Variational approximation, residual certificates and propagation errors

Fix $V$ with $m=\dim V\ge k$. Choose a basis ${\varphi }_1,\dots ,{\varphi }_m$ of $V$ and use [F6], writing $M_{ij}=q_A({\varphi }_j,{\varphi }_i)$ to distinguish the form matrix from the operator $A$. The Hermitian identity for $q_A$ follows from its defining inner products; [F3] verifies the remaining closed-form input conditions. For $z\mathrm{\ne }0$, $z^{\ast }Gz=\mathrm{\parallel }{\sum }_i{z}_i{\varphi }_i{\mathrm{\parallel }}^2>0$. By [F4], diagonalizing $G$ gives the positive square root $G^{1\mathrm{/}2}$ and its inverse $G^{-1\mathrm{/}2}$, both Hermitian, and $G^{1\mathrm{/}2}{G}^{-1\mathrm{/}2}=I$. Multiplying $G^{-1\mathrm{/}2}M{G}^{-1\mathrm{/}2}{y}_j={\rho }_j{y}_j$ by $G^{1\mathrm{/}2}$ gives $M{c}_j={\rho }_j{G}^{1\mathrm{/}2}{y}_j={\rho }_{j}G{c}_j$, and $c_j^{\ast }G{c}_l=y_j^{\ast }{G}^{-1\mathrm{/}2}G{G}^{-1\mathrm{/}2}{y}_l=y_j^{\ast }{y}_l={\delta }_{jl}$. Conversely $Mc=\rho Gc$ implies $B(G^{1\mathrm{/}2}c)=\rho G^{1\mathrm{/}2}c$, so these are all generalized eigenvalues with multiplicities. Their linear combinations $e_j:={\sum }_{i=1}^m(c_j{)}_i{\varphi }_i$ form an orthonormal eigenbasis of $V$: Gram normalization gives orthonormality, and the coefficient eigenvalue equation gives $q_A(e_j,e_i)={\rho }_j{\delta }_{ij}$. For any $k$-dimensional $L\subseteq V$, the dimension formula forces $L\cap \mathrm{span}\{e_k,\dots ,e_m\}\mathrm{\ne }\{0\}$; a unit vector in this intersection has form value at least ${\rho }_k$. The subspace $\mathrm{span}\{e_1,\dots ,e_k\}$ has supremum exactly ${\rho }_k$. Hence ${\theta }_k(V)={\rho }_k$, the $k$th generalized Ritz eigenvalue.

Suppose ${\lambda }_k(A)<\alpha$. By [F2], ${\lambda }_1\le \cdots \le {\lambda }_k={\lambda }_k(A)$ are the first $k$ isolated eigenvalues below $\alpha$, counted with multiplicity. Each corresponding eigenspace is finite-dimensional by [F1]. If $x\mathrm{\ne }0$ is an eigenvector with eigenvalue $\lambda$, self-adjointness gives $\lambda \mathrm{\parallel }x{\mathrm{\parallel }}^2=⟨Ax,x⟩=⟨x,Ax⟩=\bar{\lambda }\mathrm{\parallel }x{\mathrm{\parallel }}^2$, so $\lambda$ is real. For $x\in \ker (A-{\lambda }_{i}I)$ and $y\in \ker (A-{\lambda }_{j}I)$ with ${\lambda }_i\mathrm{\ne }{\lambda }_j$, this reality and self-adjointness give ${\lambda }_i⟨x,y⟩=⟨Ax,y⟩=⟨x,Ay⟩={\lambda }_j⟨x,y⟩$, so the eigenspaces are orthogonal. Within each finite-dimensional eigenspace choose an orthonormal basis and take as many vectors as its multiplicity among ${\lambda }_1,\dots ,{\lambda }_k$; concatenating these finite lists gives orthonormal eigenvectors $u_1,\dots ,u_k\in D(A)$ with $A{u}_j={\lambda }_j{u}_j$.

Every $k$-dimensional subspace of $V$ is a $k$-dimensional subspace of $Q(A)$, so the infimum defining ${\theta }_k(V)$ is over a subfamily of the one defining ${\lambda }_k(A)$. Therefore ${\theta }_k(V)\ge {\lambda }_k(A)$.

For $u={\sum }_{j=1}^k{c}_j{u}_j$ with ${\sum }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2=1$, orthonormality gives $\mathrm{\parallel }u\mathrm{\parallel }=1$, while [F3] gives $q_A(u_j,u_i)=⟨A{u}_j,u_i⟩={\lambda }_j{\delta }_{ij}$. Hence $q_A[u]={\sum }_j{\lambda }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\le {\lambda }_k(A)$.

