Axis fellow travelling controls the centralizer

Integers $J$ as stated exist. Indeed $2^J\ge J^2$ for integers $J\ge 4$: equality holds at four, and $2J^2\ge (J+1{)}^2$ for $J\ge 3$ propagates the inequality. A sufficiently large integer $J$ satisfies $J^2\mathrm{/}\lambda >\lambda +2c+4\kappa J$, by choosing $J>8\lambda \kappa$ and $J^2>2\lambda (\lambda +2c)$. Set $a_n=d(o,g^n)$. If $a_{2^{j+1}}-a_{2^j}\le 4\kappa$ for every $0\le j<J$, summing gives $a_{2^J}\le a_1+4\kappa J\le \lambda +c+4\kappa J$, contrary to the lower orbit bound. Thus some $N=2^j$ has $a_{2N}-a_N>4\kappa$. Subadditivity gives $K=a_N-a_{2N}\mathrm{/}2\ge 0$, and $L-2K=a_{2N}-a_N>4\kappa$. The upper orbit bound gives $L\le \lambda 2^J+c$.

Here is the asserted point-to-segment bound for arbitrary $v,a,b$. Write $p=(a\mathrm{\mid }b{)}_v$ and set $t=p+2\delta +2\mathrm{/}3$. If $t>d(v,a)$, the endpoint $a$ has distance less than $p+3\delta +1$ from $v$. Otherwise choose the point $w\in [v,a]$ at distance $t$ from $v$, using any radial sides together with the specified $[a,b]$. Slimness supplies a witness within distance less than $\delta +1\mathrm{/}3$ on $[v,b]\cup [a,b]$. If such a witness $u$ lies on $[v,b]$, put $s=d(v,u)$. Then $s>t-\delta -1\mathrm{/}3$ and the route through $w,u$ gives $d(a,b)<d(v,a)-t+\delta +1\mathrm{/}3+d(v,b)-s$. Thus $p>(t+s-\delta -1\mathrm{/}3)\mathrm{/}2>t-\delta -1\mathrm{/}3>p$, an impossibility. A witness $z$ must therefore lie on $[a,b]$, with $d(v,z)<t+\delta +1\mathrm{/}3=p+3\delta +1$. This includes repeated vertices and $\delta =0$.

Isometry gives consecutive distances $L$ and local turns $(x_{i-1}\mathrm{\mid }{x}_{i+1}{)}_{x_i}=K$. Write $A=K+\kappa$. For all $i<j$, the endpoint turn $(x_i\mathrm{\mid }{x}_{j+1}{)}_{x_j}\le A$. To prove this by induction on $j-i$, the first case is $K$. If the previous bound holds, the product identity gives $(x_i\mathrm{\mid }{x}_{j-1}{)}_{x_j}\ge L-A>A$. But F2 gives $K\ge \min \{(x_{j-1}\mathrm{\mid }{x}_i{)}_{x_j},(x_i\mathrm{\mid }{x}_{j+1}{)}_{x_j}\}-\kappa$. Its first minimum entry exceeds $K+\kappa$, so the second is at most $A$, completing the induction. Reversal of any finite subchain proves the reversed endpoint bound as well. Expanding these bounds successively gives $d(x_i,x_j)\ge (j-i)(L-2A)$ for $i<j$.

For $i<m<j$, one has $(x_i\mathrm{\mid }{x}_j{)}_{x_m}\le K+2\kappa$. If $j=m+1$, step 2.1 gives the stronger bound $A$. Otherwise its reversed version gives $(x_m\mathrm{\mid }{x}_j{)}_{x_{m+1}}\le A$, whence $(x_{m+1}\mathrm{\mid }{x}_j{)}_{x_m}\ge L-A>K+2\kappa$. Applying F2 with bridge $x_j$ to the product $(x_i\mathrm{\mid }{x}_{m+1}{)}_{x_m}\le A$ shows that the minimum of $(x_i\mathrm{\mid }{x}_j{)}_{x_m}$ and $(x_j\mathrm{\mid }{x}_{m+1}{)}_{x_m}$ is at most $A+\kappa$. The second entry is larger, proving the claim.

