2. Model description and main results

The $\beta$ -model is formally defined as follows [Reference Chatterjee, Diaconis and Sly10]. Consider a random graph with vertex set [n], where $n\ge 2$ is an integer. The presence of an edge between vertices i and j is determined independently with probability

(1) \begin{equation} p_{ij}= \frac{{\mathrm{e}}^{\beta_i + \beta_j}}{1+{\mathrm{e}}^{\beta_i + \beta_j}}, \quad 1\le i \neq j\le n,\end{equation}

where $\beta_{i}\in \mathbb{R}$ is the degree heterogeneity parameter for vertex i. We note that $\beta_i$ can be negative and may depend on n. These parameters quantify the connectivity propensity of vertices: a larger $\beta_{i}$ corresponds to a higher likelihood of vertex i forming edges. Notably, when $\beta_{i}=\beta$ is a constant for all $i\in [n]$ , the $\beta$ -model reduces to the ER random graph model with edge probability $p={\mathrm{e}}^{2\beta}/(1+\textrm{e}^{2\beta})$ .

Let $I_{ij}$ denote the indicator for the presence of an edge between vertices i and j. By construction, $I_{ii}=0$ , $I_{ij}=I_{ji}$ , and the collection $\{I_{ij},1\le i < j \le n\}$ consists of independent Bernoulli random variables with success rates $p_{ij}$ . Clearly, the adjacency matrix ${{\textit{A}}}=(I_{ij})_{n\times n}$ is symmetric with zero diagonal. In terms of ${{\textit{A}}}$ , the number of triangles $T_n$ in the $\beta$ -model on n vertices is given by

(2) \begin{equation} T_n = \frac16 \mathrm{tr}({{\textit{A}}}^3)= \sum_{1\le i < j < k\le n} I_{ij} I_{jk} I_{ki},\end{equation}

where $\mathrm{tr}({\cdot})$ denotes the matrix trace.

For simplicity of notation, we define $\mu_i={\mathrm{e}}^{\beta_i}>0$ for each $i\in [n]$ and let

\[\boldsymbol{\mu}=(\mu_1,\mu_2,\ldots,\mu_n)^\top,\]

omitting the subscript n. The edge probabilities in (1) then become

(3) \begin{equation}p_{ij}= \frac{\mu_i\mu_j}{1+ \mu_i\mu_j}.\end{equation}

Deriving asymptotic properties of $T_n$ under general parameters $\beta_i$ presents significant challenges due to degree heterogeneity. To address this, we impose the following sparsity conditions:

(4) \begin{equation} \mu_{\max}\to 0 \quad \text{and} \quad \|\boldsymbol\mu\|_2 \rightarrow \infty.\end{equation}

The first condition ensures network sparsity, while the second guarantees a non-degenerate graph structure with a substantial number of triangles (see Proposition 1). Note that, for any $s>t>0$ , $\|\boldsymbol\mu\|_s^s\le \mu_{\max}^{s-t}\|\boldsymbol\mu\|_t^t$ , since each $\mu_i$ is positive. It thus follows by the assumption $\mu_{\max}\to0$ in (4) that

(5) \begin{equation} \|\boldsymbol\mu\|_s^s = o\big(\|\boldsymbol\mu\|_t^t\big), \quad s>t>0.\end{equation}

The Kolmogorov distance between random variables X and Y is defined as

\[d_K(X,Y)=\sup_{z\in\mathbb R}|{\mathbb P}(X\leq z)-{\mathbb P}(Y\leq z)|.\]

Define the normalized triangle count $F_n$ as

(6) \begin{equation} F_n= \frac{T_n - \mathbb{E}[T_n]}{\sqrt{\mathrm{Var}[T_n]}}.\end{equation}

The following theorem provides an upper bound for $d_{K}(F_n, \mathcal{N})$ , where $\mathcal{N}$ denotes a standard normal random variable.

Theorem 1. Under the sparsity condition (4) we have

\begin{align&#x002A;} d_{K}(F_n, \mathcal{N}) \le \frac{C}{\|\boldsymbol\mu\|_2^{{5}/{2}}\big(\|\boldsymbol\mu\|_3^6+\|\boldsymbol\mu\|_2^2\big)}\sum_{\ell=1}^5A_{\ell}, \end{align&#x002A;}

where $C>0$ is a constant, and

(7) \begin{equation} \begin{alignedat}{3} A_1 & = \|\boldsymbol\mu\|_{{3}/{2}}^{{3}/{4}}\|\boldsymbol\mu\|_{{5}/{2}}^{{5}/{4}}, \qquad & A_2 & = \|\boldsymbol\mu\|_{{3}/{2}}^{{3}/{4}}\|\boldsymbol\mu\|_{2}^{1/2}\|\boldsymbol\mu\|_{{7}/{2}}^{{7}/{4}}\|\boldsymbol\mu\|_{4}^{2}, \qquad & & \\[3pt] A_3 & = \|\boldsymbol\mu\|_2^{{5}/{2}}\|\boldsymbol\mu\|_5^{5}, \qquad & A_4 & = \|\boldsymbol\mu\|_{2}^{{3}/{2}}\|\boldsymbol\mu\|^{{5}/{2}}_{{5}/{2}}\|\boldsymbol\mu\|_{5}^{{5}/{2}}, & A_5 & = \|\boldsymbol\mu\|_{{7}/{4}}^{{7}/{4}}\|\boldsymbol\mu\|_{{7}/{2}}^{{7}/{4}}. \end{alignedat} \end{equation}

The bound in Theorem 1 is complicated due to the generality of $\boldsymbol\mu$ . The following example demonstrates that, in specific cases, this bound matches the performance of existing results.

Example 1. Fix a positive integer K and a sequence of positive constants $\{\pi_r, 1\le r\le K\}$ satisfying $\sum_{r=1}^K\pi_r=1$ . Let $\{V_r, 1\le r\le K\}$ be a partition of [n] with the cardinality $|V_r|=\pi_r n+o(n)$ . Assume that, for each $i\in V_r$ with $1\le r\le K$ , $\mu_i=\theta_rn^{-\alpha/2}$ , where the constants $\theta_r>0$ and $0<\alpha<1$ . Then $\mu_{\max}$ is of order $n^{-\alpha/2}$ and, for any fixed $s>0$ ,

\begin{align&#x002A;} \|\boldsymbol\mu\|_s^s=\sum_{r=1}^K |V_r|\,\theta_r^s n^{-\alpha s/2} = n^{1-\alpha s/2}\Bigg(\sum_{r=1}^K \pi_r\theta_r^s\Bigg)(1+o(1)). \end{align&#x002A;}

In particular, when $s=2$ , the quantity $\|\boldsymbol\mu\|_2^2$ is of order $n^{1-\alpha}$ .

