[proofplan] We reduce the lower semicontinuity inequality to a subsequence realizing the liminf. The strong $L^p(U)$ convergence supplies the strong convergence of the value variable, while [weak convergence](/page/Weak%20Convergence) in $W^{1,p}(U)$ supplies weak convergence of the gradient variable in $L^p(U;\mathbb R^n)$. The Caratheodory and convexity hypotheses identify the integrand as a nonnegative normal integrand that is convex in the gradient variable, so the [Basic Ioffe lower semicontinuity theorem](/theorems/8733) applies directly. [/proofplan]

[step:Reduce the argument to a subsequence realizing the liminf]For every $v\in W^{1,p}(U)$, the [[Measurability of Caratheodory Compositions](/theorems/8735)][citetheorem:8735] gives that $x\mapsto f(x,v(x),\nabla v(x))$ is Lebesgue measurable on $U$. The upper $p$-growth bound and $v\in W^{1,p}(U)$ give \begin{align*} 0\le f(x,v(x),\nabla v(x))\le C(1+|v(x)|^p+|\nabla v(x)|^p) \end{align*} for $\mathcal L^n$-a.e. $x\in U$, and the right-hand side belongs to $L^1(U)$ because $U$ is bounded and $v,\nabla v\in L^p(U)$. Hence $I[v]\in[0,\infty)$ for every $v\in W^{1,p}(U)$. Define \begin{align*} \alpha:=\liminf_{k\to\infty} I[u_k]\in[0,\infty]. \end{align*} If $\alpha=+\infty$, then the desired inequality follows from $I[u]\ge 0$. Assume therefore that $\alpha<+\infty$. By the definition of the limit inferior, there exists a strictly increasing sequence $(k_j)_{j=1}^{\infty}$ of natural numbers such that \begin{align*} \lim_{j\to\infty} I[u_{k_j}]=\alpha. \end{align*} It is enough to prove \begin{align*} I[u]\le \liminf_{j\to\infty} I[u_{k_j}], \end{align*} because the right-hand side is then equal to $\alpha$.[/step]

[guided]First we check that the energies are well-defined finite numbers. Let $v\in W^{1,p}(U)$. The [Measurability of Caratheodory Compositions][citetheorem:8735] gives that $x\mapsto f(x,v(x),\nabla v(x))$ is Lebesgue measurable on $U$. The upper growth hypothesis gives \begin{align*} 0\le f(x,v(x),\nabla v(x))\le C(1+|v(x)|^p+|\nabla v(x)|^p) \end{align*} for $\mathcal L^n$-a.e. $x\in U$. Since $U$ is bounded, $\mathcal L^n(U)<\infty$; since $v\in W^{1,p}(U)$, both $v$ and $\nabla v$ belong to the corresponding $L^p$ spaces. Therefore the right-hand side is integrable over $U$, and $I[v]\in[0,\infty)$. The quantity we must compare with $I[u]$ is the limit inferior of the sequence of energies. Since every energy is nonnegative, the limit inferior belongs to $[0,\infty]$. Define \begin{align*} \alpha:=\liminf_{k\to\infty} I[u_k]. \end{align*} If $\alpha=+\infty$, then there is no estimate to prove beyond nonnegativity: the integrand $f$ takes values in $[0,\infty)$, so \begin{align*} I[u]=\int_U f(x,u(x),\nabla u(x))\,d\mathcal L^n(x)\ge 0, \end{align*} and hence $I[u]\le+\infty$. We may therefore assume $\alpha<+\infty$. The definition of $\liminf$ gives a subsequence whose energy converges to $\alpha$. More explicitly, there exists a strictly increasing sequence $(k_j)_{j=1}^{\infty}$ of natural numbers such that \begin{align*} \lim_{j\to\infty} I[u_{k_j}]=\alpha. \end{align*} Thus, if we prove the lower semicontinuity inequality along this subsequence, \begin{align*} I[u]\le \liminf_{j\to\infty} I[u_{k_j}], \end{align*} then the right-hand side equals $\alpha$, and we obtain exactly \begin{align*} I[u]\le \liminf_{k\to\infty} I[u_k]. \end{align*}[/guided]

[step:Identify the convergences needed for Ioffe lower semicontinuity] Because $u_k\rightharpoonup u$ in $W^{1,p}(U)$, for each $i\in\{1,\dots,n\}$ one has \begin{align*} \partial_{x_i}u_k\rightharpoonup \partial_{x_i}u\quad\text{in }L^p(U). \end{align*} Equivalently, \begin{align*} \nabla u_k\rightharpoonup \nabla u\quad\text{in }L^p(U;\mathbb R^n). \end{align*} Passing to the subsequence preserves convergence, so \begin{align*} u_{k_j}\to u\quad\text{in }L^p(U) \end{align*} and \begin{align*} \nabla u_{k_j}\rightharpoonup \nabla u\quad\text{in }L^p(U;\mathbb R^n). \end{align*} [/step]

