---
author:
- |
    [Hu Xiaohong$^{1,2}$ ]{} and [^1] [^2]\
    [$^{1}$Department of Mathematics, Sichuan University,]{}\
    [Chengdu 610064, PR China]{}\
    [$^{2}$ Department of Mathematics and Physics,]{}\
    [Chongqing University of Posts and Telecommunications,]{}\
    [Chongqing 400065, PR China]{}\
title: ' A New Extension of Serrin’s Lower Semicontinuity Theorem '
---

In this paper, we present a new extension of the famous Serrin’s lower semicontinuity theorem for the variational functional $\int_{\Omega}f(x,u,u')dx$,we prove its lower semicontinuity in $W_{loc}^{1,1}(\Omega)$ with respect to the strong $L_{loc}^{1}$ topology assuming that the integrand $f(x,s,\xi)$ has the usual continuity on all the three variables and the convexity property on the variable $\xi$ and the local absolute continuity on the variable $x$.\
[**Keyword:** ]{} Lower semicontinuity, Serrin’s theorem, strong convergence in $L^{1}$, convex function, local absolute continuity.\
[**2002 Mathematics Subject Classification:**]{} Primary 49J45, Secondary 52A41. 

Introduction and Main Results
=============================

The aim of this paper is to give some new sufficient conditions for lower semicontinuity with respect to the strong convergence in $L_{loc}^{1}$ for functionals of integral type $$F(u,\Omega)=\int_{\Omega}f(x,u(x),Du(x))dx,\label{(1.1)}$$ where $\Omega$ is an open set of $R^{n}$, $u$ is in the Sobolev Space$^{[1]}$ $W_{loc}^{1,1}(\Omega)=\{u:u\in L^{1}(K),Du\in
L^{1}(K), \forall K\subset\subset\Omega\}$, $Du$ denotes the generalized gradient of $u$, and the integrand $f(x,s,\xi):\Omega
\times R \times R^{n}\rightarrow
[0,\infty)$ satisfies the following conditions:\
(H1) $f$ is continuous in $\Omega \times R \times R^{n}$ and $f(x,s,\xi)$ is convex in $\xi\in R^{n}$ for all $(x,s)\in \Omega
\times R$.\
The integral functional $F$ is called lower semicontinuous in $W_{loc}^{1,1}(\Omega)$ with respect to the strong convergence in $L_{loc}^{1}$, if for every $u_{m},u \in W_{loc}^{1,1}(\Omega)$ such that $u_{m}\rightarrow u $ in $L_{loc}^{1}$ (where $u_{m}\rightarrow u $ in $L_{loc}^{1}$ means $\|u_{m}-u\|_{L^{1}(K)}\rightarrow 0 \ as \ m\rightarrow +\infty, \forall K\subset\subset\Omega$), then $$\liminf_{m\rightarrow+\infty}F(u_{m},\Omega)\geq
F(u,\Omega).\label{(1.2)}$$ It is well known that condition (H1) alone is not sufficient for strong lower semicontinuity of the integral $F$ in (1.1) (see book \[12\]).  In addition to (H1),Serrin published in 1961 an article$^{[13]}$ proposing some sufficient conditions for strong lower semicontinuity.  One of the most known and celebrated Serrin’s theorem on this subject is the following one.\
[**Theorem 1.1$^{[13]}$** ]{} Let f satisfy, in addition to (H1), one of the following conditions:\
(a) $f(x, s, \xi)\rightarrow +\infty$ when $|\xi|\rightarrow
+\infty$, for all $(x,s)\in \Omega\times R $;\
(b) $f(x, s, \xi)$ is strictly convex in $\xi \in R^{n}$ for all $(x,s)\in \Omega\times R $;\
(c) the derivatives $f_{x}(x,s,\xi)$, $ f_{\xi}(x,s,\xi)$ and $f_{\xi
x}(x,s,\xi)$ exist and are continuous.\
Then $F(u,\Omega)$ is lower semicontinuous in $W_{loc}^{1,1}(\Omega)$ with respect to the strong convergence in $L_{loc}^{1}$.\
The conditions (a), (b) and (c) quoted above are clearly independent, in the sense that we can find a continuous function $f$ satisfying just one of them, but none of the others .  However, the proof of Theorem 1.1 is essentially the same for every condition quoted above; indeed, the proof is based on an approximation theorem for convex functions depending continuously on parameters that can be applied, in particular, when f satisfies one of conditions (a), (b) and (c).  This fact suggests the possibility to find a suitable condition weaker than one of conditions (a), (b) and (c).  Many attempts have been made to weaken the assumptions on the integrand $f$, such as L. Ambrosio in paper \[2\], V. De Cicco in his paper \[3\] and I. Fonseca in his papers \[6\] and \[7\] proposed several generalizations of Theorem 1.1.  In the papers \[9\] and \[10\], Gori prove the following theorems:\
[**Theorem 1.2$^{[9]}$**]{} Let us assume that $f$ satisfies (H1) and also assume that, for every compact set $K \subset \Omega \times R
\times R^{n}$, there exists a constant $L = L(K)$ such that $$|f_{\xi}(x_{1},s,\xi)-f_{\xi}(x_{2},s,\xi)|\leq L|x_{1}-x_{2}|, \
\forall (x_{1},s,\xi),(x_{2},s,\xi)\in K,\label{(1.3)}$$ and, for every compact set $K_{1} \subset \Omega \times R$, there exists a constant $L_{1} = L_{1}(K_{1})$ such that\
$$|f_{\xi}(x,s,\xi)|\leq L_{1}, \ \forall (x,s)\in K_{1}, \ \forall
\xi \in R^{n},\label{(1.4)}$$ $$|f_{\xi}(x,s,\xi_{1})-f_{\xi}(x,s,\xi_{2})|\leq
L_{1}|\xi_{1}-\xi_{2}|, \ \forall (x,s)\in K_{1}, \ \forall \xi_{1},
\xi_{2}\in R^{n}.\label{(1.5)}$$ Then $F(u,\Omega)$ is lower semicontinuous in $W_{loc}^{1,1}(\Omega)$ with respect to the strong convergence in $L_{loc}^{1}$.\
[**Theorem 1.3$^{[9]}$**]{} Let f satisfy (H1) and such that, for every open set $\Omega'\times H \times K \subset \subset \Omega\times R
\times R^{n}$, there exists a constant $L=L_{\Omega'\times H \times
K}$ such that, for every $x_{1},x_{2}
\in \Omega'$, $s\in H$ and $\xi \in K$ ,\
$$|f(x_{1},s,\xi)-f(x_{2},s,\xi)|\leq L|x_{1}-x_{2}|.\label{(1.6)}$$ Then the functional $F(u,\Omega)$ is lower semicontinuous on $W_{loc}^{1,1}(\Omega)$ with respect to the $L_{loc}^{1}$ convergence.\
Condition (1.6) means that $f$ is locally Lipschitz continuous with respect to $x$, locally with respect to $(s,\xi)$ and not necessarily globally, that is, the Lipschitz constant is not uniform for $(s,\xi)\in R \times R^{n}$. This is an improvement of (c) of Serrin’s Theorem 1.1 since, when only the gradient $\nabla_{x}f$ exists and is continuous, this implies the Lipschitz continuity of $f$ with respect to $x$ on the compact subsets of $\Omega\times R \times R^{n}$.\
Then a question arises that whether there are weaker enough conditions more than locally Lipschitz continuous condition?  In this paper, we consider absolutely continuous condition.  Obviously, absolute continuity is weaker than Lipschitz continuity.  Is the local absolute continuity condition on the integrand enough for the lower semicontinuity of the variational functional?  The following theorems give a confirmed answer.\
[**Theorem 1.4** ]{} Let $\Omega \subset R$ be an open set, $f(x,s,\xi):\Omega\times R\times R\longrightarrow[0,+\infty)$ satisfy the following conditions:\
(H1) $f(x,s,\xi) $ is continuous on $\Omega\times R\times R$, $f(x,s,\xi)$ is convex in $\xi \in R$ for all $(x,s)\in \Omega\times R$;\
(H2) $f_{\xi}(x,s,\xi) $ is continuous on $\Omega\times R\times R$, and for every compact set of $\Omega\times R\times R$, $f_{\xi}(x,s,\xi) $ is absolutely continuous about $x$;\
(H3) for every compact set $K_{1}\subseteq\Omega\times R$, there exists a constant $L_{1}=L_{1}(K_{1})$, such that\
$$|f_{\xi}|\leq L_{1}, \ \forall(x,s)\in K_{1},\ \ \forall \xi\in
R,\label{(1.7)}$$ $$|f_{\xi}(x,s,\xi_{1})-f_{\xi}(x,s,\xi_{2})|\leq
L_{1}|\xi_{1}-\xi_{2}|, \ \forall(x,s)\in K_{1},\ \ \forall
\xi_{1},\xi_{2}\in R.\label{(1.8)}$$ Then the functional $F(u,\Omega)=\int_{\Omega}f(x,u(x),u'(x))dx$ is lower semicontinuous on $W_{loc}^{1,1}(\Omega)$ with respect to the strong convergence in $L_{loc}^{1}(\Omega)$.\
[**Theorem 1.5** ]{} Let $\Omega \subset R$ be an open set, $f(x,s,\xi):\Omega\times R\times R\longrightarrow[0,+\infty)$ satisfy (H1) and the following conditions:\
(H4) for every compact set $\Omega'\times H \times K\subseteq \Omega\times R \times R$, $f(x,s,\xi) $ is absolutely continuous about $x$;\
Then the functional $F(u,\Omega)$ is lower semicontinuous on $W_{loc}^{1,1}(\Omega)$ with respect to the strong convergence in $L_{loc}^{1}(\Omega)$.\

