[proofplan] We first check that the obstacle constraint defines a nonempty closed convex subset of the [Hilbert space](/page/Hilbert%20Space) $H^1_0(\Omega)$. We then minimize the strictly convex energy associated to the Dirichlet form and the linear functional $f$ on this admissible set; coercivity follows from the [Poincare Inequality with Zero Trace](/theorems/76), and weak lower semicontinuity gives existence. The Euler condition for this constrained minimizer is exactly the variational inequality. Finally, strict convexity of the Dirichlet energy gives uniqueness, and the $L^2$ representation of $f$ gives the displayed integral form. [/proofplan]

[step:Verify that the admissible obstacle set is closed and convex]Define \begin{align*} K_\psi := \{v \in H^1_0(\Omega) : v \geq \psi \ \mathcal{L}^n\text{-a.e. in } \Omega\}. \end{align*} By hypothesis, $K_\psi \neq \varnothing$. Let $v_0,v_1 \in K_\psi$ and let $\lambda \in [0,1]$. Since $K_\psi\neq \varnothing$, choose $v_* \in K_\psi$; then the set where $\psi=+\infty$ is $\mathcal{L}^n$-null, because $v_*\geq \psi$ $\mathcal{L}^n$-a.e. and $v_*$ is finite $\mathcal{L}^n$-a.e. Since $v_0 \geq \psi$ and $v_1 \geq \psi$ $\mathcal{L}^n$-a.e. in $\Omega$, the convex combination \begin{align*} v_\lambda := (1-\lambda)v_0+\lambda v_1 \end{align*} satisfies $v_\lambda(x)\geq \psi(x)$ on the full-measure set where both inequalities hold and $\psi(x)<+\infty$: if $\psi(x)\in\mathbb{R}$, then \begin{align*} v_\lambda(x) \geq (1-\lambda)\psi(x)+\lambda\psi(x) = \psi(x), \end{align*} while if $\psi(x)=-\infty$, the inequality $v_\lambda(x)\geq \psi(x)$ is automatic. Hence $v_\lambda \in K_\psi$, so $K_\psi$ is convex. To prove closedness, let $(v_k)_k$ be a sequence in $K_\psi$ such that $v_k \to v$ in $H^1_0(\Omega)$ for some $v \in H^1_0(\Omega)$. Since the $H^1_0(\Omega)$ norm controls the $L^2(\Omega)$ norm, we have $v_k \to v$ in $L^2(\Omega)$. Therefore there is a subsequence $(v_{k_j})_j$ such that $v_{k_j}(x) \to v(x)$ for $\mathcal{L}^n$-a.e. $x \in \Omega$. For each $j$, $v_{k_j} \geq \psi$ $\mathcal{L}^n$-a.e.; after discarding the union of the corresponding null sets and the null set on which pointwise convergence fails, we obtain \begin{align*} v(x) = \lim_{j \to \infty} v_{k_j}(x) \geq \psi(x) \end{align*} for $\mathcal{L}^n$-a.e. $x \in \Omega$. Thus $v \in K_\psi$, and $K_\psi$ is closed in $H^1_0(\Omega)$.[/step]

[guided]The admissible functions are precisely those Sobolev functions that stay above the obstacle almost everywhere. We must verify two structural facts because the minimization argument will take place on $K_\psi$: convexity and closedness. First take $v_0,v_1 \in K_\psi$ and $\lambda \in [0,1]$. Define the convex combination \begin{align*} v_\lambda := (1-\lambda)v_0+\lambda v_1. \end{align*} Because $K_\psi$ is nonempty, choose $v_*\in K_\psi$. The inequality $v_*\geq \psi$ almost everywhere implies that the set where $\psi=+\infty$ has $\mathcal{L}^n$-measure zero, since $v_*$ is finite almost everywhere as an $H^1_0(\Omega)$ function. This does not exclude $\psi=-\infty$, and it does not need to: the inequality $v\geq -\infty$ is automatic for finite-valued Sobolev representatives. Since $v_0 \geq \psi$ and $v_1 \geq \psi$ outside null subsets of $\Omega$, there is a measurable set $N \subset \Omega$ with $\mathcal{L}^n(N)=0$ such that $\psi(x)<+\infty$ and both inequalities hold for every $x \in \Omega \setminus N$. If $x\in\Omega\setminus N$ and $\psi(x)\in\mathbb{R}$, then \begin{align*} v_\lambda(x) = (1-\lambda)v_0(x)+\lambda v_1(x) \geq (1-\lambda)\psi(x)+\lambda\psi(x)=\psi(x). \end{align*} If instead $x\in\Omega\setminus N$ and $\psi(x)=-\infty$, then $v_\lambda(x)\geq\psi(x)$ holds automatically. Thus $v_\lambda \geq \psi$ $\mathcal{L}^n$-a.e. in $\Omega$, so $v_\lambda \in K_\psi$. This proves convexity. Now let $(v_k)_k$ be a sequence in $K_\psi$ with $v_k \to v$ in $H^1_0(\Omega)$. The obstacle condition is an almost-everywhere pointwise inequality, while the convergence is a Sobolev norm convergence. To pass the inequality to the limit, we first use the continuous embedding of $H^1_0(\Omega)$ into $L^2(\Omega)$: strong convergence in $H^1_0(\Omega)$ implies $v_k \to v$ in $L^2(\Omega)$. From $L^2$ convergence, there is a subsequence $(v_{k_j})_j$ converging to $v$ pointwise $\mathcal{L}^n$-a.e. in $\Omega$. For each $j$, the condition $v_{k_j} \in K_\psi$ means that $v_{k_j} \geq \psi$ outside some null set. The countable union of these null sets, together with the null set where pointwise convergence fails, still has $\mathcal{L}^n$-measure zero. Therefore, for every $x$ outside this single null set, \begin{align*} v(x) = \lim_{j \to \infty} v_{k_j}(x) \geq \psi(x). \end{align*} Hence $v \geq \psi$ $\mathcal{L}^n$-a.e. in $\Omega$, so $v \in K_\psi$. This proves that $K_\psi$ is closed.[/guided]

[step:Construct a minimizer of the Dirichlet energy over the obstacle set]Define the functional \begin{align*} J: H^1_0(\Omega) \to \mathbb{R}, \qquad J[v] := \frac{1}{2}\int_\Omega |\nabla v(x)|^2\,d\mathcal{L}^n(x)-f(v). \end{align*} The assumed [Poincare Inequality with Zero Trace](/theorems/76) gives a constant $C_P>0$ such that \begin{align*} \|w\|_{L^2(\Omega)} \leq C_P\|\nabla w\|_{L^2(\Omega)} \end{align*} for every $w \in H^1_0(\Omega)$. Thus $\|\nabla w\|_{L^2(\Omega)}$ is an equivalent Hilbert norm on $H^1_0(\Omega)$. Since $f \in H^{-1}(\Omega)=(H^1_0(\Omega))^*$, define its operator norm by \begin{align*} \|f\|_{H^{-1}(\Omega)} := \sup\{|f(w)| : w \in H^1_0(\Omega), \ \|w\|_{H^1_0(\Omega)} \leq 1\}. \end{align*} Using boundedness of $f$ and the equivalence of norms, there is a constant $C_f>0$ such that \begin{align*} |f(w)| \leq C_f\|\nabla w\|_{L^2(\Omega)} \end{align*} for every $w \in H^1_0(\Omega)$. Therefore \begin{align*} J[w] \geq \frac{1}{2}\|\nabla w\|_{L^2(\Omega)}^2-C_f\|\nabla w\|_{L^2(\Omega)}. \end{align*} It follows that $J[w] \to \infty$ whenever $\|\nabla w\|_{L^2(\Omega)} \to \infty$, so $J$ is coercive on $K_\psi$. More explicitly, define the polynomial $q:[0,\infty)\to\mathbb{R}$ by \begin{align*} q(r):=\frac{1}{2}r^2-C_f r. \end{align*} Since $q$ is bounded below on $[0,\infty)$, $J$ is bounded below on $K_\psi$. Since $K_\psi\neq\varnothing$, choosing any $v_*\in K_\psi$ gives the finite comparison value $J[v_*]\in\mathbb{R}$, and therefore the infimum $\inf_{v\in K_\psi}J[v]$ is finite. Let $(u_k)_k$ be a minimizing sequence in $K_\psi$, so \begin{align*} \lim_{k \to \infty} J[u_k] = \inf_{v \in K_\psi}J[v]. \end{align*} The coercivity estimate implies that $(u_k)_k$ is bounded in $H^1_0(\Omega)$. Since $H^1_0(\Omega)$ is a Hilbert space and hence reflexive, the weak [compactness theorem](/theorems/2748) for bounded sequences in reflexive Banach spaces gives a subsequence, still denoted $(u_k)_k$, and an element $u \in H^1_0(\Omega)$ such that $u_k \rightharpoonup u$ weakly in $H^1_0(\Omega)$. Because $K_\psi$ is norm-closed and convex, the Hahn-Banach separation argument for closed convex subsets of Hilbert spaces implies that $K_\psi$ is weakly closed; hence $u \in K_\psi$. These are standard functional-analytic consequences of weak compactness in reflexive spaces, Hahn-Banach separation, and weak lower semicontinuity of Hilbert norms. The map \begin{align*} w \mapsto \frac{1}{2}\int_\Omega |\nabla w(x)|^2\,d\mathcal{L}^n(x) \end{align*} is weakly lower semicontinuous on $H^1_0(\Omega)$ by the weak lower semicontinuity of the squared Hilbert norm, applied to the Hilbert norm $w\mapsto\|\nabla w\|_{L^2(\Omega)}$. The functional $f$ is weakly continuous because it is a bounded linear functional, so $u_k\rightharpoonup u$ implies $f(u_k)\to f(u)$. Hence \begin{align*} J[u] \leq \liminf_{k \to \infty}J[u_k] = \inf_{v \in K_\psi}J[v]. \end{align*} Since $u \in K_\psi$, equality holds, and $u$ minimizes $J$ over $K_\psi$.[/step]

[guided]The existence argument is the direct method in a Hilbert space. Define \begin{align*} J: H^1_0(\Omega) \to \mathbb{R}, \qquad J[v] := \frac{1}{2}\int_\Omega |\nabla v(x)|^2\,d\mathcal{L}^n(x)-f(v). \end{align*} The assumed Poincare inequality gives a constant $C_P>0$ with \begin{align*} \|w\|_{L^2(\Omega)} \leq C_P\|\nabla w\|_{L^2(\Omega)} \end{align*} for every $w\in H^1_0(\Omega)$, so the gradient norm is equivalent to the usual $H^1_0(\Omega)$ norm. Since $f\in H^{-1}(\Omega)=(H^1_0(\Omega))^*$, boundedness of $f$ with respect to this equivalent norm gives a constant $C_f>0$ such that \begin{align*} |f(w)| \leq C_f\|\nabla w\|_{L^2(\Omega)} \end{align*} for every $w\in H^1_0(\Omega)$. Therefore \begin{align*} J[w] \geq \frac{1}{2}\|\nabla w\|_{L^2(\Omega)}^2-C_f\|\nabla w\|_{L^2(\Omega)}. \end{align*} The right-hand side is a quadratic polynomial in $\|\nabla w\|_{L^2(\Omega)}$ with positive leading coefficient. Hence $J$ is coercive and bounded below on $K_\psi$. Because $K_\psi$ is nonempty, choosing $v_*\in K_\psi$ gives a finite number $J[v_*]$, so the infimum over $K_\psi$ is finite and a minimizing sequence exists. Let $(u_k)_k\subset K_\psi$ satisfy \begin{align*} \lim_{k\to\infty}J[u_k]=\inf_{v\in K_\psi}J[v]. \end{align*} Coercivity implies