This blockwise-constant scaling is standard in sparse stochastic block models and their degree-corrected variants; see, e.g., [Reference Abbe1, Reference Holland and Leinhardt17, Reference Karrer and Newman19, Reference Mossel, Neeman and Sly24] for more background. An application of Theorem 1 together with some basic calculations yields $d_K(F_n,\mathcal N)\le C\, n^{-\eta(\alpha)}$ , where

\begin{align&#x002A;} \eta(\alpha)= \begin{cases} 1-\alpha, & 0<\alpha\le \tfrac12, \\[0.2em] \tfrac34-\tfrac12{\alpha}, & \tfrac12 < \alpha\le \tfrac23, \\[0.2em] \tfrac54(1-\alpha), & \tfrac23<\alpha<1. \end{cases} \end{align&#x002A;}

In the special case $K=1$ , we have $\mu_i\equiv\mu=cn^{-\alpha/2}$ for all $i\in [n]$ , where $c>0$ and $\alpha\in (0,1)$ . Then our model reduces to the ER random graph model with edge probability

\begin{equation&#x002A;} p= \frac{\mu^2}{1+ \mu^2}\sim c^2 n^{-\alpha} \end{equation&#x002A;}

as $n\rightarrow\infty$ . In this case, the order $\eta(\alpha)$ coincides with that established in [Reference Krokowski, Reichenbachs and Thäle22, Theorem 1.1].

Notably, the upper bound in Theorem 1 may not vanish as $n\to\infty$ under the sparsity condition (4), primarily due to the $L_{3/2}$ -norm of the $\boldsymbol\mu$ term in $A_1$ and $A_2$ , which can grow very fast. To establish asymptotic normality for $T_n$ , we introduce two additional heterogeneity conditions:

(8) \begin{align} \|\boldsymbol\mu\|_{{3}/{2}}^{{3}/{2}} & = O\big(\|\boldsymbol{\mu}\|_2^6\big), \\[-10pt] \nonumber \end{align}

(9) \begin{align} \frac{\mu_{\max}}{\mu_{\min}} & = O\big(\|\boldsymbol{\mu}\|_2^{{3}/{2}}\big). \end{align}

Condition (8) directly restricts the growth rate of the $L_{3/2}$ -norm of $\boldsymbol\mu$ . Condition (9) accommodates considerable degree heterogeneity in the $\beta$ -model, thereby permitting the ratio $\mu_{\max}/\mu_{\min}$ to diverge at an appropriate rate. This is in sharp contrast to the ER random graph model, where this ratio is invariably 1. Such regularity conditions are widely adopted in statistical network analysis. For instance, in community detection problems, several constraints on the extreme values of degree heterogeneity parameters are imposed to bound the estimation error (see, e.g., [Reference Jin18]).

The following result establishes the asymptotic normality of $T_n$ under these conditions.

3. Mean and variance

In this section, we derive the asymptotic mean and variance of $T_n$ , the number of triangles in the $\beta$ -model, under the sparsity condition (4).

Proposition 1. Suppose that the sparsity condition (4) holds. As $n\to\infty$ , we have

(10) \begin{align} {\mathbb E}[{T_n}] & = \frac16\|\boldsymbol\mu\|_2^6(1+o(1)), \\[-10pt] \nonumber \end{align}

(11) \begin{align} \mathrm{Var}[{T_n}] & = \frac16\|\boldsymbol\mu\|_2^4\big(3\|\boldsymbol\mu\|_3^6+\|\boldsymbol\mu\|_2^2\big)(1+o(1)). \\[8pt] \nonumber \end{align}

Proof. We first note that, by (3) and (4), the edge probability $p_{ij}$ satisfies

(12) \begin{equation} p_{ij} = (1 + o(1)) \mu_i \mu_j, \quad 1\le i\neq j\le n. \end{equation}

For distinct vertices $i,j,k\in[n]$ , by the definition of the $\beta$ -model, the indicators $I_{ij}$ , $I_{jk}$ , and $I_{ki}$ are independent Bernoulli random variables. Using (2) and (12), the expectation of $T_n$ is equal to

(13) \begin{equation} {\mathbb E}[{T_n}] = \sum_{1\le i < j < k \le n}p_{ij}p_{jk}p_{ki} = (1+o(1))\sum_{1\le i < j < k \le n}\mu_i^2\mu_j^2\mu_k^2. \end{equation}

In terms of the norms of $\boldsymbol\mu$ , the summation on the right-hand side of (13) can be rewritten as

\begin{align&#x002A;} \sum_{1\le i < j < k \le n} \mu_i^2\mu_j^2\mu_k^2 & = \frac16\sum_{1\le i,j,k\le n} \mu_i^2\mu_j^2\mu_k^2 - \frac12\sum_{1\le i,j\le n} \mu_i^2\mu_j^4 + \frac13\sum_{i=1}^{n} \mu_i^{6} \\[3pt] & = \frac16\|\boldsymbol\mu\|_2^6 - \frac12\|\boldsymbol\mu\|_2^2\|\boldsymbol\mu\|_4^4 + \frac13\|\boldsymbol\mu\|_6^6. \end{align&#x002A;}

Combining this with (5) and (13) yields (10).

We now turn to the variance of $T_n$ . To this end, we define $X_{ij} = I_{ij} - p_{ij}$ , $1\le i\neq j\le n$ . Then, for any $1\le i\neq j\le n$ , we have ${\mathbb E}[X_{ij} ]=0$ , and by (12),

(14) \begin{equation} \mathrm{Var}[X_{ij}] = {\mathbb E}\big[X_{ij}^2\big] = p_{ij}(1 - p_{ij}) = (1 + o(1))\mu_i\mu_j. \end{equation}

By (13), we can now reformulate (2) as

(15) \begin{align} T_n & = \sum_{1\le i < j < k \le n} (X_{ij} + p_{ij})(X_{jk} + p_{jk})(X_{ki} + p_{ki}) \notag \\[3pt] & = {\mathbb E}[{T_n}] + \sum_{1\le i < j \le n}c_{ij}X_{ij} + \sum_{\substack{1\le i < j \le n\\[3pt] k\neq i,j}}p_{ij}X_{jk} X_{ki} + \sum_{1\le i < j < k \le n}X_{ij} X_{jk} X_{ki}, \end{align}

where, by (12),

\begin{align&#x002A;} c_{ij}=\sum_{k\ne i,j}p_{jk} p_{ki}=(1+o(1))\mu_i\mu_j\sum_{k\ne i,j}\mu_k^2=(1+o(1))\|\boldsymbol\mu\|_2^2\mu_i\mu_j. \end{align&#x002A;}

