Uniform local Gibbs strip limits

The finite spectral calculus [F6] defines $Z_n={\mathrm{Tr}}_{{\mathrm{\Lambda }}_n}(e^{-\beta H_{{\mathrm{\Lambda }}_n}})>0$ and the positive trace-one matrix ${\rho }_n=e^{-\beta H_{{\mathrm{\Lambda }}_n}}\mathrm{/}{Z}_n$. For any finite $X\subseteq \mathrm{\Gamma }$, use the tensor factorization of [F1] on $X\cup {\mathrm{\Lambda }}_n$ and set ${\widehat{\rho }}_X^{(n)}:={\mathrm{Tr}}_{(X\cup {\mathrm{\Lambda }}_n)\setminus X}\text{\hspace{0.17em}⁣}({\rho }_n\otimes q^{-\mathrm{\mid }X\setminus {\mathrm{\Lambda }}_n\mathrm{\mid }}{I}_{X\setminus {\mathrm{\Lambda }}_n})$, where ${\rho }_n$ is placed on ${\mathrm{\Lambda }}_n$ and each added site has density $q^{-1}I$. Tensor trace factorization shows that if $X\subseteq Y$, then tracing ${\widehat{\rho }}_Y^{(n)}$ over $Y\setminus X$ gives ${\widehat{\rho }}_X^{(n)}$: each removed added site contributes $\mathrm{Tr}(q^{-1}I)=1$, while any removed site in ${\mathrm{\Lambda }}_n$ is traced from the same ${\rho }_n$ before or after adjoining the identity densities. By [F3], these local algebras sit in the norm-completed $\mathcal{A}$; thus the consistent family [F8] gives an algebraic state ${\omega }_n$ whose restriction to ${\mathcal{A}}_{{\mathrm{\Lambda }}_n}$ is the finite Gibbs state.

Applying [F9] to $({\omega }_n)$ gives at least one pointwise-convergent subsequence. Now fix any pointwise-convergent subsequence, write it as $({\omega }_{n_k})$, and denote its state limit by $\omega$. The rest of the proof applies to this arbitrary convergent subsequence. This extraction is conditional on the exact interface [A2], not on the draft supplier's status label.

Fix local $A\in {\mathcal{A}}_X$ and $B\in {\mathcal{A}}_Y$. If $X=\mathrm{\varnothing }$ or $Y=\mathrm{\varnothing }$, the corresponding local algebra is $\mathbb{C}I$ by [F1], so embed that scalar observable in any one-site local algebra; the identity-preserving inclusion leaves both correlation functions unchanged. We may therefore take $X,Y\mathrm{\ne }\mathrm{\varnothing }$. Since $R<\mathrm{\infty }$, the set $X^{+R}=\{y:{\mathrm{dist}}_{\mathrm{\infty }}(X,\{y\})\le R\}$ is finite, so for all sufficiently large $k$, $X\cup Y\cup X^{+R}\subseteq {\mathrm{\Lambda }}_{n_k}$. For those $k$, define the finite-volume entire function $F_k(z):={\omega }_{n_k}\text{\hspace{0.17em}⁣}(A{\alpha }_z^{{\mathrm{\Lambda }}_{n_k}}(B))$, where ${\alpha }_z^{\mathrm{\Lambda }}(C):=e^{iz{H}_{\mathrm{\Lambda }}\mathrm{/}\mathrm{\hslash }}C{e}^{-iz{H}_{\mathrm{\Lambda }}\mathrm{/}\mathrm{\hslash }}$. Because the restrictions of ${\omega }_{n_k}$ to ${\mathcal{A}}_{{\mathrm{\Lambda }}_{n_k}}$ are the Gibbs states, [F7] gives both KMS boundary identities. On either boundary the finite trace state has norm one by [A2], so $\mathrm{\mid }{F}_k\mathrm{\mid }\le \mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }B\mathrm{\parallel }$ there. For any finite $\mathrm{\Lambda }$, put $H=H_{\mathrm{\Lambda }}=H^{\ast }$; diagonalizing this finite matrix shows that for $C\in {\mathcal{A}}_{\mathrm{\Lambda }}$ and $z=t+is\in S_b$, $0\le s\le b$, $\mathrm{\parallel }{\alpha }_z^{\mathrm{\Lambda }}(C)\mathrm{\parallel }\le \mathrm{\parallel }{e}^{izH\mathrm{/}\mathrm{\hslash }}\mathrm{\parallel }\mathrm{\parallel }C\mathrm{\parallel }\mathrm{\parallel }{e}^{-izH\mathrm{/}\mathrm{\hslash }}\mathrm{\parallel }\le e^{2s\mathrm{\parallel }H\mathrm{\parallel }\mathrm{/}\mathrm{\hslash }}\mathrm{\parallel }C\mathrm{\parallel }\le e^{2\beta \mathrm{\parallel }H\mathrm{\parallel }}\mathrm{\parallel }C\mathrm{\parallel }$. Consequently each $F_k$ is bounded on $S_b$, with the crude bound $e^{2\beta \mathrm{\parallel }{H}_{{\mathrm{\Lambda }}_{n_k}}\mathrm{\parallel }}\mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }B\mathrm{\parallel }$, which need not be uniform in $k$. This establishes the bounded-strip hypothesis for [F10]; its three-lines estimate uses the uniform boundary bounds and gives $\mathrm{\mid }{F}_k(z)\mathrm{\mid }\le \mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }B\mathrm{\parallel }$ throughout $S_b$.