that $(u_k)_k$ is bounded in $H^1_0(\Omega)$. We use weak compactness for bounded sequences in reflexive Banach spaces. The hypothesis needed here is reflexivity; it is satisfied because $H^1_0(\Omega)$ is a Hilbert space, and every Hilbert space is reflexive. Therefore there is a subsequence, still denoted $(u_k)_k$, and $u\in H^1_0(\Omega)$ such that $u_k\rightharpoonup u$ weakly in $H^1_0(\Omega)$. The previous step proved that $K_\psi$ is closed and convex. We next use the Hilbert-space fact that every norm-closed convex subset is weakly closed; this follows from the [Hahn-Banach separation theorem](/theorems/974), because any point outside a closed convex set can be separated by a continuous linear functional, and continuous linear functionals are weakly continuous. Its hypotheses are satisfied: $K_\psi$ is convex, norm-closed in $H^1_0(\Omega)$, and $H^1_0(\Omega)$ is a Hilbert space. Hence the weak limit still satisfies $u\in K_\psi$. The squared Hilbert norm is weakly lower semicontinuous: if $w_k\rightharpoonup w$ in a Hilbert space, then $\|w\|^2\leq\liminf_k\|w_k\|^2$. We apply this standard theorem to the equivalent Hilbert norm $w\mapsto \|\nabla w\|_{L^2(\Omega)}$, which is valid by the Poincare inequality already verified. Also, $f$ is weakly continuous because it is a bounded linear functional; this means $u_k\rightharpoonup u$ implies $f(u_k)\to f(u)$. Combining these two facts gives \begin{align*} J[u] \leq \liminf_{k\to\infty}J[u_k]=\inf_{v\in K_\psi}J[v]. \end{align*} Since $u\in K_\psi$, the opposite inequality follows from the definition of the infimum. Thus $u$ minimizes $J$ over $K_\psi$.[/guided]

[step:Derive the variational inequality from the minimizing property]Let $v \in K_\psi$ be arbitrary. For each $t \in [0,1]$, define \begin{align*} u_t := u+t(v-u). \end{align*} Since $K_\psi$ is convex and both $u$ and $v$ belong to $K_\psi$, we have $u_t \in K_\psi$. The minimizing property of $u$ gives \begin{align*} J[u_t]-J[u] \geq 0. \end{align*} Expanding the quadratic part, \begin{align*} \frac{1}{2}\int_\Omega |\nabla u_t(x)|^2\,d\mathcal{L}^n(x)-\frac{1}{2}\int_\Omega |\nabla u(x)|^2\,d\mathcal{L}^n(x)=t\int_\Omega \nabla u(x)\cdot \nabla(v-u)(x)\,d\mathcal{L}^n(x)+\frac{t^2}{2}\int_\Omega |\nabla(v-u)(x)|^2\,d\mathcal{L}^n(x). \end{align*} Since $f$ is linear, \begin{align*} f(u_t)-f(u)=t f(v-u). \end{align*} Therefore, for every $t \in (0,1]$, \begin{align*} 0 \leq t\int_\Omega \nabla u(x)\cdot \nabla(v-u)(x)\,d\mathcal{L}^n(x)+\frac{t^2}{2}\int_\Omega |\nabla(v-u)(x)|^2\,d\mathcal{L}^n(x)-t f(v-u). \end{align*} Dividing by $t>0$ gives \begin{align*} 0 \leq \int_\Omega \nabla u(x)\cdot \nabla(v-u)(x)\,d\mathcal{L}^n(x)+\frac{t}{2}\int_\Omega |\nabla(v-u)(x)|^2\,d\mathcal{L}^n(x)-f(v-u). \end{align*} Letting $t \downarrow 0$ yields \begin{align*} \int_\Omega \nabla u(x)\cdot \nabla(v-u)(x)\,d\mathcal{L}^n(x) \geq f(v-u). \end{align*} Since $v \in K_\psi$ was arbitrary, $u$ satisfies the desired variational inequality.[/step]