The $1+\binom{n}{2}+(n-2)\binom{n}{2}+\binom{n}{3}$ terms on the right-hand side of (15) are pairwise uncorrelated. By (12) and (14), we have

\begin{align&#x002A;} \mathrm{Var}[T_n] & = \sum_{1\le i < j \le n}c_{ij}^2{\mathbb E}\big[X_{ij}^2\big] + \sum_{\substack{1\le i < j \le n \\ k\neq i,j}}p_{ij}^2{\mathbb E}\big[X_{jk}^2\big]{\mathbb E}\big[X_{ki}^2\big] + \sum_{1\le i < j < k \le n}{\mathbb E}\big[X_{ij}^2\big]{\mathbb E}\big[X_{jk}^2\big]{\mathbb E}\big[X_{ki}^2\big] \\[3pt] & = (1+o(1))\Bigg(\|\boldsymbol\mu\|_2^4\sum_{1\le i < j \le n} \mu_i^3\mu_j^3 + \|\boldsymbol\mu\|_2^2\sum_{1\le i < j \le n} \mu_i^3\mu_j^3 + \sum_{1\le i < j < k \le n} \mu_i^2\mu_j^2\mu_k^2\Bigg). \end{align&#x002A;}

Since $\|\boldsymbol\mu\|_2^2=o(\|\boldsymbol\mu\|_2^4)$ and $\sum_{1\le i < j \le n} \mu_i^3\mu_j^3=\frac12\big(\|\boldsymbol\mu\|_3^6-\|\boldsymbol\mu\|_6^6\big)$ , by (10) and (13) we can proceed with

\begin{align&#x002A;} \mathrm{Var}[T_n] & = \bigg(\frac12\|\boldsymbol\mu\|_2^4\big(\|\boldsymbol\mu\|_3^6-\|\boldsymbol\mu\|_6^6\big) + \frac{1}{6}\|\boldsymbol\mu\|_2^6\bigg)(1+o(1)) \\[3pt] & = \frac16\|\boldsymbol\mu\|_2^4\big(3\|\boldsymbol\mu\|_3^6-3\|\boldsymbol\mu\|_6^6+\|\boldsymbol\mu\|_2^2\big)(1+o(1)), \end{align&#x002A;}

which, together with (5), completes the proof of (11).

An immediate consequence of Proposition 1 is the following.

Proof. Applying Proposition 1 and Chebyshev’s inequality, for any $\varepsilon>0$ , by (5) we have

\begin{align&#x002A;} {\mathbb P}\bigg(\bigg|\frac{T_n}{{\mathbb E}[T_n]}-1\bigg|>\varepsilon\bigg) \le \frac{\mathrm{Var}[T_n]}{(\varepsilon{\mathbb E}[T_n])^2} = \frac{6\big(3\|\boldsymbol\mu\|_3^6+\|\boldsymbol\mu\|_2^2\big)}{\varepsilon^2\|\boldsymbol\mu\|_2^8}(1+o(1)) \to 0, \end{align&#x002A;}

which implies that $T_n/{\mathbb E}[T_n]$ converges to 1 in probability. Thus, the desired result follows via Slutsky’s theorem and (10).

4. Proofs

This section is devoted to the proofs of our main results stated in Section 2. We first introduce several auxiliary lemmas essential to our approach.

Our primary tool for establishing the asymptotic normality of triangle counts in the $\beta$ -model is the Malliavin–Stein method [Reference Nourdin and Peccati28], a powerful synthesis of Malliavin calculus [Reference Nualart30] and Stein’s method (see, e.g., [Reference Barbour and Chen4]). This method is particularly effective for analyzing normal approximations when applied to independent Rademacher or Bernoulli random variables [Reference Krokowski, Reichenbachs and Thäle21, Reference Nourdin, Peccati and Reinert29]. In particular, by applying this method Berry–Esseen bounds for triangle counts in ER random graphs have been established [Reference Krokowski, Reichenbachs and Thäle22], while non-uniform Berry–Esseen bounds for counts of any fixed subgraph have been further derived in [Reference Butzek and Eichelsbacher8]. For more background and related bounds in the Malliavin–Stein framework, see, e.g., [Reference Eichelsbacher, Rednoß, Thäle and Zheng14] and references therein.

To state the normal approximation bound used in our approach (Lemma 1), we introduce, following [Reference Eichelsbacher, Rednoß, Thäle and Zheng14], the discrete gradient operator for functionals of independent Bernoulli random variables. For our purposes, only the first- and second-order discrete gradients are needed.

Let $m \ge 3$ be an integer, and ${{\textit{X}}}=(X_1,\ldots,X_m)$ a random vector of independent Bernoulli random variables with ${\mathbb P}(X_a = 1)=p_a$ and ${\mathbb P}(X_a = 0)=q_a$ , $a\in[m]$ , where $0<p_a<1$ and $q_a=1-p_a$ . For a measurable function $f\colon\{0,1\}^m\to\mathbb{R}$ and $F=f({{\textit{X}}})$ , define the discrete gradient of F with respect to $X_a$ by

(16) \begin{equation} D_aF=\sqrt{p_aq_a}\big(f({{\textit{X}}}_a^+)-f({{\textit{X}}}_a^-)\big),\end{equation}

where

\[{{\textit{X}}}_a^+=(X_1,\ldots,X_{a-1},1,X_{a+1},\ldots,X_m),\qquad{{\textit{X}}}_a^-=(X_1,\ldots,X_{a-1},0,X_{a+1},\ldots,X_m).\]

Using (16), we can further define the second-order discrete gradient $D_bD_aF\,:\!=\,D_b(D_aF)$ . Note that for any $a,b\in [m]$ , $D_bD_aF=D_aD_bF$ .

The following lemma plays an important role in our proof of Theorem 1.

Lemma 1. Let $F=f({{\textit{X}}})$ be a random variable with ${\mathbb E}[F]=0$ and $\mathrm{Var}[F]=1$ . Then

\begin{equation&#x002A;} d_{K}(F, \mathcal{N}) \leq C\sum_{k=1}^5 \sqrt{B_k}, \end{equation&#x002A;}

where $\mathcal{N}$ is a standard normal random variable, and

\begin{alignat&#x002A;}{2} B_1 & = \sum_{a,b,c\in [m]}\sqrt{{\mathbb E}\big[D_a^2 D_b^2\big]{\mathbb E}\big[D_{ca}^2 D_{cb}^2 \big] }, \qquad & & \\[3pt] B_2 & = \sum_{a,b,c\in [m]}\frac{1}{p_c q_c}{\mathbb E}\big[D_{ca}^2 D_{cb}^2 \big], & B_3 & = \sum_{a=1}^m \frac{1}{p_aq_a} {\mathbb E}\big[ D_a^4 \big], \\[3pt] B_4 & = \sum_{a,b\in [m]}\frac{1}{p_aq_a}\sqrt{{\mathbb E}\big[D_a^4\big]{\mathbb E}\big[D_{ab}^4\big]}, & B_5 & = \sum_{a,b\in [m]} \frac{1}{p_aq_a p_b q_b} {\mathbb E}\big[ D_{ab}^4 \big], \end{alignat&#x002A;}

with $D_a\,:\!=\,D_aF$ and $D_{ab}\,:\!=\,D_aD_bF$ .