On every compact real-time interval, [F4] gives ${\alpha }_t^{{\mathrm{\Lambda }}_{n_k}}(B)\to {\alpha }_t(B)$ uniformly in norm. The orbit $K_T=\{A{\alpha }_t(B):\mathrm{\mid }t\mathrm{\mid }\le T\}$ is compact because $t\mapsto {\alpha }_t(B)$ is norm-continuous. For any $\epsilon >0$, cover $K_T$ by finitely many norm-balls of radius $\epsilon \mathrm{/}4$, with centers $C_1,\dots ,C_m\in K_T$. Pointwise convergence in [F9] makes ${\max }_j\mathrm{\mid }{\omega }_{n_k}(C_j)-\omega (C_j)\mathrm{\mid }<\epsilon \mathrm{/}2$ for large $k$; the norm-one bounds in [A2] make the difference at any point of $K_T$ at most this maximum plus $2(\epsilon \mathrm{/}4)$. Combining this with uniform convergence of the dynamics proves $F_k(t)\to \omega (A{\alpha }_t(B))$ uniformly for $\mathrm{\mid }t\mathrm{\mid }\le T$. The same finite-net estimate applied to $\{{\alpha }_t(B)A:\mathrm{\mid }t\mathrm{\mid }\le T\}$ proves uniform convergence of the upper edges to $\omega ({\alpha }_t(B)A)$.

For $B\in {\mathcal{A}}_Y$, [F5] gives the local generator value $\delta (B)$. Every interaction term meeting $Y$ is supported in the finite neighborhood $Y^{+R}$, so when ${\mathrm{\Lambda }}_{n_k}$ contains $Y^{+R}$ the finite-volume generator satisfies ${\delta }_{n_k}(B)=\frac{i}{\mathrm{\hslash }}[H_{{\mathrm{\Lambda }}_{n_k}},B]=\delta (B)$. Differentiating the finite matrix exponential then gives $F_k^{\mathrm{\prime }}(z)={\omega }_{n_k}\text{\hspace{0.17em}⁣}(A{\alpha }_z^{{\mathrm{\Lambda }}_{n_k}}(\delta (B)))$. Taking $C=\delta (B)$ in the finite-volume estimate of step 2.2 shows that for each $k$, $F_k^{\mathrm{\prime }}$ is bounded on $S_b$ by $e^{2\beta \mathrm{\parallel }{H}_{{\mathrm{\Lambda }}_{n_k}}\mathrm{\parallel }}\mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }\delta (B)\mathrm{\parallel }$; this crude bound is finite but need not be uniform in $k$. The same finite Gibbs boundary identity [F7] applies to the pair $A,\delta (B)$. Its two real-time boundaries have modulus at most $\mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }\delta (B)\mathrm{\parallel }$ by the finite state norm bound [A2]. Since the required boundedness is now established, [F10] gives the uniform estimate $\mathrm{\mid }{F}_k^{\mathrm{\prime }}(z)\mathrm{\mid }\le L:=\mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }\delta (B)\mathrm{\parallel }$ on $S_b$. Integrating along the straight segment from $z$ to $w$ (which stays in the convex strip) yields $\mathrm{\mid }{F}_k(w)-F_k(z)\mathrm{\mid }\le L\mathrm{\mid }w-z\mathrm{\mid }$. Thus the functions are uniformly Lipschitz on the whole strip; if $L=0$, they are constant there.