[guided]The minimizer $u$ gives an inequality for all admissible variations, but the admissible set is not a linear space: if $u \in K_\psi$, then $u+\varepsilon w$ need not remain above the obstacle. The correct variations are one-sided segments from $u$ toward another admissible point. Fix $v \in K_\psi$. For $t \in [0,1]$, define \begin{align*} u_t := u+t(v-u). \end{align*} This is the point at parameter $t$ on the line segment from $u$ to $v$. Since $K_\psi$ is convex and $u,v \in K_\psi$, we have $u_t \in K_\psi$ for every $t \in [0,1]$. Because $u$ minimizes $J$ over $K_\psi$, \begin{align*} J[u_t]-J[u] \geq 0. \end{align*} We now compute this difference. Since \begin{align*} \nabla u_t = \nabla u+t\nabla(v-u), \end{align*} the Euclidean identity $|a+tb|^2=|a|^2+2t\,a\cdot b+t^2|b|^2$ gives \begin{align*} \frac{1}{2}\int_\Omega |\nabla u_t(x)|^2\,d\mathcal{L}^n(x)-\frac{1}{2}\int_\Omega |\nabla u(x)|^2\,d\mathcal{L}^n(x)=t\int_\Omega \nabla u(x)\cdot \nabla(v-u)(x)\,d\mathcal{L}^n(x)+\frac{t^2}{2}\int_\Omega |\nabla(v-u)(x)|^2\,d\mathcal{L}^n(x). \end{align*} The linear part is simpler: because $f: H^1_0(\Omega)\to \mathbb{R}$ is linear, \begin{align*} f(u_t)-f(u)=f(t(v-u))=t f(v-u). \end{align*} Substituting both identities into $J[u_t]-J[u]\geq 0$ gives, for $t \in (0,1]$, \begin{align*} 0 \leq t\int_\Omega \nabla u(x)\cdot \nabla(v-u)(x)\,d\mathcal{L}^n(x)+\frac{t^2}{2}\int_\Omega |\nabla(v-u)(x)|^2\,d\mathcal{L}^n(x)-t f(v-u). \end{align*} We divide by $t>0$: \begin{align*} 0 \leq \int_\Omega \nabla u(x)\cdot \nabla(v-u)(x)\,d\mathcal{L}^n(x)+\frac{t}{2}\int_\Omega |\nabla(v-u)(x)|^2\,d\mathcal{L}^n(x)-f(v-u). \end{align*} The only term still depending on $t$ is nonnegative and multiplied by $t$. Letting $t \downarrow 0$ removes this second-order term and leaves \begin{align*} \int_\Omega \nabla u(x)\cdot \nabla(v-u)(x)\,d\mathcal{L}^n(x) \geq f(v-u). \end{align*} Since $v \in K_\psi$ was arbitrary, this is exactly the obstacle problem variational inequality.[/guided]

[step:Prove uniqueness by testing two solutions against each other]Suppose $u_1,u_2 \in K_\psi$ both satisfy the variational inequality. Applying the inequality for $u_1$ with $v=u_2$ gives \begin{align*} \int_\Omega \nabla u_1(x)\cdot \nabla(u_2-u_1)(x)\,d\mathcal{L}^n(x) \geq f(u_2-u_1). \end{align*} Applying the inequality for $u_2$ with $v=u_1$ gives \begin{align*} \int_\Omega \nabla u_2(x)\cdot \nabla(u_1-u_2)(x)\,d\mathcal{L}^n(x) \geq f(u_1-u_2). \end{align*} Adding the two inequalities and using the linearity of $f$ gives \begin{align*} \int_\Omega \nabla u_1(x)\cdot \nabla(u_2-u_1)(x)\,d\mathcal{L}^n(x)+\int_\Omega \nabla u_2(x)\cdot \nabla(u_1-u_2)(x)\,d\mathcal{L}^n(x) \geq 0. \end{align*} The left-hand side is \begin{align*} -\int_\Omega |\nabla(u_1-u_2)(x)|^2\,d\mathcal{L}^n(x). \end{align*} Hence \begin{align*} \int_\Omega |\nabla(u_1-u_2)(x)|^2\,d\mathcal{L}^n(x) \leq 0. \end{align*} Since the integrand is nonnegative, $\nabla(u_1-u_2)=0$ in $L^2(\Omega;\mathbb{R}^n)$. Because $u_1-u_2 \in H^1_0(\Omega)$, the Poincare inequality implies \begin{align*} \|u_1-u_2\|_{L^2(\Omega)} \leq C_P\|\nabla(u_1-u_2)\|_{L^2(\Omega)}=0. \end{align*} Thus $u_1=u_2$ in $H^1_0(\Omega)$, proving uniqueness.[/step]