Some Lemmas
===========

In this section, we collect some preliminary definitions and lemmas (see papers \[1,4,8,11,14\]) which will be used in the sequel.\
[**Definition 2.1**]{}  Let $\Omega\subset R^n$ be on open set, we denote\
$$L^p_{loc}(\Omega )=\{u: \Omega\rightarrow R|\ \int_{\Omega^{\prime}}|u|^pdx<+\infty
,\forall \Omega^{\prime}\subset\subset\Omega\},$$ and\
$$W^{1,1}_{loc}(\Omega )=\{u|\ u\in L^1_{loc}(\Omega ), Du\in
L^1_{loc}(\Omega )\}$$ [**Remark 2.1**]{}  Notice that $u_m\rightarrow u$ in $L^1_{loc}(\Omega )$ implies that $u_m$ converges to $u$ in measure.\
[**Definition 2.2**]{}  Let $f: [a,b]\rightarrow R$ be a real function, if $\forall\varepsilon >0, \exists\delta >0$, such that for any finite disjoint open interval $\{(a_i,b_i)\}^n_{i=1}$ on $[a,b]$, when $\sum\limits^n_{i=1}(b_i-a_i)<\delta$, we have $$\sum\limits^n_{i=1}|f(b_i)-f(a_i)|<\varepsilon ,$$ then we call $f(x)$ is a absolutely continuous function on $[a,b]$.\
[**Remark 2.2**]{}  If $f(x)$ is Lipschitz continuous on $[a,b]$, then $f(x)$ is absolutely continuous on $[a,b]$.\
[**Lemma 2.1**]{}  Let $f(x)$ is a absolutely continuous function on $[a,b]$, then $f(x)$ is almost everywhere differentiable on $[a,b]$ and $f^{\prime}(x)$ is a integrable function on $[a,b]$.\
[**Lemma 2.2**]{}(Lebesgue Dominated convergence theorem)  Let $(X,\mathcal{R},\mu)$ be a measure space, $f$ and $\{f_n\}(n\geq
1)$ be measurable functions on $E\in\mathcal{R}$, if\
(1) $\{f_n\}$ is convergence in measure to $f$ on $E$;\
(2) there exists a integrable function $h(x)$ on $E$, such that $$|f_n(x)|\leq h(x),\ \ \ \ a.e.  x\in E,$$ then $f(x), f_{n}(x)(n\geq 1)$ are integrable on $E$, and $$\lim\limits_{n\rightarrow +\infty}\int_Ef_n(x)d\mu =\int_Ef(x)d\mu
.$$ [**Lemma 2.3**]{}  Let $f(x)$ be a measurable function on $E$, the $f(x)$ is Lebesgue integrable on $E$ if and only if $|f(x)|$ is Lebesgue integrable on $E$, and $$|\int_Ef(x)dx|\leq\int_E|f(x)|dx$$ [**Lemma 2.4**]{}  Let $f(x)$ be Lebesgue integrable on $E, then
\forall\varepsilon >0, \exists\delta >0$, for every measurable subset $A$ of $E$, when $m(A)<\delta$, we have $$|\int_Af(x)dx|\leq\int_A|f(x)|dx<\varepsilon .$$ [**Definition 2.3**]{}  Let function $\eta (x): R^n\rightarrow
[0,1]$ satisfies $\eta\in C^{\infty}_c(R^n), {\rm supp}(\eta
)\subseteq B(0,1), \eta (-x)=\eta (x)$ and $\int_{R^n}\eta (x)dx=1$. Giving a function $v: \Omega\rightarrow R$ and $\varepsilon>0$, we define the convolution of $v$ with step $\varepsilon$ as $$v_{\varepsilon}(x)=\eta_{\varepsilon}*v(x)=\int_{R^n}\eta_{\varepsilon}(x-y)v(y)dy=\int_{R^n}\eta_{\varepsilon}(y)v(x-y)dy,$$ where $\eta_{\varepsilon}(x)=\varepsilon^{-n}\eta\left(\frac{x}{\varepsilon}\right),
{\rm supp}(\eta )=\overline{\{x\in R^n,|\eta (x)\neq 0\}}.$\
We have the following approximations about the convolution of $v$:\
[**Lemma 2.5$^{[1]}$**]{}  Let $\Omega\subset R^n$ be an open set, and a function $v: \Omega\rightarrow R$, then\
(i) If $v\in L^1_{loc}(\Omega )$ and ${\rm
supp}(v)\subset\subset\Omega $, then $v_{\varepsilon}\in C^{\infty}_0(\Omega )$ provided $\varepsilon<dist({\rm
supp}(v),\partial\Omega)$, and $$v_{\varepsilon}\rightarrow v\ {\rm in}\ L^1_{loc}(\Omega )\ {\rm
as}\ \varepsilon\rightarrow 0^{+}.$$ (ii) If $v\in L^p(\Omega )$ where $1\leq p<+\infty $, then $v_{\varepsilon}\in L^{p}(\Omega )$.  Also $$v_{\varepsilon}\rightarrow v\ {\rm in}\ L^p_{loc}(\Omega)\ {\rm
as}\ \varepsilon\rightarrow 0^{+}.$$ A function $f: R^n\rightarrow R$ is called convex if for every $x,y\in R^n, \lambda\in (0,1)$, we have $$f(\lambda x+(1-\lambda
)y)\leq\lambda f(x)+(1-\lambda )f(y)$$.