To apply Lemma 1, we denote the set of all possible edges in the $\beta$ -model on n vertices by $\{e_1,e_2,\ldots,e_m\}$ , where $m=\binom{n}{2}$ . For each edge $e_a$ with $a\in[m]$ , let $1\le i_a < j_a\le n$ denote its endpoints. Note that the edge indicators $\{I_{i_a j_a}, a\in [m]\}$ are independent Bernoulli random variables with ${\mathbb P}(I_{i_a j_a}=1)=p_{i_a j_a}$ and ${\mathbb P}(I_{i_a j_a}=0)=1-p_{i_a j_a}$ . Hence, the normalized triangle count $F_n$ in (6) is a measurable function of these variables and is thus amenable to analysis via Lemma 1.

By (16), we have

(17) \begin{equation} D_a\,:\!=\, D_aF_n=\frac{\sqrt{h_a}}{\sqrt{\mathrm{Var}[T_n]}\,}V_a, \quad a\in[m],\end{equation}

where

(18) \begin{align} h_a & = p_{i_aj_a}(1-p_{i_aj_a}), \\[-10pt] \nonumber \end{align}

(19) \begin{align} V_a & = \sum_{k\neq i_a,j_a} I_{i_ak}I_{j_ak}. \\[8pt] \nonumber \end{align}

In other words, $V_a$ counts the number of wedges (i.e. paths of length two) with endpoints $i_a$ and $j_a$ . Graphically, the gradient $D_aF_n$ measures the sensitivity of triangle counts to flipping the status of $e_a$ .

For the second-order gradient, consider any two given edges $e_a=\{i_a,j_a\}$ and $e_b=\{i_b,j_b\}$ . By definition, we can immediately obtain that $D_{ab}=0$ if either $e_a$ and $e_b$ are an identical edge or disjoint (i.e. the intersection $\{i_a,j_a\} \cap \{i_b,j_b\}$ is the empty set $\varnothing$ ). If $|\{i_a,j_a\}\cap\{i_b,j_b\}|=1$ , then (16) implies that

(20) \begin{equation} D_bV_a=\sqrt{h_b}\,I_{\ell_1\ell_2},\end{equation}

where $\{\ell_1,\ell_2\}=(\{i_a,j_a\}\cup\{i_b,j_b\})\setminus(\{i_a,j_a\}\cap\{i_b,j_b\})$ , i.e., $\ell_1$ and $\ell_2$ are the endpoints of $e_a$ and $e_b$ excluding their common vertex. Consequently, it follows by (17) and (20) that, for any $a,b\in [m]$ ,

(21) \begin{equation} D_{ab}\,:\!=\,D_aD_bF_n=D_bD_aF_n=\frac{\sqrt{h_ah_b}}{\sqrt{\mathrm{Var}[T_n]}\,}\Delta_{ab},\end{equation}

with

(22) \begin{equation} \Delta_{ab} = \begin{cases} 0 & \text{if}\ \{i_a,j_a\}=\{i_b,j_b\}\ \text{or}\ \{i_a,j_a\}\cap\{i_b,j_b\}=\varnothing, \\[3pt] I_{\ell_1\ell_2} & \text{if}\ |\{i_a,j_a\}\cap\{i_b,j_b\}|=1. \end{cases} \end{equation}

In addition to Lemma 1, we further require the following auxiliary lemmas.

Proof. We begin with the inequality ${\mathbb E}[|X+Y|^s]\le 2^{s-1}{\mathbb E}[|X|^s+|Y|^s]$ , which holds for any random variables X and Y. Applying this to $V_a$ given in (19), we have

(23) \begin{equation} {\mathbb E}[V_a^{s}] \le C{\mathbb E}\Bigg[\Bigg|\sum_{k\neq i_a,j_a}(I_{i_ak}I_{j_ak}-p_{i_ak}p_{j_ak})\Bigg|^s + \Bigg(\sum_{k\neq i_a,j_a} p_{i_ak}p_{j_ak}\Bigg)^s\Bigg]. \end{equation}

To bound the first term on the right-hand side of (23), observe that $I_{i_ak}I_{j_ak}$ are independent Bernoulli variables with success rate $p_{i_ak}p_{j_ak}$ for $k\neq i_a,j_a$ . By Rosenthal’s inequality we have

\begin{align&#x002A;} {\mathbb E}\Bigg|\sum_{k\neq i_a,j_a}(I_{i_ak}I_{j_ak}-p_{i_ak}p_{j_ak})\Bigg|^s & \le C\Bigg[\sum_{k\neq i_a,j_a}{\mathbb E}|I_{i_ak}I_{j_ak}-p_{i_ak}p_{j_ak}|^{s} \\[3pt] & \qquad + \Bigg(\sum_{k\neq i_a,j_a}\mathrm{Var}[I_{i_ak}I_{j_ak}]\Bigg)^{s/2}\Bigg] \\[3pt] & \le C\Bigg[\sum_{k\neq i_a,j_a}p_{i_ak}p_{j_ak} + \Bigg(\sum_{k\neq i_a,j_a}p_{i_ak}p_{j_ak}\Bigg)^{s/2}\Bigg]. \end{align&#x002A;}

Substituting this into (23) and using the inequality $x^{s/2}\le x+x^s$ for $x>0$ and $s\ge2$ , by (12) we thus have

\begin{align&#x002A;} {\mathbb E}[V_a^{s}] & \le C\Bigg[\sum_{k\neq i_a,j_a}p_{i_ak}p_{j_ak} + \Bigg(\sum_{k\neq i_a,j_a}p_{i_ak}p_{j_ak}\Bigg)^{s}\Bigg] \\[3pt] & \le C\Bigg[\mu_{i_a}\mu_{j_a}\sum_{k\neq i_a,j_a}\mu_k^2 + \Bigg(\mu_{i_a}\mu_{j_a}\sum_{k\neq i_a,j_a}\mu_k^2\Bigg)^{s}\Bigg] \le C\big(\|\boldsymbol\mu\|_2^2\mu_{i_a} \mu_{j_a} +\|\boldsymbol\mu\|_2^{2s}\mu_{i_a}^s \mu_{j_a}^s \big), \end{align&#x002A;}

which completes the proof of Lemma 2.

