Proof. It is enough to check non-degenerate triangles. Triangles contained entirely in \(\mathcal{A}\) have all side lengths equal to \(\alpha\), and triangles contained entirely in \(\mathcal{B}\) have all side lengths equal to \(\beta\). Hence those triangles automatically satisfy the triangle inequality.

Consider next a triangle with two vertices in \(\mathcal{A}\) and one vertex in \(\mathcal{B}\), say \(a_i,a_{i'}\in\mathcal{A}\) with \(i\neq i'\), and \(b_j\in\mathcal{B}\). One side has length \(\alpha\). For a vertex \(a_\ell\in\mathcal{A}\), the cross-block side \(d(a_\ell,b_j)\) has length \(\gamma\) exactly when \[\label{eq:gamma-congruence-A} j\equiv \ell+s \pmod{n} \quad\text{for some }s\in\{0,1,\ldots,k-1\},\tag{29}\] and otherwise it has length \(\delta\). Equivalently, for fixed \(j\), the \(\gamma\)-neighbours of \(b_j\) inside \(\mathcal{A}\) are precisely \[\label{eq:gamma-neighbours-A} a_j,a_{j-1},\ldots,a_{j-k+1},\tag{30}\] where subscripts are understood modulo \(n\). Hence exactly \(k\) vertices of \(\mathcal{A}\) have distance \(\gamma\) from \(b_j\), and the remaining \(n-k\) vertices have distance \(\delta\) from \(b_j\). Therefore the two cross-block side lengths are either \((\gamma,\gamma)\), \((\delta,\delta)\), or \((\gamma,\delta)\). Consequently the side lengths of such a triangle are one of \[\label{eq:triangle-types-A} \alpha,\gamma,\gamma;\qquad \alpha,\delta,\delta;\qquad \alpha,\gamma,\delta.\tag{31}\] The type \((\alpha,\gamma,\gamma)\) occurs if and only if \(k\geq2\), because a non-degenerate triangle must use two distinct vertices of \(\mathcal{A}\). The type \((\alpha,\delta,\delta)\) occurs because \(n-k\geq2\), which follows from \(k\leq\lfloor n/2\rfloor\) and \(n\geq3\). The mixed type \((\alpha,\gamma,\delta)\) occurs because \(k\geq1\) and \(n-k\geq1\).

The same argument for a triangle with two vertices in \(\mathcal{B}\) and one vertex in \(\mathcal{A}\) gives the same three possibilities with \(\alpha\) replaced by \(\beta\): \[\label{eq:triangle-types-B} \beta,\gamma,\gamma;\qquad \beta,\delta,\delta;\qquad \beta,\gamma,\delta.\tag{32}\] Hence the triangle inequality is equivalent to the following conditions. For the triples \((s,\gamma,\gamma)\), with \(s\in\{\alpha,\beta\}\), one needs \(s\leq2\gamma\), and this condition is relevant precisely when \(k\geq2\). For the triples \((s,\delta,\delta)\), one needs \(s\leq2\delta\). For the mixed triples \((s,\gamma,\delta)\), one needs \[\label{eq:mixed-triangle-equivalent} s\leq\gamma+\delta, \qquad |\gamma-\delta|\leq s.\tag{33}\] Combining these conditions for \(s=\alpha\) and \(s=\beta\) gives exactly 25 –28 . ◻