For nested $V_n$, every $k$-subspace of $V_n$ is a $k$-subspace of $V_{n+1}$; thus ${\theta }_k(V_{n+1})\le {\theta }_k(V_n)$. By step 2.1 this sequence is bounded below by ${\lambda }_k(A)$, so it has a limit at least ${\lambda }_k(A)$.

Let $\epsilon >0$ and put $M=({\sum }_{j=1}^k\mathrm{\parallel }{u}_j{\mathrm{\parallel }}_{Q(A)}^2{)}^{1\mathrm{/}2}$. For each $j$, density gives an approximation from some $V_{n_j}$; taking $N={\max }_j{n}_j$ and using nesting, choose all $w_j\in V_N$ with $\mathrm{\parallel }{w}_j-u_j{\mathrm{\parallel }}_{Q(A)}<\delta$. For coefficients $c$ with ${\sum }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2=1$, put $w={\sum }_j{c}_j{w}_j$ and $u={\sum }_j{c}_j{u}_j$. Then $\mathrm{\parallel }w-u{\mathrm{\parallel }}_{Q(A)}\le \sqrt{k}\delta$ and $\mathrm{\parallel }u\mathrm{\parallel }=1$, so $\mathrm{\parallel }w\mathrm{\parallel }\ge 1-\sqrt{k}\delta$; taking $\sqrt{k}\delta <1\mathrm{/}2$ ensures the $w_j$ are linearly independent. Writing $a(x,y)=q_A(x,y)+(c+1)⟨x,y⟩$, [F5] and $a[x]\ge \mathrm{\parallel }x{\mathrm{\parallel }}^2$ give $\mathrm{\mid }{q}_A(x,y)\mathrm{\mid }\le (c+2)\mathrm{\parallel }x{\mathrm{\parallel }}_{Q(A)}\mathrm{\parallel }y{\mathrm{\parallel }}_{Q(A)}$. Since $\mathrm{\parallel }u{\mathrm{\parallel }}_{Q(A)}\le M$ and $\mathrm{\parallel }w{\mathrm{\parallel }}_{Q(A)}\le M+\sqrt{k}\delta$, this yields $\mathrm{\mid }{q}_A[w]-q_A[u]\mathrm{\mid }\le (c+2)\sqrt{k}\delta (2M+\sqrt{k}\delta )$ uniformly in $c$. Also $\mathrm{\mid }\mathrm{\parallel }w{\mathrm{\parallel }}^2-1\mathrm{\mid }\le \sqrt{k}\delta (2+\sqrt{k}\delta )$ uniformly, so for sufficiently small $\delta$, $\mathrm{\parallel }w{\mathrm{\parallel }}^2\ge 1\mathrm{/}2$. The bound just proved for $q_A[w]-q_A[u]$ and the uniform bound $\mathrm{\mid }{q}_A[u]\mathrm{\mid }\le {\max }_j\mathrm{\mid }{\lambda }_j\mathrm{\mid }$ then show that $q_A[w]\mathrm{/}\mathrm{\parallel }w{\mathrm{\parallel }}^2$ converges uniformly to $q_A[u]$. The values $q_A[u]={\sum }_j{\lambda }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2$ are at most ${\lambda }_k(A)$ by step 2.2; hence choose $\delta$ small enough that every such $w$ satisfies $q_A[w]\mathrm{/}\mathrm{\parallel }w{\mathrm{\parallel }}^2\le {\lambda }_k(A)+\epsilon$.

The span of $w_1,\dots ,w_k$ is a $k$-dimensional subspace of $V_N$, so step 3.2 gives ${\theta }_k(V_N)\le {\lambda }_k(A)+\epsilon$. By monotonicity from step 3.1 the same upper bound holds for all $n\ge N$. As $\epsilon >0$ is arbitrary, the limit in step 3.1 is at most ${\lambda }_k(A)$; combined with its lower bound from step 2.1 it equals ${\lambda }_k(A)$.

Apply the Borel sum rule (Unbounded Borel functional calculus: domains, products, spectral mapping, Statement, clause 3) to $f(x)=x$ and $h(x)=-\lambda$. By [F1], $f(T)=T$, while $h(T)=-\lambda I$ has domain $H$; thus $u\in D(f(T))\cap D(h(T))$ and $r=(x\mapsto x-\lambda )(T)u$. The norm identity in [F3] therefore gives $\mathrm{\parallel }r{\mathrm{\parallel }}^2={\int }_{\mathbb{R}}\mathrm{\mid }x-\lambda {\mathrm{\mid }}^{2}d{\mu }_u(x)$. The measure ${\mu }_u$ has total mass one by [F2].

