[proofplan] We compare the penalized minimizer $u_\varepsilon$ with fixed admissible functions, for which the penalty term vanishes. This gives a uniform $H^1_0(\Omega)$ bound and also forces the constraint violation $(u_\varepsilon-\psi)^-$ to converge to zero in $L^2(\Omega)$. Compactness then yields a subsequential limit $\bar u$, and the vanishing negative part shows $\bar u \in K_\psi$. Passing to the limit in the variational inequality for the energies identifies $\bar u$ as the obstacle minimizer $u$, and a matching limsup-liminf argument for the Dirichlet energy upgrades [weak convergence](/page/Weak%20Convergence) to strong convergence. [/proofplan] [step:Define the energies and fix an admissible comparison function] Define the obstacle energy functional \begin{align*} J: H^1_0(\Omega) \to \mathbb{R} \end{align*} by \begin{align*} J[v] := \frac{1}{2}\int_\Omega |\nabla v|^2 \, d\mathcal{L}^n(x) - \int_\Omega f v \, d\mathcal{L}^n(x). \end{align*} For each $\varepsilon > 0$, define the penalty functional \begin{align*} P_\varepsilon: H^1_0(\Omega) \to [0,\infty) \end{align*} by \begin{align*} P_\varepsilon[v] := \int_\Omega B_\varepsilon(v-\psi) \, d\mathcal{L}^n(x). \end{align*} This integral is finite for every $v \in H^1_0(\Omega)$. Indeed, if $L_\varepsilon$ is a global Lipschitz constant for $\beta_\varepsilon$, then $B_\varepsilon(0)=0$ and $\beta_\varepsilon(0)=0$ imply $|B_\varepsilon(s)| \leq \frac{L_\varepsilon}{2}|s|^2$ for all $s \in \mathbb{R}$, while $v-\psi \in L^2(\Omega)$. Thus $I_\varepsilon[v] = J[v] + P_\varepsilon[v]$. Since $K_\psi \neq \varnothing$, fix an admissible comparison function $w_0 \in K_\psi$. Because $w_0-\psi \geq 0$ for $\mathcal{L}^n$-a.e. $x \in \Omega$ and $B_\varepsilon(s)=0$ for $s \geq 0$, we have \begin{align*} P_\varepsilon[w_0] = 0. \end{align*} Thus minimality of $u_\varepsilon$ gives \begin{align*} J[u_\varepsilon] + P_\varepsilon[u_\varepsilon] = I_\varepsilon[u_\varepsilon] \leq I_\varepsilon[w_0] = J[w_0]. \end{align*} [/step] [step:Derive a uniform $H^1_0(\Omega)$ bound for the penalized minimizers] Let $C_P > 0$ be a [Poincare inequality](/theorems/75) constant for $\Omega$, so that \begin{align*} \|v\|_{L^2(\Omega)} \leq C_P \|\nabla v\|_{L^2(\Omega)} \end{align*} for every $v \in H^1_0(\Omega)$. Since $P_\varepsilon[u_\varepsilon] \geq 0$, the comparison inequality gives \begin{align*} \frac{1}{2}\|\nabla u_\varepsilon\|_{L^2(\Omega)}^2 - \int_\Omega f u_\varepsilon \, d\mathcal{L}^n(x) \leq J[w_0]. \end{align*} By the [Cauchy-Schwarz inequality](/theorems/432) in $L^2(\Omega)$ and the [Poincare inequality](/theorems/75), \begin{align*} \left|\int_\Omega f u_\varepsilon \, d\mathcal{L}^n(x)\right| \leq \|f\|_{L^2(\Omega)}\|u_\varepsilon\|_{L^2(\Omega)} \leq C_P\|f\|_{L^2(\Omega)}\|\nabla u_\varepsilon\|_{L^2(\Omega)}. \end{align*} Set $A_\varepsilon := \|\nabla u_\varepsilon\|_{L^2(\Omega)}$ and $M := C_P\|f\|_{L^2(\Omega)}$. Then \begin{align*} \frac{1}{2}A_\varepsilon^2 - M A_\varepsilon \leq J[w_0]. \end{align*} The scalar quadratic inequality implies that $(A_\varepsilon)_{\varepsilon>0}$ is bounded. Applying Poincare again, $(u_\varepsilon)_{\varepsilon>0}$ is bounded in $H^1_0(\Omega)$. [/step] [step:Show that the violation of the obstacle constraint vanishes in $L^2(\Omega)$] From \begin{align*} J[u_\varepsilon] + P_\varepsilon[u_\varepsilon] \leq J[w_0] \end{align*} and the uniform $H^1_0(\Omega)$ bound, there exists a constant $C_1 > 0$, independent of $\varepsilon$, such that \begin{align*} P_\varepsilon[u_\varepsilon] \leq C_1. \end{align*} Indeed, the uniform bound controls $|J[u_\varepsilon]|$ by the [Cauchy-Schwarz inequality](/theorems/432) and the [Poincare inequality](/theorems/75). By the lower bound on $B_\varepsilon$, \begin{align*} \frac{c}{\varepsilon}\int_\Omega ((u_\varepsilon-\psi)^-)^2 \, d\mathcal{L}^n(x) \leq \int_\Omega B_\varepsilon(u_\varepsilon-\psi) \, d\mathcal{L}^n(x) \leq C_1. \end{align*} Therefore \begin{align*} \|(u_\varepsilon-\psi)^-\|_{L^2(\Omega)}^2 \leq \frac{C_1}{c}\varepsilon. \end{align*} In particular, \begin{align*} (u_\varepsilon-\psi)^- \to 0 \quad \text{strongly in } L^2(\Omega) \end{align*} as $\varepsilon \downarrow 0$. [/step] [step:Extract a compactness subsequence and prove that its limit is admissible] Let $(\varepsilon_k)_{k=1}^\infty$ be any sequence with $\varepsilon_k \downarrow 0$. Since $(u_{\varepsilon_k})_{k=1}^\infty$ is bounded in the [Hilbert space](/page/Hilbert%20Space) $H^1_0(\Omega)$, the Hilbert-space [weak sequential compactness](/theorems/214) theorem for bounded sequences applies. Hence there exist a subsequence, not relabelled, and an element $\bar u \in H^1_0(\Omega)$ such that \begin{align*} u_{\varepsilon_k} \rightharpoonup \bar u \quad \text{weakly in } H^1_0(\Omega). \end{align*} By the assumed compact embedding $H^1_0(\Omega) \hookrightarrow L^2(\Omega)$, equivalently the compactness property supplied in standard domains by the [Rellich-Kondrachov Theorem](/theorems/64), \begin{align*} u_{\varepsilon_k} \to \bar u \quad \text{strongly in } L^2(\Omega). \end{align*} The map $N: L^2(\Omega) \to L^2(\Omega)$ defined by $N(v) := (v-\psi)^-$ is Lipschitz with constant $1$, because $|a^- - b^-| \leq |a-b|$ for all $a,b \in \mathbb{R}$. Therefore \begin{align*} (u_{\varepsilon_k}-\psi)^- \to (\bar u-\psi)^- \quad \text{strongly in } L^2(\Omega). \end{align*} The previous step also gives $(u_{\varepsilon_k}-\psi)^- \to 0$ strongly in $L^2(\Omega)$, so [uniqueness of limits](/theorems/625) in $L^2(\Omega)$ gives $(\bar u-\psi)^-=0$ in $L^2(\Omega)$. Hence $\bar u \geq \psi$ for $\mathcal{L}^n$-a.e. $x \in \Omega$, and therefore $\bar u \in K_\psi$. [guided] Fix an arbitrary sequence $(\varepsilon_k)_{k=1}^\infty$ with $\varepsilon_k \downarrow 0$. The purpose of passing to sequences is to prove convergence as $\varepsilon \downarrow 0$ by the usual sequential criterion: every sequence of parameters tending to zero must have the corresponding functions tending to the same limit. The uniform estimate from the previous steps says that $(u_{\varepsilon_k})_{k=1}^\infty$ is bounded in $H^1_0(\Omega)$. The space $H^1_0(\Omega)$ is a Hilbert space with its