[proofplan] We localize the distributional equation between nested open sets and mollify $u$ and $f$ there, obtaining smooth functions $u_\varepsilon$ satisfying the classical Poisson equation $-\Delta u_\varepsilon=f_\varepsilon$ on an intermediate set. The smooth interior Calderon-Zygmund estimate, together with the local interpolation inequality, gives a uniform $W^{2,p}$ bound for $u_\varepsilon$ on $V$. Since $1<p<\infty$, boundedness in $W^{2,p}(V)$ gives weak compactness; the strong $L^p$ convergence of $u_\varepsilon$ and [weak convergence](/page/Weak%20Convergence) of its derivatives identify the weak limits as the distributional derivatives of $u$. Lower semicontinuity of the Sobolev norm then gives the stated estimate. [/proofplan]

[step:Choose nested open sets with room for mollification] Let $V,U' \subset \mathbb{R}^n$ be open sets with $V \subset\subset U' \subset\subset U$. Let $\mathcal{L}^n$ denote $n$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}^n$. For $x\in\mathbb{R}^n$ and $r>0$, write $B(x,r):=\{y\in\mathbb{R}^n: |y-x|<r\}$ for the open Euclidean ball. Choose open sets $W,W' \subset \mathbb{R}^n$ such that \begin{align*} V \subset\subset W \subset\subset W' \subset\subset U'. \end{align*} Define \begin{align*} \rho := \operatorname{dist}(W,\mathbb{R}^n \setminus W') > 0. \end{align*} Let $\eta \in C_c^\infty(B(0,1))$ be a nonnegative [standard mollifier](/page/Standard%20Mollifier) satisfying \begin{align*} \int_{\mathbb{R}^n} \eta(z)\,d\mathcal{L}^n(z)=1. \end{align*} For $0<\varepsilon<\rho$, define \begin{align*} \eta_\varepsilon: \mathbb{R}^n \to \mathbb{R}, \quad z \mapsto \varepsilon^{-n}\eta(z/\varepsilon). \end{align*} Extend $u|_{W'}$ and $f|_{W'}$ by zero to $\mathbb{R}^n$, and define \begin{align*} u_\varepsilon: W \to \mathbb{R}, \quad x \mapsto \int_{\mathbb{R}^n}\eta_\varepsilon(x-y)u(y)\,d\mathcal{L}^n(y), \end{align*} and \begin{align*} f_\varepsilon: W \to \mathbb{R}, \quad x \mapsto \int_{\mathbb{R}^n}\eta_\varepsilon(x-y)f(y)\,d\mathcal{L}^n(y). \end{align*} Because $0<\varepsilon<\rho$ and $x \in W$ imply $B(x,\varepsilon) \subset W'$, these convolutions use only the original values of $u$ and $f$ on $W'$ when evaluated on $W$. [/step]

[step:Mollify the distributional equation inside $W$]For every $0<\varepsilon<\rho$, the functions $u_\varepsilon$ and $f_\varepsilon$ belong to $C^\infty(W)$, and \begin{align*} -\Delta u_\varepsilon = f_\varepsilon \quad \text{pointwise on } W. \end{align*} This is the standard local commutation of convolution with constant-coefficient distributional derivatives, applied to the identity $-\Delta u=f$ on $W'$; the support condition $B(x,\varepsilon)\subset W'$ ensures that no boundary term from the zero extension is seen on $W$. Moreover, by the local approximation property of mollifiers in $L^p$, applied on $W \subset\subset W'$, we have \begin{align*} u_\varepsilon \to u \quad \text{in } L^p(W) \end{align*} and \begin{align*} f_\varepsilon \to f \quad \text{in } L^p(W) \end{align*} as $\varepsilon \downarrow 0$.[/step]