Proof. Consider a vector that is constant on each of the two blocks, and write it as \[\label{eq:block-weighting} w(a_i)=w_{\mathcal{A}}(t),\qquad w(b_i)=w_{\mathcal{B}}(t)\tag{42}\] for all \(i\). For such a vector, every weighting equation attached to a point of \(\mathcal{A}\) has left-hand side \(A(t)w_{\mathcal{A}}(t)+C(t)w_{\mathcal{B}}(t)\), and every weighting equation attached to a point of \(\mathcal{B}\) has left-hand side \(C(t)w_{\mathcal{A}}(t)+B(t)w_{\mathcal{B}}(t)\). Thus this block-constant vector is a weighting exactly when it satisfies the \(2\times 2\) system \[\label{eq:block-weighting-system} \begin{pmatrix} A(t)&C(t)\\ C(t)&B(t) \end{pmatrix} \begin{pmatrix} w_{\mathcal{A}}(t)\\ w_{\mathcal{B}}(t) \end{pmatrix} = \begin{pmatrix} 1\\ 1 \end{pmatrix}.\tag{43}\] Solving 43 gives \[\begin{align} w_{\mathcal{A}}(t)&=\frac{B(t)-C(t)}{A(t)B(t)-C(t)^2}, \tag{44}\\ w_{\mathcal{B}}(t)&=\frac{A(t)-C(t)}{A(t)B(t)-C(t)^2}. \tag{45} \end{align}\] Therefore \[\label{eq:mag-sum-weights} \mathop{\mathrm{Mag}}(tX)=n\bigl(w_{\mathcal{A}}(t)+w_{\mathcal{B}}(t)\bigr) =\frac{n\bigl(A(t)+B(t)-2C(t)\bigr)}{A(t)B(t)-C(t)^2}.\tag{46}\]  ◻

Proof. Since \[\label{eq:A-B-C-at-zero} A(0)=n, \qquad B(0)=n, \qquad C(0)=n,\tag{50}\] we have \[\label{eq:N-D-at-zero} N(0)=n(n+n-2n)=0, \qquad D(0)=n^2-n^2=0.\tag{51}\]  ◻

Proof. Put \[\label{eq:sigma-one} \sigma_1=k\gamma+(n-k)\delta.\tag{53}\] Then \[\begin{align} A'(0)&=-(n-1)\alpha, & B'(0)&=-(n-1)\beta, & C'(0)&=-\sigma_1. \label{eq:first-derivatives-A-B-C} \end{align}\tag{54}\] Therefore \[\begin{align} N'(0)&=n\bigl(-(n-1)(\alpha+\beta)+2\sigma_1\bigr), \tag{55}\\ D'(0)&=n\bigl(-(n-1)(\alpha+\beta)+2\sigma_1\bigr). \tag{56} \end{align}\] Both quantities vanish exactly under 52 . ◻

Proof. By Proposition 3, both \(N(t)\) and \(D(t)\) vanish at \(t=0\). By Proposition 4, their first derivatives also vanish at \(t=0\). Since \(D''(0)\neq 0\), the functions \(D(t)\) and \(D'(t)\) are nonzero for all sufficiently small positive \(t\). Since \(N(t)\) and \(D(t)\) are real analytic functions of \(t\), l’Hopital’s rule applied twice gives \[\label{eq:lhopital-proof} \lim_{t\to 0+}\frac{N(t)}{D(t)} = \lim_{t\to 0+}\frac{N'(t)}{D'(t)} = \lim_{t\to 0+}\frac{N''(t)}{D''(t)} = \frac{N''(0)}{D''(0)}.\tag{57}\] The last equality follows from the continuity of \(N''\) and \(D''\) at \(0\), together with \(D''(0)\neq0\). Using 49 proves ?? . ◻

Proof. Put \[\label{eq:sigma-two} \sigma_2=k\gamma^2+(n-k)\delta^2.\tag{60}\] A direct calculation gives \[\begin{align} N''(0)&=n\bigl((n-1)(\alpha^2+\beta^2)-2\sigma_2\bigr), \tag{61}\\ D''(0)&=n(n-1)(\alpha^2+\beta^2)+2(n-1)^2\alpha\beta -2\sigma_1^2-2n\sigma_2. \tag{62} \end{align}\] Under the balanced condition, \[\label{eq:sigma-one-balanced} \sigma_1=\frac{(n-1)(\alpha+\beta)}{2}.\tag{63}\] Moreover, \[\label{eq:sigma-two-variance} \sigma_2=\frac{\sigma_1^2}{n}+\frac{k(n-k)}{n}(\gamma-\delta)^2.\tag{64}\] Substituting 63 and 64 into 61 and 62 yields \[\begin{align} D''(0)&=(n-1)(\alpha^2+\beta^2)-2k(n-k)(\gamma-\delta)^2, \tag{65}\\ N''(0)&=D''(0)+\frac{(n-1)^2}{2}(\alpha-\beta)^2. \tag{66} \end{align}\] Since \(\Omega_{n,k}=2D''(0)\), Proposition 5 gives \[\label{eq:one-plus-proof} \lim_{t\to 0+}\mathop{\mathrm{Mag}}(tX) =1+\frac{\frac{(n-1)^2}{2}(\alpha-\beta)^2}{D''(0)} =1+\frac{(n-1)^2(\alpha-\beta)^2}{\Omega_{n,k}(\alpha,\beta,\gamma,\delta)}.\tag{67}\]  ◻