Fix integers $a<b$ and the specified segment $\gamma :[0,D]\to X$ from $x_a$ to $x_b$, parametrized by length. Steps 3.1 and 1.2 give for each interior index $i$ a point $z_i=\gamma (t_i)$ with $d(x_i,z_i)<E$. Use $z_a=x_a$, $z_b=x_b$ and $t_a=0,t_b=D$ at the endpoints. Only finitely many witnesses are selected. Consecutive parameters satisfy $\mathrm{\mid }{t}_{i+1}-t_i\mathrm{\mid }=d(z_i,z_{i+1})<L+2E$. For $0<t<D$, take the first index $i$ with $t_i\ge t$. Then $i>a$, $t_{i-1}<t$, and $\mathrm{\mid }{t}_i-t\mathrm{\mid }<L+2E$. Consequently $d(\gamma (t),x_i)<L+3E$. The endpoints satisfy the same conclusion with distance zero. This proves both orbit-chord bounds; no monotonicity of the parameters was assumed.

F3 identifies the sequences $(x_{-i}{)}_{i\ge 1}$ and $(x_j{)}_{j\ge 1}$ as pole representatives. Step 3.1 bounds their mixed products at $o=x_0$ by $K+2\kappa$. Therefore F4 gives $B_o(g^{-},g^{+})\le K+4\kappa$. For every bijective isometry $h$, the exact finite-product identity $(hy\mathrm{\mid }hz{)}_{ho}=(y\mathrm{\mid }z{)}_o$ passes first to mixed liminfs and then to the supremum over representatives. This passage is exact because applying $h$ is a bijection between the representing sequences of the respective classes, by F3. Thus $B_{ho}(h\xi ,h\eta )=B_o(\xi ,\eta )$. If $h\in H$, symmetry gives $B_{ho}(g^{-},g^{+})=B_o(g^{-},g^{+})\le K+4\kappa$. F4 implies that the mixed liminf of our same two fixed orbit sequences at $ho$ is at most this number. Some $i,j\ge 1$ consequently satisfy $(x_{-i}\mathrm{\mid }{x}_j{)}_{ho}<K+4\kappa +1$; otherwise every tail infimum would be at least that larger threshold.

Apply step 1.2 at $v=ho$ to any segment $[x_{-i},x_j]$. There is $z$ on it with $d(ho,z)<K+4\kappa +3\delta +2$. Step 4.1 supplies $m\in [-i,j]\cap \mathbb{Z}$ with $d(z,x_m)<L+3E$. Addition gives $d(ho,x_m)<R$. Since $o=e$, left invariance identifies this with $\mathrm{\mid }{g}^{-mN}h\mathrm{\mid }<R$. F3 shows that $g\in H$; composition and inverses of pole-pair preserving actions show directly that $H$ is a subgroup. Hence $g^{-mN}h\in ⟨g⟩h$ and belongs to $H$. F5 makes the set of these possible representatives finite, so there are finitely many right cosets. Inversion bijects right and left cosets and preserves word length. Finally $K\le L$ yields the displayed uniform upper bound for $R$.

If $h$ commutes with $g$, then $h{g}^{\pm n}=g^{\pm n}h$ and $d(h{g}^{\pm n},g^{\pm n})=\mathrm{\mid }h\mathrm{\mid }$. Bounded perturbation of sequence terms preserves boundary classes by the product formula, so $h$ fixes each pole and belongs to $H$. Thus step 5.1 supplies a representative $g^{-mN}h$ in the same centralizer coset, with the same radius bound. The subgroup $⟨g⟩$ is central in $C_G(g)$, so either coset convention gives the stated quotient. All selections were finite for each specified element or segment; no family of geodesics, nearest-point attainment, properness theorem or AC is used. The inequalities remain strict and valid at $\delta =0$, $c=0$ and $N=1$.