Now, we give some properties about convex functions:\
[**Lemma 2.6$^{[4]}$.**]{}  Let $f_i: \Omega\rightarrow R,
\{f_i\}_{i\in N}$ be a sequence of functions ,\
(i) if $f_i$ is convex, then $f=\sup_{i\in N}f_i$ is also convex;\
(ii) if $f_i$ is lower semicontinuous, then $f=\sup_{i\in N}f_i$ is also lower semicontinuous.  The following approximation result was proved by De Giorgi$^{[8]}$.\
[**Lemma 2.7$^{[8]}$**]{}  Let $U\subseteq R^d$ be an open set and $f: U\times R^n\rightarrow [0,+\infty )$ be a continuous function with compact support on $U$, such that, for every $t\in U,
f(t,\cdot )$ is convex on $R^n$. Then there exists a sequence $\{\alpha_{q}\}^{\infty}_{q=1}\subseteq C^{\infty}_c(R^n),
\alpha_{q}\geq 0, \int_{R^n}\alpha_{q}dx=1, {\rm
supp}(\alpha_{q})\subseteq B(0,1)$. Let $$a_{q}(t)=\int_{R^n}f(t,\xi )\{(n+1)\alpha_{q}(\xi )+<\nabla \alpha_{q}(\xi ),\xi
>\}d\xi ,$$ and $$b_q(t)=-\int_{R^n}f(t,\xi )\nabla \alpha_q(\xi )d\xi .$$ Then $$f_j(t,\xi )=\max\limits_{1\leq q\leq j}\{0,a_q(t)+<b_q(t),\xi
>\},j\in N,$$ satisfy the following results:\
(i) for every $j\in N, f_j: U\times R^n\rightarrow [0,+\infty )$ is a continuous function with compact support on $U$ such that, $\forall t\in U, f_j(t,\cdot )$ is convex on $R^n$. Moreover, $\forall(t,\xi )\in U\times R^n, f_j(t,\xi )\leq f_{j+1}(t,\xi )$ and $$f(t,\xi )=\sup\limits_{j\in N}f_j(t,\xi )$$\
(ii) for every $j\in N$, there exists a constant $M_j>0$, such that, $\forall (t,\xi )\in U\times R^n$, $$|f_j(t,\xi )|\leq M_j(1+|\xi |),$$ and, $\forall t\in U, \forall\xi_1,\xi_2\in R^n.$ $$|f_j(t,\xi_1)-f_j(t,\xi_2)|\leq M_j|\xi_1-\xi_2|.$$