Proof. The possible pairs \((n,k)\) with \(3\leq n\leq 5\) and \(1\leq k\leq \lfloor n/2\rfloor\) are \[\label{eq:small-n-all-pairs} (3,1),\qquad (4,1),\qquad (4,2),\qquad (5,1),\qquad (5,2).\tag{68}\] We check these cases one by one, using only the triangle inequalities and the balanced condition.

Set \[\label{eq:u-v-eta} u=\max\{\alpha,\beta\}, \qquad v=\min\{\alpha,\beta\}, \qquad \eta=|\gamma-\delta|.\tag{69}\] The mixed triangle inequalities give \[\label{eq:eta-leq-v} \eta\leq v.\tag{70}\] Recall that \[\label{eq:Omega-small-proof} \Omega_{n,k} = 2(n-1)(u^2+v^2)-4k(n-k)\eta^2.\tag{71}\] By Corollary 1, it is enough to show that \(\Omega_{n,k}\) cannot be negative. Indeed, if \(\Omega_{n,k}>0\), then \[\lim_{t\to 0+}\mathop{\mathrm{Mag}}(tX) = 1+\frac{(n-1)^2(\alpha-\beta)^2}{\Omega_{n,k}} \geq 1.\]

We first treat the three cases with \(k=1\).

For \((n,k)=(3,1)\), we have \[\label{eq:n3k1-Omega} \Omega_{3,1}=4(u^2+v^2)-8\eta^2.\tag{72}\] If \(\Omega_{3,1}<0\), then \[\label{eq:n3k1-contradiction} \eta^2>\frac{u^2+v^2}{2}\geq v^2,\tag{73}\] hence \(\eta>v\), contradicting 70 . Thus \(\Omega_{3,1}\geq 0\).

For \((n,k)=(4,1)\), we have \[\label{eq:n4k1-Omega} \Omega_{4,1}=6(u^2+v^2)-12\eta^2.\tag{74}\] If \(\Omega_{4,1}<0\), then again \[\label{eq:n4k1-contradiction} \eta^2>\frac{u^2+v^2}{2}\geq v^2,\tag{75}\] so \(\eta>v\), contradicting 70 . Hence \(\Omega_{4,1}\geq 0\).

For \((n,k)=(5,1)\), we have \[\label{eq:n5k1-Omega} \Omega_{5,1}=8(u^2+v^2)-16\eta^2.\tag{76}\] If \(\Omega_{5,1}<0\), then \[\label{eq:n5k1-contradiction} \eta^2>\frac{u^2+v^2}{2}\geq v^2,\tag{77}\] so \(\eta>v\), contradicting 70 . Thus \(\Omega_{5,1}\geq 0\).

Next consider \((n,k)=(4,2)\). The balanced condition is \[\label{eq:n4k2-balanced-sum} \gamma+\delta=\frac{3}{4}(u+v).\tag{78}\] Since \(k=2\) and \(n-k=2\), both the \(\gamma\)-edges and the \(\delta\)-edges occur in pairs. Therefore the triangle inequalities force \[\label{eq:n4k2-lower-bounds} u\leq 2\gamma, \qquad u\leq 2\delta,\tag{79}\] or equivalently \[\label{eq:n4k2-gamma-delta-lower} \gamma,\delta\geq \frac{u}{2}.\tag{80}\] In particular, \[\label{eq:n4k2-necessary} \frac{3}{4}(u+v)=\gamma+\delta\geq u,\tag{81}\] so a metric space can exist only if \(u\leq 3v\). Moreover, under the lower bounds 80 , the largest possible value of \(|\gamma-\delta|\) occurs when the smaller of \(\gamma,\delta\) is \(u/2\). Thus \[\label{eq:n4k2-eta-bound} \eta \leq \frac{3}{4}(u+v)-u = \frac{3v-u}{4}.\tag{82}\] Now \[\label{eq:n4k2-Omega} \Omega_{4,2}=6(u^2+v^2)-16\eta^2.\tag{83}\] If \(\Omega_{4,2}<0\), then \[\label{eq:n4k2-negative-condition} \eta^2>\frac{3}{8}(u^2+v^2).\tag{84}\] On the other hand, 82 gives \[\label{eq:n4k2-upper-square} \eta^2\leq \frac{(3v-u)^2}{16}.\tag{85}\] But, since \(u\geq v>0\), \[\label{eq:n4k2-elementary-positive} 6(u^2+v^2)-(3v-u)^2 = 5u^2+6uv-3v^2 >0.\tag{86}\] Equivalently, \[\label{eq:n4k2-contradiction} \frac{(3v-u)^2}{16} < \frac{3}{8}(u^2+v^2),\tag{87}\] which contradicts 84 . Hence \(\Omega_{4,2}>0\).