For $m\ge 1$ and $r\ge 1$, let $G_{m,r}:=\{-m+\frac{2mj}{2^r}+i\frac{b\mathrm{\ell }}{2^r}:0\le j,\mathrm{\ell }\le 2^r\}\subset [-m,m]+i[0,b]$. The union of these grids over $m,r$ is countable and dense in the closed strip: for any point in a fixed rectangle, choosing $r$ large makes each coordinate mesh $2m\mathrm{/}{2}^r$ and $b\mathrm{/}{2}^r$ arbitrarily small. Enumerate the grid points by increasing $m+r+j+\mathrm{\ell }$, breaking ties lexicographically. For a bounded complex sequence, enclose its values in a fixed square $[-M,M{]}^2$ with $M>0$; subdivide the current square into four closed subsquares in a fixed lexicographic order and take the first containing infinitely many terms. Choose at each stage the least natural index larger than the previous one whose value lies in that square. At least one subsquare contains infinitely many terms because there are only four. The two coordinate intervals of the nested squares are closed, nested, and have lengths tending to zero; [F11] gives their unique common coordinates, and the selected values converge to that complex point. This defines a deterministic subsequence rule $E$ for each bounded complex sequence. Starting with $i_0(n)=n$, at grid point $g_j$ apply $E$ to $F_{i_{j-1}(n)}(g_j)$, obtaining a strictly increasing index map $s_j$, and set $i_j=i_{j-1}\circ s_j$. The deterministic recursion [F12] defines all $i_j$. The diagonal indices $k_j:=i_j(j)$ are strictly increasing; for each fixed $r$, the tail $(k_j{)}_{j\ge r}$ is a subsequence of $i_r$, so $F_{k_j}(g_r)$ converges. Thus one subsequence converges at every grid point without an additional choice principle.

Fix $m$ and $\epsilon >0$. If $L=0$, convergence at a single grid point and constancy give uniform convergence on the rectangle $K_m=[-m,m]+i[0,b]$. If $L>0$, choose $r$ so that every point of $K_m$ is within $\epsilon \mathrm{/}(8L)$ of some point of $G_{m,r}$. The grid is finite, so the diagonal subsequence is uniformly Cauchy on that grid: for sufficiently large $p,q$, the values $F_{k_p},F_{k_q}$ differ by less than $\epsilon \mathrm{/}2$ at every grid point. At an arbitrary $z\in K_m$, choose a grid point $g$ within $\epsilon \mathrm{/}(8L)$; the two Lipschitz estimates from step 3.2 and the grid difference give $\mathrm{\mid }{F}_{k_p}(z)-F_{k_q}(z)\mathrm{\mid }\le L\mathrm{\mid }z-g\mathrm{\mid }+\mathrm{\mid }{F}_{k_p}(g)-F_{k_q}(g)\mathrm{\mid }+L\mathrm{\mid }g-z\mathrm{\mid }<\epsilon$. Thus the subsequence is uniformly Cauchy on each $K_m$. Completeness of $\mathbb{C}$ gives a limit $F_{A,B}$ locally uniformly on the closed strip; step 2.2 gives $\mathrm{\mid }{F}_{A,B}(z)\mathrm{\mid }\le \mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }B\mathrm{\parallel }$.

Each $F_k$ is entire, and step 4.1 gives locally uniform convergence on every compact subset of $S_b^{\circ }$. By [F13], $F_{A,B}$ is holomorphic there; uniform convergence on compact rectangles including their horizontal edges makes it continuous on $S_b$. Step 3.1 gives uniform convergence of the two boundary sequences on every compact real interval, so the lower and upper boundaries of $F_{A,B}$ are respectively $\omega (A{\alpha }_t(B))$ and $\omega ({\alpha }_t(B)A)$ for every real $t$. Hence the stated local-pair strip conclusion holds.