usual Sobolev norm, so bounded sequences have weakly convergent subsequences by the Hilbert-space weak [sequential compactness](/page/Sequential%20Compactness) theorem. Thus, after passing to a subsequence and keeping the same notation, there is some $\bar u \in H^1_0(\Omega)$ such that \begin{align*} u_{\varepsilon_k} \rightharpoonup \bar u \quad \text{weakly in } H^1_0(\Omega). \end{align*} Weak convergence alone is not enough to pass the pointwise inequality $u_{\varepsilon_k} \geq \psi$ to the limit, because the penalized functions need not actually satisfy that inequality. This is why we used the penalty estimate: it says the negative part of $u_{\varepsilon_k}-\psi$ tends to zero in $L^2(\Omega)$. The compact embedding hypothesis supplies the missing strong convergence; on standard Sobolev domains this compactness is the content of the [Rellich-Kondrachov Theorem](/theorems/64): \begin{align*} u_{\varepsilon_k} \to \bar u \quad \text{strongly in } L^2(\Omega). \end{align*} Now define the negative-part map $N: L^2(\Omega) \to L^2(\Omega)$ by $N(v) := (v-\psi)^-$. This map is well-defined because $\psi \in H^1(\Omega) \subset L^2(\Omega)$ on the bounded domain $\Omega$, and it is Lipschitz with constant $1$. The scalar inequality behind this is \begin{align*} |a^- - b^-| \leq |a-b| \end{align*} for all $a,b \in \mathbb{R}$. Applying this pointwise with $a=v(x)-\psi(x)$ and $b=z(x)-\psi(x)$ and then integrating gives \begin{align*} \|N(v)-N(z)\|_{L^2(\Omega)} \leq \|v-z\|_{L^2(\Omega)}. \end{align*} Therefore the strong $L^2(\Omega)$ convergence of $u_{\varepsilon_k}$ implies \begin{align*} (u_{\varepsilon_k}-\psi)^- \to (\bar u-\psi)^- \quad \text{strongly in } L^2(\Omega). \end{align*} But the penalty estimate already gives \begin{align*} (u_{\varepsilon_k}-\psi)^- \to 0 \quad \text{strongly in } L^2(\Omega). \end{align*} Since limits in $L^2(\Omega)$ are unique, we obtain $(\bar u-\psi)^-=0$ in $L^2(\Omega)$. This means $\bar u-\psi \geq 0$ for $\mathcal{L}^n$-a.e. $x \in \Omega$, so $\bar u \in K_\psi$. [/guided] [/step] [step:Pass to the limit in the minimality inequality to identify the weak limit] Let $w \in K_\psi$ be arbitrary. Since $w-\psi \geq 0$ for $\mathcal{L}^n$-a.e. $x \in \Omega$, the penalty vanishes on $w$: \begin{align*} P_{\varepsilon_k}[w] = 0. \end{align*} Minimality of $u_{\varepsilon_k}$ gives \begin{align*} J[u_{\varepsilon_k}] + P_{\varepsilon_k}[u_{\varepsilon_k}] \leq J[w]. \end{align*} Since $P_{\varepsilon_k}[u_{\varepsilon_k}] \geq 0$, \begin{align*} J[u_{\varepsilon_k}] \leq J[w]. \end{align*} The weak convergence $u_{\varepsilon_k} \rightharpoonup \bar u$ in $H^1_0(\Omega)$ gives weak convergence of gradients in $L^2(\Omega;\mathbb{R}^n)$. The squared $L^2$ norm is weakly lower semicontinuous, hence \begin{align*} \int_\Omega |\nabla \bar u|^2 \, d\mathcal{L}^n(x) \leq \liminf_{k\to\infty}\int_\Omega |\nabla u_{\varepsilon_k}|^2 \, d\mathcal{L}^n(x). \end{align*} The strong convergence $u_{\varepsilon_k}\to \bar u$ in $L^2(\Omega)$ and $f\in L^2(\Omega)$ imply, by the [Cauchy-Schwarz