[proofplan] The proof is a direct verification of what the Green kernel is designed to do. We test the distribution $Lu$ against an arbitrary compactly supported smooth function and use the assumed passage of $L_x$ through the $y$-integration. The Green identity $L_xG_y=\delta_y$ then collapses the integral to the [regular distribution](/page/Regular%20Distribution) induced by $f$. The boundary condition follows in the same way from the homogeneous trace condition $\mathcal{T}G_y=0$ for each pole $y$. [/proofplan] [step:Test the represented function against a compactly supported smooth function] Let $\mathcal{D}'(\Omega)$ denote the space of distributions on $\Omega$, the [topological dual](/page/Topological%20Dual) of $C_c^\infty(\Omega)$. Let $\phi \in C_c^\infty(\Omega)$ be an arbitrary [test function](/page/Test%20Function). Since $f \in L^1_{\mathrm{loc}}(\Omega)$ and $\operatorname{supp}\phi$ is compact in $\Omega$, the product $f\phi$ is Lebesgue integrable over $\Omega$. Let $T_f \in \mathcal{D}'(\Omega)$ denote the regular distribution induced by $f$, defined by \begin{align*} T_f(\phi)=\int_\Omega f(x)\phi(x)\,d\mathcal{L}^n(x). \end{align*} By hypothesis, the represented function $u$ belongs to the domain of $L$, so $Lu$ is well-defined as a distribution. To prove $Lu=f$ in $\mathcal{D}'(\Omega)$, it is enough to prove \begin{align*} (Lu)(\phi)=T_f(\phi) \end{align*} for every $\phi \in C_c^\infty(\Omega)$. [/step] [step:Use the Green identity to collapse the distributional integral] For each $y \in \Omega$, define $G_y \in \mathcal{D}'(\Omega)$ by $G_y := G(\cdot,y)$, and let $\delta_y \in \mathcal{D}'(\Omega)$ denote the Dirac distribution at $y$, defined by $\delta_y(\psi)=\psi(y)$ for every $\psi \in C_c^\infty(\Omega)$. By the assumed distributional passage-through identity, \begin{align*} Lu=\int_\Omega L_xG_y\,f(y)\,d\mathcal{L}^n(y) \end{align*} in $\mathcal{D}'(\Omega)$. This identity is part of the hypothesis and includes the legitimacy of testing the distribution-valued integral under the $y$-integration. Evaluating both sides on $\phi$ gives \begin{align*} (Lu)(\phi)=\int_\Omega (L_xG_y)(\phi)f(y)\,d\mathcal{L}^n(y). \end{align*} For each $y \in \Omega$, the defining property of the Green kernel gives $L_xG_y=\delta_y$, so \begin{align*} (L_xG_y)(\phi)=\delta_y(\phi)=\phi(y). \end{align*} Therefore \begin{align*} (Lu)(\phi)=\int_\Omega \phi(y)f(y)\,d\mathcal{L}^n(y)=T_f(\phi). \end{align*} Since $\phi \in C_c^\infty(\Omega)$ was arbitrary, $Lu=T_f$, that is, $Lu=f$ in $\mathcal{D}'(\Omega)$. [guided] We want to prove an equality of distributions, so we do not compare values pointwise. Instead, we compare the two distributions after applying them to an arbitrary test function. Let $\phi \in C_c^\infty(\Omega)$ be fixed. For each $y \in \Omega$, define $G_y \in \mathcal{D}'(\Omega)$ by $G_y := G(\cdot,y)$, and let $\delta_y \in \mathcal{D}'(\Omega)$ denote the Dirac distribution at $y$, defined by $\delta_y(\psi)=\psi(y)$ for every $\psi \in C_c^\infty(\Omega)$. Because $f \in L^1_{\mathrm{loc}}(\Omega)$ and $\operatorname{supp}\phi$ is compact in $\Omega$, the product $f\phi$ is Lebesgue integrable over $\Omega$. The distribution associated to the forcing term $f$ is therefore the regular distribution $T_f \in \mathcal{D}'(\Omega)$ defined by \begin{align*} T_f(\phi)=\int_\Omega f(x)\phi(x)\,d\mathcal{L}^n(x). \end{align*} The hypothesis says that applying $L$ to the represented function $u$ may be passed through the $y$-integration in the distributional sense. Since the statement also assumes that $u$ lies in the domain of $L$, the distribution $Lu$ is defined. The passage-through assumption includes the operation of testing the resulting distribution-valued integral against $\phi$, and therefore \begin{align*} (Lu)(\phi)=\int_\Omega (L_xG_y)(\phi)f(y)\,d\mathcal{L}^n(y). \end{align*} Now the Green kernel property enters. For each fixed $y \in \Omega$, the distribution $G_y$ satisfies $L_xG_y=\delta_y$ in $\mathcal{D}'(\Omega)$. Applying both sides to $\phi$ gives \begin{align*} (L_xG_y)(\phi)=\delta_y(\phi)=\phi(y). \end{align*} Substituting this identity into the previous integral yields \begin{align*} (Lu)(\phi)=\int_\Omega \phi(y)f(y)\,d\mathcal{L}^n(y). \end{align*} This is exactly $T_f(\phi)$. Since the test function $\phi$ was arbitrary, the distributions $Lu$ and $T_f$ agree on all of $C_c^\infty(\Omega)$, and hence $Lu=f$ in $\mathcal{D}'(\Omega)$. [/guided] [/step] [step:Pass the boundary trace through the representation] The operator $\mathcal{T}$ is the boundary trace operator named in the theorem statement. By hypothesis, the represented function $u$ belongs to the domain of $\mathcal{T}$, so $\mathcal{T}u$ is well-defined. The assumed trace passage-through identity gives \begin{align*} \mathcal{T}u=\int_\Omega \mathcal{T}G_y\,f(y)\,d\mathcal{L}^n(y). \end{align*} For every $y \in \Omega$, the Green kernel satisfies the homogeneous boundary condition $\mathcal{T}G_y=0$. Hence \begin{align*} \mathcal{T}u=\int_\Omega 0\cdot f(y)\,d\mathcal{L}^n(y)=0. \end{align*} Thus $u$ satisfies the prescribed homogeneous boundary condition. [guided] We now verify the boundary condition, using the trace operator $\mathcal{T}$ that is part of the theorem statement. The statement assumes that $u$ lies in the domain of $\mathcal{T}$, so the boundary trace $\mathcal{T}u$ is defined. It also assumes that the trace may be passed through the Green representation in the appropriate boundary trace sense. Therefore \begin{align*} \mathcal{T}u=\int_\Omega \mathcal{T}G_y\,f(y)\,d\mathcal{L}^n(y). \end{align*} The defining homogeneous boundary property of the Green kernel says that, for each pole $y \in \Omega$, the trace of $G_y := G(\cdot,y)$ vanishes: \begin{align*} \mathcal{T}G_y=0. \end{align*} Substituting this identity inside the trace representation gives \begin{align*} \mathcal{T}u=\int_\Omega 0\cdot f(y)\,d\mathcal{L}^n(y)=0. \end{align*} Thus the represented function satisfies exactly the homogeneous boundary condition encoded by $\mathcal{T}$. [/guided] [/step] [step:Conclude the representation solves the forced boundary value problem] The preceding steps prove both required assertions: $Lu=f$ in $\mathcal{D}'(\Omega)$ and $\mathcal{T}u=0$ in the prescribed boundary trace sense. Therefore the Green [representation formula](/theorems/39) produces a solution of the forced equation with the corresponding homogeneous boundary condition. [/step]