Proof of Theorem 1.4
====================

We will divide into four steps to complete the proof of theorem 1.4.\
[**Step 1:**]{}  Let $\{\beta_i(x,s)\}_{i\in N}$ be a sequence of smooth functions satisfying\
(1) there exists a compact set $\Omega^{\prime}\times
H\subset\subset\Omega\times R$, such that $\beta_i(x,s)=0, \forall
(x,s)\in (\Omega\backslash\Omega^{\prime})\times (R\backslash H),$\
(2) for every $i\in N, \beta_i(x,s)\leq\beta_{i+1}(x,s),\forall
(x,s)\in\Omega'\times H$,\
(3) $\lim\limits_{i\rightarrow +\infty}\beta_i(x,s)=1,\forall
(x,s)\in\Omega'\times H$.\
Let $$f_i(x,s,\xi )=\beta_i(x,s)f(x,s,\xi ),\ \ \ \ \ i=1,2,\cdots .$$ It is clear that, for each $i\in N, f_i$ satisfies all the hypothesis in theorem 1.4 and also vanishes if $(x,s)$ is outside $\Omega'\times H$. Thus $$\lim\limits_{i\rightarrow +\infty}f_i(x,s,\xi )=f(x,s,\xi ),\ \ \
\ \forall (x,s,\xi )\in\Omega'\times H\times R,$$ and $$f_i(x,s,\xi )\leq f_{i+1}(x,s,\xi )\leq f(x,s,\xi ),\forall i\in
N, \forall (x,s,\xi )\in\Omega'\times H\times R.$$ By Levi lemma, we have $$\lim\limits_{i\rightarrow
+\infty}\int_{\Omega'}f_i(x,s,\xi
)dx=\int_{\Omega'}f(x,s,\xi )d\xi .$$ Thus, without loss of generality, we can assume that there exists an open set $\Omega'\times H\subset\subset\Omega\times R$, such that $$f(x,s,\xi )=0,\ \ \ \ \forall (x,s,\xi )\in (\Omega
\backslash\Omega')\times (R\backslash H)\times R.\label{(3.1)}$$ Let $u_m, u\in W^{1,1}_{loc}(\Omega )$ such that $u_m\rightarrow u$ in $L^1_{loc}(\Omega )$.  We will prove that $$\liminf\limits_{m\rightarrow +\infty} F(u_m,\Omega )\geq
F(u,\Omega ).$$ Without loss of generality, we can assume that $$\liminf\limits_{m\rightarrow +\infty} F(u_m,\Omega
)=\lim\limits_{m\rightarrow +\infty}F(u_m,\Omega )<+\infty .$$ By (3.1), we have $F(u_m,\Omega )=F(u_m,\Omega'), F(u,\Omega
)=F(u,\Omega')$, thus we will only prove the following inequality: $$\lim\limits_{m\rightarrow +\infty}F(u_m,\Omega')\geq
F(u,\Omega').\label{(3.2)}$$ [**Step 2:**]{}   Let $\eta_{\varepsilon}\in C^{\infty}_c(R)$ be a mollifier and for $\epsilon>0$,define $$v_{\varepsilon}(x)=\eta_{\varepsilon}*u(x)=\int_{\Omega}\eta_{\varepsilon}(x-y)u(y)dy,\
\ \ \ x\in [\Omega_{\varepsilon}],\label{(3.3)}$$ where $\left[\Omega_{\varepsilon}\right]\triangleq \{x\in\Omega
:{\rm dist}(x,\partial\Omega )>\varepsilon\}$. We have $$\begin{aligned}
[u_{\varepsilon}(x)]'&=&[\eta_{\varepsilon}*u(x)]_x=[\int_{\Omega}\eta_{\varepsilon}(x-y)u(y)dy]_x\nonumber\\
&=&\int_{\Omega}[\eta_{\varepsilon}(x-y)]_xu(y)dy=\int_{\Omega}-[\eta_{\varepsilon}(x-y)]_yu(y)dy\label{(3.4)}\\
&=&\int_{B(x,\varepsilon
)}\eta_{\varepsilon}(x-y)[u(y)]_ydy=[u']_{\varepsilon}(x),\
\ \ \ x\in\Omega_{\varepsilon}.\nonumber\end{aligned}$$ In the following, we denote the derivative of $u_{\varepsilon}$ as $u'_{\varepsilon}$. When $u\in W^{1,1}_{loc}(\Omega )$, we know $u'\in L^1_{loc}(\Omega )$.  By Lemma 2.5, we know $u'_{\varepsilon}\in C^{\infty}_0(\Omega )$ and $$u'_{\varepsilon}\rightarrow u'\ {\rm in}\ L^1_{loc}(\Omega )\ {\rm
as}\ \varepsilon\rightarrow 0^{+},\label{(3.5)}$$ i.e., $\forall\delta >0, \ \exists\epsilon>0$, such that $$\int_{\Omega'}|u'_{\varepsilon}-u'|dx<\delta .\label{(3.6)}$$ New we estimate the difference for the integrand values on different points: $$\begin{aligned}
f(x,u_m,u'_m)-f(x,u,u')&=&f(x,u_m,u'_m)-f(x,u_m,u'_{\varepsilon})\nonumber\\
&+&f(x,u_m,u'_{\varepsilon})-f(x,u,u'_{\varepsilon})\label{(3.7)}\\
&+&f(x,u,u'_{\varepsilon})-f(x,u,u').\nonumber\end{aligned}$$ By the convexity of $f(x,s,\xi )$ with respect to $\xi$, we have $$f(x,u_m,u'_m)-f(x,u_m,u'_{\varepsilon})\geq
f_{\xi}(x,u_m,u'_{\varepsilon})\cdot(u'_m-u'_{\varepsilon}).\label{(3.8)}$$ By (\[(3.8)\]), we have $$\begin{aligned}
f(x,u_m,u'_m)-f(x,u_m,u'_{\varepsilon})&\geq & f_{\xi}(x,u_m,u'_{\varepsilon})\cdot
u'_m-f_{\xi}(x,u_m,u'_{\varepsilon})\cdot u'_{\varepsilon}\nonumber\\
