Finite-element eigenvalue approximation

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.

Choose an $a$-orthonormal basis ${\psi }_1,\dots ,{\psi }_N$ of $V_h$ by [F3]. For $u_h\in V_h$, define $K_h{u}_h:={\sum }_{i=1}^{N}b(u_h,{\psi }_i){\psi }_i\mathrm{.}$ If $v_h={\sum }_{i=1}^N{c}_i{\psi }_i$, then $a(K_h{u}_h,v_h)={\sum }_{i}b(u_h,{\psi }_i)\overline{c_i}=b(u_h,v_h)$, since $b$ is conjugate-linear in its second argument and the basis is orthonormal. This finite coordinate formula also makes $K_h$ linear in $u_h$.

For $u_h,v_h\in V_h$, $a(K_h{u}_h,v_h)=b(u_h,v_h)=\overline{b(v_h,u_h)}=\overline{a(K_h{v}_h,u_h)}=a(u_h,K_h{v}_h),$ so $K_h$ is self-adjoint in the energy inner product. Also $a(K_h{u}_h,u_h)=b(u_h,u_h)>0$ when $u_h\mathrm{\ne }0$, so $K_h$ is positive and injective. To verify compactness from the compact-closure definition, use the $a$-orthonormal basis ${\psi }_1,\dots ,{\psi }_N$ from step 1.1. The matrix entries $m_{ij}:=a(K_h{\psi }_j,{\psi }_i)$ give, for $x={\sum }_j{c}_j{\psi }_j$, $\mathrm{\parallel }{K}_{h}x{\mathrm{\parallel }}_a^2={\sum }_i\mathrm{\mid }{\sum }_j{m}_{ij}{c}_j{\mathrm{\mid }}^2\le ({\sum }_{i,j}\mathrm{\mid }{m}_{ij}{\mathrm{\mid }}^2){\sum }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2$, so $K_h$ is bounded. The coordinate map identifies $V_h$ isometrically with ${\mathbb{R}}^N$ in the real case and ${\mathbb{R}}^{2N}$ in the complex case. The closure of the image of its closed unit ball is therefore closed and bounded in finite-dimensional Euclidean coordinates, hence compact by Heine--Borel [F7]; the unit-ball criterion in [F7] makes $K_h$ compact.

The inner product norm on $V_h$ is complete by [F8], so the compact self-adjoint spectral theorem [F4] applies to $K_h$. It gives an $a$-orthonormal eigenbasis ${\varphi }_{h,1},\dots ,{\varphi }_{h,N}$ with positive eigenvalues ordered ${\mu }_{h,1}\ge \cdots \ge {\mu }_{h,N}>0$. The equation $K_h{\varphi }_{h,j}={\mu }_{h,j}{\varphi }_{h,j}$ is equivalent to the Ritz equation with ${\lambda }_{h,j}={\mu }_{h,j}^{-1}$, so these are the positive Ritz eigenvalues in increasing order. For $v_h={\sum }_{j=1}^N{c}_j{\varphi }_{h,j}$, $a(v_h,v_h)={\sum }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2,b(v_h,v_h)=a(K_h{v}_h,v_h)={\sum }_j{\mu }_{h,j}\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\mathrm{.}$

On $W_h:=\mathrm{span}\{{\varphi }_{h,1},\dots ,{\varphi }_{h,m}\}$, ${\sum }_j{\mu }_{h,j}\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\ge {\mu }_{h,m}{\sum }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2$, with equality for $v_h={\varphi }_{h,m}$. Thus ${\max }_{v_h\in W_h\setminus \{0\}}a(v_h,v_h)\mathrm{/}b(v_h,v_h)={\mu }_{h,m}^{-1}$, so the minimum in the statement is at most ${\lambda }_{h,m}$.

Let $W_h$ be any $m$-dimensional subspace of $V_h$. The linear map $v_h\mapsto (a(v_h,{\varphi }_{h,1}),\dots ,a(v_h,{\varphi }_{h,m-1}))$ has a nonzero kernel by [F5]; when $m=1$ its target is the zero-dimensional space and the kernel is all of $W_h$. A nonzero vector in that kernel has only coefficients $c_j$ with $j\ge m$ in step 3.1. Since ${\mu }_{h,j}\le {\mu }_{h,m}$ for those indices, its quotient is at least ${\mu }_{h,m}^{-1}$. Hence the maximum on every $W_h$ is at least ${\lambda }_{h,m}$. That maximum is attained by [F6], and the span in step 4.1 attains the outer minimum.