The following inequality is used repeatedly in our analysis.

With the above preparation, we proceed to prove Theorem 1.

Proof of Theorem 1. Recall the second-order discrete gradients $D_{ab}$ given in (21) for all $a,b\in[m]$ , where $m=\binom{n}{2}$ represents the number of all possible edges in the $\beta$ -model on n vertices. Observe that $\Delta_{ab}^s=\Delta_{ab}=\Delta_{ba}$ for all $s>0$ , and $\Delta_{ca}$ and $\Delta_{cb}$ are independent for distinct edges $e_a,e_b,e_c$ . Combining Lemma 1 with (17) and (21), we obtain

(24) \begin{equation} d_{K}(F_n, \mathcal{N}) \leq \frac{C}{\mathrm{Var}[T_n]}\sum_{k=1}^5 \sqrt{\widetilde{B}_k}, \end{equation}

where

\begin{align&#x002A;} \widetilde{B}_1 & \,:\!=\, \sum_{a,b,c\in [m]}h_ah_bh_c\sqrt{{\mathbb E}\big[V_a^2 V_b^2\big]{\mathbb E}[\Delta_{ca}\Delta_{cb}]}, \\ \widetilde{B}_2 & \,:\!=\, \sum_{a,b,c\in [m]}h_ah_bh_c{\mathbb E}[\Delta_{ca}\Delta_{cb}], & \!\!\!\!\!\!\!\!\!\!\!\widetilde{B}_3 \,:\!=\, \sum_{a=1}^m h_a{\mathbb E}\big[V_a^4\big],\qquad \ \ \\ \widetilde{B}_4 & \,:\!=\, \sum_{a,b\in [m]}h_ah_b\sqrt{{\mathbb E}\big[V_a^4\big]{\mathbb E}[\Delta_{ab}]}, & \widetilde{B}_5 \,:\!=\, \sum_{a,b\in [m]} h_ah_b {\mathbb E}[\Delta_{ab}]. \end{align&#x002A;}

To prove Theorem 1, by (11) and (24) it suffices to show that $\sum_{k=1}^5 \sqrt{\widetilde{B}_k}\le C\|\boldsymbol\mu\|_{2}^{3/2}\sum_{\ell=1}^5A_{\ell}$ , where $A_{\ell}>0$ $(\ell=1,2,\ldots, 5)$ are defined in (7). Applying the Cauchy–Schwarz inequality, a sufficient condition for this to hold is

(25) \begin{equation} \sum_{k=1}^5\widetilde{B}_k \le C\|\boldsymbol\mu\|_{2}^3\sum_{\ell=1}^5A_{\ell}^2. \end{equation}

Thus, the remainder of the proof is devoted to verifying (25).

We now analyze the terms $\widetilde{B}_k$ individually, beginning with $\widetilde{B}_3$ . From (12) and (18), we deduce that $h_a=(1 + o(1)) \mu_{i_a} \mu_{j_a}$ , $a\in [m]$ . Applying Lemma 2 with $s = 4$ yields

\begin{align&#x002A;} \widetilde{B}_3 \le C\sum_{a=1}^m\big(\|\boldsymbol\mu\|_2^2\mu_{i_a}^2\mu_{j_a}^2 + \|\boldsymbol\mu\|_2^{8}\mu_{i_a}^5 \mu_{j_a}^5\big). \end{align&#x002A;}

Since, for any $t>0$ ,

\begin{align&#x002A;} \sum_{a=1}^m\mu_{i_a}^t\mu_{j_a}^t = \sum_{1\le i < j \le n}\mu_i^t\mu_j^t = \frac12\Bigg[\Bigg(\sum_{i=1}^n\mu_i^t\Bigg)^2 - \sum_{i=1}^n\mu_i^{2t}\Bigg] \le \frac12\|\boldsymbol\mu\|_t^{2t}, \end{align&#x002A;}

we have

(26) \begin{equation} \widetilde{B}_3\le C\big(\|\boldsymbol\mu\|_2^6+\|\boldsymbol\mu\|_2^8\|\boldsymbol\mu\|_5^{10}\big) = C\|\boldsymbol\mu\|_2^3\big(\|\boldsymbol\mu\|_2^3+A_3^2\big). \end{equation}

Next, consider $\widetilde{B}_5$ , which involves pairs of edges $e_a=(i_a,j_a)$ and $e_b=(i_a,j_b)$ sharing a common vertex $i_a$ . By (12) and (22), we have

(27) \begin{align} \widetilde{B}_5 & = \sum_{i_a=1}^n\sum_{j_a\neq i_a}\sum_{j_b\neq i_a,j_a}p_{i_aj_b}(1-p_{i_aj_a})p_{i_aj_b}(1-p_{i_aj_b})p_{j_aj_b} \notag \\[3pt] & = \sum_{i=1}^n\sum_{j\neq i}\sum_{k\neq i,j}p_{ij}(1-p_{ij})p_{ik}(1-p_{ik})p_{jk} \notag \\[3pt] & \le \sum_{i=1}^n\sum_{j\neq i}\sum_{k\neq i,j}p_{ij}p_{ik}p_{jk} \le \sum_{i,j,k\in[n]}\mu_i^2\mu_j^2\mu_k^2 = \|\boldsymbol\mu\|_2^6, \end{align}

where in the second equality we relabeled the indices $i_a,j_a, j_b$ as i, j, k for simplicity, and in the last inequality we applied the bound $p_{ij}\le \mu_{i}\mu_j$ .

