) \nonumber \\ &\qquad\qquad-\eta_{h-1}a(x,T_h(u^m),dT_h(u^m)),dT_k(u)\bigr)\eta_R(\rho(x))\,d\nu \nonumber \\ &=I_{5}^m+I_{6}^{mh}. \end{aligned}
\end{equation}
\tag{5.32}
$$
In view of (5.18) we can establish the inequality
$$
\begin{equation}
\begin{gathered} \, |I^{mh}_{6}|\leqslant C_{9}(h,k)\|\chi_{\{\mathfrak{B}_{R+1}\colon |u^m|\geqslant k\}}\, d T_k(u)\|_{M}. \end{gathered}
\end{equation}
\tag{5.33}
$$
Using Lemma 2 we select $k$ such that $\operatorname{meas}\{x\in\mathfrak{B}_{R+1}\colon |u(x)|= k\}=0$. Then by (5.11) we have
$$
\begin{equation*}
\chi_{\{\mathfrak{B}_{R+1}\colon |u^m|\geqslant k\}}\, dT_k(u)\to \chi_{\{\mathfrak{B}_{R+1}\colon |u|\geqslant k\}}\, dT_k(u)= 0 \quad \text{$\nu$-a.e. in } \mathfrak{B}_{R+1}, \quad m\to\infty.
\end{equation*}
\notag
$$
In view of (5.13) we have $M(x,|d T_k(u)|_g)\in L_1(\mathbb{H}^n)$, and from Lebesgue’s theorem we deduce that
$$
\begin{equation*}
\chi_{\{\mathfrak{B}_{R+1}\colon |u^m|\geqslant k\}}\, dT_k(u)\to 0 \quad \text{in } L_M(\mathbb{H}^n), \qquad m\to\infty.
\end{equation*}
\notag
$$
Hence from (5.33) we obtain
It follows from (5.31), (5.32) and (5.34) that
$$
\begin{equation}
|I^m_{5}|\leqslant \varepsilon_9(m,h)+\varepsilon(h).
\end{equation}
\tag{5.35}
$$
Using the notation
$$
\begin{equation*}
q^m(x)=\Lambda\bigl(x,T_k(u^m),dT_k(u^m),dT_k(u)\bigr),
\end{equation*}
\notag
$$
from Lemma 6 we obtain
$$
\begin{equation*}
\begin{aligned} \, 0 &\leqslant \int_{\mathbb{H}^n}q^m(x) \eta_R(\rho(x))\,d\nu=I^{m}_{5}-I_{54}^m \\ &=I^{m}_{5}-\int_{\mathbb{H}^n}\eta_R(\rho(x)) \bigl(a(x,T_k(u^m),dT_k(u)),d(T_k(u^m)-T_k(u))\bigr) \,d\nu. \end{aligned}
\end{equation*}
\notag
$$
Similarly to how we deduced (3.19), using (5.12) we can show that
$$
\begin{equation*}
\begin{aligned} \, &\eta_R(\rho(x))a(x,T_k(u^m),dT_k (u)) \\ &\qquad \to \eta_R(\rho(x)) a(x,T_k(u),dT_k(u)) \quad\text{in } L_{\overline{M}}(\mathbb{H}^n), \qquad m\to \infty. \end{aligned}
\end{equation*}
\notag
$$
In combination with (5.13), this yields
$$
\begin{equation}
I^m_{54}=\int_{\mathbb{H}^n}\bigl(a(x,T_k(u^m),dT_k(u)),d(T_k(u^m)-T_k(u))\bigr) \eta_R(\rho(x))\,d\nu=\varepsilon_{10}(m,h).
\end{equation}
\tag{5.36}
$$
From (5.35) and (5.36) we obtain
$$
\begin{equation*}
\int_{\mathbb{H}^n}q^m(x)\eta_R(\rho(x)) \,d\nu\leqslant \varepsilon_{11}(m,h)+\varepsilon(h).
\end{equation*}
\notag
$$
Using (5.26) and taking the limit in the last inequality, first as $m\to\infty$ and then as $h\to\infty$, we establish the relation
$$
\begin{equation*}
\lim_{m\to \infty}\int_{\mathfrak{B}_R}q^m(x) \,d\nu=0,
\end{equation*}
\notag
$$
where positive $R$ and $ k$ can be arbitrary.
By Lemma 7 for $w^j=T_{k}(u^j)$ and $w=T_{k}(u)$, taking (5.12) into account we have (5.17). In view of (3.18),
$$
\begin{equation}
dT_k(u^m)\to dT_k(u) \quad \text{$\nu$-a.e. in } \mathbb{H}^n, \qquad m\to\infty.
