Conforming Galerkin invariant subspaces converge across an isolated cluster

Since $E_{r-1}\subseteq E_s$ when $r>1$, its directed best-approximation gap is at most ${\delta }_s$. Choose ${\delta }_s$ below the smallness thresholds for the upper eigenvalue theorem at $s$ and, if $r>1$, at $r-1$. At index $s$, [F2] gives ${\lambda }_{h,s}\le \lambda +C_s{\delta }_s^2$. For $r\le j\le s$, Ritz ordering gives ${\lambda }_{h,j}\le {\lambda }_{h,s}$, while the conforming lower bound in [F2] gives ${\lambda }_{h,j}\ge {\lambda }_j=\lambda$. Hence $\lambda \le {\lambda }_{h,j}\le \lambda +C_s{\delta }_s^2$. When $r>1$, Ritz ordering and the upper bound at $r-1$ give ${\lambda }_{h,j}\le {\lambda }_{r-1}+C_{r-1}{\delta }_s^2<\lambda -\frac{1}{2}(\lambda -{\lambda }_{r-1})$ for $j<r$. For each discrete index $j>s$ that exists, [F2] gives ${\lambda }_{h,j}\ge {\lambda }_j\ge {\lambda }_{s+1}>\lambda$. Thus exactly the $s-r+1$ indexed Ritz eigenvalues in the cluster are separated from all other discrete Ritz eigenvalues.

Boundedness and coercivity give $\alpha \mathrm{\parallel }v{\mathrm{\parallel }}_V^2\le a(v,v)\le M\mathrm{\parallel }v{\mathrm{\parallel }}_V^2$, so the energy norm is equivalent to the Hilbert norm of $V$ and $V$ is complete in the energy norm. Thus $(V,a)$ is a Hilbert space. For fixed $u$, the map $v\mapsto \overline{(Ju,Jv{)}_H}=(Jv,Ju{)}_H$ is a bounded linear functional on $(V,a)$. By [F5] there is a unique $Ku\in V$ with $a(v,Ku)=(Jv,Ju{)}_H$ for every $v$; conjugating gives $a(Ku,v)=(Ju,Jv{)}_H$. Uniqueness in this identity makes $K$ linear. If $C_J=\mathrm{\parallel }J{\mathrm{\parallel }}_{V\to H}$, then coercivity and Cauchy–Schwarz give $\mathrm{\mid }a(Ku,v)\mathrm{\mid }=\mathrm{\mid }(Ju,Jv{)}_H\mathrm{\mid }\le (C_J^2\mathrm{/}\alpha )\mathrm{\parallel }u{\mathrm{\parallel }}_a\mathrm{\parallel }v{\mathrm{\parallel }}_a$. Taking the supremum over $\mathrm{\parallel }v{\mathrm{\parallel }}_a=1$ and using Riesz norm equality yields $\mathrm{\parallel }Ku{\mathrm{\parallel }}_a\le (C_J^2\mathrm{/}\alpha )\mathrm{\parallel }u{\mathrm{\parallel }}_a$. Hermitian symmetry shows that $K$ is self-adjoint. The continuous eigenrelation in [F1] gives $K{e}_j={\lambda }_j^{-1}{e}_j$ and $K(e_j\mathrm{/}\sqrt{{\lambda }_j})={\lambda }_j^{-1}(e_j\mathrm{/}\sqrt{{\lambda }_j})$. In particular, on $E$ the operator $K$ acts as $\mu I$ for $\mu :=1\mathrm{/}\lambda$.