&=&f_{\xi}(x,u_m,u'_{\varepsilon})\cdot
u'_m-f_{\xi}(x,u,u'_{\varepsilon})\cdot u'\nonumber\\
&+&f_{\xi}(x,u,u'_{\varepsilon})\cdot
u'-f_{\xi}(x,u_m,u'_{\varepsilon})\cdot u'_{\varepsilon}\label{(3.9)}\\
&=&f_{\xi}(x,u_m,u'_{\varepsilon})\cdot
u'_m-f_{\xi}(x,u,u'_{\varepsilon})\cdot u'\nonumber\\
&+& f_{\xi}(x,u,u'_{\varepsilon})\cdot
(u'-u'_{\varepsilon})\nonumber\\
&+& [f_{\xi}(x,u,u'_{\varepsilon})-f_{\xi}(x,u_m,u'_{\varepsilon})]\cdot
u'_{\varepsilon}.\nonumber\end{aligned}$$ By (\[(3.7)\]) and (\[(3.9)\]), we have $$\begin{aligned}
f(x,u_m,u'_m)-f(x,u,u')&\geq & f_{\xi}(x,u_m,u'_{\varepsilon}) \cdot
u'_m-f_{\xi}(x,u,u'_{\varepsilon})\cdot u'\nonumber\\
& +&f_{\xi}(x,u,u'_{\varepsilon})\cdot
(u'-u'_{\varepsilon})\nonumber\\
&+& [f_{\xi}(x,u,u'_{\varepsilon})-f_{\xi}(x,u_m,u'_{\varepsilon})]\cdot
u'_{\varepsilon}\label{(3.10)}\\
&+&f(x,u_m,u'_{\varepsilon})-f(x,u,u'_{\varepsilon})\nonumber\\
&+& f(x,u,u'_{\varepsilon})-f(x,u,u').\nonumber\end{aligned}$$ This implies $$\begin{aligned}
&&\int_{\Omega'}[f(x,u_m,u'_m)-f(x,u,u')]dx\nonumber\\
&\geq &\int_{\Omega'}[f_{\xi}(x,u_m,u'_{\varepsilon})\cdot
u'_m-f_{\xi}(x,u,u'_{\xi})\cdot
u']dx\nonumber\\
&+&\int_{\Omega'}[f_{\xi}(x,u,u'_{\varepsilon})\cdot
(u'-u'_{\varepsilon})]dx \nonumber\\
&+&\int_{\Omega'}[f_{\xi}(x,u,u'_{\varepsilon})-f_{\xi}(x,u_m,u'_{\varepsilon})]\cdot
u'_{\varepsilon}dx\label{(3.11)}\\
&+&\int_{\Omega'}[f(x,u_m,u'_{\varepsilon})-f(x,u,u'_{\varepsilon})]dx\nonumber\\
&+&\int_{\Omega'}[f(x,u,u'_{\varepsilon})-f(x,u,u')]dx.\nonumber\end{aligned}$$ [**Step 3:**]{}  Now, we estimate the right side of inequality (\[(3.11)\]).\
By (\[(1.7)\]) and (\[(3.6)\]), we have $$\int_{\Omega'}[f_{\xi}(x,u,u'_{\varepsilon})\cdot
(u'-u'_{\varepsilon})]dx\geq
-L_1\int_{\Omega'}|u'-u'_{\varepsilon}|dx\geq
-L_1\delta .\label{(3.12)}$$ Thus $$\lim\limits_{\varepsilon\rightarrow
0}\int_{\Omega'}[f_{\xi}(x,u,u'_{\varepsilon})\cdot
(u'-u'_{\varepsilon})]dx\geq 0\label{(3.13)}$$ Since $f(x,s,\xi )$ and $f_{\xi}(x,s,\xi )$ are continuous functions , they are bounded functions on compact subset.  By remark 2.1 and Lebesgue dominated convergence theorem, we obtain $$\lim\limits_{m\rightarrow
+\infty}\int_{\Omega'}[f_{\xi}(x,u,u'_{\varepsilon})-f_{\xi}(x,u_m,u'_{\varepsilon})]\cdot
u'_{\varepsilon}dx=0,\label{(3.14)}$$ and $$\lim\limits_{m\rightarrow
+\infty}\int_{\Omega'}[f(x,u_m,u'_{\varepsilon})-f(x,u,u'_{\varepsilon})]dx=0.\label{(3.15)}$$ Now, we will prove $$\lim\limits_{\varepsilon\rightarrow
0}\int_{\Omega'}[f(x,u,u'_{\varepsilon})-f(x,u,u')]dx\geq
0\label{(3.16)}$$ By lemma 2.7, there exists a sequence of non-negative continuous functions $f_j(x,s,\xi )\ (j\in N)$, such that $f_j(x,s,\xi )$ is convex on $\xi$, and $\forall (x,s,\xi )\in\Omega'\times
H\times R$, $$\begin{aligned}
&&f_j(x,s,\xi )\leq f_{j+1}(x,s,\xi ),\label{(3.17)}\\
&&f(x,s,\xi )=\sup\limits_{j\in N}f_j(x,s,\xi ),\label{(3.18)}\\
&&|f_j(x,s,\xi_1)-f_j(x,s,\xi_2)|\leq
M_j|\xi_1-\xi_2|.\label{(3.19)}\end{aligned}$$ By Levi lemma, we obtain $$\lim\limits_{j\rightarrow
+\infty}\int_{\Omega'}f_j(x,u,u'_{\varepsilon})dx=\int_{\Omega'}f(x,u,u'_{\varepsilon})dx,\label{(3.20)}$$ and $$\lim\limits_{j\rightarrow
+\infty}\int_{\Omega'}f_j(x,u,u')dx=\int_{\Omega'}f(x,u,u')dx.\label{(3.21)}$$ In order to prove (\[(3.16)\]), we only need to prove $$\lim\limits_{\varepsilon\rightarrow
0}\int_{\Omega'}[f_j(x,u,u'_{\varepsilon})-f_j(x,u,u')]dx\geq
0,\ \forall j\in N.\label{(3.22)}$$ By (\[(3.20)\]) and (\[(3.21)\]), we have $$\int_{\Omega'}[f_j(x,u,u'_{\varepsilon})-f_j(x,u,u')]dx\geq
-M_j\int_{\Omega'}|u'_{\varepsilon}-u'|dx\geq
-M_j \delta .$$ Thus (\[(3.16)\]) holds.\
[**Step 4:**]{} Now, we need to prove $$\lim\limits_{m\rightarrow
+\infty}\int_{\Omega'}[f_{\xi}(x,u_m,u'_{\varepsilon})\cdot
u'_m-f_{\xi}(x,u,u'_{\varepsilon})\cdot
u']dx=0.  \label{(3.23)}$$

