[proofplan] We reduce the PDE $Lu = f$ to an abstract operator equation $(I - \gamma K)u = Kf$ on $L^2(U)$, where $K$ is a compact operator constructed by inverting the shifted operator $L_\gamma = L + \gamma I$ (which is an isomorphism by the [First Existence Theorem](/theorems/93)) and composing with the compact Sobolev embedding. The abstract Fredholm alternative for the compact operator $\gamma K$ then yields the dichotomy. The main work is translating back: identifying $\ker(I - \gamma K)$ with $N$, identifying $\ker(I - \gamma K^*)$ with $N^*$, and showing that the abstract solvability condition $Kf \perp \ker(I - \gamma K^*)$ is equivalent to $f \perp N^*$. [/proofplan] [step:Construct the compact resolvent operator $K: L^2(U) \to L^2(U)$] Fix $\gamma \ge 0$ large enough so that $L_\gamma := L + \gamma I: H^1_0(U) \to H^{-1}(U)$ is an isomorphism, as guaranteed by the [First Existence Theorem](/theorems/93). Define the resolvent operator as the composition: \begin{align*} K: L^2(U) \xrightarrow{\iota} H^{-1}(U) \xrightarrow{L_\gamma^{-1}} H^1_0(U) \xrightarrow{j} L^2(U), \end{align*} where $\iota: L^2(U) \hookrightarrow H^{-1}(U)$ is the canonical embedding (each $g \in L^2$ acts on $v \in H^1_0$ by $v \mapsto \int_U g\, v \, d\mathcal{L}^n$) and $j: H^1_0(U) \hookrightarrow L^2(U)$ is the Sobolev inclusion. Since $\iota$ and $L_\gamma^{-1}$ are bounded and $j$ is compact by the [Rellich-Kondrachov Theorem](/theorems/64), the composition $K$ is compact. [guided] Why construct $K$? We want to apply the Fredholm alternative, which requires a compact operator. The idea is to "solve away" the invertible part of $L$ and package what remains into a compact perturbation of the identity. The shifted operator $L_\gamma$ is an isomorphism for $\gamma$ large enough (First Existence Theorem). So $L_\gamma^{-1}: H^{-1}(U) \to H^1_0(U)$ is a bounded linear map. To get an operator on $L^2(U)$, we embed $L^2$ into $H^{-1}$ on the input side and $H^1_0$ into $L^2$ on the output side: \begin{align*} K: L^2(U) \xrightarrow{\iota} H^{-1}(U) \xrightarrow{L_\gamma^{-1}} H^1_0(U) \xrightarrow{j} L^2(U). \end{align*} The embedding $\iota$ is bounded (by Cauchy–Schwarz). The map $L_\gamma^{-1}$ is bounded by the open mapping theorem. The embedding $j: H^1_0(U) \hookrightarrow L^2(U)$ is compact by the [Rellich-Kondrachov Theorem](/theorems/64), since $U$ is bounded. A composition of bounded operators with one compact factor is compact, so $K$ is compact. [/guided] [/step] [step:Reduce the PDE $Lu = f$ to the operator equation $(I - \gamma K)u = Kf$] Write $L = L_\gamma - \gamma I$. Then $Lu = f$ becomes $L_\gamma u = f + \gamma u$. Applying $K = j \circ L_\gamma^{-1} \circ \iota$ to $f + \gamma u \in L^2(U) \subset H^{-1}(U)$: \begin{align*} u = K(f + \gamma u) = Kf + \gamma Ku. \end{align*} Rearranging gives $(I - \gamma K)u = Kf$. Conversely, if $(I - \gamma K)u = Kf$ in $L^2(U)$, then $u = K(f + \gamma u) = j \circ L_\gamma^{-1} \circ \iota(f + \gamma u)$. In particular, $u \in H^1_0(U)$ and $L_\gamma u = f + \gamma u$, i.e., $Lu = f$. [/step] [step:Apply the abstract Fredholm alternative to $I - \gamma K$] Since $K: L^2(U) \to L^2(U)$ is compact, the abstract Fredholm alternative for the operator $I - \gamma K$ on the [Hilbert space](/page/Hilbert%20Space) $L^2(U)$ gives: - $\ker(I - \gamma K)$ is finite-dimensional. - $\operatorname{Range}(I - \gamma K)$ is closed and equals $\ker(I - \gamma K^*)^\perp$. - $\dim \ker(I - \gamma K) = \dim \ker(I - \gamma K^*)$. - Either $\ker(I - \gamma K) = \{0\}$ (and $I - \gamma K$ is bijective), or $\ker(I - \gamma K) \ne \{0\}$. It remains to identify the abstract kernels and solvability condition with the PDE quantities $N$, $N^*$, and $f \perp N^*$. [/step] [step:Identify $\ker(I - \gamma K)$ with $N = \{u \in H^1_0(U) : Lu = 0\}$] A function $u \in L^2(U)$ satisfies $(I - \gamma K)u = 0$ if and only if $u = \gamma Ku$. By the equivalence established earlier (with $f = 0$), this holds if and only if $u \in H^1_0(U)$ and $Lu = 0$. Note that $u = \gamma Ku \in \operatorname{Range}(K) \subset H^1_0(U)$, so elements of the kernel automatically have $H^1_0$ regularity. Therefore $\ker(I - \gamma K) = N$. [/step] [step:Identify $\ker(I - \gamma K^*)$ with $N^* = \{v \in H^1_0(U) : L^*v = 0\}$] [claim:Structure of the adjoint $K^*$] $K^* = j \circ (L_\gamma^*)^{-1} \circ \iota$, where $L_\gamma^* = L^* + \gamma I$. [/claim] [proof] Let $\phi, \psi \in L^2(U)$. Set $u = K\phi \in H^1_0(U)$, so $B_\gamma[u, v] = (\phi, v)_{L^2(U)}$ for all $v \in H^1_0(U)$, where $B_\gamma$ is the bilinear form of $L_\gamma$. Set $w = j \circ (L_\gamma^*)^{-1} \circ \iota(\psi) \in H^1_0(U)$, so $B_\gamma^*[w, v] = (\psi, v)_{L^2(U)}$ for all $v \in H^1_0(U)$. Since $B_\gamma^*[w, v] = B_\gamma[v, w]$ by definition of the adjoint form: \begin{align*} (K\phi, \psi)_{L^2(U)} = (u, \psi)_{L^2(U)} = B_\gamma[u, w] = (\phi, w)_{L^2(U)}, \end{align*} where the second equality uses $B_\gamma[v, w] = (\psi, v)_{L^2(U)}$ with $v = u$, and the third uses $B_\gamma[u, v] = (\phi, v)_{L^2(U)}$ with $v = w$. Since this holds for all $\phi$, we conclude $K^*\psi = w = j \circ (L_\gamma^*)^{-1} \circ \iota(\psi)$. [/proof] Since $K^*$ has the same structure as $K$ with $L_\gamma$ replaced by $L_\gamma^*$, the argument of the previous step applies: $w \in \ker(I - \gamma K^*)$ if and only if $w = \gamma K^* w$, which holds if and only if $w \in H^1_0(U)$ and $L^*w = 0$. Therefore $\ker(I - \gamma K^*) = N^*$. [/step] [step:Show the abstract solvability condition $Kf \perp \ker(I - \gamma K^*)$ equals $f \perp N^*$] Let $w \in N^* = \ker(I - \gamma K^*)$. Then $L^*w = 0$, so $L_\gamma^* w = \gamma w$, which means $B_\gamma[v, w] = \gamma (v, w)_{L^2(U)}$ for all $v \in H^1_0(U)$. Set $u_0 = Kf \in H^1_0(U)$, so $B_\gamma[u_0, v] = (f, v)_{L^2(U)}$ for all $v \in H^1_0(U)$. Then: \begin{align*} \gamma (Kf, w)_{L^2(U)} = \gamma (u_0, w)_{L^2(U)} = B_\gamma[u_0, w] = (f, w)_{L^2(U)}, \end{align*} where the second equality uses $B_\gamma[v, w] = \gamma (v, w)_{L^2(U)}$ with $v = u_0$, and the third uses $B_\gamma[u_0, v] = (f, v)_{L^2(U)}$ with $v = w$. Since $\gamma > 0$: \begin{align*} (Kf, w)_{L^2(U)} = 0 \iff (f, w)_{L^2(U)} = 0. \end{align*} [/step] [step:Assemble the two cases of the Fredholm alternative] **Case 1: $N = \{0\}$.** By Fredholm, $\dim N^* = \dim N = 0$, so $N^* = \{0\}$. The operator $I - \gamma K$ is bijective, so for every $f \in L^2(U)$ the equation $(I - \gamma K)u = Kf$ has a unique solution, and hence $Lu = f$ has a unique weak solution $u \in H^1_0(U)$. **Case 2: $N \ne \{0\}$.** By Fredholm, $\dim N = \dim N^* < \infty$. The equation $(I - \gamma K)u = Kf$ is solvable if and only if $Kf \perp \ker(I - \gamma K^*)$. By the previous step, this is equivalent to $(f, v)_{L^2(U)} = 0$ for all $v \in N^*$. When a solution $u_0$ exists, the full solution set is $u_0 + N$, since $(I - \gamma K)(u_0 + w) = Kf$ for any $w \in N = \ker(I - \gamma K)$. [/step]