Finally consider \((n,k)=(5,2)\). The balanced condition is \[\label{eq:n5k2-balanced} 2\gamma+3\delta=2(u+v).\tag{88}\] Again, since \(k=2\) and \(n-k=3\), the triangle inequalities force \[\label{eq:n5k2-lower-bounds} u\leq 2\gamma, \qquad u\leq 2\delta,\tag{89}\] so \[\label{eq:n5k2-gamma-delta-lower} \gamma,\delta\geq \frac{u}{2}.\tag{90}\] Therefore \[\label{eq:n5k2-necessary} 2(u+v)=2\gamma+3\delta\geq \frac{5u}{2},\tag{91}\] and hence a metric space can exist only if \[\label{eq:n5k2-u-bound} u\leq 4v.\tag{92}\]

We now bound \(\eta=|\gamma-\delta|\) directly. If \(\gamma\geq \delta\), then \(\gamma=\delta+\eta\), and 88 gives \[\label{eq:n5k2-case-one-delta} 5\delta+2\eta=2(u+v).\tag{93}\] Since \(\delta\geq u/2\), we obtain \[\label{eq:n5k2-case-one-eta} \eta\leq v-\frac{u}{4} = \frac{4v-u}{4}.\tag{94}\] If \(\delta\geq \gamma\), then \(\delta=\gamma+\eta\), and 88 gives \[\label{eq:n5k2-case-two-gamma} 5\gamma+3\eta=2(u+v).\tag{95}\] Since \(\gamma\geq u/2\), we obtain \[\label{eq:n5k2-case-two-eta} \eta\leq \frac{4v-u}{6} \leq \frac{4v-u}{4},\tag{96}\] where the last inequality uses 92 . Hence, in all cases, \[\label{eq:n5k2-eta-bound} \eta\leq \frac{4v-u}{4}.\tag{97}\]

Now \[\label{eq:n5k2-Omega} \Omega_{5,2}=8(u^2+v^2)-24\eta^2.\tag{98}\] If \(\Omega_{5,2}<0\), then \[\label{eq:n5k2-negative-condition} \eta^2>\frac{u^2+v^2}{3}.\tag{99}\] But 97 implies \[\label{eq:n5k2-upper-square} \eta^2\leq \frac{(4v-u)^2}{16}.\tag{100}\] Since \(u\geq v>0\), \[\label{eq:n5k2-elementary-positive} 16(u^2+v^2)-3(4v-u)^2 = 13u^2+24uv-32v^2 >0.\tag{101}\] Equivalently, \[\label{eq:n5k2-contradiction} \frac{(4v-u)^2}{16} < \frac{u^2+v^2}{3},\tag{102}\] contradicting 99 . Hence \(\Omega_{5,2}>0\).

