> [!theoremb] Theorem > > Let $(E, \inp_E)$, $(H, \inp_H)$ be [[Hilbert Space|Hilbert spaces]] such that $H \subset E$ is [[Compact Operator|compactly]] embedded as a [[Dense|dense]] subspace. > > Let $\inp_L: H^2 \to \complex$ be a bounded [[Sesquilinear Form|sesquilinear form]]. Suppose that there exists $\gamma > 0$ such that $\inp_{L, \gamma} = \inp_L + \mu\inp_E$ is coercive with respect to $H$ for all $\mu \ge \gamma$, then: > 1. **First Existence Theorem**: For every $\mu \ge \gamma$, there exists a [[Bounded Linear Map|bounded linear map]] $L_\mu^{-1}: E \to H$ such that $\angles{L_\mu^{-1}y, \phi}_{L, \mu} = \angles{y, \phi}_{E}$ for all $\phi \in H$. > 2. **Second Existence Theorem**: The spaces $\ker(L) = \bracs{x \in H: \angles{x, \phi}_L = 0 \forall \phi \in H}$ and $\ker(L^*) = \bracs{x \in H: \angles{\phi, x}_L = 0 \forall \phi \in H}$ are both finite-dimensional and have the same dimension. For any $y \in E$, there exists $x \in H$ such that $\angles{x, \phi}_L = \angles{y, \phi}_E$ for all $\phi \in H$ if and only if $y \perp \ker(L^*)$ with respect to $E$. In particular, either for every $y \in E$, there exists a unique $x \in H$ such that $\angles{x, \phi}_L = \angles{y, \phi}_E$ for all $\phi \in H$, or $\ker(L) \ne \bracs{0}$. > 3. **Third Existence Theorem**: Let $\sigma(L)$ be the set of $\lambda \in \complex$ such that $\bracs{x \in H: \angles{x, \phi}_L = \angles{\lambda x, \phi}_E \forall \phi \in H} \ne \bracs{0}$, then $\sigma(L)$ is at most countable. If $\sigma(L) \subset \real$ is infinite, then it accumulates at $\infty$. For each $\lambda \not\in \sigma(L)$, there exists a unique $x \in H$ such that $\angles{x, \phi}_L = \angles{y, \phi}_E$ for all $\phi \in H$. > > > *Proof*. $(1)$: Let $\mu \ge \gamma$, then by the [[Lax-Milgram Theorem]], there exists a toplinear isomorphism $A \in \laut{H}$ such that > $ > \angles{x, \phi}_{L, \mu} = \angles{Ax, \phi}_{H} \quad \forall x, \phi \in H > $ > Let $y \in E$, then since $H$ is continuously embedded in $E$, by the [[Riesz Representation Theorem]], there exists a unique $y^* \in H$ such that > $ > \angles{y, \phi}_E = \angles{y^*, \phi}_H \quad \forall \phi \in H > $ > Since $L_\mu$ is an isomorphism, > $ > \anglesn{A^{-1} y^*, \phi}_{L, \mu} = \angles{y^*, \phi}_H = \angles{y, \phi}_E \quad \forall \phi \in H > $ > and $A^{-1}y^* \in H$ is the unique element of $H$ satisfying the above. The inversion map $L_\mu^{-1}$ can be expressed as the following composition > $ > \begin{CD} > E @>>> E^* @>\iota>> H^* @>\text{Riesz}>> H @>A^{-1}>> H > \end{CD} > $ > of bounded linear maps. > > $(2)$: Let $y \in E$ and $x \in H$, then > $ > \angles{x, \phi}_{L} = \angles{y, \phi}_E \quad \forall \phi \in H > $ > if and only if > $ > \angles{x, \phi}_{L, \gamma} = \angles{y + \gamma x, \phi}_E \quad \forall \phi \in H > $ > Since $H \subset E$ is dense, this is equivalent to $x = L_\gamma^{-1}(y + \gamma x)$ and $x - \gamma L_\gamma^{-1}x = L_\gamma^{-1}y$. > > As the embedding $\iota: H \to E$ is compact, $\gamma L_\gamma^{-1}$ viewed as an operator $E \to E$ is compact, and $(I - \gamma L_{\gamma}^{-1}): E \to E$ is [[Fredholm Operator|Fredholm]] with index $0$. By the [[Fredholm Alternative]], either $(I - \gamma L^{-1}_{\gamma}) \in \laut{E}$, or $\ker(I - \gamma L_\gamma^{-1})$ is non-trivial and finite-dimensional. From the above chain of reasoning, > $ > \ker(L) = \ker{(I - \gamma L_\gamma^{-1})} \quad \ker((I - \gamma L_\gamma^{-1})^*) = \ker(L^*) > $ > with > $ > \dim \ker(L) = \dim \ker{(I - \gamma L_\gamma^{-1})}= \dim \ker((I - \gamma L_\gamma^{-1})^*) = \dim \ker(L^*) > $ > > Lastly, $L_\gamma^{-1}y \in \im{I - \gamma L_\gamma^{-1}}$ if and only if $L_\gamma^{-1}y \perp \ker((I - \gamma L_\gamma^{-1})^*)$, if and only if > $ > \angles{L_\gamma^{-1}y, \phi}_E = \angles{y, (L_\gamma^{-1})^*\phi} = 0 \quad \forall \phi \in \ker((I - \gamma L_\gamma^{-1})^*) > $ > if and only if > $ > \angles{y, \phi} = 0 \quad \forall \phi \in \ker((I - \gamma L_\gamma^{-1})^*) = \ker(L^*) > $ > since $(\gamma L_\gamma^{-1})^*\phi = \phi$ for all $\phi \in \ker((I - \gamma L_\gamma^{-1})^*)$. > > $(3)$: Let $x \in E$, $\lambda \in \complex$, then > $ > \begin{align*} > \angles{x, \phi}_{L} &= \angles{\lambda x, \phi}_E &\forall \phi \in H \\ > \angles{x, \phi}_{L, \gamma} &= \angles{(\lambda + \gamma)x, \phi}_E &\forall \phi \in H \\ > x &= L_{\gamma}^{-1}[(\lambda + \gamma)x] \\ > x &= (\lambda + \gamma)L_\gamma^{-1}x \\ > [L_\gamma^{-1} - (\lambda + \gamma)^{-1}]x&= 0 > \end{align*} > $ > Therefore $\sigma(L)$ is in bijection with $\sigma(L_\gamma^{-1})$ via the map > $ > \sigma(L) \to \sigma(L_\gamma^{-1}) \quad \begin{align*} > \lambda \mapsto \begin{cases} > (\lambda + \gamma)^{-1} &\lambda \ne -\gamma \\ > 0 &\lambda = -\gamma > \end{cases} > \end{align*} > $ > By $(1)$, $\mu \not\in \sigma(L)$ for every $\mu \ge \gamma$. Therefore if $\sigma(L) \subset \real$, then it accumulates at $\infty$.