Fix $u\in E$. Decompose $u=P_{F}u+(I-P_F)u$ using [F2]. The spectral projection $P_F$ commutes with $B_h$ because $F$ is a sum of whole eigenspaces. By [F1], the orthogonal spectral expansion of $(B_h-\mu I)u$ on $F^{\perp }$ contains only nonzero eigenvalues outside $\mathrm{\Lambda }$ and the kernel component; all their multipliers belong to ${\mathrm{\Sigma }}_h\setminus \mathrm{\Lambda }$ and are at distance at least $\mathrm{\Delta }$ from $\mu$. Parseval therefore gives $\mathrm{\parallel }(B_h-\mu I)u{\mathrm{\parallel }}^2\ge {\mathrm{\Delta }}^2\mathrm{\parallel }(I-P_F)u{\mathrm{\parallel }}^2\mathrm{.}$

If $u\mathrm{\ne }0$, divide step 1.1 by ${\mathrm{\Delta }}^2\mathrm{\parallel }u{\mathrm{\parallel }}^2$ and take square roots to obtain $\mathrm{dist}(u,F)\mathrm{/}\mathrm{\parallel }u\mathrm{\parallel }\le \epsilon \mathrm{/}\mathrm{\Delta }$. The same bound is vacuous for $u=0$. Taking the supremum over nonzero $u\in E$ proves the stated directed gap estimate.

Choose an orthonormal basis $f_1,\dots ,f_N$ by [F3]. For $x={\sum }_j{c}_j{f}_j$ with $\mathrm{\parallel }x\mathrm{\parallel }\le 1$, orthonormality gives $\mathrm{\mid }{c}_j\mathrm{\mid }\le 1$, so the triangle inequality yields $\mathrm{\parallel }Ax\mathrm{\parallel }\le {\sum }_j\mathrm{\parallel }A{f}_j\mathrm{\parallel }$. Thus $A$ maps the closed unit ball to a bounded set. Its closure is closed and bounded in finite-dimensional coordinates, hence compact by [F3]. The compact-operator unit-ball criterion in [F3] makes $A$ compact; Hermitian symmetry makes it self-adjoint.

Apply [F1] to the compact self-adjoint operator from step 1.1 to obtain an orthonormal eigenbasis $(e_j{)}_{j=1}^N$ with real eigenvalues ${\lambda }_j$. Expand $u={\sum }_{j=1}^N{c}_j{e}_j$, where $c_j=(u,e_j)$ and ${\sum }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2=\mathrm{\parallel }u{\mathrm{\parallel }}^2=1$ by [F2]. Hence $\mu ={\sum }_j{\lambda }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\in \mathbb{R}$.

Orthogonality gives ${\rho }^2=\mathrm{\parallel }(A-\mu I)u{\mathrm{\parallel }}^2={\sum }_{j=1}^N\mathrm{\mid }{\lambda }_j-\mu {\mathrm{\mid }}^2\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\ge \mathrm{dist}(\mu ,\mathrm{\Sigma }(A){)}^2{\sum }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2$. Since the coefficient squares sum to one by step 2.1, taking nonnegative square roots proves the spectral-proximity bound.

If there is no eigenvalue other than the simple target, then $N=1$ and $u$ is a unit multiple of $e$, so the angle is zero. Otherwise choose the indexing so $e_k=e$ and ${\lambda }_k=\lambda$. The phase-invariant angle obeys ${\sin }^2\mathrm{\angle }(u,e)=1-\mathrm{\mid }{c}_k{\mathrm{\mid }}^2={\sum }_{j\mathrm{\ne }k}\mathrm{\mid }{c}_j{\mathrm{\mid }}^2$. For each $j\mathrm{\ne }k$, $\mathrm{\mid }{\lambda }_j-\mu \mathrm{\mid }\ge \mathrm{\Delta }$ by definition. Therefore ${\rho }^2={\sum }_j\mathrm{\mid }{\lambda }_j-\mu {\mathrm{\mid }}^2\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\ge {\mathrm{\Delta }}^2{\sum }_{j\mathrm{\ne }k}\mathrm{\mid }{c}_j{\mathrm{\mid }}^2={\mathrm{\Delta }}^2{\sin }^2\mathrm{\angle }(u,e)$. Taking square roots proves the directional estimate; if the target is the only eigenvalue, simplicity gives the zero-angle branch already handled.

The proximity estimate minimizes over the entire spectrum and contains no ordering or target label. It only says that some eigenvalue lies within $\rho$ of $\mu$; without an additional spectral-location certificate it does not identify that eigenvalue as the lowest one.

Taking adjoints of $L^{-1}L=I$ and $L{L}^{-1}=I$ by [F3] gives $L^{\ast }(L^{-1}{)}^{\ast }=I$ and $(L^{-1}{)}^{\ast }{L}^{\ast }=I$. Hence $L^{\ast }$ is invertible with inverse $(L^{-1}{)}^{\ast }=L^{-\ast }$; taking adjoints again gives $(L^{-\ast }{)}^{\ast }=L^{-1}$. The matrix $\tilde{H}=L^{-1}H{L}^{-\ast }$ is Hermitian, since ${\tilde{H}}^{\ast }=(L^{-\ast }{)}^{\ast }{H}^{\ast }(L^{-1}{)}^{\ast }=L^{-1}H{L}^{-\ast }=\tilde{H}\mathrm{.}$ The coordinate map $c\mapsto y=L^{\ast }c$ is invertible by [F1].

