[proofplan] We first isolate the coercive part of the operator. The [Lax-Milgram theorem](/theorems/4946) identifies the operator induced by $B_0$ as an isomorphism $A_0:H \to H^*$, so for each $F \in H^*$ the equation $A(u)=F$ can be transferred to the equation $(I_H+C)u=A_0^{-1}F$ on $H$, where $C$ is compact. [The Fredholm alternative](/page/The%20Fredholm%20Alternative) for compact perturbations of the identity gives finite-dimensional kernel, closed range, and the injective-if-and-only-if-surjective dichotomy. Finally, we translate the range condition back from the transformed operator to $A$ and identify the obstruction space with $\ker A^*$ by the closed-range annihilator identity. [/proofplan] [step:Represent the coercive part by an isomorphism] Define the [bounded linear operator](/page/Bounded%20Linear%20Operator) $A_0:H \to H^*$ by $A_0(u)(v)=B_0[u,v]$ for $u,v \in H$. The boundedness of $B_0$ gives boundedness of $A_0$, and coercivity gives $A_0(u)(u)=B_0[u,u]\geq \alpha \|u\|_H^2$ for every $u \in H$. By the [Lax-Milgram Theorem](/theorems/91), $A_0:H \to H^*$ is a bounded linear isomorphism. [guided] The purpose of this step is to turn the coercive [bilinear form](/page/Bilinear%20Form) into an invertible operator. Define \begin{align*} A_0:H \to H^* \end{align*} by \begin{align*} A_0(u)(v)=B_0[u,v] \end{align*} for $u,v \in H$. This is the operator associated to the unperturbed bilinear form. We verify the hypotheses needed for the Lax-Milgram theorem. The form $B_0$ is bilinear by hypothesis. It is bounded by hypothesis, so there is a constant $M_0>0$ such that \begin{align*} |B_0[u,v]|\leq M_0\|u\|_H\|v\|_H \end{align*} for all $u,v \in H$. Hence, for fixed $u \in H$, the map $v \mapsto B_0[u,v]$ is a bounded linear functional on $H$, and $A_0(u)\in H^*$. The same inequality gives \begin{align*} \|A_0(u)\|_{H^*}\leq M_0\|u\|_H \end{align*} so $A_0$ is bounded. Coercivity is exactly the estimate \begin{align*} A_0(u)(u)=B_0[u,u]\geq \alpha \|u\|_H^2 \end{align*} for all $u \in H$. Therefore the [Lax-Milgram Theorem](/theorems/91) applies, and it gives that \begin{align*} A_0:H \to H^* \end{align*} is a bounded linear isomorphism. This is the step where coercivity is used: it supplies the invertible reference operator against which the compact perturbation will be measured. [/guided] [/step] [step:Convert the equation to a compact perturbation of the identity] Define $C:H \to H$ by $C=A_0^{-1}K$. Since $K:H \to H^*$ is compact and $A_0^{-1}:H^* \to H$ is bounded, the composition $C$ is compact. Define $T:H \to H$ by $T=I_H+C$. For $u \in H$ and $F \in H^*$, the equation $A(u)=F$ is equivalent to $Tu=A_0^{-1}F$. Indeed, $A(u)=A_0(u)+K(u)=A_0(u)+A_0(Cu)=A_0(Tu)$, and applying $A_0^{-1}$ gives the equivalence. [guided] Now we rewrite the original equation on the [Hilbert space](/page/Hilbert%20Space) $H$ rather than on its dual $H^*$. Define the operator \begin{align*} C:H \to H \end{align*} by $C=A_0^{-1}K$. This definition is meaningful because $K:H \to H^*$ and $A_0^{-1}:H^* \to H$. The operator $K$ is compact by hypothesis, and $A_0^{-1}$ is bounded by the previous step. A bounded operator composed after a [compact operator](/page/Compact%20Operator) is compact, so $C:H \to H$ is compact. Define \begin{align*} T:H \to H \end{align*} by $T=I_H+C$, where $I_H:H \to H$ is the identity map. For $u \in H$, the definition of $C$ gives $A_0(Cu)=K(u)$. Therefore \begin{align*} A(u)=A_0(u)+K(u)=A_0(u)+A_0(Cu)=A_0(Tu). \end{align*} Since $A_0$ is