[guided]Assume that $u_1,u_2\in K_\psi$ both solve the variational inequality. The strategy is to use each solution as the comparison function for the other, which is legitimate because both functions are admissible. For $u_1$, choose $v=u_2$ and obtain \begin{align*} \int_\Omega \nabla u_1(x)\cdot \nabla(u_2-u_1)(x)\,d\mathcal{L}^n(x) \geq f(u_2-u_1). \end{align*} For $u_2$, choose $v=u_1$ and obtain \begin{align*} \int_\Omega \nabla u_2(x)\cdot \nabla(u_1-u_2)(x)\,d\mathcal{L}^n(x) \geq f(u_1-u_2). \end{align*} Adding the inequalities cancels the right-hand side because $f$ is linear: \begin{align*} f(u_2-u_1)+f(u_1-u_2)=f(0)=0. \end{align*} Thus \begin{align*} \int_\Omega \nabla u_1(x)\cdot \nabla(u_2-u_1)(x)\,d\mathcal{L}^n(x)+\int_\Omega \nabla u_2(x)\cdot \nabla(u_1-u_2)(x)\,d\mathcal{L}^n(x) \geq 0. \end{align*} Let $z:=u_1-u_2\in H^1_0(\Omega)$. Since $\nabla(u_2-u_1)=-\nabla z$ and $\nabla(u_1-u_2)=\nabla z$, the left-hand side equals \begin{align*} -\int_\Omega |\nabla z(x)|^2\,d\mathcal{L}^n(x). \end{align*} Therefore \begin{align*} \int_\Omega |\nabla z(x)|^2\,d\mathcal{L}^n(x) \leq 0. \end{align*} The integrand is nonnegative, so $\nabla z=0$ in $L^2(\Omega;\mathbb{R}^n)$. Applying the Poincare inequality to $z\in H^1_0(\Omega)$ gives \begin{align*} \|z\|_{L^2(\Omega)} \leq C_P\|\nabla z\|_{L^2(\Omega)}=0. \end{align*} Hence $z=0$ in $H^1_0(\Omega)$, meaning $u_1=u_2$. This proves uniqueness.[/guided]

[step:Rewrite the right-hand side when the functional is represented by an $L^2$ function]Assume that there exists $F \in L^2(\Omega)$ such that \begin{align*} f(w)=\int_\Omega F(x)w(x)\,d\mathcal{L}^n(x) \end{align*} for every $w \in H^1_0(\Omega)$. Taking $w=v-u$ in this representation gives \begin{align*} f(v-u)=\int_\Omega F(x)(v-u)(x)\,d\mathcal{L}^n(x). \end{align*} Substituting this identity into the variational inequality proves the asserted $L^2$ form: \begin{align*} \int_\Omega \nabla u(x)\cdot \nabla(v-u)(x)\,d\mathcal{L}^n(x) \geq \int_\Omega F(x)(v-u)(x)\,d\mathcal{L}^n(x). \end{align*} This completes the proof.[/step]

[guided]Assume that the functional $f\in H^{-1}(\Omega)$ is represented by a function $F\in L^2(\Omega)$, meaning that for every $w\in H^1_0(\Omega)$, \begin{align*} f(w)=\int_\Omega F(x)w(x)\,d\mathcal{L}^n(x). \end{align*} This representation applies to the admissible difference $w=v-u$, since $u,v\in H^1_0(\Omega)$ and $H^1_0(\Omega)$ is a [vector space](/page/Vector%20Space). Therefore \begin{align*} f(v-u)=\int_\Omega F(x)(v-u)(x)\,d\mathcal{L}^n(x). \end{align*} Substituting this identity into \begin{align*} \int_\Omega \nabla u(x)\cdot \nabla(v-u)(x)\,d\mathcal{L}^n(x) \geq f(v-u) \end{align*} gives \begin{align*} \int_\Omega \nabla u(x)\cdot \nabla(v-u)(x)\,d\mathcal{L}^n(x) \geq \int_\Omega F(x)(v-u)(x)\,d\mathcal{L}^n(x). \end{align*} This is exactly the asserted $L^2$ form of the obstacle problem inequality.[/guided]