Substituting $c=L^{-\ast }y$ into $Hc=\lambda Sc$ and multiplying by $L^{-1}$ gives $L^{-1}H{L}^{-\ast }y=\lambda L^{-1}L{L}^{\ast }{L}^{-\ast }y=\lambda y\mathrm{.}$ Every operation is reversible because $L$ is invertible, so the generalized and whitened eigenproblems have exactly the same eigenvalues, with eigenvectors related by $y=L^{\ast }c$.

With $c=L^{-\ast }y$, matrix multiplication gives $c^{\ast }Hc=y^{\ast }{L}^{-1}H{L}^{-\ast }y=y^{\ast }\tilde{H}y,c^{\ast }Sc=y^{\ast }{L}^{-1}L{L}^{\ast }{L}^{-\ast }y=y^{\ast }y\mathrm{.}$ Since $c\mathrm{\ne }0$ iff $y\mathrm{\ne }0$, and $S>0$ makes $c^{\ast }Sc>0$, both equal denominators are positive. Division proves equality of the two Rayleigh quotients.

For fixed $v$, the functional $w\mapsto a(v,w)$ is bounded and conjugate-linear. Apply [F1] to the continuous datum $F(w)=a(v,w)$; its continuous solution is $v$, and the resulting discrete solution is the unique $P_{h}v\in V_h$ in the statement. If $v_1,v_2\in V$ and $c_1,c_2\in \mathbb{K}$, then $c_1{P}_h{v}_1+c_2{P}_h{v}_2$ satisfies the defining equation for $P_h(c_1{v}_1+c_2{v}_2)$, so uniqueness proves linearity.

The continuous equation for $v$ and the defining discrete equation for $P_{h}v$ agree on each $w_h\in V_h$. By [F2], $a(v-P_{h}v,w_h)=0$.

Fix any $w_h\in V_h$. Since $v-w_h=(v-P_{h}v)+(P_{h}v-w_h)$ and $P_{h}v-w_h\in V_h$, step 2.1 and Hermitian symmetry eliminate both cross terms in the squared energy norm. Therefore $\mathrm{\parallel }v-w_h{\mathrm{\parallel }}_a^2=\mathrm{\parallel }v-P_{h}v{\mathrm{\parallel }}_a^2+\mathrm{\parallel }{P}_{h}v-w_h{\mathrm{\parallel }}_a^2\ge \mathrm{\parallel }v-P_{h}v{\mathrm{\parallel }}_a^2\mathrm{.}$ The choice $w_h=P_{h}v$ attains equality, proving the best-approximation identity (the exact supplier interface [F3]).

If $v\in E$ is nonzero, put $\widehat{v}=v\mathrm{/}\mathrm{\parallel }v{\mathrm{\parallel }}_a$. By linearity in step 1.1 and the definition [F4], $\mathrm{\parallel }v-P_{h}v{\mathrm{\parallel }}_a=\mathrm{\parallel }v{\mathrm{\parallel }}_a\mathrm{\parallel }\widehat{v}-P_h\widehat{v}{\mathrm{\parallel }}_a\le {\delta }_a(E,V_h)\mathrm{\parallel }v{\mathrm{\parallel }}_a$; the same inequality is immediate for $v=0$. If $P_{h}v=0$, step 3.1 makes the left side $\mathrm{\parallel }v{\mathrm{\parallel }}_a$. Since ${\delta }_a(E,V_h)<1$, this forces $v=0$. Thus $P_h{\mathrm{\mid }}_E$ is injective.

Put ${\varphi }_i:=e_i\mathrm{/}\sqrt{{\lambda }_i}$, $p_i:=P_h{\varphi }_i$, and $r_i:={\varphi }_i-p_i$. By [F1], the ${\varphi }_i$ are $a$-orthonormal. For any coefficient vector $c\in {\mathbb{K}}^m$, linearity of $P_h$ and [F2]–[F3] give ${\parallel {\sum }_{i=1}^m{c}_i{r}_i\parallel }_a={\parallel v-P_{h}v\parallel }_a\le {\delta }_h\mathrm{\parallel }v{\mathrm{\parallel }}_a={\delta }_h\mathrm{\parallel }c{\mathrm{\parallel }}_2,v:={\sum }_{i=1}^m{c}_i{\varphi }_i\mathrm{.}$ In particular, the error Gram matrix $G_{ij}:=a(r_j,r_i)$ is positive semidefinite and $\mathrm{\parallel }G{\mathrm{\parallel }}_2\le {\delta }_h^2$.

Since each $r_i$ is energy-orthogonal to $V_h$, it is orthogonal to every $p_j$. Therefore the stiffness Gram matrix $(A_h{)}_{ij}:=a(p_j,p_i)$ satisfies $(A_h{)}_{ij}=a({\varphi }_j-r_j,{\varphi }_i-r_i)={\delta }_{ij}-G_{ij}\mathrm{.}$ It follows that $\mathrm{\parallel }{A}_h-I{\mathrm{\parallel }}_2=\mathrm{\parallel }G{\mathrm{\parallel }}_2\le {\delta }_h^2$.