[proofplan] The proof is a direct specialization of the [Green representation formula for Poisson's equation](/theorems/42) for the Dirichlet Green function. The cited theorem is applied only because the present statement explicitly assumes that its remaining approximation or boundary-regularity hypotheses hold for $u$; the bounded $C^2$ domain, the Dirichlet Green function, and the boundary trace supply the geometric and boundary objects appearing in the formula. The continuity hypotheses on $f$ and $g$ make the interior and boundary integrands meaningful after the substitutions $-\Delta u=f$ in $\Omega$ and $u|_{\partial\Omega}=g$ on $\partial\Omega$. The definition of the Poisson kernel then rewrites the boundary term in the stated form. [/proofplan] [step:Apply the Green representation formula to $u$] Fix $x \in \Omega$. The [Green representation formula for Poisson's equation](/theorems/42) requires a bounded domain with a Dirichlet Green function and enough regularity or approximation control for the integration-by-parts identity defining the boundary term. The bounded $C^2$ domain $\Omega$ and its Dirichlet Green function $G$ are hypotheses of the theorem statement, while the remaining applicability condition for $u$ is exactly the stated approximation or $C^2(\overline{\Omega})$ alternative. Therefore the formula gives \begin{align*} u(x)= -\int_\Omega G(y,x)\Delta u(y)\,d\mathcal L^n(y)-\int_{\partial\Omega}\frac{\partial G}{\partial\nu_y}(y,x)u(y)\,d\mathcal H^{n-1}(y). \end{align*} Here the first integral is taken with respect to $n$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\Omega$, and the boundary integral is taken with respect to $(n-1)$-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) on $\partial\Omega$. [guided] Fix a point $x \in \Omega$. We want to invoke the [Green representation formula for Poisson's equation](/theorems/42), so we check what is needed in the present argument. The formula requires the Dirichlet Green function on a sufficiently regular bounded domain and enough regularity or approximation control on the represented function so that the integration-by-parts formula, including the boundary normal derivative term, is valid. The domain hypothesis supplies a bounded $C^2$ domain $\Omega$, the theorem statement supplies its Dirichlet Green function $G$, and the remaining regularity requirement for $u$ is not being reproved here because it is exactly the statement's assumption that the formula applies by approximation or by $u\in C^2(\overline{\Omega})$. Under these verified or explicitly assumed hypotheses, the cited formula expresses the value of $u$ at $x$ as \begin{align*} u(x)= -\int_\Omega G(y,x)\Delta u(y)\,d\mathcal L^n(y)-\int_{\partial\Omega}\frac{\partial G}{\partial\nu_y}(y,x)u(y)\,d\mathcal H^{n-1}(y). \end{align*} The variable $y$ is the integration variable. The first integral is with respect to $n$-dimensional Lebesgue measure on $\Omega$, and the second is with respect to $(n-1)$-dimensional Hausdorff measure on $\partial\Omega$. The derivative $\frac{\partial G}{\partial\nu_y}(y,x)$ is the outward normal derivative in the first variable of $G$, evaluated at the boundary point $y\in\partial\Omega$. The continuity of the boundary trace and the stated applicability of the Green formula are what make this boundary expression legitimate in this proof. [/guided] [/step] [step:Substitute the equation and boundary trace] Because $-\Delta u=f$ in $\Omega$, we have $-\Delta u(y)=f(y)$ for every $y \in \Omega$. Because $u|_{\partial \Omega}=g$, we have $u(y)=g(y)$ for every $y \in \partial\Omega$. Substituting these two identities into the [representation formula](/theorems/39) gives \begin{align*} u(x)=\int_\Omega G(y,x)f(y)\,d\mathcal L^n(y)-\int_{\partial\Omega}\frac{\partial G}{\partial\nu_y}(y,x)g(y)\,d\mathcal H^{n-1}(y). \end{align*} [/step] [step:Rewrite the boundary term using the Poisson kernel] By definition, \begin{align*} P(x,y)=-\frac{\partial G}{\partial\nu_y}(y,x) \end{align*} for $x \in \Omega$ and $y \in \partial\Omega$. Therefore \begin{align*} -\int_{\partial\Omega}\frac{\partial G}{\partial\nu_y}(y,x)g(y)\,d\mathcal H^{n-1}(y)=\int_{\partial\Omega}P(x,y)g(y)\,d\mathcal H^{n-1}(y). \end{align*} Combining this identity with the preceding step yields, for the fixed point $x \in \Omega$, \begin{align*} u(x)=\int_\Omega G(y,x)f(y)\,d\mathcal L^n(y)+\int_{\partial\Omega}P(x,y)g(y)\,d\mathcal H^{n-1}(y). \end{align*} Since $x \in \Omega$ was arbitrary, the representation holds for every $x \in \Omega$. [/step]