Invoking Lemma 3 with $s = \frac{5}{2}$ and $t = \frac{3}{2}$ gives $\|\boldsymbol\mu\|_2^4 \le \|\boldsymbol\mu\|_{{3}/{2}} ^ {{3}/{2}} \|\boldsymbol\mu\|_{{5}/{2}}^{{5}/{2}}=A_1^2$ . Combining this with (26) and (27), and noting that $\|\boldsymbol\mu\|_2^3=o(\|\boldsymbol\mu\|_2^4)$ under our assumptions, we have

(28) \begin{equation} \widetilde{B}_3+\widetilde{B}_5\le C \|\boldsymbol\mu\|_{2}^3\big(A_1^2+A_3^2\big). \end{equation}

For $\widetilde{B}_4$ , Lemma 2 with $s=4$ provides

(29) \begin{equation} \sqrt{{\mathbb E}[V_a^{4}]} \le C \Big(\|\boldsymbol\mu\|_2\mu_{i_a}^{{1}/{2}}\mu_{j_a}^{{1}/{2}} + \|\boldsymbol\mu\|_2^{4}\mu_{i_a}^2\mu_{j_a}^2\Big). \end{equation}

Following the approach for $\widetilde{B}_5$ , by (29) we bound $\widetilde{B}_4$ as

(30) \begin{align} \widetilde{B}_4 & \le C\sum_{i=1}^n\sum_{j\neq i}\sum_{k\neq i,j}p_{ij}p_{ik} \Big(\|\boldsymbol\mu\|_2\mu_{i}^{{1}/{2}}\mu_{j}^{{1}/{2}} + \|\boldsymbol\mu\|_2^{4}\mu_{i}^2\mu_{j}^2\Big)p_{jk}^{1/2} \notag \\[3pt] & \le C\sum_{i,j,k\in[n]}\Big(\|\boldsymbol\mu\|_2\mu_{i}^{{5}/{2}}\mu_{j}^{2}\mu_{k}^{{3}/{2}} + \|\boldsymbol\mu\|_2^{4}\mu_{i}^{4}\mu_{j}^{{7}/{2}}\mu_{k}^{{3}/{2}}\Big) = C\|\boldsymbol\mu\|_{2}^{3}\big(A_1^2+A_2^2\big), \end{align}

which, together with (28), implies that

(31) \begin{equation} \widetilde{B}_3 + \widetilde{B}_4 + \widetilde{B}_5 \le C\|\boldsymbol\mu\|_{2}^3\big(A_1^2+A_2^2+A_3^2\big). \end{equation}

We now turn to $\widetilde{B}_2$ . By distinguishing cases where edges $e_a$ and $e_b$ are identical or distinct, we decompose this term as

(32) \begin{equation} \widetilde{B}_2 = \sum_{a,c\in[m]}h^2_ah_c{\mathbb E}[\Delta_{ca}] + \sum_{\{a,b,c\}\subset[m]}h_ah_bh_c{\mathbb E}[\Delta_{ca}]{\mathbb E}[\Delta_{cb}] \,=\!:\, \widetilde{B}_{21} + \widetilde{B}_{22}, \end{equation}

where the simple fact ${\mathbb E}[\Delta_{ca}^2]={\mathbb E}[\Delta_{ca}]$ is used.

For $\widetilde{B}_{21}$ , analogous to the analysis of $\widetilde{B}_5$ and $\widetilde{B}_4$ , we obtain

(33) \begin{align} \widetilde{B}_{21} & = \sum_{i=1}^n\sum_{j\neq i}\sum_{k\neq i,j}p_{ij}^2(1-p_{ij})^2p_{ik}(1-p_{ik})p_{jk} \notag \\[3pt] & \le \sum_{i,j,k\in[n]}p_{ij}^2p_{ik}p_{jk} \le \sum_{i,j,k\in[n]}\mu_{i}^3\mu_{j}^3\mu_k^2 =\|\boldsymbol\mu\|_2^2\|\boldsymbol\mu\|_3^6. \end{align}

To bound $\widetilde{B}_{22}$ , we only need to consider triples of edges $(e_a,e_b,e_c)$ where $e_c$ shares a common vertex with both $e_a$ and $e_b$ . Two configurations exist: either $e_c$ connects to $e_a$ and $e_b$ via distinct common vertices (Figure 1(a)), or all three edges share a unique common vertex (Figure 1(b)). This yields

(34) \begin{align} \widetilde{B}_{22} & = \sum_{1\le i\ne j\le n}\sum_{k\neq i,j}\sum_{\ell\neq i,j,k} \big(p_{ij}(1-p_{ij})p_{ik}(1-p_{ik})p_{j\ell}(1-p_{j\ell})p_{i\ell}p_{jk} \notag \\[3pt] & \qquad\qquad\qquad\qquad\quad + p_{ij}(1-p_{ij})p_{ik}(1-p_{ik})p_{i\ell}(1-p_{i\ell})p_{jk}p_{j\ell}\big) \notag \\[3pt] & \le 2\sum_{i,j,k,\ell\in[n]}p_{ij}p_{ik}p_{j\ell}p_{i\ell}p_{jk} \le 2\sum_{i,j,k,\ell\in[n]}\mu_{i}^3\mu_{j}^3\mu_k^2\mu_{\ell}^2 = 2\|\boldsymbol\mu\|_2^4\|\boldsymbol\mu\|_3^6. \end{align}

Figure 1.

Edge $e_c$ shares a common vertex with two other edges, $e_a$ and $e_b$ .

Substituting (33) and (34) into (32) and applying the sparsity condition (4), we obtain

(35) \begin{equation} \widetilde{B}_2 \le C\big(\|\boldsymbol\mu\|_2^2\|\boldsymbol\mu\|_3^6 + \|\boldsymbol\mu\|_2^4\|\boldsymbol\mu\|_3^6\big) \le C\|\boldsymbol\mu\|_2^4\|\boldsymbol\mu\|_3^6. \end{equation}

Invoking Lemma 3 with $s = 2$ and $t = 4$ gives $\|\boldsymbol\mu\|_3^6 \le \|\boldsymbol\mu\|_{2} ^2 \|\boldsymbol\mu\|_4^4$ . Utilizing the asymptotic relations $\|\boldsymbol\mu\|_2 = o\big(\|\boldsymbol\mu\|_2^2\big)$ , $\|\boldsymbol\mu\|_4^4 = o\big(\|\boldsymbol\mu\|_{{7}/{2}}^{{7}/{2}}\big)$ , and $\|\boldsymbol\mu\|_2^4 = o\big(\|\boldsymbol\mu\|_{{7}/{4}}^{{7}/{2}}\big)$ , we have

\begin{equation&#x002A;} \|\boldsymbol\mu\|_2\|\boldsymbol\mu\|_3^6 \le \|\boldsymbol\mu\|_{2}^3\|\boldsymbol\mu\|_4^4 = o\big(\|\boldsymbol\mu\|_{2}^4\|\boldsymbol\mu\|_{{7}/{2}}^{{7}/{2}}\big) = o\big(A_5^2\big). \end{equation&#x002A;}

Thus, from (35) we can conclude that

(36) \begin{equation} \widetilde{B}_2= o\big(\|\boldsymbol\mu\|_{2}^3A_5^2\big). \end{equation}

Consider $\widetilde{B}_1 $ . Following a similar decomposition to (32), we can rewrite it as