\end{equation}
\tag{5.37}
$$
To prove (2.21) consider the test function ${v = T_k(u^m-\xi)}$ in (4.23), where $\xi\in C_0^1(\mathbb{H}^n)$; then we obtain
$$
\begin{equation}
\begin{aligned} \, \nonumber &\int_{\mathbb{H}^n} \bigl(a(x,T_m(u^m),du^m),dT_k(u^m-\xi)\bigr)\,d\nu \\ \nonumber &\qquad\qquad +\int_{\mathbb{H}^n} \bigl(b_0^m(x,u^m)-f^m\bigr)T_k(u^m-\xi)\,d\nu +\int_{\mathbb{H}^n} b_1^m(x,u^m)T_k(u^m-\xi)\,d\mu \\ &\qquad=I^m+J^m=0. \end{aligned}
\end{equation}
\tag{5.38}
$$
Set $\mathbf{M}=k+\|\xi\|_{\infty}$. If $|u^m|\geqslant \mathbf{M}$, then $|u^m-\xi|\geqslant |u^m|-\|\xi\|_{\infty}\geqslant k$, and therefore $\{\mathbb{H}^n\colon |u^m-\xi|< k\}\subseteq \{\mathbb{H}^n\colon |u^m|< \mathbf{M}\}$. Hence
$$
\begin{equation}
\begin{aligned} \, I^m &=\int_{\mathbb{H}^n} \bigl(a(x,T_m(u^m),du^m),dT_k(u^m-\xi)\bigr)\,d\nu \nonumber \\ &=\int_{\mathbb{H}^n} \bigl(a(x,T_{\mathbf{M}}(u^m),d T_{\mathbf{M}}(u^m)),dT_k(u^m-\xi)\bigr)\,d\nu \nonumber \\ &=\int_{\mathbb{H}^n} \bigl(a(x,T_{\mathbf{M}}(u^m),dT_{\mathbf{M}}(u^m)),dT_{\mathbf{M}}(u^m)\,{-}\,d\xi\bigr) \chi_{\{\mathbb{H}^n\colon |u^m-\xi|<k\}}\,d\nu, \quad m\geqslant \mathbf{M}. \end{aligned}
\end{equation}
\tag{5.39}
$$
Let $w^m=u^m-\xi$ and $w=u-\xi$. Since on the set where $|w^m|\to k$ as $m\to\infty$ we have $|w|= k$, it follows that $d w=0$. Therefore,
$$
\begin{equation}
\begin{aligned} \, &dT_k(w^m)-dT_k(w) =\chi_{\{\mathbb{H}^n\colon |w^m|<k\}}(dw^m-d w) \nonumber \\ &\qquad +\bigl(\chi_{\{\mathbb{H}^n\colon |w^m|<k\}}-\chi_{\{\mathbb{H}^n\colon |w|<k\}}\bigr)\,dw\to 0 \quad \text{$\nu$-a.e. in } \mathbb{H}^n, \qquad m\to\infty. \end{aligned}
\end{equation}
\tag{5.40}
$$
Using inequality (2.9) and conditions (2.15) and (2.16) for $\varepsilon\in (0,1)$ we find that
$$
\begin{equation*}
\begin{aligned} \, &\bigl(a(x,T_{\mathbf{M}}(u^m),dT_{\mathbf{M}}(u^m)), dT_{\mathbf{M}}(u^m)-d\xi\bigr)\chi_{\{\mathbb{H}^n\colon |u^m-\xi|< k\}} \\ &\qquad \geqslant(c_0-\varepsilon \operatorname{g}(\mathbf{M}))M(x,|dT_{\mathbf{M}}(u^m)|_g)-C_1 G(x)-M(x,\varepsilon^{-1} |d\xi|_g). \end{aligned}
\end{equation*}
\notag
$$
Taking $\varepsilon<c_0/\operatorname{g}(\mathbf{M})$ we obtain the inequality
$$
\begin{equation*}
\begin{aligned} \, &\bigl(a(x,T_{\mathbf{M}}(u^m),dT_{\mathbf{M}}(u^m)),d( T_M(u^m)-\xi)\bigr)\chi_{\{\mathbb{H}^n\colon |u^m-\xi|<k\}} \\ &\qquad \geqslant -C_{10}(G(x)+M (x,|d\xi|_g))\in L_1(\mathbb{H}^n). \end{aligned}
\end{equation*}
\notag
$$
From (5.40), (5.12) and (5.37), since $a(x,r,y)$ is continuous in $(r,y)$, by Fatou’s lemma we obtain
$$
\begin{equation}
\begin{aligned} \, \lim_{m\to \infty}\inf I^m &\geqslant \int_{\mathbb{H}^n}\bigl(a(x,T_{\mathbf{M}}(u),dT_{\mathbf{M}}(u)),dT_k(u-\xi)\bigr)\,d\nu \nonumber \\ &=\int_{\mathbb{H}^n}\bigl(a(x,u,du),dT_k( u-\xi)\bigr)\,d\nu\geqslant C_I. \end{aligned}
\end{equation}
\tag{5.41}
$$
By (5.11), using Lemma 8 we have
$$
\begin{equation}
T_k(u^m-\xi)\stackrel {*}\rightharpoonup T_k(u-\xi) \quad \text{in } L_{\infty}(\mathbb{H}^n), \qquad m\to \infty.