inequality](/theorems/432), \begin{align*} \int_\Omega f u_{\varepsilon_k} \, d\mathcal{L}^n(x) \to \int_\Omega f \bar u \, d\mathcal{L}^n(x). \end{align*} Therefore \begin{align*} J[\bar u] \leq \liminf_{k\to\infty}J[u_{\varepsilon_k}] \leq J[w]. \end{align*} Since $w \in K_\psi$ was arbitrary and $\bar u \in K_\psi$, the function $\bar u$ minimizes $J$ over $K_\psi$. By uniqueness of the obstacle solution, $\bar u = u$. [/step] [step:Recover convergence of the Dirichlet energies] Taking $w=u$ in the minimality inequality and using $P_{\varepsilon_k}[u]=0$ gives \begin{align*} J[u_{\varepsilon_k}] + P_{\varepsilon_k}[u_{\varepsilon_k}] \leq J[u]. \end{align*} Since $P_{\varepsilon_k}[u_{\varepsilon_k}] \geq 0$, \begin{align*} \limsup_{k\to\infty}J[u_{\varepsilon_k}] \leq J[u]. \end{align*} On the other hand, weak lower semicontinuity of the Dirichlet term and strong $L^2(\Omega)$ convergence in the load term give \begin{align*} J[u] \leq \liminf_{k\to\infty}J[u_{\varepsilon_k}]. \end{align*} Thus \begin{align*} J[u_{\varepsilon_k}] \to J[u]. \end{align*} Since \begin{align*} \int_\Omega f u_{\varepsilon_k} \, d\mathcal{L}^n(x) \to \int_\Omega f u \, d\mathcal{L}^n(x), \end{align*} we conclude that \begin{align*} \int_\Omega |\nabla u_{\varepsilon_k}|^2 \, d\mathcal{L}^n(x) \to \int_\Omega |\nabla u|^2 \, d\mathcal{L}^n(x). \end{align*} [/step] [step:Upgrade weak convergence and norm convergence to strong convergence] We already know that \begin{align*} \nabla u_{\varepsilon_k} \rightharpoonup \nabla u \quad \text{weakly in } L^2(\Omega;\mathbb{R}^n) \end{align*} and that \begin{align*} \|\nabla u_{\varepsilon_k}\|_{L^2(\Omega)} \to \|\nabla u\|_{L^2(\Omega)}. \end{align*} Using the Hilbert space identity in $L^2(\Omega;\mathbb{R}^n)$, \begin{align*} \|\nabla u_{\varepsilon_k}-\nabla u\|_{L^2(\Omega)}^2 = \|\nabla u_{\varepsilon_k}\|_{L^2(\Omega)}^2 + \|\nabla u\|_{L^2(\Omega)}^2 - 2\int_\Omega \nabla u_{\varepsilon_k}\cdot \nabla u \, d\mathcal{L}^n(x). \end{align*} The weak convergence gives convergence of the [inner product](/page/Inner%20Product) term, and the norm convergence gives convergence of the first term. Hence \begin{align*} \|\nabla u_{\varepsilon_k}-\nabla u\|_{L^2(\Omega)} \to 0. \end{align*} By the [Poincare inequality](/theorems/75), \begin{align*} \|u_{\varepsilon_k}-u\|_{L^2(\Omega)} \leq C_P\|\nabla u_{\varepsilon_k}-\nabla u\|_{L^2(\Omega)} \to 0. \end{align*} Therefore \begin{align*} u_{\varepsilon_k} \to u \quad \text{strongly in } H^1_0(\Omega). \end{align*} It remains to pass from subsequences to the full parameter limit. Suppose strong convergence failed. Then there would exist a number $\delta>0$ and a sequence $\varepsilon_k \downarrow 0$ such that \begin{align*} \|u_{\varepsilon_k}-u\|_{H^1_0(\Omega)} \geq \delta \end{align*} for every $k$. The argument above applies to this sequence and produces a further subsequence that converges strongly to $u$ in $H^1_0(\Omega)$, contradicting the displayed lower bound. Hence \begin{align*} u_\varepsilon \to u \quad \text{strongly in } H^1_0(\Omega) \end{align*} as $\varepsilon \downarrow 0$. [/step]