By [F3], ${\mu }_u$ is supported on $\sigma (T)$. Set $d=\mathrm{dist}(\lambda ,\sigma (T))$. For every $x\in \sigma (T)$, $\mathrm{\mid }x-\lambda \mathrm{\mid }\ge d$ by the definition of infimum, so step 1.1 yields $\mathrm{\parallel }r{\mathrm{\parallel }}^2\ge d^2{\mu }_u(\sigma (T))=d^2$. Both sides are nonnegative, hence $d\le \mathrm{\parallel }r\mathrm{\parallel }$.

Since $I-P_{\mathrm{\Delta }}=E_T(\mathbb{R}\setminus \mathrm{\Delta })$, the PVM norm identity gives $\mathrm{\parallel }(I-P_{\mathrm{\Delta }})u{\mathrm{\parallel }}^2={\mu }_u(\mathbb{R}\setminus \mathrm{\Delta })={\mu }_u(\sigma (T)\setminus \mathrm{\Delta })$ by support from [F3]. On this set $\mathrm{\mid }x-\lambda \mathrm{\mid }\ge g$, so step 1.1 implies $\mathrm{\parallel }r{\mathrm{\parallel }}^2\ge g^2\mathrm{\parallel }(I-P_{\mathrm{\Delta }})u{\mathrm{\parallel }}^2$. Taking nonnegative square roots and dividing by $g>0$ proves the cluster estimate. If $\sigma (T)\setminus \mathrm{\Delta }$ is empty, [F3] directly gives ${\mu }_u(\mathbb{R}\setminus \mathrm{\Delta })=0$, yielding the stated zero bound.

Fix $s$ in the compact segment joining $t_0$ to $t$, put $W(\tau )=e^{i\tau T}$ and $U(s)=W((s-t_0)\mathrm{/}\mathrm{\hslash })$, and define $w(s)=U(s)v(s)$. For $h$ with $s+h$ in the interval, the group law gives $\frac{w(s+h)-w(s)}{h}=U(s+h)\frac{v(s+h)-v(s)}{h}+U(s)\frac{W(h\mathrm{/}\mathrm{\hslash })-I}{h}v(s)$. By [F1], $U$ is strongly continuous and unitary, the first term tends to $U(s)v^{\mathrm{\prime }}(s)$, and since $v(s)\in D(T)$ the second tends to $\frac{i}{\mathrm{\hslash }}U(s)Tv(s)$ by the derivative of $W$ at zero and the scalar chain rule. Thus $w$ is differentiable and $w^{\mathrm{\prime }}(s)=U(s)v^{\mathrm{\prime }}(s)+\frac{i}{\mathrm{\hslash }}U(s)Tv(s)=-\frac{i}{\mathrm{\hslash }}U(s)r(s)$, after substituting $i\mathrm{\hslash }{v}^{\mathrm{\prime }}=Tv+r$. The derivative is continuous because $v^{\mathrm{\prime }}$, $Tv$, and hence $r$ are continuous and $U$ is strongly continuous.

If $t\ge t_0$, apply [F3] on $[t_0,t]$; if $t<t_0$, apply it on $[t,t_0]$ and negate the resulting identity. In either case $w(t)-w(t_0)=-\frac{i}{\mathrm{\hslash }}{\int }_{t_0}^{t}U(s)r(s)ds$, using step 1.1. Then [F4] gives $\mathrm{\parallel }w(t)-w(t_0)\mathrm{\parallel }\le \frac{1}{\mathrm{\hslash }}{\int }_{\min (t,t_0)}^{\max (t,t_0)}\mathrm{\parallel }U(s)r(s)\mathrm{\parallel }ds=\frac{1}{\mathrm{\hslash }}{\int }_{\min (t,t_0)}^{\max (t,t_0)}\mathrm{\parallel }r(s)\mathrm{\parallel }ds$; the equality uses unitarity of $U(s)$. Since $w(t_0)=v(t_0)$ and $w(t)=U_T(t,t_0{)}^{\ast }v(t)$, unitarity also gives $\mathrm{\parallel }w(t)-w(t_0)\mathrm{\parallel }=\mathrm{\parallel }v(t)-U_T(t,t_0)v(t_0)\mathrm{\parallel }$. This is the required estimate, including $t=t_0$ and either time orientation.