[guided]The reason for introducing $W'$ is to mollify without touching the boundary of $U'$. Since $W \subset\subset W'$, the number \begin{align*} \rho := \operatorname{dist}(W,\mathbb{R}^n \setminus W') \end{align*} is positive. Thus, whenever $0<\varepsilon<\rho$ and $x \in W$, the support of the function $y \mapsto \eta_\varepsilon(x-y)$ is contained in $B(x,\varepsilon)\subset W'$. Therefore the convolution defining $u_\varepsilon(x)$ and $f_\varepsilon(x)$ depends only on the values of $u$ and $f$ on $W'$. The distributional equation says that for every [test function](/page/Test%20Function) $\psi \in C_c^\infty(W')$, \begin{align*} -\int_{W'} u(y)\Delta\psi(y)\,d\mathcal{L}^n(y)=\int_{W'} f(y)\psi(y)\,d\mathcal{L}^n(y). \end{align*} Convolution with a smooth compactly supported kernel commutes with the constant-coefficient operator $\Delta$ on the smaller set $W$, precisely because the translated kernel remains supported in $W'$. Hence \begin{align*} -\Delta u_\varepsilon = \eta_\varepsilon * (-\Delta u)=\eta_\varepsilon * f=f_\varepsilon \quad \text{on } W. \end{align*} Here the convolution is understood locally on $W'$ and evaluated only on $W$. Finally, since $u,f \in L^p(W')$ and $1<p<\infty$, the standard $L^p$ approximation theorem for mollifiers gives \begin{align*} u_\varepsilon \to u \quad \text{in } L^p(W) \end{align*} and \begin{align*} f_\varepsilon \to f \quad \text{in } L^p(W). \end{align*} This strong convergence is the mechanism that will later identify the weak Sobolev limit with the original distribution $u$.[/guided]

[step:Apply the smooth interior estimate uniformly] By the smooth interior Calderon-Zygmund estimate for the Laplacian on nested open sets, applied with $V\subset\subset W$ and exponent $1<p<\infty$, and by the local [Sobolev interpolation inequality](/theorems/906) for first derivatives on the same pair of sets, there exists a constant $C_0=C_0(n,p,V,W)>0$ such that every smooth function $v \in C^\infty(W)$ satisfies \begin{align*} \|v\|_{W^{2,p}(V)} \le C_0\left(\|-\Delta v\|_{L^p(W)}+\|v\|_{L^p(W)}\right). \end{align*} Applying this estimate to $v=u_\varepsilon$ and using $-\Delta u_\varepsilon=f_\varepsilon$ on $W$ gives \begin{align*} \|u_\varepsilon\|_{W^{2,p}(V)} \le C_0\left(\|f_\varepsilon\|_{L^p(W)}+\|u_\varepsilon\|_{L^p(W)}\right). \end{align*} [Young's convolution inequality](/theorems/463) applied to the zero extensions of $u|_{W'}$ and $f|_{W'}$ gives \begin{align*} \|u_\varepsilon\|_{L^p(W)} \le \|u\|_{L^p(W')} \end{align*} and \begin{align*} \|f_\varepsilon\|_{L^p(W)} \le \|f\|_{L^p(W')}. \end{align*} Since $W' \subset U'$, we obtain the uniform estimate \begin{align*} \|u_\varepsilon\|_{W^{2,p}(V)} \le C_0\left(\|f\|_{L^p(U')}+\|u\|_{L^p(U')}\right). \end{align*} The two external ingredients used here are the smooth interior Calderon-Zygmund estimate for the Laplacian and the local Sobolev interpolation inequality for first derivatives; they are invoked exactly in the nested-open-set form stated above. [/step]

[step:Extract weak limits for all first and second derivatives] Because $1<p<\infty$, the space $L^p(V)$ is reflexive. The uniform $W^{2,p}(V)$ bound implies that the families \begin{align*} (u_\varepsilon)_{\varepsilon>0}, \quad (\partial_{x_i}u_\varepsilon)_{\varepsilon>0}, \quad (\partial_{x_i}\partial_{x_j}u_\varepsilon)_{\varepsilon>0} \end{align*} are bounded in $L^p(V)$ for all $1\le i,j\le n$. Choose a sequence $\varepsilon_k \downarrow 0$. By weak compactness in $L^p(V)$ and a diagonal extraction over the finite index set $\{1,\dots,n\}\cup \{1,\dots,n\}^2$, there exist functions $h_i,g_{ij}\in L^p(V)$ and a subsequence, still denoted by $\varepsilon_k$, such that \begin{align*} \partial_{x_i}u_{\varepsilon_k} \rightharpoonup h_i \quad \text{weakly in } L^p(V) \end{align*} for every $1\le i\le n$, and \begin{align*} \partial_{x_i}\partial_{x_j}u_{\varepsilon_k} \rightharpoonup g_{ij} \quad \text{weakly in } L^p(V) \end{align*} for every $1\le i,j\le n$. Also, since $u_{\varepsilon_k}\to u$ strongly in $L^p(W)$ and $V\subset W$, we have \begin{align*} u_{\varepsilon_k}\to u \quad \text{strongly in } L^p(V). \end{align*} [/step]

