[proofplan] We prove existence by the direct method in the [calculus of variations](/page/Calculus%20of%20Variations). The energy functional is coercive on $W^{1,p}_0(\Omega)$ by [Poincare Inequality with Zero Trace](/theorems/76) and the boundedness of $f$, so a minimizing sequence is bounded; reflexivity gives a weakly convergent subsequence, and weak lower semicontinuity gives a minimizer. The Euler equation of this minimizer is obtained by differentiating the energy along affine lines. Uniqueness follows by testing the difference of two weak equations against their difference and using the strict monotonicity of the map $\xi \mapsto |\xi|^{p-2}\xi$. [/proofplan] [step:Define the variational functional on $W^{1,p}_0(\Omega)$] Let $X := W^{1,p}_0(\Omega)$, equipped with the norm \begin{align*} \|v\|_X := \|\nabla v\|_{L^p(\Omega)} \end{align*} for $v \in X$. This is a norm on $X$ equivalent to the usual $W^{1,p}$ norm by [Poincare Inequality with Zero Trace](/theorems/76), applied because $\Omega$ is bounded and $1<p<\infty$. Define the functional $J: X \to \mathbb{R}$ by \begin{align*} J[u] := \frac{1}{p}\int_\Omega |\nabla u|^p \, d\mathcal{L}^n - f(u). \end{align*} The integral is finite because $\nabla u \in L^p(\Omega;\mathbb{R}^n)$, and $f(u)$ is finite because $f \in X^*$. [/step] [step:Prove coercivity and choose a bounded minimizing sequence] Let $\|f\|_{X^*}$ denote the operator norm of $f: X \to \mathbb{R}$. For every $u \in X$, the definition of the dual norm gives \begin{align*} |f(u)| \leq \|f\|_{X^*}\|u\|_X. \end{align*} Therefore \begin{align*} J[u] \geq \frac{1}{p}\|u\|_X^p - \|f\|_{X^*}\|u\|_X. \end{align*} Since $p>1$, the right-hand side tends to $+\infty$ as $\|u\|_X \to \infty$. Hence $J$ is coercive. Let \begin{align*} m := \inf_{u \in X} J[u]. \end{align*} Since $J[0]=0$, we have $m \leq 0$, and coercivity implies $m > -\infty$. Choose a minimizing sequence $(u_k)_{k=1}^\infty$ in $X$ such that $J[u_k] \to m$. Coercivity implies that $(u_k)_{k=1}^\infty$ is bounded in $X$. [guided] The purpose of this step is to show that minimizing the energy cannot run off to infinity in the Sobolev norm. Let $\|f\|_{X^*}$ be the operator norm of the bounded [linear map](/page/Linear%20Map) $f: X \to \mathbb{R}$. By definition of this norm, every $u \in X$ satisfies \begin{align*} |f(u)| \leq \|f\|_{X^*}\|u\|_X. \end{align*} Substituting this into the definition of $J$ gives the lower bound \begin{align*} J[u] = \frac{1}{p}\|u\|_X^p - f(u) \geq \frac{1}{p}\|u\|_X^p - |f(u)|. \end{align*} Using the dual norm estimate, we obtain \begin{align*} J[u] \geq \frac{1}{p}\|u\|_X^p - \|f\|_{X^*}\|u\|_X. \end{align*} The expression on the right is a degree-$p$ polynomial in $\|u\|_X$ with positive leading coefficient and $p>1$, so it tends to $+\infty$ as $\|u\|_X \to \infty$. This proves coercivity. Now define \begin{align*} m := \inf_{u \in X} J[u]. \end{align*} The value $J[0]=0$ shows $m \leq 0$. The explicit lower bound also shows that $J$ is bounded below. Indeed, define the scalar function $g:[0,\infty)\to\mathbb{R}$ by \begin{align*} g(r):=\frac{1}{p}r^p-\|f\|_{X^*}r. \end{align*} Since $p>1$, this function has a finite global minimum on $[0,\infty)$. Thus $m>-\infty$. By the definition of an infimum, there exists a sequence $(u_k)_{k=1}^\infty$ in $X$ such that $J[u_k]\to m$. If this sequence were unbounded in $X$, coercivity would force $J[u_k]\to+\infty$, contradicting $J[u_k]\to m<+\infty$. Therefore $(u_k)_{k=1}^\infty$ is bounded in $X$. [/guided] [/step] [step:Extract a weak limit and prove that it minimizes $J$] Since $1<p<\infty$, the standard reflexivity theorem for Sobolev spaces applies to the closed subspace $X=W^{1,p}_0(\Omega)$ of $W^{1,p}(\Omega)$, so $X$ is reflexive. Because $(u_k)_{k=1}^\infty$ is bounded in $X$, there exist a subsequence $(u_{k_j})_{j=1}^\infty$ and a function $u \in X$ such that \begin{align*} u_{k_j} \rightharpoonup u \quad \text{weakly in } X. \end{align*} Define $E: X \to \mathbb{R}$ by \begin{align*} E[w] := \frac{1}{p}\int_\Omega |\nabla w|^p \, d\mathcal{L}^n. \end{align*} Define the integrand $G:\mathbb{R}^n\to\mathbb{R}$ by \begin{align*} G(\xi):=\frac{1}{p}|\xi|^p. \end{align*} The function $G$ is convex, continuous, and nonnegative. The weak lower semicontinuity theorem for convex integral functionals requires precisely these hypotheses on $G$ and [weak convergence](/page/Weak%20Convergence) in $L^p(\Omega;\mathbb{R}^n)$ of the vector fields entering the integral. Since weak convergence in $X$ means weak convergence of gradients in $L^p(\Omega;\mathbb{R}^n)$ under the isometric embedding $w \mapsto \nabla w$, the theorem applies and gives that $E$ is weakly lower semicontinuous on $X$. Since $f:X\to\mathbb{R}$ is bounded and linear, it is weakly continuous. Hence \begin{align*} J[u] \leq \liminf_{j\to\infty} J[u_{k_j}] = m. \end{align*} Since $m$ is the infimum of $J$ over $X$, we also have $m \leq J[u]$. Therefore $J[u]=m$, so $u$ minimizes $J$ on $X$. [/step] [step:Differentiate the energy along affine lines to obtain the weak equation] Fix $v \in X$. Define $\phi_v: \mathbb{R} \to \mathbb{R}$ by \begin{align*} \phi_v(t) := J[u+tv]. \end{align*} Since $u$ minimizes $J$ on $X$ and $u+tv \in X$ for every $t \in \mathbb{R}$, the function $\phi_v$ has a minimum at $t=0$. Let $A: \Omega \to \mathbb{R}^n$ and $B: \Omega \to \mathbb{R}^n$ be the measurable maps $A(x):=\nabla u(x)$ and $B(x):=\nabla v(x)$. For each $t \in \mathbb{R}$, define $H_t: \Omega \to \mathbb{R}$ by \begin{align*} H_t(x) := \frac{|A(x)+tB(x)|^p-|A(x)|^p}{pt}. \end{align*} For $\mathcal{L}^n$-a.e. $x\in\Omega$, the map $s \mapsto |A(x)+sB(x)|^p/p$ is differentiable at $s=0$ with derivative $|A(x)|^{p-2}A(x)\cdot B(x)$; this remains valid when $1<p<2$ and $A(x)=0$, because then $|sB(x)|^p/s \to 0$ as $s\to0$. Moreover, the mean-value estimate for $G:\mathbb{R}^n\to\mathbb{R}$, $G(\xi)=|\xi|^p/p$, states that there is a constant $C_p>0$, depending only on $p$, such that for all $\xi,\zeta\in\mathbb{R}^n$, \begin{align*} |G(\xi+\zeta)-G(\xi)|\leq C_p\left(|\xi|^{p-1}|\zeta|+|\zeta|^p\right). \end{align*} Applying this estimate with $\xi=A(x)$ and $\zeta=tB(x)$, and using $|t|\leq 1$, gives \begin{align*} |H_t(x)| \leq C_p\left(|A(x)|^{p-1}|B(x)|+|B(x)|^p\right). \end{align*} The right-hand side is integrable over $\Omega$: $|A|^{p-1}\in L^{p'}(\Omega)$, $|B|\in L^p(\Omega)$, and Holder's