Restricted history-path gap

On legal clock $c_i$, $H_{\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{c}\mathrm{k}0}$ has eigenvalue $i$; the propagation terms have diagonal entries $1\mathrm{/}2$ at the endpoints, $1$ internally, and adjacent entries $-1\mathrm{/}2$. Therefore the clock matrix is the displayed $H_0(s)=aD+t{L}_N$, where $a=1-s$, $t=s\mathrm{/}2$, and $D=\mathrm{diag}(0,1,\dots ,N)$.

If $0\le s\le 1\mathrm{/}6$, the Rayleigh quotient at $e_0$ gives the ground eigenvalue $E_0\le t$. On the codimension-one subspace $e_0^{\perp }$, $D\ge I$ and $L_N\ge 0$, so every Rayleigh quotient there is at least $a$. If the second eigenvalue $E_1$ were below $a$, the span of eigenvectors for the two lowest eigenvalues would have dimension at least two and would contain a nonzero vector orthogonal to $e_0$; its quotient, as an eigenbasis expansion supported on eigenvalues at most $E_1$, would be at most $E_1<a$, a contradiction. Hence $E_1\ge a$, and $E_1-E_0\ge a-t=1-3s\mathrm{/}2\ge 3\mathrm{/}4$.

Now suppose $s\ge 1\mathrm{/}6$, so $t\ge 1\mathrm{/}12$, and let $v$ be the normalized positive ground vector from [F1], with eigenvalue $E_0$. Set ${\pi }_i=v_i^2$, so ${\sum }_i{\pi }_i=1$ and ${\pi }_0\ge {\pi }_1\ge \cdots \ge {\pi }_N>0$. For a vector $f=(f_i)$ with ${\sum }_i{\pi }_i{f}_i=0$, set $x_i=v_i{f}_i$; then $x\perp v$ and $\mathrm{\parallel }x{\mathrm{\parallel }}^2={\sum }_i{\pi }_i\mathrm{\mid }{f}_i{\mathrm{\mid }}^2$. The eigenvalue equation $H_{0}v=E_{0}v$, expanded edge by edge, gives the ground-state identity ${\displaystyle ⟨x,(H_0-E_{0}I)x⟩=t\sum _{i=0}^{N-1}{v}_i{v}_{i+1}\mathrm{\mid }{f}_i-f_{i+1}{\mathrm{\mid }}^2\mathrm{.}}$ Indeed, the diagonal coefficient at vertex $i$ after subtracting $E_0$ is $t$ times the sum of neighboring ratios $v_j\mathrm{/}{v}_i$; multiplying by $v_i^2\mathrm{\mid }{f}_i{\mathrm{\mid }}^2$ assigns $t{v}_i{v}_j\mathrm{\mid }{f}_i{\mathrm{\mid }}^2$ to each incident edge, and the off-diagonal term supplies $-2t{v}_i{v}_j\mathrm{Re}({\bar{f}}_i{f}_j)$.

Since $v_i\ge v_{i+1}$, each edge weight satisfies $v_i{v}_{i+1}\ge {\pi }_{i+1}$. Define a continuous-time reversible path chain with stationary masses ${\pi }_i$ and symmetric edge conductances $c_i={\pi }_{i+1}$: its rate from $i$ to $i+1$ is $c_i\mathrm{/}{\pi }_i={\pi }_{i+1}\mathrm{/}{\pi }_i\le 1$, and its reverse rate is $c_i\mathrm{/}{\pi }_{i+1}=1$. Each total exit rate is at most $2$. Let $Q$ be its generator and $P=I+Q\mathrm{/}2$. Then $P$ is row-stochastic, reversible with stationary law $\pi$, and irreducible, and its Dirichlet form is $⟨f,(I-P)f{⟩}_{\pi }=\frac{1}{2}{\sum }_i{c}_i\mathrm{\mid }{f}_i-f_{i+1}{\mathrm{\mid }}^2$.

For any set $S$ with $0<\pi (S)\le 1\mathrm{/}2$, if $0\in S$, choose the largest $k$ for which $\{0,\dots ,k\}\subseteq S$. Then $k<N$, edge $(k,k+1)$ crosses the cut, and the tail mass is at least $1\mathrm{/}2$. Monotonicity gives ${\sum }_{j=k+1}^N{\pi }_j\le N{\pi }_{k+1}$, so ${\pi }_{k+1}\ge 1\mathrm{/}(2N)$. The chain flow across the edge is $c_k\mathrm{/}2={\pi }_{k+1}\mathrm{/}2$, hence $Q_P(S,S^c)\mathrm{/}\pi (S)\ge 1\mathrm{/}(2N)$. If $0\mathrm{\notin }S$, put $k=\min S-1$; then $S\subseteq \{k+1,\dots ,N\}$ and $\pi (S)\le N{\pi }_{k+1}$, while the crossing flow is again ${\pi }_{k+1}\mathrm{/}2$, giving the same ratio bound. Thus $\varphi (P)\ge 1\mathrm{/}(2N)$.

By [F2], $1-{\lambda }_2(P)\ge 1\mathrm{/}(8N^2)$. The generator gap is twice this number because $P=I+Q\mathrm{/}2$, so its Dirichlet-form Poincaré constant is at least $1\mathrm{/}(4N^2)$. For any $x\perp v$, set $f_i=x_i\mathrm{/}{v}_i$; then ${\sum }_i{\pi }_i{f}_i=0$. The ground-state identity in step 2.2 and $v_i{v}_{i+1}\ge c_i$ give $⟨x,(H_0-E_{0}I)x⟩\ge t\mathrm{/}(4N^2)\mathrm{\parallel }x{\mathrm{\parallel }}^2$. By [F3], expand $x$ in an orthonormal eigenbasis of $H_0$ with $v$ as its simple ground vector; choose $x$ to be a unit eigenvector for $E_1$, orthogonal to $v$. Its quotient is exactly $E_1-E_0$, so the established lower bound applies to this $x$. Thus $E_1-E_0\ge t\mathrm{/}(4N^2)\ge 1\mathrm{/}(48N^2)$ for $s\ge 1\mathrm{/}6$. Together with step 2.1, this proves the uniform bound.