Let $$\begin{aligned}
&&g(x,s)\triangleq
f_{\xi}(x,s,u'_{\varepsilon}),\label{(3.24)}\\
&&G_m(x)\triangleq\int^{u_m(x)}_{u(x)}g(x,s)ds.\label{(3.25)}\end{aligned}$$ By condition $(H_2)$, $ g(x,s)$ is a absolutely continuous function on $x$. By Lemma 2.1, $g(x,s)$ is almost everywhere differentiable, i.e. $\frac{\partial g}{\partial x}$ exists almost everywhere.  Derivating the both sides of (\[(3.25)\]), we obtain $$G^{\prime}_m(x)=g(x,u_m)\cdot u'_m-g(x,u)\cdot
u'+\int^{u_m(x)}_{u(x)}\frac{\partial g}{\partial x}dx,\ \ \
\ a.e.  x\in\Omega'.\label{(3.26)}$$ Because $G_m(x)$ vanishes outside $\Omega'$, we obtain $$\int_{\Omega'}G^{\prime}_m(x)dx=0.  \label{(3.27)}$$ By (\[(3.26)\]), we have $$\begin{aligned}
&&|\int_{\Omega'}[f_{\xi}(x,u_m,u'_{\varepsilon})\cdot
u'_m-f_{\xi}(x,u,u'_{\varepsilon})\cdot
u']dx|\nonumber\\
&=&\left|\int_{\Omega'}[g(x,u_m)\cdot
u'_m-g(x,u)\cdot
u']dx\right|\label{(3.28)}\\
&=&|-\int_{\Omega'}\int^{u_m(x)}_{u(x)}\frac{\partial
g}{\partial x}dsdx|\leq\int_{D_m}\left|\frac{\partial g}{\partial
x}\right|dxds, \nonumber\end{aligned}$$ where $$D_m=\{(x,s)\in\Omega'\times H|\ \min\{u_m(x),u(x)\}\leq
s(x)\leq\max\{u_m(x),u(x)\}\}.$$ We note $$|D_m|=|\int_{\Omega'}\int^{u_m}_udsdx|\leq\int_{\Omega'}|u_m-u|dx\rightarrow
0\ (m\rightarrow +\infty ).\label{(3.29)}$$ By Fubini theorem, we have $$\int_{\Omega\times R}|\frac{\partial g}{\partial
x}|dxds=\int_Hds\int_{\Omega'}\left|\frac{\partial
g}{\partial x}\right|dx\label{(3.30)}$$ Since $g(x,s)$ is absolutely continuous about $x$, $\frac{\partial
g}{\partial x}$ is integrable and absolutely integrable, i.e. $$\int_{\Omega'}\left|\frac{\partial g}{\partial
x}\right|dx<+\infty .\label{(3.31)}$$ Thus $$\int_{\Omega\times R}\left|\frac{\partial g}{\partial
x}\right|dxds<+\infty\label{(3.32)}$$ Because of the absolute continuity of integral, we have $$\lim\limits_{m\rightarrow +\infty}\int_{D_m}\left|\frac{\partial
g}{\partial x}\right|dxds=0\label{(3.33)}$$ By (\[(3.28)\]), we obtain $$\lim\limits_{m\rightarrow
+\infty}|\int_{\Omega'}[f_{\xi}(x,u_m,u'_{\varepsilon})\cdot
u'_m-f_{\xi}(x,u,u'_{\varepsilon})\cdot
u']dx|=0\label{(3.34)}$$ Thus we have proved (\[(3.23)\]).  By (\[(3.13)\])-(\[(3.16)\]) and (\[(3.23)\]), we have $$\lim\limits_{m\rightarrow
+\infty}\int_{\Omega'}[f(x,u_m,u'_m)-f(x,u,u')]dx\geq
0\label{(3.35)}$$ Thus we deduce that the functional $F(u,\Omega )$ is lower semicontinuous on $W^{1,1}_{loc}(\Omega )$ with respect to the strong convergence in $L^1_{loc}(\Omega )$, which does complete the proof.\

