A repeated eigenspace can be stable while chosen eigenvectors oscillate

Let $(x^{(n)})$ be Cauchy in $H$. For each $j\in I$, $x_j^{(n)}$ is Cauchy in $\mathbb{R}$ because $\mathrm{\mid }{x}_j^{(n)}-x_j^{(m)}\mathrm{\mid }\le \mathrm{\parallel }{x}^{(n)}-x^{(m)}{\mathrm{\parallel }}_H$; let $x_j$ be its limit by [F2]. The norms $\mathrm{\parallel }{x}^{(n)}{\mathrm{\parallel }}_H$ are bounded: after some index every term is within $1$ of one fixed term, and the finitely many earlier norms have a maximum. Thus they are bounded by some $C$. For every finite $F\subseteq I$, passage to coordinate limits gives ${\sum }_{j\in F}\mathrm{\mid }{x}_j{\mathrm{\mid }}^2\le C^2$, so $x=(x_j)\in H$ by [F1]. Given $\epsilon >0$, choose $N$ so $\mathrm{\parallel }{x}^{(n)}-x^{(m)}{\mathrm{\parallel }}_H<\epsilon \mathrm{/}2$ for $n,m\ge N$. For fixed $n\ge N$ and finite $F$, let $m\to \mathrm{\infty }$ in ${\sum }_{j\in F}\mathrm{\mid }{x}_j^{(n)}-x_j^{(m)}{\mathrm{\mid }}^2<{\epsilon }^2\mathrm{/}4$; taking the supremum over finite $F$ gives $\mathrm{\parallel }{x}^{(n)}-x{\mathrm{\parallel }}_H\le \epsilon \mathrm{/}2<\epsilon$. Thus $H$ is complete.

Define $U:V\to H$ by $(Uu{)}_j=\sqrt{{\lambda }_j}{u}_j$. The estimate ${\sum }_j{\lambda }_j\mathrm{\mid }\alpha u_j+\beta v_j{\mathrm{\mid }}^2\le 2\mathrm{\mid }\alpha {\mathrm{\mid }}^2\mathrm{\parallel }u{\mathrm{\parallel }}_V^2+2\mathrm{\mid }\beta {\mathrm{\mid }}^2\mathrm{\parallel }v{\mathrm{\parallel }}_V^2$ shows $V$ is a vector space, and the coordinate formula makes $U$ linear. It is onto, since $y\in H$ has preimage $u_j=y_j\mathrm{/}\sqrt{{\lambda }_j}\in V$, and it is an isometry from $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_V$ to $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_H$. Moreover $a(u,v)=(Uu,Uv{)}_H$, so the series defining $a$ is well-defined and $a$ is the inner product inducing $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_V$. Step 1.1 and this isometry make $V$ a Hilbert space. Since ${\lambda }_j\ge 1$, $a(u,u)\ge \mathrm{\parallel }Ju{\mathrm{\parallel }}_H^2$; [F9] gives $\mathrm{\mid }a(u,v)\mathrm{\mid }\le \mathrm{\parallel }u{\mathrm{\parallel }}_V\mathrm{\parallel }v{\mathrm{\parallel }}_V$, so $a$ is bounded and coercive. Finite truncations are dense in $H$ by the tail property in [F1], so $J$ is injective with dense range.

Let $B_V:=\{u\in V:\mathrm{\parallel }u{\mathrm{\parallel }}_V\le 1\}$. For any $\epsilon >0$, choose $N\ge 2$ with $1\mathrm{/}(N+1)<{\epsilon }^2\mathrm{/}4$ by [F8]. If $x=Ju$ with $u\in B_V$, then ${\sum }_{j>N}\mathrm{\mid }{x}_j{\mathrm{\mid }}^2={\sum }_{j>N}\mathrm{\mid }{u}_j{\mathrm{\mid }}^2\le (N+1{)}^{-1}{\sum }_{j>N}{\lambda }_j\mathrm{\mid }{u}_j{\mathrm{\mid }}^2\le 1\mathrm{/}(N+1)$. The ellipsoid $C_N:=\{(z_1,\dots ,z_N)\in {\mathbb{R}}^N:{\sum }_{j\le N}{\lambda }_j\mathrm{\mid }{z}_j{\mathrm{\mid }}^2\le 1\}$ is closed and bounded, hence compact by [F7]. Its open balls of radius $\epsilon \mathrm{/}2$ centered at its points cover it, so compactness gives a finite $\epsilon \mathrm{/}2$-net with centers in $C_N$. Each center $z$ lifts by zero padding to $\tilde{z}=(z_1,\dots ,z_N,0,\dots )\in J(B_V)$. For $x=Ju\in J(B_V)$, its prefix lies in $C_N$ and the tail norm is less than $\epsilon \mathrm{/}2$, so some lift satisfies $\mathrm{\parallel }x-\tilde{z}{\mathrm{\parallel }}_H<\epsilon$. Thus $J(B_V)$ is totally bounded. Its closure $C:=\overline{J(B_V)}$ is totally bounded as well: a finite $\epsilon \mathrm{/}2$-net for $J(B_V)$ is an $\epsilon$-net for $C$ by the triangle inequality and the definition of closure. The set $C$ is complete because it is closed in the complete space $H$. By [F6], $C$ is compact; the unit-ball criterion in Compact linear operator now shows that $J$ is compact.

For every coordinate vector $e_j$, $a(e_j,v)={\lambda }_j(J{e}_j,Jv{)}_H$. For nonzero $u\in V$, $\frac{a(u,u)}{\mathrm{\parallel }Ju{\mathrm{\parallel }}_H^2}=\frac{{\sum }_j{\lambda }_j\mathrm{\mid }{u}_j{\mathrm{\mid }}^2}{{\sum }_j\mathrm{\mid }{u}_j{\mathrm{\mid }}^2}\ge 1$, with equality exactly when $u_j=0$ for every $j\ge 3$. Conversely, testing any weak eigenpair equation at $v=e_j$ gives $({\lambda }_j-\lambda )u_j=0$ for each $j$, so every eigenvalue is one of the ${\lambda }_j$ and its eigenspace is the span of the corresponding coordinate vectors. Hence the first eigenvalue is $1$ with eigenspace $E$, and the next is ${\lambda }_3=3$.

The restriction of the continuous form to each $V_n=E$ is the identity form, so its two Ritz values are $1$ and the full cluster eigenspace $E_n=E$. For every $u\in E$, $u\in V_n$, so [F4] gives ${\delta }_a(E,V_n)=0$; likewise the gap from $E$ to $E_n$ is zero. The cluster theorem [F5] applies with $r=1,s=2$ and its gap conclusion is attained exactly.

For every positive integer $n$, set $u_n=e_1$ when $n$ is even and $u_n=e_2$ when $n$ is odd. Both are unit Ritz eigenvectors for the same repeated eigenvalue $1$. Their successive energy distances are $\mathrm{\parallel }{e}_1-e_2{\mathrm{\parallel }}_a=\sqrt{2}$, so $(u_n)$ is not Cauchy and cannot converge. This does not contradict step 4.1: the invariant subspace remains exactly $E$, while the repeated eigenspace supplies no preferred basis vector.