[proofplan] Subtract the prescribed lifting $G$ from $u$ to reduce the boundary condition to the homogeneous Dirichlet case. The difference $v := u - G$ belongs to $H^1_0(U)$ and satisfies a Poisson equation with right-hand side $f + \Delta G \in L^2(U)$. We then apply the global homogeneous Dirichlet $H^2$ estimate on bounded $C^2$ domains and recover the asserted estimate for $u = v + G$ by the triangle inequality. [/proofplan] [step:Subtract the lifting to obtain homogeneous boundary data] Define the map $v: U \to \mathbb{R}$ by \begin{align*} v(x) := u(x) - G(x) \quad \text{for } x \in U. \end{align*} Since $u \in H^1(U)$ and $G \in H^2(U) \subset H^1(U)$, we have $v \in H^1(U)$. The trace operator is linear, and the hypotheses give $\operatorname{Tr}u = g$ and $\operatorname{Tr}G = g$. Hence \begin{align*} \operatorname{Tr}v = \operatorname{Tr}u - \operatorname{Tr}G = g - g = 0. \end{align*} Therefore $v \in H^1_0(U)$. [guided] The purpose of introducing $v$ is to remove the inhomogeneous boundary condition. Define the map $v: U \to \mathbb{R}$ by \begin{align*} v(x) := u(x) - G(x) \quad \text{for } x \in U. \end{align*} Because $u \in H^1(U)$ and $G \in H^2(U)$, and because $H^2(U) \subset H^1(U)$, the difference $v$ belongs to $H^1(U)$. Now we check the boundary condition carefully. The trace operator is linear on Sobolev spaces of the relevant orders. The hypotheses say that $\operatorname{Tr}u = g$ and $\operatorname{Tr}G = g$ on $\partial U$. Therefore \begin{align*} \operatorname{Tr}v = \operatorname{Tr}(u - G) = \operatorname{Tr}u - \operatorname{Tr}G = g - g = 0. \end{align*} Thus $v$ has zero trace. By the definition of $H^1_0(U)$ as the [Sobolev space](/page/Sobolev%20Space) of $H^1$ functions with zero trace on a bounded Lipschitz domain, and since a bounded $C^2$ domain is Lipschitz, we conclude that $v \in H^1_0(U)$. [/guided] [/step] [step:Compute the weak equation solved by the homogeneous unknown] Since $G \in H^2(U)$, its weak second derivatives belong to $L^2(U)$, so \begin{align*} \Delta G := \sum_{i=1}^n \partial_{x_i}^2 G \end{align*} defines an element of $L^2(U)$. Let $\varphi \in H^1_0(U)$. Using the weak equation for $u$ and the integration-by-parts identity for $G \in H^2(U)$ against $H^1_0(U)$ test functions, we get \begin{align*} \int_U \nabla v \cdot \nabla \varphi \, d\mathcal{L}^n(x) = \int_U \nabla u \cdot \nabla \varphi \, d\mathcal{L}^n(x) - \int_U \nabla G \cdot \nabla \varphi \, d\mathcal{L}^n(x). \end{align*} The first term equals $\int_U f\varphi \, d\mathcal{L}^n(x)$. For the second term, \begin{align*} \int_U \nabla G \cdot \nabla \varphi \, d\mathcal{L}^n(x) = -\int_U (\Delta G)\varphi \, d\mathcal{L}^n(x). \end{align*} Therefore \begin{align*} \int_U \nabla v \cdot \nabla \varphi \, d\mathcal{L}^n(x) = \int_U (f + \Delta G)\varphi \, d\mathcal{L}^n(x). \end{align*} Thus $v$ solves $-\Delta v = f + \Delta G$ weakly in $U$ with homogeneous Dirichlet boundary data. [/step] [step:Apply the homogeneous global $H^2$ estimate] Set $F: U \to \mathbb{R}$ by \begin{align*} F(x) := f(x) + \Delta G(x) \quad \text{for } x \in U. \end{align*} Since $f \in L^2(U)$ and $\Delta G \in L^2(U)$, we have $F \in L^2(U)$. The domain $U$ is bounded with $C^2$ boundary, and $v \in H^1_0(U)$ solves $-\Delta v = F$ weakly in $U$. By the homogeneous global Dirichlet $H^2$ estimate for the Poisson equation on bounded $C^2$ domains, applied to $v$ and $F$, there exists a constant $C_0 = C_0(U) > 0$ such that $v \in H^2(U)$ and \begin{align*} \|v\|_{H^2(U)} \leq C_0\|F\|_{L^2(U)}. \end{align*} Equivalently, \begin{align*} \|v\|_{H^2(U)} \leq C_0\|f + \Delta G\|_{L^2(U)}. \end{align*} Here we are citing a result not yet verified in the wiki: the homogeneous global Dirichlet $H^2$ estimate for the Poisson equation on bounded $C^2$ domains. [/step] [step:Estimate the lifted solution in $H^2(U)$] Since $u = v + G$, and since both $v$ and $G$ belong to $H^2(U)$, we have $u \in H^2(U)$. By the triangle inequality in $H^2(U)$, \begin{align*} \|u\|_{H^2(U)} \leq \|v\|_{H^2(U)} + \|G\|_{H^2(U)}. \end{align*} Using the estimate for $v$ gives \begin{align*} \|u\|_{H^2(U)} \leq C_0\|f + \Delta G\|_{L^2(U)} + \|G\|_{H^2(U)}. \end{align*} The triangle inequality in $L^2(U)$ gives \begin{align*} \|f + \Delta G\|_{L^2(U)} \leq \|f\|_{L^2(U)} + \|\Delta G\|_{L^2(U)}. \end{align*} Because the triangle inequality in $L^2(U)$ and the [Cauchy-Schwarz inequality](/theorems/432) in $\mathbb{R}^n$ give \begin{align*} \|\Delta G\|_{L^2(U)} \leq \sum_{i=1}^n \|\partial_{x_i}^2 G\|_{L^2(U)} \leq \sqrt n\left(\sum_{i=1}^n \|\partial_{x_i}^2 G\|_{L^2(U)}^2\right)^{1/2} \leq \sqrt n\|G\|_{H^2(U)}, \end{align*} we obtain \begin{align*} \|u\|_{H^2(U)} \leq C_0\|f\|_{L^2(U)} + (C_0\sqrt n + 1)\|G\|_{H^2(U)}. \end{align*} Defining $C := \max\{C_0, C_0\sqrt n + 1\}$, which depends only on $U$ because $n$ is determined by $U \subset \mathbb{R}^n$, yields \begin{align*} \|u\|_{H^2(U)} \leq C\left(\|f\|_{L^2(U)} + \|G\|_{H^2(U)}\right). \end{align*} This is the desired estimate. [/step]