Stable continuum Mayer coefficients obey an integrated tree bound

Fix $n\ge 2$ and positions $x_1,\dots ,x_n$, and put $f_{ij}=f_{\beta }(x_i-x_j)$. For each nonempty label set $I$, let $U_I={\sum }_{i<j,i,j\in I}v(x_i-x_j)$. If $U_I=+\mathrm{\infty }$, the adopted convention gives $e^{-2\beta U_I}=0$; otherwise all pair values in the sum are finite. In either case the exponential definition and addition law [F2], together with evenness of $v$, give ${\prod }_{i\in I}{\prod }_{j\in I\setminus \{i\}}\mathrm{\mid }1+f_{ij}\mathrm{\mid }=e^{-2\beta U_I}$. Stability and monotonicity [F1, F2] imply this product is at most $e^{2\beta B\mathrm{\mid }I\mathrm{\mid }}$. Writing $q_i={\prod }_{j\in I\setminus \{i\}}\mathrm{\mid }1+f_{ij}\mathrm{\mid }$, all $q_i$ are nonnegative and their product is at most $e^{2\beta B\mathrm{\mid }I\mathrm{\mid }}$; hence some $r\in I$ has $q_r\le e^{2\beta B}$.

The maps $(x_1,\dots ,x_n)\mapsto x_i-x_j$ are continuous and hence Borel under [A1, F4]. Since $v$ is Borel [F1], composition with $v$ is Borel [F4]. The adopted map $u\mapsto e^{-\beta u}$ on $\mathbb{R}\cup \{+\mathrm{\infty }\}$ is Borel: on $\mathbb{R}$ it is continuous by [F2], and its value at the Borel singleton $+\mathrm{\infty }$ is $0$ by [F4]. Thus $f_{\beta }$ is Borel, and $U_n$ is a finite sum of finite products of Borel factors [F4]. The same composition and arithmetic facts on ${\mathbb{R}}^{dn}$ make each tree weight Borel. Repeated application of the two-factor identity in [F5] identifies the finite product Borel sigma-algebra, and the Borel-to-Lebesgue clause in [F5] makes the tree weight measurable for product Lebesgue measure.

For disjoint label sets $I,J$, define $g(I,J)$ as the finite sum of ${\prod }_{\{i,j\}\in E(G)}{f}_{ij}$ over simple graphs on $I\cup J$ in which every connected component meets $I$; set $g(\mathrm{\varnothing },\mathrm{\varnothing })=1$ and $g(\mathrm{\varnothing },J)=0$ for nonempty $J$. Evenness in [F1] makes each unordered-edge weight independent of which endpoint is written first. Choose $r\in I$ as in step 1.1 and put $I^{\mathrm{\prime }}=I\setminus \{r\}$. In each graph let $K\subseteq J$ be the neighbors of $r$ in $J$. Edges from $r$ to $I^{\mathrm{\prime }}$ are independently present or absent, and summing their two weights gives ${\prod }_{j\in I^{\mathrm{\prime }}}(1+f_{rj})$. Deleting $r$ leaves a graph all of whose components meet $I^{\mathrm{\prime }}\cup K$, and conversely any such graph reconstructs exactly one original graph after adjoining the chosen $r$-edges. Thus $g(I,J)=({\prod }_{j\in I^{\mathrm{\prime }}}(1+f_{rj})){\sum }_{K\subseteq J}\left({\prod }_{k\in K}{f}_{rk}\right)g(I^{\mathrm{\prime }}\cup K,J\setminus K)$.

Under Countable Choice [A1], the product Lebesgue spaces are sigma-finite by [F3], so Tonelli [F6] applies. Root a labelled tree $T$ at $1$. A finite tree with at least two vertices has at least two leaves: a longest simple path exists in a finite graph, and each endpoint has exactly one incident edge, since another incident edge would either extend the path or create a cycle. At least one of the two leaves differs from the root. Deleting a leaf preserves connectedness and acyclicity on the remaining vertices, so the remaining graph is a tree unless only the root remains. Delete such a leaf and integrate its coordinate first. Translation invariance [F7] and evenness of $f_{\beta }$ give ${\int }_{{\mathbb{R}}^d}\mathrm{\mid }{f}_{\beta }(x_{\mathrm{\ell }}-x_j)\mathrm{\mid }d{x}_{\mathrm{\ell }}=C_{\beta }$. Repeating after each leaf deletion proves that the pinned tree integral is $C_{\beta }^{n-1}$.