inequality applies with conjugate exponents $p'$ and $p$, while $|B|^p\in L^1(\Omega)$. Here $p'$ is defined by \begin{align*} \frac{1}{p}+\frac{1}{p'}=1. \end{align*} The [dominated convergence theorem](/theorems/4) therefore gives \begin{align*} \phi_v'(0)=\int_\Omega |\nabla u|^{p-2}\nabla u\cdot\nabla v \, d\mathcal{L}^n - f(v). \end{align*} Because $\phi_v$ has a minimum at $0$, $\phi_v'(0)=0$. Therefore \begin{align*} \int_\Omega |\nabla u|^{p-2}\nabla u\cdot\nabla v \, d\mathcal{L}^n = f(v). \end{align*} Since $v\in X$ was arbitrary, $u$ is a weak solution. [/step] [step:Prove the strict monotonicity needed for uniqueness] [claim:Strict monotonicity of the $p$-Laplace vector field] For all $\xi,\eta \in \mathbb{R}^n$ with $\xi\neq \eta$, \begin{align*} \left(|\xi|^{p-2}\xi-|\eta|^{p-2}\eta\right)\cdot(\xi-\eta)>0. \end{align*} [/claim] [proof] Define $F:\mathbb{R}^n \to \mathbb{R}$ by \begin{align*} F(\xi) := \frac{1}{p}|\xi|^p. \end{align*} The function $F$ is convex for $p>1$ and strictly convex on every line segment with distinct endpoints. Its gradient is \begin{align*} \nabla F(\xi)=|\xi|^{p-2}\xi. \end{align*} For $\xi,\eta\in\mathbb{R}^n$, convexity gives \begin{align*} F(\eta)\geq F(\xi)+\nabla F(\xi)\cdot(\eta-\xi) \end{align*} and \begin{align*} F(\xi)\geq F(\eta)+\nabla F(\eta)\cdot(\xi-\eta). \end{align*} Adding these inequalities yields \begin{align*} \left(\nabla F(\xi)-\nabla F(\eta)\right)\cdot(\xi-\eta)\geq 0. \end{align*} If $\xi\neq\eta$, strict convexity of $F$ on the segment joining $\xi$ and $\eta$ makes the combined inequality strict. Substituting $\nabla F(\zeta)=|\zeta|^{p-2}\zeta$ gives the claimed inequality. [/proof] [/step] [step:Compare two weak solutions and conclude uniqueness] Let $u_1,u_2\in X$ be weak solutions. Define \begin{align*} w := u_1-u_2 \in X. \end{align*} Using $v=w$ in the weak formulation for $u_1$ and $u_2$, and subtracting the two identities, gives \begin{align*} \int_\Omega \left(|\nabla u_1|^{p-2}\nabla u_1-|\nabla u_2|^{p-2}\nabla u_2\right)\cdot\nabla w \, d\mathcal{L}^n = 0. \end{align*} The integrand is measurable because it is a composition of measurable gradients with continuous maps on $\mathbb{R}^n$. It is integrable since $|\nabla u_i|^{p-1}\in L^{p'}(\Omega)$ for $i=1,2$, $\nabla w\in L^p(\Omega;\mathbb{R}^n)$, and Holder's inequality with conjugate exponents $p'$ and $p$ bounds the absolute value of the integral, where \begin{align*} \frac{1}{p}+\frac{1}{p'}=1. \end{align*} Since $\nabla w=\nabla u_1-\nabla u_2$, the integrand is nonnegative $\mathcal{L}^n$-a.e. by the strict monotonicity claim, and it is positive wherever $\nabla u_1\neq\nabla u_2$. Hence \begin{align*} \nabla u_1=\nabla u_2 \quad \mathcal{L}^n\text{-a.e. in }\Omega. \end{align*} Thus $\nabla w=0$ $\mathcal{L}^n$-a.e. in $\Omega$. Since $w\in W^{1,p}_0(\Omega)$ and $\Omega$ is bounded, [Poincare Inequality with Zero Trace](/theorems/76) implies \begin{align*} \|w\|_{L^p(\Omega)} \leq C_\Omega \|\nabla w\|_{L^p(\Omega)}=0, \end{align*} where $C_\Omega>0$ is the Poincare constant of $\Omega$. Therefore $w=0$ in $L^p(\Omega)$, so $u_1=u_2$ as elements of $W^{1,p}_0(\Omega)$. This proves uniqueness and completes the proof. [/step]