Positive-reference expanders have only synchronized critical PSD phases

Since ${\mathrm{\Delta }}_L\mathbf{1}=0$ and $d>0$, testing the two order inequalities at $\mathbf{1}$ gives $c_{-}\le 0\le c_{+}$. For $v\perp \mathbf{1}$, $v^{T}Lv\ge (1+c_{-})d\mathrm{\parallel }v{\mathrm{\parallel }}^2>0$. If the graph were disconnected, subtracting the mean from the indicator of a component would give a nonzero mean-zero vector with $Lv=0$, a contradiction. Thus the graph is connected. A one-vertex graph already satisfies the conclusion.

First assume $w=c_{+}-c_{-}>0$, and use $s,B,q,K$ from the numerical-budget lemma. If a nonsynchronized critical PSD state existed on the phase torus, use the mass lemma's common rotation and choose angular representatives in $(-\pi ,\pi ]$ with real nonnegative phasor average. The PSD tests and mass lemma applies because $a<1\mathrm{/}32$. The kernel/half-circle lemma gives $m_0=\mathrm{\mid }{C}_{\pi \mathrm{/}2}\mathrm{\mid }>0$, while the mass bound gives $m_0\le \sum H<5a^{2}n\mathrm{/}2<an$. Start at ${\beta }_0=\pi \mathrm{/}2$, $b_0=m_0$. As long as the cap $an$ has not been reached and the index is below $K$, decrease the angle by $\eta =\arcsin (8a\mathrm{/}s)$ and set $b_{j+1}=\min (q{b}_j,an)$. The tail-growth lemma (a), with $ϵ=s\mathrm{/}2$ and $u=1$, proves $\mathrm{\mid }{C}_{{\beta }_{j+1}}\mathrm{\mid }\ge b_{j+1}$ by induction. Every prospective angle remains above $3\pi \mathrm{/}8$ by the first numerical budget, so its hypotheses hold before each update.

If the cap is reached by index $K$, stop there. Then $b_M=an$ and ${\beta }_M>3\pi \mathrm{/}8>\pi \mathrm{/}4$, whence $b_M{\sin }^2{\beta }_M>an\mathrm{/}2>5a^{2}n\mathrm{/}2$. If no cap is reached, $b_K=q^K{m}_0$ and $q^K\ge B\mathrm{/}(2a)$. For each subsequent index, set ${\eta }_j=\arcsin (4B{m}_0\mathrm{/}(s{b}_j))$, ${\beta }_{j+1}={\beta }_j-{\eta }_j$, $b_{j+1}=\min (q{b}_j,n\mathrm{/}2)$, and stop at the first $b_M=n\mathrm{/}2$. Before the final cap, $b_j=q^j{m}_0$, so the inverse-sine arguments are at most $8a\mathrm{/}s<1$. By the inverse-sine bound, ${\eta }_j\le (\pi \mathrm{/}2)4B\mathrm{/}(s{q}^j)$. Their total decrement, including every prospective step, is therefore bounded by the second numerical budget; thus all angles exceed $\pi \mathrm{/}4$. Tail-growth (b) gives the cardinality induction because its required sine is $4B{m}_0\mathrm{/}(s\mathrm{\mid }{C}_{{\beta }_j}\mathrm{\mid })\le 4B{m}_0\mathrm{/}(s{b}_j)$. The geometric-growth stopping bound applies to this positive sequence with cap $n\mathrm{/}2$ and $q>1$, so the cap is reached in finitely many steps. At the end, $b_M{\sin }^2{\beta }_M>n\mathrm{/}4>5a^{2}n\mathrm{/}2$.

In either stopping case, $H(t)\ge {\sin }^2{\beta }_M$ whenever $\mathrm{\mid }t\mathrm{\mid }\ge {\beta }_M$ and $0\le {\beta }_M\le \pi \mathrm{/}2$, from the piecewise definition of $H$, oddness of sine, its range bound and its monotonicity on $[0,\pi \mathrm{/}2]$. Consequently ${\sum }_{x}H({\theta }_x)\ge \mathrm{\mid }{C}_{{\beta }_M}\mathrm{\mid }{\sin }^2{\beta }_M\ge b_M{\sin }^2{\beta }_M>5a^{2}n\mathrm{/}2$, contradicting the mass lemma. Hence no nonsynchronized critical PSD state exists when $w>0$.

If $w=0$, step 1.1 forces $c_{-}=c_{+}=0$. Replace these two bounds by $c_{-}^{\mathrm{\prime }}=-t$, $c_{+}^{\mathrm{\prime }}=t$ with $t>0$ sufficiently small. The same matrix inequalities still hold, $c_{-}^{\mathrm{\prime }}>-1$, and the strict numerical inequality persists by continuity at $t=0$: its left side becomes $f(t)=32a(1+7t)\log ((1+t+a)\mathrm{/}(2a))\mathrm{/}(1-t{)}^2$, which is continuous on $(-1\mathrm{/}2,1\mathrm{/}2)$ by [F5], since $a>0$, $1+t+a>0$ and $1-t\mathrm{\ne }0$. As $f(0)<1$, continuity with tolerance $(1-f(0))\mathrm{/}2$ gives $f(t)<1$ for all sufficiently small $t>0$. The perturbed width is $2t>0$, so steps 2.1–1.4 apply without changing $G,d,a,\theta$. This proves the conclusion also in the zero-width case.