[step:Identify the weak limits as distributional derivatives of $u$]Let $\phi \in C_c^\infty(V)$ be arbitrary. Since $u_{\varepsilon_k}\in C^\infty(V)$, [integration by parts](/theorems/210) gives \begin{align*} \int_V u_{\varepsilon_k}(x)\partial_{x_i}\phi(x)\,d\mathcal{L}^n(x)=-\int_V \partial_{x_i}u_{\varepsilon_k}(x)\phi(x)\,d\mathcal{L}^n(x). \end{align*} Define the conjugate exponent $p'\in(1,\infty)$ by the displayed identity \begin{align*} p' = \frac{p}{p-1}. \end{align*} The left-hand side converges to \begin{align*} \int_V u(x)\partial_{x_i}\phi(x)\,d\mathcal{L}^n(x) \end{align*} by strong $L^p(V)$ convergence and the fact that $\partial_{x_i}\phi\in L^{p'}(V)$. The right-hand side converges to \begin{align*} -\int_V h_i(x)\phi(x)\,d\mathcal{L}^n(x) \end{align*} by weak convergence in $L^p(V)$ and $\phi\in L^{p'}(V)$. Hence $h_i=\partial_{x_i}u$ in $\mathcal{D}'(V)$. Similarly, [integration by parts](/theorems/2098) twice gives \begin{align*} \int_V u_{\varepsilon_k}(x)\partial_{x_i}\partial_{x_j}\phi(x)\,d\mathcal{L}^n(x)=\int_V \partial_{x_i}\partial_{x_j}u_{\varepsilon_k}(x)\phi(x)\,d\mathcal{L}^n(x). \end{align*} Passing to the limit as before yields \begin{align*} \int_V u(x)\partial_{x_i}\partial_{x_j}\phi(x)\,d\mathcal{L}^n(x)=\int_V g_{ij}(x)\phi(x)\,d\mathcal{L}^n(x). \end{align*} Therefore $g_{ij}=\partial_{x_i}\partial_{x_j}u$ in $\mathcal{D}'(V)$ for every $1\le i,j\le n$. Since $u$, all first weak derivatives, and all second weak derivatives belong to $L^p(V)$, we conclude that $u\in W^{2,p}(V)$.[/step]