[step:Verify the integrand hypotheses for the Ioffe theorem] By the Caratheodory hypothesis, for each $(s,\xi)\in\mathbb R\times\mathbb R^n$ the section $x\mapsto f(x,s,\xi)$ from $U$ to $[0,\infty)$ is Lebesgue measurable, and for $\mathcal L^n$-a.e. $x\in U$ the section $(s,\xi)\mapsto f(x,s,\xi)$ from $\mathbb R\times\mathbb R^n$ to $[0,\infty)$ is continuous. Hence $f$ is a nonnegative normal integrand. The additional hypothesis says that, for $\mathcal L^n$-a.e. $x\in U$ and every $s\in\mathbb R$, the gradient section $\xi\mapsto f(x,s,\xi)$ from $\mathbb R^n$ to $[0,\infty)$ is convex. These are precisely the normality, nonnegativity, and convexity-in-gradient hypotheses required by [citetheorem:8733] with $m=1$. [/step]

[step:Apply Ioffe lower semicontinuity to the value and gradient pair]We apply [citetheorem:8733] to the sequence \begin{align*} (u_{k_j},\nabla u_{k_j})_{j=1}^{\infty} \end{align*} and the limit pair $(u,\nabla u)$. The domain $U$ is bounded and open, $1<p<\infty$, the integrand hypotheses were verified above, the value component satisfies strong convergence in $L^p(U)$, and the gradient component satisfies weak convergence in $L^p(U;\mathbb R^n)$. Therefore \begin{align*} \int_U f(x,u(x),\nabla u(x))\,d\mathcal L^n(x)\le \liminf_{j\to\infty}\int_U f(x,u_{k_j}(x),\nabla u_{k_j}(x))\,d\mathcal L^n(x). \end{align*} By the definition of $I$, this is \begin{align*} I[u]\le \liminf_{j\to\infty} I[u_{k_j}]. \end{align*}[/step]

[guided]We now use the main lower semicontinuity input. The theorem we invoke is [citetheorem:8733], the Basic Ioffe Lower Semicontinuity Theorem. In the present scalar case, the value variable has dimension $m=1$, and the gradient variable lies in $\mathbb R^n$. We verify its hypotheses one by one. The set $U$ is bounded and open because it is a bounded Lipschitz domain. The exponent satisfies $1<p<\infty$ by hypothesis. The integrand \begin{align*} f:U\times\mathbb R\times\mathbb R^n\to[0,\infty) \end{align*} is nonnegative. From the Caratheodory assumption, it is measurable in the spatial variable and continuous, hence lower semicontinuous, in the finite-dimensional variables for $\mathcal L^n$-a.e. $x\in U$; this is the normal-integrand condition. The separate convexity hypothesis gives that for $\mathcal L^n$-a.e. $x\in U$ and every $s\in\mathbb R$, the map $\xi\mapsto f(x,s,\xi)$ from $\mathbb R^n$ to $[0,\infty)$ is convex. The convergence hypotheses also match the theorem. The value component converges strongly: \begin{align*} u_{k_j}\to u\quad\text{in }L^p(U). \end{align*} The gradient component converges weakly: \begin{align*} \nabla u_{k_j}\rightharpoonup \nabla u\quad\text{in }L^p(U;\mathbb R^n). \end{align*} This is exactly the mixed convergence structure for Ioffe lower semicontinuity: strong convergence in the variable where the integrand is merely continuous, and weak convergence in the variable where the integrand is convex. Applying [citetheorem:8733] gives \begin{align*} \int_U f(x,u(x),\nabla u(x))\,d\mathcal L^n(x)\le \liminf_{j\to\infty}\int_U f(x,u_{k_j}(x),\nabla u_{k_j}(x))\,d\mathcal L^n(x). \end{align*} Using the definition of the functional $I$, this becomes \begin{align*} I[u]\le \liminf_{j\to\infty} I[u_{k_j}]. \end{align*}[/guided]

[step:Return from the subsequence to the original sequence] From the first step, \begin{align*} \liminf_{j\to\infty}I[u_{k_j}]=\lim_{j\to\infty}I[u_{k_j}]=\liminf_{k\to\infty}I[u_k]. \end{align*} Combining this identity with the inequality obtained from [citetheorem:8733] yields \begin{align*} I[u]\le \liminf_{k\to\infty}I[u_k]. \end{align*} This is the asserted sequential lower semicontinuity of $I$ under weak $W^{1,p}(U)$ convergence together with strong $L^p(U)$ convergence. [/step]