Proof of Theorem 1.5
====================

In order to prove Theorem 1.5, we will verify all the conditions in Theorem 1.4 from the assumptions in Theorem 1.5.  Now we will divide into four steps to complete the proof of Theorem 1.5:\
[**Step 1:**]{}  Similar to the first step of the proof in Theorem 1.4, without loss of generality, we assume that the integrand $f(x,s,\xi)$ vanishes outside a compact subset of $\Omega\times
R$. Thus we assume that there exists an open set $\Omega'\times
H\subset\subset\Omega\times R$, such that $$f(x,s,\xi )\equiv 0,\ \ \ \ \forall (x,s,\xi )\in
(\Omega\backslash\Omega')\times (R\backslash H)\times
R.\label{(4.1)}$$ Let $u_m,u\in W^{1,1}_{loc}(\Omega )$, such that $u_m\rightarrow u$ in $L^1_{loc}(\Omega )$, we need to prove $$\lim\limits_{m\rightarrow +\infty}F(u_m,\Omega')\geq
F(u,\Omega').\label{(4.2)}$$ By lemma 2.7, there exists a functions sequence $\{f_j(x,s,\xi
)\}_{j\in N}$, such that $\forall j\in N$, $f_j$ is a continuous function on $\Omega'\times H\subset\subset\Omega\times R$; $\forall (x,s)\in\Omega'\times H$, $f_j(x,s,\cdot )$ is convex on $R$, and $\forall (x,s,\xi )\in\Omega'\times H\times R$, $$\begin{aligned}
&&f_j(x,s,\xi )\leq f_{j+1}(x,s,\xi )\label{(4.3)}\\
&&f(x,s,\xi )=\sup\limits_{j\in N}f_j(x,s,\xi )\label{(4.4)}\\
&&|f_j(x,s,\xi_1)-f_j(x,s,\xi_2)|\leq
M_j|\xi_1-\xi_2|,(x,s)\in\Omega'\times H, \xi_1,\xi_2\in
R.\label{(4.5)}\end{aligned}$$

Let $\eta_{\varepsilon}\in C^{\infty}_c(R)(0<\varepsilon <<1)$ be a mollifier and define the $f_{j,\varepsilon}=f_j*\eta_{\varepsilon}$, i.e. $$f_{j,\varepsilon}(x,s,\xi )=\int_Rf_j(x,s,\xi
-z)\eta_{\varepsilon}(z)dz.\label{(4.6)}$$ By (\[(4.5)\]), we have $$\begin{aligned}
&&|f_{j,\varepsilon}(x,s,\xi )-f_j(x,s,\xi )|\nonumber\\
&=&|\int_Rf_j(x,s,\xi -z)\eta_{\varepsilon}(z)dz-\int_Rf_j(x,s,\xi
)\eta_{\varepsilon}(z)dz|\nonumber\\
&\leq&\int_R|f_j(x,s,\xi -z)-f_j(x,s,\xi
)|\eta_{\varepsilon}(z)dz\label{(4.7)}\\
&\leq&\int_{{\rm
supp}\eta_{\varepsilon}}M_j|z|\cdot\eta_{\varepsilon}(z)dz\leq
M_j\cdot\varepsilon.  \nonumber\end{aligned}$$ Choosing $\varepsilon =\varepsilon_j=\frac{1}{jM_j}\rightarrow 0$. By (\[(4.7)\]), we have $$|f_{j,\varepsilon_j}(x,s,\xi
)-f_j(x,s,\xi )|\leq M_j\varepsilon_j=\frac{1}{j}.\label{(4.8)}$$ So $$f_j(x,s,\xi )-\frac{2}{j}\leq f_{j,\varepsilon_j}(x,s,\xi
)-\frac{1}{j}\leq f_j(x,s,\xi )\leq f(x,s,\xi )\label{(4.9)}$$ By (\[(4.3)\]), (\[(4.4)\]) and Levi lemma, we have $$\lim\limits_{i\rightarrow
+\infty}\int_{\Omega'}f_j(x,u(x),u'(x))dx=\int_{\Omega'}f(x,u(x),u'(x))dx.\label{(4.10)}$$ Denote $$F_j(u,\Omega')=\int_{\Omega'}[f_{j,\varepsilon_j}(x,u(x),u'(x))-\frac{1}{j}]dx\label{(4.11)}$$ By (\[(4.9)\])-(\[(4.11)\]), we have $$\lim\limits_{j\rightarrow +\infty}F_j(u,\Omega')=F(u,\Omega'
)=\int_{\Omega'}f(x,u(x),u'(x))dx.\label{(4.12)}$$ Obviously, $$F_j(u,\Omega')\leq F(u,\Omega'),\ \ \ \ \forall
j\in N.$$ Thus $$\sup\limits_{j\in
N}F_j(u,\Omega')=F(u,\Omega').\label{(4.13)}$$ By Lemma 2.6, in order to prove that $F(u,\Omega')$ is lower semicontinuous on $W^{1,1}_{loc}(\Omega )$ with respect to the strong convergence in $L^{\prime}_{loc}(\Omega )$, it’s enough that we prove lower semicontinuity with respect to the strong convergence in $L^1_{loc}(\Omega )$ for the functional sequence $\{F_j(u,\Omega')\}_{j\in N}$ on $W^{1,1}_{loc}(\Omega )$ .\
[**Step 2:**]{}  In order to prove that $\forall j\in N,
F_j(u,\Omega')$ is lower semicontinuous on $W^{1,1}_{loc}(\Omega )$ with respect to the strong convergence in $L^1_{loc}(\Omega )$, we will prove that $\forall j\in N$, the integrand $f_{j,\varepsilon_j}(x,u(x),u'(x))$ satisfy all conditions of theorem 1.4.\
$\forall\xi_1,\xi_2\in R, 0<\lambda <1$, by the convexity of $f_j(x,s,\cdot )$ on $R$, we have $$f_j(x,s,\lambda\xi_1+(1-\lambda )\xi_2)\leq\lambda
f_j(x,s,\xi_1)+(1-\lambda )f_j(x,s,\xi_2).\label{(4.14)}$$ Thus $$\begin{aligned}
&&f_{j,\varepsilon_j}(x,s,\lambda\xi_1+(1-\lambda )\xi_2)\nonumber\\
&=&\int_Rf_j(x,s,\lambda\xi_1+(1-\lambda
)\xi_2-z)\eta_{\varepsilon_j}(z)dz\nonumber\\
&=&\int_Rf_j(x,s,\lambda (\xi_1-z)+(1-\lambda
)(\xi_2-z))\cdot\eta_{\varepsilon_j}(z)dz\label{(4.15)}\\
&\leq&\lambda\int_Rf_j(x,s,(\xi_1-z))\cdot\eta_{\varepsilon_j}(z)dz+(1-\lambda
)\int_Rf_j(x,s,(\xi_2-z))\cdot\eta_{\varepsilon_j}(z)dz\nonumber\\
&=&\lambda f_{j,\varepsilon_j}(x,s,\xi_1)+(1-\lambda
)f_{j,\varepsilon_j}(x,s,\xi_2).\nonumber\end{aligned}$$ Thus $f_{j,\varepsilon_j}$ satisfy $(H_1)$.\
[**Step 3:**]{}  $\forall (x,s)\in\Omega'\times H, \xi_1,\xi_2\in
R$, by (\[(4.5)\]), we have $$\begin{aligned}
&&|f_{j,\varepsilon_j}(x,s,\xi_1)-f_{j,\varepsilon_j}(x,s,\xi_2)|\nonumber\\
&=&|\int_R[f_j(x,s,\xi_1-z)\eta_{\varepsilon_j}(z)-f_j(x,s,\xi_{2}-z)\eta_{\varepsilon_j}(z)]dz|\nonumber\\
&\leq&\int_R|f_j(x,s,\xi_1-z)-f_j(x,s,\xi_2-z)|\cdot\eta_{\varepsilon_j}(z)dz\label{(4.16)}\\
&\leq&\int_{{\rm
supp}\eta_{\varepsilon}}M_j|\xi_1-\xi_2|\eta_{\varepsilon_j}(z)dz\leq
M_j|\xi_1-\xi_2|.\nonumber\end{aligned}$$ Thus $$\left|\frac{\partial f_{j,\varepsilon_j}}{\partial\xi}\right|\leq
M_j.\label{(4.17)}$$ So $f_{j,\varepsilon_j}$ satisfies (1.7) in the condition (H3) of Theorem 1.4.\
Now, we will prove $f_{j,\varepsilon_j}$ satisfies (1.8) in the condition (H3) of Theorem 1.4.  By ${\rm
supp}(\eta_{\varepsilon_j})\subseteq B(0,\varepsilon_j)$,we have