(37) \begin{align} \widetilde{B}_1 & = \sum_{a,c\in [m]} h^2_a h_c\sqrt{{\mathbb E}\big[V_a^4\big]{\mathbb E}[\Delta_{ca}]} + \sum_{\{a,b,c\}\subset[m]}h_ah_bh_c\sqrt{{\mathbb E}\big[V_a^2V_b^2\big]{\mathbb E}[\Delta_{ca}]{\mathbb E}[\Delta_{cb}]} \notag \\[3pt] & \,=\!:\, \widetilde{B}_{11}+\widetilde{B}_{12}. \end{align}

Comparing $\widetilde{B}_{11}$ with the expression for $\widetilde{B}_4$ , by (30) we obtain

(38) \begin{equation} \widetilde{B}_{11} = o(\widetilde{B}_4)=o\big(\|\boldsymbol\mu\|_{2}^{3}\big(A_1^2+A_2^2\big)\big). \end{equation}

For $\widetilde{B}_{12}$ , using the Cauchy–Schwarz inequality and (29) gives

\begin{align&#x002A;} \widetilde{B}_{12} & \le \sum_{\{a,b,c\}\subset[m]}h_ah_bh_c\big({\mathbb E}\big[V_a^4\big]\big)^{{1}/{4}}\big({\mathbb E}\big[V_b^4\big]\big)^{{1}/{4}} \sqrt{{\mathbb E}[\Delta_{ca}]{\mathbb E}[\Delta_{cb}]} \\[3pt] & \le C\|\boldsymbol\mu\|_2\sum_{\{a,b,c\}\subset[m]}h_ah_bh_c\Big(\mu_{i_a}^{{1}/{4}}\mu_{j_a}^{{1}/{4}} + \|\boldsymbol\mu\|_2^{3/2}\mu_{i_a}\mu_{j_a}\Big)\Big(\mu_{i_b}^{{1}/{4}}\mu_{j_b}^{{1}/{4}} + \|\boldsymbol\mu\|_2^{3/2}\mu_{i_b}\mu_{j_b}\Big) \\[3pt] & \qquad\qquad\qquad\qquad\quad \times \sqrt{{\mathbb E}[\Delta_{ca}]{\mathbb E}[\Delta_{cb}]}. \end{align&#x002A;}

Then, analogously to (34), we can proceed with

(39) \begin{align} \widetilde{B}_{12} & \le C\|\boldsymbol\mu\|_2\sum_{1\le i\neq j\le n}\sum_{k\neq i,j}\sum_{\ell\neq i,j,k} \Big[p_{ij}p_{ik}p_{j\ell}\Big(\mu_{i}^{{1}/{4}}\mu_{k}^{{1}/{4}} + \|\boldsymbol\mu\|_2^{3/2}\mu_{i}\mu_{k}\Big) \notag \\[4pt] & \qquad\qquad\qquad\qquad\qquad\qquad\qquad \times \Big(\mu_{j}^{{1}/{4}}\mu_{\ell}^{{1}/{4}} + \|\boldsymbol\mu\|_2^{3/2}\mu_{j}\mu_{\ell}\Big)p_{i\ell}^{1/2}p_{jk}^{1/2} \notag \\[4pt] & \qquad\qquad\qquad\qquad\qquad\qquad + p_{ij}p_{ik}p_{i\ell}\Big(\mu_{i}^{{1}/{4}}\mu_{k}^{{1}/{4}} + \|\boldsymbol\mu\|_2^{3/2}\mu_{i}\mu_{k}\Big) \notag \\[4pt] & \qquad\qquad\qquad\qquad\qquad\qquad\qquad \times \Big(\mu_{i}^{{1}/{4}}\mu_{\ell}^{{1}/{4}} + \|\boldsymbol\mu\|_2^{3/2}\mu_{i}\mu_{\ell}\Big)p_{jk}^{1/2}p_{j\ell}^{1/2}\Big] \notag \\[4pt] & \le C\|\boldsymbol\mu\|_2 \notag \\[4pt] & \quad \times \sum_{i,j,k,\ell\in[n]}\Big[\mu_i^{5/2}\mu_j^{5/2}\mu_k^{3/2}\mu_{\ell}^{3/2} \notag \\[4pt] & \qquad\qquad\qquad\qquad\quad \times \Big(\mu_i^{1/4}\mu_j^{1/4}\mu_k^{1/4}\mu_{\ell}^{1/4} + \|\boldsymbol\mu\|_2^{3/2}\mu_i\mu_j^{1/4}\mu_k\mu_{\ell}^{1/4} + \|\boldsymbol\mu\|_2^{3}\mu_i\mu_j\mu_k\mu_{\ell}\Big) \notag \\[4pt] & \qquad\qquad\qquad\quad\!\! + \mu_i^3\mu_j^2\mu_k^{3/2}\mu_{\ell}^{3/2}\Big(\mu_{i}^{{/1}{2}}\mu_{k}^{{/1}{4}}\mu_{\ell}^{{/1}{4}} + \|\boldsymbol\mu\|_2^{3/2}\mu_{i}^{5/4}\mu_{k}\mu_{\ell}^{1/4} + \|\boldsymbol\mu\|_2^{3}\mu_{i}^2\mu_k\mu_{\ell}\Big)\Big] \notag \end{align}

(39) \begin{align} & = C\|\boldsymbol\mu\|_{2}\Big(\|\boldsymbol\mu\|_{{7}/{4}}^{{7}/{2}}\|\boldsymbol\mu\|^{{11}/{2}}_{{11}/{4}} + \|\boldsymbol\mu\|^{{7}/{4}}_{{7}/{4}}\|\boldsymbol\mu\|_{2}^{{3}/{2}}\|\boldsymbol\mu\|_{{5}/{2}}^{{5}/{2}}\|\boldsymbol\mu\|^{{11}/{4}}_{{11}/{4}} \|\boldsymbol\mu\|_{{7}/{2}}^{{7}/{2}} + \|\boldsymbol\mu\|_{2}^{3}\|\boldsymbol\mu\|^{5}_{{5}/{2}}\|\boldsymbol\mu\|^{7}_{{7}/{2}} \notag \\[3pt] & \qquad\qquad\quad + \|\boldsymbol\mu\|_{{7}/{4}}^{{7}/{2}}\|\boldsymbol\mu\|^{2}_2\|\boldsymbol\mu\|^{{7}/{2}}_{{7}/{2}} + \|\boldsymbol\mu\|_{{7}/{4}}^{{7}/{4}}\|\boldsymbol\mu\|_{2}^{{7}/{2}}\|\boldsymbol\mu\|_{{5}/{2}}^{{5}/{2}}\|\boldsymbol\mu\|_{{17}/{4}}^{{17}/{4}} + \|\boldsymbol\mu\|_{2}^{5}\|\boldsymbol\mu\|_{{5}/{2}}^{5}\|\boldsymbol\mu\|_{5}^{5}\Big), \end{align}

where the second inequality follows from $p_{ij}\le \mu_i\mu_j$ for all $i,j\in[n]$ , together with symmetry. We denote the six terms in the final parentheses above in turn by $\widetilde{B}_{12}^{(1)},\widetilde{B}_{12}^{(2)},\ldots,\widetilde{B}_{12}^{(6)}$ . Clearly, from the relation $\big(\widetilde{B}_{12}^{(2)}\big)^2=\widetilde{B}_{12}^{(1)}\widetilde{B}_{12}^{(3)}$ , we have

