**Idea of the proof.** (1) If $u$ solves (DP), compare $I[w]$ with $I[u]$ for any $w \in \mathcal{A}$. Expanding $I[w]-I[u]$ and using Green’s identity with the boundary term vanishing (since $w-u=0$ on $\partial U$) shows \begin{align*} I[w]-I[u] = \tfrac{1}{2}\int_U |\nabla(w-u)|^2 \, d\mathcal{L}^n \ge 0. \end{align*} (2) Conversely, if $u$ minimizes $I$ over $\mathcal{A}$, take interior variations $u+\tau v$ with $v \in C_c^\infty(U)$ so that the boundary condition remains $g$. Differentiating $I[u+\tau v]$ at $\tau=0$ (justified below) yields the Euler–Lagrange identity \begin{align*} \int_U \nabla u \cdot \nabla v \, d\mathcal{L}^n = \int_U f\,v \, d\mathcal{L}^n \quad \text{for all } v \in C_c^\infty(U), \end{align*} which means $-\,\Delta u = f$ in distributions, hence classically since $u \in C^2(U)$.

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**Step 1 (from PDE to minimality: compute $I[w]-I[u]$).** Fix $w \in \mathcal{A}$ and set $\varphi := w-u \in C^2(U)\cap C(\overline{U})$ with $\varphi|_{\partial U}=0$. Then \begin{align*} I[w]-I[u] &= \int_U \left( \tfrac{1}{2}|\nabla(u+\varphi)|^2 - f(u+\varphi) - \tfrac{1}{2}|\nabla u|^2 + f u \right) d\mathcal{L}^n \\ &= \int_U \left( \nabla u \cdot \nabla \varphi + \tfrac{1}{2}|\nabla \varphi|^2 - f\,\varphi \right) d\mathcal{L}^n. \tag{A1} \end{align*} Since $u$ solves (DP), $-\,\Delta u = f$ in $U$ and $u=g$ on $\partial U$. By Gauss–Green on the $C^1$ domain $U$, \begin{align*} \int_U \nabla u \cdot \nabla \varphi \, d\mathcal{L}^n = - \int_U (\Delta u)\,\varphi \, d\mathcal{L}^n + \int_{\partial U} \varphi\, \partial_\nu u \, d\mathcal{H}^{n-1} = \int_U f\,\varphi \, d\mathcal{L}^n, \tag{A2} \end{align*} because $\varphi=0$ on $\partial U$. Inserting (A2) into (A1) gives \begin{align*} I[w]-I[u] = \tfrac{1}{2} \int_U |\nabla \varphi|^2 \, d\mathcal{L}^n \ge 0, \end{align*} hence $I[u] \le I[w]$ for all $w \in \mathcal{A}$, i.e. $u$ minimizes $I$.

**Step 2 (from minimality to PDE: first variation and its justification).** Assume $u \in \mathcal{A}$ minimizes $I$ over $\mathcal{A}$. Let $v \in C_c^\infty(U)$. For each $\tau \in \mathbb{R}$, define \begin{align*} i(\tau) := I[u+\tau v] = \int_U \left( \tfrac{1}{2}|\nabla u + \tau \nabla v|^2 - f(u+\tau v) \right) d\mathcal{L}^n. \end{align*} Because $v$ has compact support in $U$, $u+\tau v \in \mathcal{A}$ for all $\tau$, so $\tau=0$ is a minimizer of $i$. We claim $i$ is differentiable at $0$ with \begin{align*} i'(0) = \int_U \nabla u \cdot \nabla v \, d\mathcal{L}^n - \int_U f\, v \, d\mathcal{L}^n. \tag{B1} \end{align*} To justify differentiability under the integral sign, note that the integrand is a polynomial in $\tau$ with continuous coefficients: \begin{align*} \tfrac{1}{2}|\nabla u + \tau \nabla v|^2 - f(u+\tau v) = \tfrac{1}{2}|\nabla u|^2 + \tau\, \nabla u \cdot \nabla v + \tfrac{1}{2}\tau^2 |\nabla v|^2 - f u - \tau f v. \end{align*} Hence \begin{align*} \frac{d}{d\tau} \Big[ \tfrac{1}{2}|\nabla u + \tau \nabla v|^2 - f(u+\tau v) \Big] = \nabla u \cdot \nabla v + \tau |\nabla v|^2 - f v, \end{align*} which is continuous in $\tau$ and integrable over $U$ (since $u, v$ are smooth and $U$ is bounded). Therefore $i$ is $C^1$ and \begin{align*} i'(\tau) = \int_U \big( \nabla u \cdot \nabla v + \tau |\nabla v|^2 - f v \big) d\mathcal{L}^n, \end{align*} so in particular (B1) holds. Since $\tau=0$ minimizes $i$, $i'(0)=0$, and hence \begin{align*} \int_U \nabla u \cdot \nabla v \, d\mathcal{L}^n = \int_U f\, v \, d\mathcal{L}^n \quad \text{for all } v \in C_c^\infty(U). \tag{B2} \end{align*} Equation (B2) is the weak formulation of $-\,\Delta u = f$ in $U$. Because $u \in C^2(U)$, this implies $-\,\Delta u = f$ pointwise in $U$. Together with $u=g$ on $\partial U$ (since $u \in \mathcal{A}$), $u$ solves (DP).