We have shown, case by case, that \(\Omega_{n,k}<0\) is impossible for every pair in 68 . Therefore, in all cases with \(\Omega_{n,k}>0\), Corollary 1 gives \[\lim_{t\to 0+}\mathop{\mathrm{Mag}}(tX)\geq 1.\]

It remains only to handle the possible equality case \(\Omega_{n,k}=0\). The above argument shows that equality can occur only in the cases \(k=1\), and then necessarily \[\label{eq:k1-equality-case} u=v, \qquad \eta=v.\tag{103}\] Thus \(\alpha=\beta=v\), and hence \(A(t)=B(t)\). The magnitude formula ?? becomes, after cancelling the common factor \(A(t)-C(t)\), \[\label{eq:k1-degenerate-cancellation} \mathop{\mathrm{Mag}}(tX) = \frac{2n(A(t)-C(t))}{(A(t)-C(t))(A(t)+C(t))} = \frac{2n}{A(t)+C(t)}.\tag{104}\] Since \(A(0)=C(0)=n\), we get \[\label{eq:k1-degenerate-limit-final} \lim_{t\to 0+}\mathop{\mathrm{Mag}}(tX) = \frac{2n}{2n} = 1.\tag{105}\] Thus every balanced \(t\)-scaled \((n,k;\alpha,\beta,\gamma,\delta)\)-two-chunk cyclic metric space with \(3\leq n\leq 5\) has finite small-scale magnitude at least \(1\). ◻

Proof. Let \(u,v,\eta\) be as in 69 .

For \(k=1\), the estimate \[\label{eq:n6k1-Omega} \Omega_{6,1}=10(u^2+v^2)-20\eta^2 \geq 10(u^2-v^2)\tag{106}\] shows that \(\Omega_{6,1}\geq 0\). If \(\Omega_{6,1}>0\), then Corollary 1 gives a limit at least \(1\). If \(\Omega_{6,1}=0\), then \(\alpha=\beta\), and the same direct expansion as in 105 gives limit \(1\).

For \(k=3\), the balanced condition gives \[\label{eq:n6k3-balanced} \gamma+\delta=\frac{5}{6}(u+v).\tag{107}\] Since \(\gamma,\delta\geq u/2\), \[\label{eq:n6k3-eta-bound} \eta\leq \gamma+\delta-u=\frac{5v-u}{6}.\tag{108}\] Thus \[\label{eq:n6k3-Omega-positive} \Omega_{6,3} =10(u^2+v^2)-36\eta^2 \geq 10(u^2+v^2)-(5v-u)^2>0.\tag{109}\] Again Corollary 1 implies that the finite small-scale limit is at least \(1\). This proves (1).

It remains to prove (2). We use two explicit one-parameter families. The particular normalizations used below are not canonical; they are chosen to make the balanced condition automatic and to keep the resulting one-variable formulae transparent.

First, we impose the simplifying condition \(\gamma=\delta\). If \(\alpha=r\) and \(\beta=1\), then the balanced condition forces \(\gamma=\delta=5(r+1)/12\). Thus, for \(1\leq r\leq 5\), define \[\label{eq:n6-family-zero} \alpha=r, \qquad \beta=1, \qquad \gamma=\delta=\frac{5(r+1)}{12}.\tag{110}\] This family is balanced, and Lemma 1 shows that it is a metric family exactly in the displayed range. Since \(\gamma=\delta\), Corollary 1 gives \[\label{eq:n6-L-zero} L_0(r) :=\lim_{t\to 0+}\mathop{\mathrm{Mag}}(tX) =1+\frac{25(r-1)^2}{10(r^2+1)}.\tag{111}\] The function \(L_0\) is continuous and increasing on \([1,5]\), and \[\label{eq:n6-L-zero-endpoints} L_0(1)=1, \qquad L_0(5)=\frac{33}{13}.\tag{112}\] Hence this family realizes every value in \([1,33/13]\).

Second, for \(1\leq r\leq 5\), define \[\label{eq:n6-family-one} \alpha=r, \qquad \beta=1, \qquad \delta=\frac{r}{2}, \qquad \gamma=\frac{r+5}{4}.\tag{113}\] This family is also balanced and metric by Lemma 1. Here \[\label{eq:n6-family-one-eta} \gamma- \delta=\frac{5-r}{4},\tag{114}\] and Corollary 1 gives \[\label{eq:n6-L-one} L_1(r) :=\lim_{t\to 0+}\mathop{\mathrm{Mag}}(tX) =1+\frac{25(r-1)^2}{4(2r^2+5r-10)}.\tag{115}\] The denominator in 115 vanishes at \[\label{eq:n6-root} r_0=\frac{\sqrt{105}-5}{4}\in(1,5).\tag{116}\] Moreover, \[\begin{align} \lim_{r\to r_0-}L_1(r)&=-\infty, \tag{117}\\ \lim_{r\to r_0+}L_1(r)&=+\infty, \tag{118}\\ L_1(1)&=1, \tag{119}\\ L_1(5)&=\frac{33}{13}. \tag{120} \end{align}\] By continuity, \(L_1\) realizes every value below \(1\) on \([1,r_0)\), and every value at least \(33/13\) on \((r_0,5]\). Together with 111 , this realizes every value in \(\mathbb{R}\setminus\{1\}\). This proves (2).

