[proofplan] The distribution $\lambda=-\Delta u-f$ rewrites the displayed variational inequality as the measure inequality $\lambda(v-u)\ge 0$ for admissible competitors. Positive test-function perturbations $u+\phi$ first force $\lambda$ to be nonnegative. Then the compatibility condition $u-\psi\in H^1_0(\Omega)$ makes the convex competitor $u-t(u-\psi)$ admissible and gives the zero-product condition with the gap $u-\psi$. Conversely, under the stated integrability and capacity hypotheses, nonnegativity of $\lambda$ together with vanishing on the nonnegative gap implies $\lambda(v-u)\ge 0$ for every admissible $v$, which is exactly the variational inequality. [/proofplan] [step:Rewrite the variational inequality using the reaction measure] Let $\mathcal{L}^n$ denote $n$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}^n$. Let $C_c^\infty(\Omega)$ denote the space of smooth real-valued functions with compact support in $\Omega$, let $\mathcal{D}'(\Omega)$ denote the space of distributions on $\Omega$, namely the continuous linear functionals on $C_c^\infty(\Omega)$ with its standard test-function topology, and let $H^{-1}(\Omega)$ denote the [dual space](/page/Dual%20Space) $(H^1_0(\Omega))^*$. For every [test function](/page/Test%20Function) $\phi \in C_c^\infty(\Omega)$, the distribution $\lambda$ acts by \begin{align*} \lambda(\phi)=(-\Delta u)(\phi)-T_f(\phi), \end{align*} where $T_f \in \mathcal{D}'(\Omega)$ is the [regular distribution](/page/Regular%20Distribution) \begin{align*} T_f(\phi):=\int_\Omega f(x)\phi(x)\,d\mathcal{L}^n(x). \end{align*} Since $u \in H^1(\Omega)$, the distribution $-\Delta u$ is defined by \begin{align*} (-\Delta u)(\phi)=\int_\Omega \nabla u(x)\cdot \nabla \phi(x)\,d\mathcal{L}^n(x). \end{align*} Therefore \begin{align*} \lambda(\phi)=\int_\Omega \nabla u(x)\cdot \nabla \phi(x)\,d\mathcal{L}^n(x)-\int_\Omega f(x)\phi(x)\,d\mathcal{L}^n(x). \end{align*} Because $K_\psi\subset H^1_0(\Omega)$, every admissible displacement $v-u$ belongs to $H^1_0(\Omega)$. Since $\Omega$ is bounded and $f\in L^2(\Omega)$, the [Cauchy-Schwarz inequality](/theorems/432) and the continuous embedding $H^1_0(\Omega)\hookrightarrow L^2(\Omega)$ show that $T_f$ acts continuously on $H^1_0(\Omega)$. Thus $\lambda=-\Delta u-f$ extends from $C_c^\infty(\Omega)$ to a bounded linear functional on $H^1_0(\Omega)$, hence to an element of $H^{-1}(\Omega)$, by \begin{align*} \lambda(h)=\int_\Omega \nabla u(x)\cdot \nabla h(x)\,d\mathcal{L}^n(x)-\int_\Omega f(x)h(x)\,d\mathcal{L}^n(x) \end{align*} for $h\in H^1_0(\Omega)$. With this extension, the variational inequality is equivalent to \begin{align*} \lambda(v-u)\ge 0 \end{align*} for every $v\in K_\psi$. [guided] The purpose of introducing $\lambda$ is to isolate the part of the Euler-Lagrange equation that is caused by the obstacle constraint. Here $\mathcal{L}^n$ denotes $n$-dimensional Lebesgue measure on $\mathbb{R}^n$, $C_c^\infty(\Omega)$ means the space of smooth real-valued functions with compact support in $\Omega$, $\mathcal{D}'(\Omega)$ means the distribution space dual to this test-function space, and $H^{-1}(\Omega)$ means the dual space $(H^1_0(\Omega))^*$. For