$$\begin{aligned}
\frac{\partial f_{j,\varepsilon_j}}{\partial\xi}(x,s,\xi
)&=&\int_R\frac{\partial f_j(x,s,\xi
-z)}{\partial\xi}\cdot\eta_{\varepsilon_j}(z)dz\nonumber\\
&=&-\int_R\frac{\partial f_j(x,s,\xi -z)}{\partial
z}\cdot\eta_{\varepsilon_j}(z)dz\label{(4.18)}\\
&=&\int_Rf_j(x,s,\xi
-z)\frac{\partial\eta_{\varepsilon_j}(z)}{\partial
z}dz.\nonumber\end{aligned}$$

By (\[(4.5)\]) and (\[(4.18)\]), we have $$\begin{aligned}
&&\left|\frac{\partial
f_{j,\varepsilon_j}}{\partial\xi}(x,s,\xi_1)-\frac{\partial
f_{j,\varepsilon_j}}{\partial\xi}(x,s,\xi_2)\right|\nonumber\\
&\leq&\int_R\left|f_j(x,s,\xi_1-z)-f_j(x,s,\xi_2-z)\right|\cdot\left|\frac{\partial\eta_{\varepsilon_j}(z)}{\partial
z}\right|dz\label{(4.19)}\\
&\leq&M_j|\xi_1-\xi_2|\cdot\int_R\left|\frac{\partial\eta_{\varepsilon_j}(z)}{\partial
z}\right|dz=L_jM_j|\xi_1-\xi_2|,\nonumber\end{aligned}$$ where $$L_j=\int_R\left|\frac{\partial\eta_{\varepsilon_j}(z)}{\partial
z}\right|dz\label{(4.20)}$$ is a constant depending on $\varepsilon_j$. Thus $f_{j,\varepsilon_j}$ satisfies (1.8).  So we have proved that $f_{j,\varepsilon_j}$ satisfies $(H_3)$.\
[**Step 4:**]{}  Next we will prove that $f_{j,\varepsilon_j}$ satisfies condition $(H2)$.\
By the condition $(H4)$, for every compact subset $\Omega'\times H\times K$, $f(x,s,\xi )$ is absolutely continuous about $x$, i.e. $\forall\varepsilon_0>0,\ \exists\delta >0$ such that for any finite disjoint open interval $\{(x_i,y_j)\}^n_{i=1}$ in $\Omega'$, when $\Sigma^n_{i=1}(y_i-x_i)<\delta$, we have $$\sum\limits^n_{i=1}|f(y_i,s,\xi )-f(x_i,s,\xi
)|<\varepsilon_0\label{(4.21)}$$ By Lemma 2.7, there exists continuous functions sequence $\{f_j(x,s,\xi )\}_{i\in N}$, $\forall j\in N$, $\forall
(x,s)\in\Omega'\times H, \ f_j(x,s,\cdot )$ is convex on $R$, and $\forall (x,s,\xi )\in\Omega'\times H\times R$, we have $$f_j(x,s,\xi )=\max\limits_{1\leq q\leq
j}\{0,a_q(x,s)+b_q(x,s)\xi\},\ j\in N.\label{(4.22)}$$ where $$\begin{aligned}
&&a_q(x,s)=\int_Rf(x,s,\xi )\left[2\eta_q(\xi
)+\xi\frac{\partial\eta_q(\xi )}{\partial\xi}\right]d\xi
,\label{(4.23)}\\
&&b_q(x,s)=-\int_Rf(x,s,\xi )\frac{\partial\eta_q(\xi
)}{\partial\xi}d\xi ,\label{(4.24)}\end{aligned}$$ and $\eta_q\in C^{\infty}_c(R)\ (q\in N)$ are mollifiers satisfying $\eta_q\geq 0, \ \int_R\eta_q(\xi )d\xi =1$ and ${\rm
supp}(\eta_q)\subseteq B(0,1)$, $\forall j\in N$. By (\[(4.22)\]), without of loss generality, we assume that there exists $l\in \{1,\cdots,j\}$, such that $$f_j(x,s,\xi )=a_l(x,s)+b_l(x,s)\cdot\xi\label{(4.25)}$$ where $a_l,b_l$ are given by (\[(4.23)\])-(\[(4.24)\]).  By (\[(4.21)\]), we obtain $$\begin{aligned}
&&\sum\limits^n_{i=1}|a_l(y_i,s)-a_l(x_i,s)|\nonumber\\
&=&\sum\limits^n_{i=1}|\int_R[f(y_i,s,\xi )-f(x_i,s,\xi )]\cdot
[2\eta_l(\xi )+\xi\frac{\partial\eta_l(\xi )}{\partial\xi}]d\xi
|\nonumber\\
&\leq&\int_R\sum\limits^n_{i=1}|f(y_i,s,\xi )-f(x_i,s,\xi )|\cdot
[2\eta_l(\xi )+|\xi\frac{\partial\eta_l(\xi
)}{\partial\xi}|]d\xi\label{(4.26)}\\
&\leq&\varepsilon_0\int_{B(0,1)}[2\eta_l(\xi
)+|\xi\frac{\partial\eta_l(\xi )}{\partial\xi}|]d\xi\leq
(2+A_l)\cdot\varepsilon_0,\nonumber\end{aligned}$$ where $$A_l=\int_{B(0,1)}|\frac{\partial\eta_l(\xi
)}{\partial\xi}|d\xi$$ is a constant.  Similar to the above proof, we have $$\begin{aligned}
&&\sum\limits^n_{i=1}|b_l(y_i,s)-b_l(x_i,s)|\nonumber\\
&=&\sum\limits^n_{i=1}|\int_R[f(y_i,s,\xi )-f(x_i,s,\xi
)]\cdot\frac{\partial\eta_l(\xi )}{\partial\xi}d\xi
|\nonumber\\
&\leq&\int_R\sum\limits^n_{i=1}|f(y_i,s,\xi )-f(x_i,s,\xi
)|\cdot\left|\frac{\partial\eta_l(\xi
)}{\partial\xi}\right|d\xi\label{(4.27)}\\
&\leq&\varepsilon_0\int_{B(0,1)}\left|\frac{\partial\eta_l(\xi
)}{\partial\xi}\right|d\xi\leq A_l\cdot\varepsilon_0.\nonumber\end{aligned}$$ Thus $$\begin{aligned}
 &&\sum\limits^n_{i=1}|f_j(y_i,s,\xi
)-f_j(x_i,s,\xi )|\nonumber\\
&=&\sum\limits^n_{i=1}|a_l(y_i,s)-a_l(x_i,s)+[b_l(y_i,s)-b_l(x_i,s)]\cdot\xi
|\nonumber\\
&\leq&\sum\limits^n_{i=1}|a_l(y_i,s)-a_l(x_i,s)|+\sum\limits^n_{i=1}|b_l(y_i,s)-b_l(x_i,s)|\cdot
|\xi |\label{(4.28)}\\
&\leq&(2+A_l)\varepsilon_0+A_l\varepsilon_0K_1=(2+A_l+A_lK_1)\varepsilon_0\triangleq\sigma.\nonumber\end{aligned}$$ Since $\xi$ varies on a compact set, then $K_1=\sup\limits_{\xi}\{|\xi |\}<+\infty$. Choosing $\varepsilon_0$ sufficient small so that $\sigma$ is enough small.  Thus $f_j(x,s,\xi )$ is absolutely continuous about $x$ on any compact subset of $\Omega\times R\times R$. By (\[(4.6)\]) and (\[(4.28)\]), we have $$\begin{aligned}
&&\sum\limits^n_{i=1}|f_{j,\varepsilon_j}(y_i,s,\xi
)-f_{j,\varepsilon_j}(x_i,s,\xi
)|\nonumber\\
&=&\sum\limits^n_{i=1}\left|\int_R[f_j(y_i,s,\xi
-z)\eta_{\varepsilon_j}(z)-f_j(x_i,s,\xi-
z)\eta_{\varepsilon_j}(z)]dz\right|\nonumber\\
&\leq&\int_R\sum\limits^n_{i=1}|f_j(y_i,s,\xi -z)-f_j(x_i,s,\xi
-z)|\cdot\eta_{\varepsilon_j}(z)dz\label{(4.29)}\\
&\leq&\sigma\cdot\int_{B(0,\varepsilon_j)}\eta_{\varepsilon_j}(z)dz=\sigma
.\nonumber \end{aligned}$$ By (\[(4.18)\]) and (\[(4.29)\]), we obtain $$\begin{aligned}
  &&\sum\limits^n_{i=1}\left|\frac{\partial
f_{j,\varepsilon_j}}{\partial\xi}(y_i,s,\xi)-\frac{\partial
f_{j,\varepsilon_j}}{\partial\xi}(x_i,s,\xi)\right|\nonumber\\
&\leq&\int_R\sum\limits^n_{i=1}|f_j(y_i,s,\xi -z)-f_j(x_i,s,\xi
-z)|\cdot\left|\frac{\partial\eta_{\varepsilon_j}(z)}{\partial
z}\right|dz\label{(4.30)}\\
&\leq&\sigma\int_R\left|\frac{\partial\eta_{\varepsilon_j(z)}}{\partial
z}\right|dz=L_j\sigma\nonumber\end{aligned}$$ where $L_j$ are constants depending on $\varepsilon_j$ and given by (\[(4.20)\]) ($\forall j\in N$).  By (\[(4.30)\]), for every compact subset on $\Omega\times R\times R, \frac{\partial
f_{j,\varepsilon_j}}{\partial\xi}$ is absolutely continuous about $x$. Thus $f_{j,\varepsilon_j}$ satisfy condition $(H_2)$.

Now,we have proved $f_{j,\varepsilon_j}$ satisfies all conditions in Theorem 1.4, so $F_j(u,\Omega')$ is lower semicontinuous in $W^{1,1}_{loc}(\Omega )$ with respect to the strong convergence in $L^1_{loc}(\Omega )$.  Thus $F(u,\Omega )$ has the same lower semicontinuity.  This completes the proof of Theorem 1.5.\

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[^1]: Corresponding author: huxh@cqupt.edu.cn and zhangshiqing@msn.com

[^2]: Supported partially by NSF of China.