Discrete spectrum and min–max principle for a compact coercive form

For $f\in H$, the functional $v\mapsto (f,Jv{)}_H$ is bounded on $V$ because $\mathrm{\mid }(f,Jv{)}_H\mathrm{\mid }\le C_J\mathrm{\parallel }f{\mathrm{\parallel }}_H\mathrm{\parallel }v{\mathrm{\parallel }}_V$. Apply [F2] to obtain a unique $Sf\in V$ satisfying $a(Sf,v)=(f,Jv{)}_H$ for every $v\in V$. Uniqueness makes $S:H\to V$ linear.

Testing with $v=Sf$ and using coercivity gives $\alpha \mathrm{\parallel }Sf{\mathrm{\parallel }}_V^2\le a(Sf,Sf)=\mathrm{Re}(f,JSf{)}_H\le \mathrm{\mid }(f,JSf{)}_H\mathrm{\mid }\le C_J\mathrm{\parallel }f{\mathrm{\parallel }}_H\mathrm{\parallel }Sf{\mathrm{\parallel }}_V\mathrm{.}$ If $Sf\mathrm{\ne }0$, divide by $\alpha \mathrm{\parallel }Sf{\mathrm{\parallel }}_V$; if $Sf=0$ the same bound is immediate. Thus $S$ is bounded from $H$ to $V$ with $\mathrm{\parallel }Sf{\mathrm{\parallel }}_V\le (C_J\mathrm{/}\alpha )\mathrm{\parallel }f{\mathrm{\parallel }}_H$.

Set $T:=JS:H\to H$. By [F3], $T$ is compact. For $f,g\in H$, testing the defining equations for $Sf$ and $Sg$ at $Sg$ and $Sf$ respectively and using Hermitian symmetry gives $(Tf,g{)}_H=(f,Tg{)}_H$; hence $T$ is self-adjoint. Also $(f,Tf{)}_H=a(Sf,Sf)\ge 0$. If $Tf=0$, injectivity of $J$ gives $Sf=0$, so $(f,Jv{)}_H=0$ for every $v\in V$ by step 1.1. Density of $J(V)$ then gives $f=0$. Thus $T$ is positive and injective.

Apply [F4] to $T$. Its nonzero eigenvalues are positive because $T$ is positive, and its kernel is zero by step 2.2; therefore the eigenspaces of its nonzero eigenvalues span $H$. There are infinitely many eigenvalue occurrences, because each eigenspace has finite dimension while $H$ is infinite-dimensional. Write $d(\nu ):=\dim E_{\nu }$. Recursively list a largest eigenvalue among those whose multiplicity has not yet been exhausted: after any finite number of entries the remaining set is nonempty, and if $\rho >0$ is any remaining value, only finitely many remaining eigenvalues are at least $\rho$, each with finite multiplicity. Iterated trichotomy gives a maximum of these finitely many occurrences, which is then maximal in the whole remainder. The resulting sequence lists every $\nu$ exactly $d(\nu )$ times, since only finitely many occurrences are at least its value. Thus ${\mu }_1\ge {\mu }_2\ge \cdots >0$ includes all multiplicities. Compactness gives ${\mu }_j\to 0$; Countable Choice selects an orthonormal basis of each finite-dimensional eigenspace, and these bases give the ordered family $(f_j)$ [F2, F4].

Since $T{f}_j=J(S{f}_j)={\mu }_j{f}_j$ and $J$ is injective, each $f_j$ belongs to the image $J(V)$. Identify it with its unique preimage $e_j\in V$. The equation $a(S{f}_j,v)=(f_j,Jv{)}_H$ and $S{f}_j={\mu }_j{e}_j$ imply $a(e_j,v)={\mu }_j^{-1}(J{e}_j,Jv{)}_H(v\in V)\mathrm{.}$ Set ${\lambda }_j:={\mu }_j^{-1}$. Then $0<{\lambda }_1\le {\lambda }_2\le \cdots$, ${\lambda }_j\to +\mathrm{\infty }$, $\mathrm{\parallel }J{e}_j{\mathrm{\parallel }}_H=1$, and the displayed form eigenrelation holds.

By [F1], the energy norm induced by $a$ is equivalent to $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_V$, so $V$ is complete in that norm. Put ${\varphi }_j:=e_j\mathrm{/}\sqrt{{\lambda }_j}$. The eigenrelation in step 4.1 gives $a({\varphi }_i,{\varphi }_j)={\delta }_{ij}$, so $({\varphi }_j)$ is an orthonormal family in $V$ with inner product $a$. If $v\in V$ is $a$-orthogonal to every ${\varphi }_j$, the same eigenrelation gives $(Jv,J{e}_j{)}_H=0$ for every $j$. The vectors $(J{e}_j)$ are an orthonormal basis of $H$ by step 3.1, so $Jv=0$; injectivity of $J$ gives $v=0$. Thus the closed span of $({\varphi }_j)$ has zero orthogonal complement in $V$, and [F6] implies it is all of $V$.

Parseval in $H$ for $(J{e}_j)$ and in $V$ for $({\varphi }_j)$ yields, for every $v\in V$, $\mathrm{\parallel }Jv{\mathrm{\parallel }}_H^2={\sum }_j\mathrm{\mid }(Jv,J{e}_j{)}_H{\mathrm{\mid }}^2,a(v,v)={\sum }_j\mathrm{\mid }a(v,{\varphi }_j){\mathrm{\mid }}^2={\sum }_j{\lambda }_j\mathrm{\mid }(Jv,J{e}_j{)}_H{\mathrm{\mid }}^2\mathrm{.}$ For $v\mathrm{\ne }0$, $\mathrm{\parallel }Jv{\mathrm{\parallel }}_H>0$ by injectivity, so $R(v)$ is the weighted average of the ${\lambda }_j$ with weights $\mathrm{\mid }(Jv,J{e}_j{)}_H{\mathrm{\mid }}^2$.

Let $E_m:=\mathrm{span}\{e_1,\dots ,e_m\}$. Step 6.1 shows $R(v)\le {\lambda }_m$ for every nonzero $v\in E_m$, with equality at $e_m$; hence ${\max }_{v\in E_m\setminus \{0\}}R(v)={\lambda }_m$. This gives one inequality in the min–max identity, attained by $E_m$.

Let $S\subseteq V$ have dimension $m$. The map $S\to {\mathbb{K}}^{m-1}$ taking $v$ to $((Jv,J{e}_1{)}_H,\dots ,(Jv,J{e}_{m-1}{)}_H)$ has a nonzero kernel by [F7]. For a nonzero vector $v$ in that kernel, the expansion of step 6.1 has no terms with index below $m$, so $R(v)\ge {\lambda }_m$. Thus the maximum Rayleigh quotient on every such $S$ is at least ${\lambda }_m$. It is attained on the unit sphere by [F8], so the outer minimum is at least ${\lambda }_m$ and equals the value from step 7.1.