[guided]This step starts from the compactness conclusions already established: after passing to a subsequence, the first derivatives $\partial_{x_i}u_{\varepsilon_k}$ converge weakly in $L^p(V)$ to functions $h_i\in L^p(V)$, the second derivatives $\partial_{x_i}\partial_{x_j}u_{\varepsilon_k}$ converge weakly in $L^p(V)$ to functions $g_{ij}\in L^p(V)$, and $u_{\varepsilon_k}\to u$ strongly in $L^p(V)$. We must prove that these functions are not artificial limits, but are exactly the weak derivatives of $u$. Fix a test function $\phi\in C_c^\infty(V)$. For the first derivatives, the smoothness of $u_{\varepsilon_k}$ permits classical integration by parts: \begin{align*} \int_V u_{\varepsilon_k}(x)\partial_{x_i}\phi(x)\,d\mathcal{L}^n(x)=-\int_V \partial_{x_i}u_{\varepsilon_k}(x)\phi(x)\,d\mathcal{L}^n(x). \end{align*} There is no boundary term because $\phi$ has compact support in $V$. Now we pass to the limit on both sides. Define the Hölder conjugate exponent $p'\in(1,\infty)$ by the displayed identity \begin{align*} p' = \frac{p}{p-1}. \end{align*} Since $\partial_{x_i}\phi\in C_c^\infty(V)\subset L^{p'}(V)$ and $u_{\varepsilon_k}\to u$ strongly in $L^p(V)$, Hölder's inequality gives convergence of the left-hand side to \begin{align*} \int_V u(x)\partial_{x_i}\phi(x)\,d\mathcal{L}^n(x). \end{align*} Since $\partial_{x_i}u_{\varepsilon_k}\rightharpoonup h_i$ weakly in $L^p(V)$ and $\phi\in L^{p'}(V)$, the right-hand side converges to \begin{align*} -\int_V h_i(x)\phi(x)\,d\mathcal{L}^n(x). \end{align*} Therefore \begin{align*} \int_V u(x)\partial_{x_i}\phi(x)\,d\mathcal{L}^n(x)=-\int_V h_i(x)\phi(x)\,d\mathcal{L}^n(x), \end{align*} which is exactly the definition of $h_i=\partial_{x_i}u$ in $\mathcal{D}'(V)$. For second derivatives, the same idea is applied with two integrations by parts. Since $u_{\varepsilon_k}$ is smooth and $\phi$ is compactly supported in $V$, \begin{align*} \int_V u_{\varepsilon_k}(x)\partial_{x_i}\partial_{x_j}\phi(x)\,d\mathcal{L}^n(x)=\int_V \partial_{x_i}\partial_{x_j}u_{\varepsilon_k}(x)\phi(x)\,d\mathcal{L}^n(x). \end{align*} The left-hand side converges by strong $L^p(V)$ convergence of $u_{\varepsilon_k}$ to $u$, because $\partial_{x_i}\partial_{x_j}\phi\in L^{p'}(V)$. The right-hand side converges by weak $L^p(V)$ convergence of $\partial_{x_i}\partial_{x_j}u_{\varepsilon_k}$ to $g_{ij}$, because $\phi\in L^{p'}(V)$. Hence \begin{align*} \int_V u(x)\partial_{x_i}\partial_{x_j}\phi(x)\,d\mathcal{L}^n(x)=\int_V g_{ij}(x)\phi(x)\,d\mathcal{L}^n(x). \end{align*} This is precisely the distributional identity $g_{ij}=\partial_{x_i}\partial_{x_j}u$ on $V$. Thus the weak compactness argument has produced all first and second weak derivatives of $u$ as actual $L^p(V)$ functions. Together with $u\in L^p(V)$, this proves $u\in W^{2,p}(V)$.[/guided]

[step:Pass the uniform estimate to the limit] The norm on $W^{2,p}(V)$ is lower semicontinuous under weak convergence of all derivatives in $L^p(V)$. Since $u_{\varepsilon_k}\to u$ strongly in $L^p(V)$, $\partial_{x_i}u_{\varepsilon_k}\rightharpoonup \partial_{x_i}u$ weakly in $L^p(V)$, and $\partial_{x_i}\partial_{x_j}u_{\varepsilon_k}\rightharpoonup \partial_{x_i}\partial_{x_j}u$ weakly in $L^p(V)$, we obtain \begin{align*} \|u\|_{W^{2,p}(V)} \le \liminf_{k\to\infty}\|u_{\varepsilon_k}\|_{W^{2,p}(V)}. \end{align*} Using the uniform estimate from the smooth approximation step gives \begin{align*} \|u\|_{W^{2,p}(V)} \le C_0\left(\|f\|_{L^p(U')}+\|u\|_{L^p(U')}\right). \end{align*} The intermediate sets $W$ and $W'$ were fixed once after $V$ and $U'$ were chosen, so the dependence of $C_0=C_0(n,p,V,W)$ may be recorded as dependence on $n$, $p$, $V$, and $U'$. Renaming $C_0$ as $C=C(n,p,V,U')$ proves the asserted estimate. Since $V\subset\subset U$ was arbitrary after choosing an intermediate $U'$, this also proves $u\in W^{2,p}_{\mathrm{loc}}(U)$. [/step]