Finally, (3) follows from (2) by choosing any \(R<1\). The concrete matrix in 2 gives the explicit value \(R=44/59\). ◻

Proof. Put \[\label{eq:m-q-definition} m=n-1, \qquad q=n-k.\tag{121}\] Since \(n\geq 7\) and \(2\leq k\leq \lfloor n/2\rfloor\), one can choose a real number \(\theta\) satisfying \[\label{eq:theta-choice} \sqrt{\frac{m}{kq}}<\theta<\min\left\{1,\frac{n-2}{2k}\right\}.\tag{122}\] Indeed, \(kq>m\), and the inequality \[\label{eq:theta-interval-nonempty} \frac{m}{kq}<\frac{(n-2)^2}{4k^2}\tag{123}\] follows from \(k\leq n/2\), \(q\geq n/2\), and \((n-2)^2>4(n-1)\) for \(n\geq 7\).

For \(r>1\) close to \(1\) and for \(0\leq h\leq \theta\), define \[\begin{align} \beta&=1, \tag{124}\\ \alpha&=r, \tag{125}\\ \delta(r,h)&=\frac{(n-1)(r+1)}{2n}-\frac{k}{n}h, \tag{126}\\ \gamma(r,h)&=\delta(r,h)+h. \tag{127} \end{align}\] Then \[\label{eq:general-family-balanced} k\gamma(r,h)+(n-k)\delta(r,h)=\frac{(n-1)(r+1)}{2},\tag{128}\] so the balanced condition holds. By the strict upper bound in 122 , the inequalities in Lemma 1 hold for all \(0\leq h\leq\theta\), after restricting \(r\) to a sufficiently small interval \((1,1+\varepsilon)\). Hence 124 –127 define balanced \((n,k;\alpha,\beta,\gamma,\delta)\)-two-chunk cyclic metric spaces.

For this family, Corollary 1 gives \[\label{eq:general-family-limit} L(r,h) :=\lim_{t\to 0+}\mathop{\mathrm{Mag}}(tX) =1+\frac{m^2(r-1)^2}{2m(r^2+1)-4kq h^2},\tag{129}\] whenever the denominator is nonzero. For \(r\) sufficiently close to \(1\), the denominator in 129 is positive at \(h=0\), while it is negative at \(h=\theta\), because \[\label{eq:general-denominator-signs} 2m(1^2+1)-4kq\theta^2=4m-4kq\theta^2<0.\tag{130}\] Thus the denominator crosses zero as \(h\) varies from \(0\) to \(\theta\). On the positive side of the crossing, \(L(r,h)\) tends to \(+\infty\); on the negative side, it tends to \(-\infty\). Also, \[\begin{align} L(r,0)&=1+\frac{m(r-1)^2}{2(r^2+1)}\longrightarrow 1 \quad \text{as } r\to 1+, \tag{131}\\ L(r,\theta)&\longrightarrow 1 \quad \text{as } r\to 1+, \tag{132} \end{align}\] with \(L(r,\theta)<1\) for \(r\) sufficiently close to \(1\).

Now let \(R>1\). Choose \(r>1\) sufficiently close to \(1\) so that \(L(r,0)<R\). Since \(L(r,h)\to +\infty\) at the positive side of the zero of the denominator, the intermediate value theorem gives an \(h\in[0,\theta]\) with \(L(r,h)=R\). Similarly, if \(R<1\), choose \(r>1\) sufficiently close to \(1\) so that \(L(r,\theta)>R\). Since \(L(r,h)\to -\infty\) at the negative side of the zero of the denominator, the intermediate value theorem again gives an \(h\in[0,\theta]\) with \(L(r,h)=R\). This proves the proposition. ◻