Let $P_h$ be the energy projection from [F3] and set $K_h:=P_{h}K{P}_h:V\to V$. It is self-adjoint because $P_h$ is an orthogonal projection and $K$ is self-adjoint. It is bounded as a composition of bounded operators, and its range lies in the finite-dimensional space $V_h$. To check compactness under the actual compact-closure definition, the image $K_h({\bar{B}}_V)$ is bounded in $V_h$. Choose a finite $a$-orthonormal basis of $V_h$ by finite Gram--Schmidt and identify its real coordinates with ${\mathbb{R}}^d$ or ${\mathbb{R}}^{2d}$. The closure of this image in $V_h$ is closed and bounded in those coordinates, hence compact by Heine--Borel. Its inclusion into $V$ is isometric, so it is compact and closed in $V$; therefore the ambient closure of $K_h({\bar{B}}_V)$ is a closed subset of this compact set and is compact. The compact-operator definition now makes $K_h$ compact [F6]. It is positive because $a(K_{h}u,u)=a(K{P}_{h}u,P_{h}u)=\mathrm{\parallel }J{P}_{h}u{\mathrm{\parallel }}_H^2\ge 0$. For $u_h,v_h\in V_h$, $a(K_h{u}_h,v_h)=a(K{u}_h,v_h)=(J{u}_h,J{v}_h{)}_H\mathrm{.}$ If $K_h{u}_h=0$, taking $v_h=u_h$ gives $\mathrm{\parallel }J{u}_h{\mathrm{\parallel }}_H^2=0$, so injectivity of $J$ gives $u_h=0$. Thus $K_h{\mathrm{\mid }}_{V_h}$ is positive definite. Moreover, $K_h{u}_h={\mu }_h{u}_h$ is equivalent by the displayed identity to $b(u_h,v_h)={\mu }_{h}a(u_h,v_h)$ for every $v_h\in V_h$, the Ritz equation with ${\lambda }_{h,j}=1\mathrm{/}{\mu }_h$. Conversely, the same identity and nondegeneracy of $a$ on $V_h$ show that every Ritz eigenvector is an eigenvector of $K_h$. Its positive eigenspaces are therefore exactly the conforming Ritz eigenspaces. Put $\mu :=1\mathrm{/}\lambda$. If $r>1$, set ${\gamma }_{-}:=\lambda -{\lambda }_{r-1}$; set ${\mathrm{\Delta }}_{+}:=1\mathrm{/}\lambda -1\mathrm{/}{\lambda }_{s+1}>0$. When $r>1$, put ${\mathrm{\Delta }}_{-}:=1\mathrm{/}(\lambda -{\gamma }_{-}\mathrm{/}2)-1\mathrm{/}\lambda >0$. Step 1.1 and smallness give a positive distance from $\mu$ to all eigenvalues outside the $r,\dots ,s$ Ritz cluster: lower-index reciprocal eigenvalues are at least $1\mathrm{/}(\lambda -{\gamma }_{-}\mathrm{/}2)$ when $r>1$, higher-index reciprocals are at most $1\mathrm{/}{\lambda }_{s+1}$ when present, and zero is separated by $\mu$. Define ${\mathrm{\Delta }}_{r,s,\lambda }:=\min \{\mu ,{\mathrm{\Delta }}_{+}\}$ if $r=1$ and ${\mathrm{\Delta }}_{r,s,\lambda }:=\min \{\mu ,{\mathrm{\Delta }}_{+},{\mathrm{\Delta }}_{-}\}$ if $r>1$. The cluster subspace $E_h$ of $K_h$ has dimension $s-r+1$ and this positive spectral separation from its complement.

For $u\in E$, the inclusion $E\subseteq E_s$ and [F3] give $\mathrm{\parallel }{P}_{h}u-u{\mathrm{\parallel }}_a\le {\delta }_s\mathrm{\parallel }u{\mathrm{\parallel }}_a$. Since $Ku=\mu u$, $K_{h}u-\mu u=P_{h}K(P_{h}u-u)+\mu (P_{h}u-u)\mathrm{.}$ The orthogonal projection $P_h$ has norm at most one, so $\mathrm{\parallel }(K_h-\mu I)u{\mathrm{\parallel }}_a\le (\mathrm{\parallel }K\mathrm{\parallel }+\mu ){\delta }_s\mathrm{\parallel }u{\mathrm{\parallel }}_a\mathrm{.}$ The constant ${\mathrm{\Delta }}_{r,s,\lambda }$ from step 2.1 is therefore independent of ${\delta }_s$ under the chosen smallness bound.

Apply the residual subspace estimate [F4] to $K_h$, its separated cluster $E_h$, and the residual bound from step 3.1. It yields ${\delta }_a(E,E_h)\le \frac{\mathrm{\parallel }K\mathrm{\parallel }+\mu }{{\mathrm{\Delta }}_{r,s,\lambda }}{\delta }_s,$ where ${\mathrm{\Delta }}_{r,s,\lambda }>0$ is the uniform spectral separation obtained in step 2.1. This is the claimed invariant-subspace estimate, including the case of multiplicity greater than one.