an isomorphism, applying $A_0^{-1}:H^* \to H$ to both sides shows that, for each $F \in H^*$, \begin{align*} A(u)=F \quad \iff \quad Tu=A_0^{-1}F. \end{align*} This is the key reduction: solvability for $A$ is exactly solvability for a compact perturbation of the identity on $H$. [/guided] [/step] [step:Apply the Fredholm alternative to the transformed operator] The space $H$ is a real Hilbert space, hence a [Banach space](/page/Banach%20Space). The operator $T=I_H+C$ is the identity on $H$ plus the compact linear operator $C:H \to H$. By the [Fredholm Alternative](/theorems/94) for compact perturbations of the identity, $\ker T$ is finite-dimensional, $\operatorname{Range}(T)$ is closed, and $T$ is injective if and only if $T$ is surjective. Since $A=A_0T$ and $A_0$ is an isomorphism, we have $\ker A=\ker T$ and $\operatorname{Range}(A)=A_0(\operatorname{Range}(T))$. Thus $\ker A$ is finite-dimensional and $\operatorname{Range}(A)$ is closed. Also, $A$ is injective if and only if $T$ is injective, and $A$ is surjective if and only if $T$ is surjective. Therefore $A$ is injective if and only if $A$ is surjective. [guided] We now apply the [Fredholm alternative](/theorems/72) to the transformed operator. The theorem applies to an operator of the form identity plus compact on a Banach space. Here $H$ is a real Hilbert space, hence a Banach space, and the previous step proved that $C:H \to H$ is compact. Therefore \begin{align*} T=I_H+C:H \to H \end{align*} has exactly the form required by the [Fredholm Alternative](/theorems/94). The theorem gives three facts: $\ker T$ is finite-dimensional, $\operatorname{Range}(T)$ is closed, and $T$ is injective if and only if $T$ is surjective. We transfer these facts back to $A$ using $A=A_0T$. Since $A_0:H \to H^*$ is an isomorphism, \begin{align*} Au=0 \quad \iff \quad A_0Tu=0 \quad \iff \quad Tu=0, \end{align*} so $\ker A=\ker T$. Also, a functional $F \in H^*$ lies in $\operatorname{Range}(A)$ exactly when $F=A_0z$ for some $z \in \operatorname{Range}(T)$, hence \begin{align*} \operatorname{Range}(A)=A_0(\operatorname{Range}(T)). \end{align*} Because $A_0$ is a homeomorphism between Banach spaces, it sends closed subspaces to closed subspaces. Thus $\operatorname{Range}(A)$ is closed, and $\ker A$ is finite-dimensional. Finally, injectivity and surjectivity are preserved under composition with the isomorphism $A_0$, so $A$ is injective if and only if $A$ is surjective. [/guided] [/step] [step:Identify the solvability condition with the adjoint kernel] We prove that $\operatorname{Range}(A)=\{F \in H^*: F(w)=0 \text{ for every } w \in \ker A^*\}$. First let $F \in \operatorname{Range}(A)$. Then $F=A(u)$ for some $u \in H$. If $w \in \ker A^*$, then $F(w)=A(u)(w)=A^*(w)(u)=0$. Hence every element of $\operatorname{Range}(A)$ annihilates $\ker A^*$. Conversely, suppose $F \in H^*$ annihilates $\ker A^*$ and suppose, for contradiction, that $F \notin \operatorname{Range}(A)$. Since $\operatorname{Range}(A)$ is a closed linear subspace of $H^*$, the [Hahn-Banach separation theorem](/page/Hahn-Banach%20Theorem) gives a nonzero functional $\Lambda \in (H^*)^*$ such that $\Lambda(G)=0$ for every $G \in \operatorname{Range}(A)$ and $\Lambda(F)\neq 0$. Since $H$ is Hilbert, it is [reflexive](/page/Reflexive%20Space), so there exists $w \in H$ such that $\Lambda(G)=G(w)$ for every $G \in H^*$. For every $v \in H$, the functional $A(v)$ belongs to $\operatorname{Range}(A)$, hence $A^*(w)(v)=A(v)(w)=\Lambda(A(v))=0$. Thus $w \in \ker A^*$. Since $F$ annihilates $\ker A^*$, this gives $F(w)=0$. But $F(w)=\Lambda(F)\neq 0$, a contradiction. Therefore $F \in \operatorname{Range}(A)$. [guided] We identify exactly which right-hand sides lie in the range of $A$. First suppose $F \in \operatorname{Range}(A)$. Then there is $u \in H$ such that $F=A(u)$. If $w \in \ker A^*$, then the definition of the adjoint gives \begin{align*} F(w)=A(u)(w)=A^*(w)(u)=0. \end{align*} Thus every element of $\operatorname{Range}(A)$ annihilates $\ker A^*$. For the reverse inclusion, suppose $F \in H^*$ annihilates $\ker A^*$, meaning $F(w)=0$ for every $w \in \ker A^*$. Assume for contradiction that $F \notin \operatorname{Range}(A)$. The previous step proved that $\operatorname{Range}(A)$ is closed. Therefore the [Hahn-Banach separation theorem](/page/Hahn-Banach%20Theorem) gives a nonzero functional $\Lambda \in (H^*)^*$ such that $\Lambda(G)=0$ for every $G \in \operatorname{Range}(A)$ and $\Lambda(F)\neq 0$. Because $H$ is a Hilbert space, it is [reflexive](/page/Reflexive%20Space), and the canonical embedding of $H$ into $(H^*)^*$ is onto. Hence there exists $w \in H$ such that \begin{align*} \Lambda(G)=G(w) \end{align*} for every $G \in H^*$. Now take any $v \in H$. Since $A(v) \in \operatorname{Range}(A)$, we have \begin{align*} A^*(w)(v)=A(v)(w)=\Lambda(A(v))=0. \end{align*} This holds for every $v \in H$, so $w \in \ker A^*$. The annihilation hypothesis on $F$ then gives $F(w)=0$. But the representing property of $w$ gives $F(w)=\Lambda(F)$, and $\Lambda(F)\neq 0$, a contradiction. Therefore $F \in \operatorname{Range}(A)$. [/guided] [/step] [step:Separate the two alternatives] If $\ker A=\{0\}$, then $A$ is injective. From the [Fredholm alternative](/theorems/219) step, $A$ is also surjective. Hence for every $F \in H^*$ there exists at least one $u \in H$ with $A(u)=F$, and injectivity gives uniqueness. If $\ker A\neq \{0\}$, then the finite-dimensionality of $\ker A$ follows from $\ker A=\ker T$ and the [Fredholm Alternative](/theorems/94) for $T=I_H+C$. The previous step shows that, for each $F \in H^*$, the equation $A(u)=F$ is solvable exactly when $F \in \operatorname{Range}(A)$, equivalently exactly when $F(w)=0$ for every $w \in \ker A^*$. The two cases are mutually exclusive because $\ker A$ is either zero or nonzero. Together they exhaust all possibilities, so the stated Fredholm alternative holds. [guided] We now split according to the kernel of $A$. If $\ker A=\{0\}$, then $A$ is injective. The Fredholm step proved that injectivity of $A$ is equivalent to surjectivity of $A$, so $A:H \to H^*$ is surjective. Hence for every $F \in H^*$ there exists $u \in H$ such that $A(u)=F$. If $u_1,u_2 \in H$ both solve the equation, then $A(u_1-u_2)=0$, so $u_1-u_2 \in \ker A=\{0\}$ and $u_1=u_2$. Thus the solution exists and is unique for every $F \in H^*$. If $\ker A\neq \{0\}$, then the Fredholm step gives $\ker A=\ker T$, and the [Fredholm Alternative](/theorems/94) gives that $\ker T$ is finite-dimensional. Hence $\ker A$ is nonzero and finite-dimensional. The range-identification step proved \begin{align*} \operatorname{Range}(A)=\{F \in H^*: F(w)=0 \text{ for every } w \in \ker A^*\}. \end{align*} Therefore $A(u)=F$ is solvable exactly when $F$ annihilates $\ker A^*$. The alternatives are mutually exclusive because $\ker A$ is either zero or nonzero, and they exhaust all cases. This proves the stated Fredholm alternative for the coercive compact perturbation $A$. [/guided] [/step]