Let ${\mathcal{F}}_I(J)$ be the forests on $I\cup J$ with exactly one vertex from $I$ in each tree, and define $h(I,J)=e^{2\beta B(\mathrm{\mid }I\mathrm{\mid }+\mathrm{\mid }J\mathrm{\mid })}{\sum }_{F\in {\mathcal{F}}_I(J)}{\prod }_{\{i,j\}\in E(F)}\mathrm{\mid }{f}_{ij}\mathrm{\mid }$, with the same empty-root convention as $g$. For nonempty $I$, delete the root $r$ selected in step 1.1. Its component cannot contain another vertex of $I$, and the set $K\subseteq J$ of its neighbors gives exactly the roots of the resulting branches; this is a bijection with forests in ${\mathcal{F}}_{I^{\mathrm{\prime }}\cup K}(J\setminus K)$. The exponential addition law [F2] factors the size weight, so $h(I,J)=e^{2\beta B}{\sum }_{K\subseteq J}({\prod }_{k\in K}\mathrm{\mid }{f}_{rk}\mathrm{\mid })h(I^{\mathrm{\prime }}\cup K,J\setminus K)$. If $I=\mathrm{\varnothing }$, both $g$ and $h$ are $1$ for $J=\mathrm{\varnothing }$ and $0$ otherwise. Induct on $\mathrm{\mid }I\mathrm{\mid }+\mathrm{\mid }J\mathrm{\mid }$ from these base cases. Step 1.1 supplies the root bound and step 2.1 supplies the graph recurrence; applying the finite-sum real triangle inequality from [F12, F13] gives $\mathrm{\mid }g(I,J)\mathrm{\mid }\le q_r{\sum }_K({\prod }_{k\in K}\mathrm{\mid }{f}_{rk}\mathrm{\mid })\mathrm{\mid }g(I^{\mathrm{\prime }}\cup K,J\setminus K)\mathrm{\mid }$. The induction hypothesis and $q_r\le e^{2\beta B}$ bound this by $h(I,J)$. At $I=\{1\}$, $J=\{2,\dots ,n\}$, $g(I,J)=U_n$ and every forest in ${\mathcal{F}}_I(J)$ is a spanning tree. This proves the pointwise tree bound.

In the Prüfer encoding, repeatedly delete the least-labelled leaf and record its neighbor; after $n-2$ deletions this gives a word of length $n-2$ over $\{1,\dots ,n\}$. Step 2.2 guarantees a leaf other than the root at every stage with at least two vertices, and deleting a leaf leaves a tree until the final vertex pair. At each stage a vertex occurs in the remaining word exactly once for each incident edge deleted before that vertex; it is recorded when a neighboring leaf is removed, and the vertex itself is deleted once it becomes a leaf, except for the final two vertices. Thus its occurrence count is its current degree minus one, and absence from the word is equivalent to being a current leaf. Conversely, from any such word, repeatedly join the least label absent from the remaining word to its first symbol and delete that label and symbol, then join the last two labels. Adding a new vertex with one edge to a tree makes the new vertex a leaf and preserves connectedness and acyclicity; induction on the word length therefore shows this constructs a tree. In the resulting tree the least absent label is a leaf and the first symbol is its neighbor, so encoding reverses every decoding step. The procedures give a bijection with all $n^{n-2}$ words. Integrating the pointwise tree bound of step 3.1 and using monotonicity, homogeneity and finite additivity [F8, F9] with the tree integral from step 2.2 yields $\int \mathrm{\mid }{U}_n(0,x_2,\dots ,x_n)\mathrm{\mid }d{x}_2\cdots d{x}_n\le e^{2\beta Bn}{n}^{n-2}{C}_{\beta }^{n-1}$.