\end{equation}
\tag{5.42}
$$
We split the integral $J^m$ into two. The first term
$$
\begin{equation*}
J^m_1=\int_{\mathbb{H}^n}b_0^m(x,u^m)T_k(u^m-\xi)\,d\nu+ \int_{\mathbb{H}^n}b_1^m(x,u^m)T_k(u^m-\xi)\,d\mu
\end{equation*}
\notag
$$
can be estimated as follows. Using (4.20) and (5.42) and taking the limit as $m\to\infty$ we see that
$$
\begin{equation}
J^m_2= \int_{\Omega}f^m T_k(u^m-\xi)\,dx\to \int_{\Omega}f T_k(u-\xi)\,dx=C_{J_2}.
\end{equation}
\tag{5.43}
$$
Now from (5.38) we obtain
Let $\operatorname{supp}\xi\subset \mathfrak{B}_{l_0}$, $l\geqslant l_0$, $\mathfrak{B}_{l,s}^m=\{x\in\mathfrak{B}_{l}\colon \ |u^m(x)|< s\}$, $s\geqslant \mathbf{M}$ and $\mathfrak{B}_{l,s}=\{x\in\mathfrak{B}_{l}\colon \ |u(x)|< s\}$. We choose $s$ so that $\operatorname{meas} \{x\in\mathfrak{B}_{l}\colon \ |u(x)|= s\}=0$. Then taking (4.19) and the inequality $u^m(x)T_k(u^m-\xi)\geqslant0$ for $|u^m(x)|>\mathbf{M}$ into account we see that
$$
\begin{equation*}
\begin{aligned} \, J^m_1 &=\int_{\mathbb{H}^n\backslash \mathfrak{B}_{l,s}^m}b_0^m(x,u^m)T_k(w^m)\,d\nu+\int_{\mathbb{H}^n\backslash \mathfrak{B}_{l,s}^m}b_1^m(x,u^m)T_k(w^m)\,d\mu \\ &\qquad +\int_{\mathfrak{B}_{l,s}^m}b_0^m(x,u^m)T_k(w^m)\,d\nu+ \int_{\mathfrak{B}_{l,s}^m}b_1^m(x,u^m)T_k(w^m)\,d\mu \\ &\geqslant\int_{\mathfrak{B}_{l,s}^m}b_0^m(x,T_s(u^m))T_k(w^m)\,d\nu+ \int_{\mathfrak{B}_{l,s}^m}b_1^m(x,T_s(u^m))T_k(w^m)\,d\mu= \overline{J}^{\,lm}_1. \end{aligned}
\end{equation*}
\notag
$$
Recalling (5.14), (5.15) and (5.42) and taking the limit as $m\to \infty$ we obtain
$$
\begin{equation*}
\begin{aligned} \, &\int_{\mathfrak{B}_{l,s}}b_1(x,T_s(u))T_k(u-\xi)\,d\mu +\int_{\mathfrak{B}_{l,s}}b_0(x,T_s(u))T_k(u-\xi)\,d\nu \\ &\qquad= \lim_{m\to \infty}\overline{J}^{\,lm}_1\leqslant \liminf_{m\to \infty} J_1^m. \end{aligned}
\end{equation*}
\notag
$$
Since by (2.19) we have
$$
\begin{equation*}
\int_{\mathfrak{B}_{l,s}\setminus \mathfrak{B}_{l_0,s}}b_0(x,T_s(u))T_k(u-\xi)\,d\nu= \int_{\mathfrak{B}_{l,s}\setminus \mathfrak{B}_{l_0,s}}|b_0(x,T_s(u))T_k(u)|\,d\nu,
\end{equation*}
\notag
$$
Beppo Levi’s theorem shows that we can take the limit as $l\to \infty$. Setting $\mathbb{H}_{s}=\{x\in\mathbb{H}^n\colon \ |u(x)|< s\}$ and taking the limit as $l\to \infty$ we obtain
$$
\begin{equation*}
\int_{\mathbb{H}_{s}}b_1(x,u)T_k(u-\xi)\,d\mu +\int_{\mathbb{H}_{s}}b_0(x,u)T_k(u-\xi)\,d\nu\leqslant \liminf_{m\to \infty} J_1^m.
\end{equation*}
\notag
$$
Since
$$
\begin{equation*}
\int_{\mathbb{H}_{s}\setminus\mathbb{H}_{M}}b_1(x,u)T_k(u-\xi)\,d\mu= \int_{\mathbb{H}_{s}\setminus\mathbb{H}_{M}}|b_1(x,u)T_k(u-\xi)|\,d\mu
\end{equation*}
\notag
$$
by (2.19), we can take the limit as $s\to \infty$. As a result,
$$
\begin{equation}
\int_{\mathbb{H}^n}b_1(x,u)T_k(u-\xi)\,d\mu +\int_{\mathbb{H}^n}b_0(x,u)T_k(u-\xi)\,d\nu\leqslant \liminf_{m\to \infty} J_1^m.
\end{equation}
\tag{5.44}
$$
Combining (5.38), (5.41), (5.44) and (5.43) we deduce (2.21). Theorem 1 is proved. 
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