a test function $\phi \in C_c^\infty(\Omega)$, the distribution $-\Delta u$ is defined through [integration by parts](/theorems/210) without boundary terms: \begin{align*} (-\Delta u)(\phi)=\int_\Omega \nabla u(x)\cdot \nabla \phi(x)\,d\mathcal{L}^n(x). \end{align*} The function $f\in L^2(\Omega)$ also defines a distribution $T_f \in \mathcal{D}'(\Omega)$ by \begin{align*} T_f(\phi)=\int_\Omega f(x)\phi(x)\,d\mathcal{L}^n(x). \end{align*} Hence \begin{align*} \lambda(\phi)=\int_\Omega \nabla u(x)\cdot \nabla \phi(x)\,d\mathcal{L}^n(x)-\int_\Omega f(x)\phi(x)\,d\mathcal{L}^n(x). \end{align*} For a general admissible competitor $v\in K_\psi$, the displacement $v-u$ is not usually a smooth compactly supported function, so we must extend the formula. The definition of $K_\psi$ gives $v-u\in H^1_0(\Omega)$. Since $\Omega$ is bounded, the embedding $H^1_0(\Omega)\hookrightarrow L^2(\Omega)$ is continuous, and therefore the map \begin{align*} h\mapsto \int_\Omega f(x)h(x)\,d\mathcal{L}^n(x) \end{align*} is a bounded linear functional on $H^1_0(\Omega)$ by the Cauchy-Schwarz inequality. The map \begin{align*} h\mapsto \int_\Omega \nabla u(x)\cdot \nabla h(x)\,d\mathcal{L}^n(x) \end{align*} is also bounded on $H^1_0(\Omega)$ because $u\in H^1(\Omega)$. Hence $\lambda$ has a well-defined action as an element of $H^{-1}(\Omega)$ on every $h\in H^1_0(\Omega)$: \begin{align*} \lambda(h)=\int_\Omega \nabla u(x)\cdot \nabla h(x)\,d\mathcal{L}^n(x)-\int_\Omega f(x)h(x)\,d\mathcal{L}^n(x). \end{align*} Substituting $h=v-u$ gives exactly the left-hand side of the variational inequality. Thus the obstacle inequality can be read as saying that the reaction functional $\lambda$ evaluates nonnegatively on every admissible displacement $v-u\in H^1_0(\Omega)$. [/guided] [/step] [step:Derive nonnegativity of the reaction from upward perturbations] Assume first that $u$ solves the variational inequality. Let $\phi \in C_c^\infty(\Omega)$ satisfy $\phi\ge 0$ in $\Omega$. Define the competitor $v_\phi:\Omega\to\mathbb{R}$ by \begin{align*} v_\phi(x)=u(x)+\phi(x) \end{align*} for $x\in\Omega$. Since $u-\psi\ge 0$ quasi-everywhere and $\phi\ge 0$ everywhere, $v_\phi-\psi\ge 0$ quasi-everywhere, so $v_\phi\in K_\psi$. The variational inequality with $v=v_\phi$ gives \begin{align*} \lambda(\phi)\ge 0. \end{align*} Since this holds for every nonnegative $\phi\in C_c^\infty(\Omega)$, the distribution $\lambda$ is positive. Under the standing hypothesis that $\lambda$ is represented by a Radon measure, this means precisely that the representing Radon measure is nonnegative on $\Omega$. [/step] [step:Use the convex competitor toward the obstacle to prove zero complementarity] The standing compatibility assumption gives $u-\psi\in H^1_0(\Omega)$. For $t\in(0,1)$, define $v_t:\Omega\to\mathbb{R}$ by \begin{align*} v_t(x)=u(x)-t(u(x)-\psi(x)) \end{align*} for $x\in\Omega$. Equivalently, $v_t=(1-t)u+t\psi$. Since $u\in H^1_0(\Omega)$ and $u-\psi\in H^1_0(\Omega)$, we have $v_t=u-t(u-\psi)\in H^1_0(\Omega)$. Also \begin{align*} v_t-\psi=(1-t)(u-\psi), \end{align*} so $v_t-\psi\ge 0$ quasi-everywhere and hence $v_t\in K_\psi$. Applying the variational inequality to $v_t$ gives \begin{align*} \lambda(v_t-u)\ge 0. \end{align*} The displacement $v_t-u=-t(u-\psi)$ belongs to $H^1_0(\Omega)$ by the same compatibility assumption, and the quasi-continuous representative of $u-\psi$ is $\lambda$-integrable by hypothesis. The $H^{-1}$ action of the Radon measure representative therefore agrees with integration against this quasi-continuous representative, so \begin{align*} \lambda(v_t-u)=-t\int_\Omega (u-\psi)\,d\lambda. \end{align*} Consequently \begin{align*} -t\int_\Omega (u-\psi)\,d\lambda \ge 0. \end{align*} Because $t>0$, \begin{align*} \int_\Omega (u-\psi)\,d\lambda \le 0. \end{align*} But $u-\psi\ge 0$ quasi-everywhere and $\lambda\ge 0$. By the stated capacity absolute-continuity hypothesis on the Radon representative of $\lambda$, every set of zero Sobolev capacity has $\lambda$-measure zero. Therefore the quasi-everywhere inequality gives $(u-\psi)\ge 0$ $\lambda$-almost everywhere. Hence \begin{align*} \int_\Omega (u-\psi)\,d\lambda \ge 0. \end{align*} Combining the two inequalities yields \begin{align*} \int_\Omega (u-\psi)\,d\lambda = 0. \end{align*} [/step] [step:Convert zero complementarity into support on the contact set] Let $w:\Omega\to[0,\infty]$ denote the chosen quasi-continuous representative of $u-\psi$. The equality \begin{align*} \int_\Omega w\,d\lambda=0 \end{align*} with $w\ge 0$ $\lambda$-almost everywhere and $\lambda\ge 0$ implies $w=0$ $\lambda$-almost everywhere. Therefore, for the quasi-open positivity set $\{w>0\}$, \begin{align*} \lambda(\{w>0\})=0. \end{align*} Equivalently, $\lambda$ is concentrated on the quasi-contact set $\{w=0\}$, which is the set where the quasi-continuous representatives satisfy $u=\psi$. [/step] [step:Recover the variational inequality from the complementarity conditions] Conversely, assume the complementarity conditions hold together with the standing identity $\lambda=-\Delta u-f$ and the stated measure-action compatibility. Let $v\in K_\psi$ be arbitrary. Define $g:\Omega\to\mathbb{R}$ by \begin{align*} g(x)=v(x)-u(x) \end{align*} for $x\in\Omega$. By the strengthened integrability hypothesis, the chosen quasi-continuous representative of $g=v-u$ is $\lambda$-integrable. Since $v-\psi\ge 0$ quasi-everywhere and $u-\psi\ge 0$ quasi-everywhere, \begin{align*} g=(v-\psi)-(u-\psi)\ge -(u-\psi) \end{align*} quasi-everywhere. By the stated capacity absolute-continuity hypothesis on the Radon representative of $\lambda$, sets of zero Sobolev capacity have $\lambda$-measure zero, so this quasi-everywhere inequality holds $\lambda$-almost everywhere. Integrating with respect to the nonnegative measure $\lambda$ gives \begin{align*} \int_\Omega g\,d\lambda \ge -\int_\Omega (u-\psi)\,d\lambda=0. \end{align*} Using the distributional identity $\lambda=-\Delta u-f$, this is \begin{align*} \int_\Omega \nabla u(x)\cdot \nabla(v-u)(x)\,d\mathcal{L}^n(x)-\int_\Omega f(x)(v-u)(x)\,d\mathcal{L}^n(x)\ge 0. \end{align*} Rearranging gives \begin{align*} \int_\Omega \nabla u(x)\cdot \nabla(v-u)(x)\,d\mathcal{L}^n(x) \ge \int_\Omega f(x)(v-u)(x)\,d\mathcal{L}^n(x). \end{align*} Since $v\in K_\psi$ was arbitrary, $u$ solves the obstacle variational inequality. [/step]