(40) \begin{equation} \widetilde{B}_{12}^{(2)} \le \frac12\big(\widetilde{B}_{12}^{(1)}+\widetilde{B}_{12}^{(3)}\big). \end{equation}

Setting $s = \frac{7}{2}$ and $t = 5$ in Lemma 3 yields $\|\boldsymbol\mu\|^{{17}/{2}}_{{17}/{4}} \le \|\boldsymbol\mu\|^{5}_5\|\boldsymbol\mu\|^{{7}/{2}}_{{7}/{2}}$ , which implies that $\big(\widetilde{B}_{12}^{(5)}\big)^2\le\widetilde{B}_{12}^{(4)}\widetilde{B}_{12}^{(6)}$ . Hence, we also have

(41) \begin{equation} \widetilde{B}_{12}^{(5)} \le \frac12\big(\widetilde{B}_{12}^{(4)}+\widetilde{B}_{12}^{(6)}\big) = \frac12\|\boldsymbol\mu\|^{2}_2\big(A_5^2+A_4^2\big). \end{equation}

Furthermore, setting $(s,t)=\big(2,\frac72\big)$ and (2,5) in Lemma 3 yields $\|\boldsymbol\mu\|^{{11}/{2}}_{{11}/{4}} \le \|\boldsymbol\mu\|^{2}_2\|\boldsymbol\mu\|^{{7}/{2}}_{{7}/{2}}$ and $\|\boldsymbol\mu\|^7_{{7}/{2}}\le \|\boldsymbol\mu\|^{2}_2\|\boldsymbol\mu\|^{5}_5$ , from which it follows that

(42) \begin{equation} \widetilde{B}_{12}^{(1)} \le \widetilde{B}_{12}^{(4)}, \qquad \widetilde{B}_{12}^{(3)}\le \widetilde{B}_{12}^{(6)}. \end{equation}

Combining (40)–(42) with (39), we thus have

(43) \begin{equation} \widetilde{B}_{12} \le C\|\boldsymbol\mu\|_{2}\big(\widetilde{B}_{12}^{(4)}+\widetilde{B}_{12}^{(6)}\big) = C\|\boldsymbol\mu\|_{2}^3\big(A_4^2+A_5^2\big). \end{equation}

Hence, by (36)–(38) and (43) we arrive at $\widetilde{B}_{1} + \widetilde{B}_{2} \le C\|\boldsymbol\mu\|_{2}^3\big(A_1^2+A_2^2+A_4^2+A_5^2\big)$ . This, together with (31), proves (25), and thus completes the proof of Theorem 1.

To prove Theorem 2, we introduce the following auxiliary lemma, which provides a reverse Cauchy–Schwarz inequality for positive vectors.

Lemma 4. Let ${{\textit{x}}}=(x_1,\ldots,x_n)$ and ${{\textit{y}}}=(y_1,\ldots,y_n)$ be positive vectors (i.e. all entries are positive). For any real $s,t>0$ ,

\begin{equation&#x002A;} \|{{\textit{x}}}\|_s^{s}\,\|{{\textit{y}}}\|_t^{t} \le \frac{x_{\max}^{\,s/2}\, y_{\max}^{\,t/2}}{x_{\min}^{\,s/2}\, y_{\min}^{\,t/2}} \Bigg(\sum_{i=1}^n x_i^{s/2} y_i^{t/2}\Bigg)^2. \end{equation&#x002A;}

Proof. Since all entries are positive, we have

\begin{align&#x002A;} x_{\min}^{s/2}y_{\min}^{t/2}\|{{\textit{x}}}\|_s^{s}\|{{\textit{y}}}\|_t^{t} = x_{\min}^{s/2}y_{\min}^{t/2}\sum_{i=1}^n\sum_{j=1}^n x_i^{s}y_j^{t} & = \sum_{1\le i,j\le n} x_{\min}^{s/2}x_i^{s/2} x_i^{s/2}\cdot y_{\min}^{t/2}y_j^{t/2} y_j^{t/2} \\[3pt] & \le \sum_{1\le i,j\le n}x_j^{s/2}x_i^{s/2}x_{\max}^{s/2}\cdot y_i^{t/2}y_j^{t/2}y_{\max}^{t/2} \\[3pt] & = x_{\max}^{s/2}y_{\max}^{t/2}\Bigg(\sum_{i=1}^n x_i^{s/2}y_i^{t/2}\Bigg)^2, \end{align&#x002A;}

which proves the inequality as claimed.

Applying Lemma 4 to ${{\textit{x}}}={{\textit{y}}}=\boldsymbol{\mu}$ yields that, for any $s,t>0$ ,

(44) \begin{equation} \|\boldsymbol\mu\|_s^s \,\|\boldsymbol\mu\|_t^t \le \bigg(\frac{\mu_{\max}}{\mu_{\min}}\bigg)^{({s+t})/{2}}\|\boldsymbol\mu\|_{({s+t})/{2}}^{s+t}.\end{equation}

Proof of Theorem 2. Since convergence in Kolmogorov distance implies convergence in distribution, Theorem 1 reduces our task to proving that under the conditions of Theorem 2,

(45) \begin{equation} A_{\ell} = o\Big(\|\boldsymbol\mu\|_2^{{5}/{2}}\|\boldsymbol\mu\|_3^6 + \|\boldsymbol\mu\|_2^{{9}/{2}}\Big), \quad \ell=1,2,\ldots,5. \end{equation}

This relation holds for $A_3$ and $A_4$ only under the sparsity condition (4). Indeed, by (5) we have

\begin{align&#x002A;} A_3 = \|\boldsymbol\mu\|_2^{{5}/{2}}\|\boldsymbol\mu\|_5^5 = o\Big(\|\boldsymbol\mu\|_2^{{9}/{2}}\Big), \qquad A_4 = \|\boldsymbol\mu\|_2^{{3}/{2}}\|\boldsymbol\mu\|_{{5}/{2}}^{{5}/{2}}\|\boldsymbol\mu\|_{5}^{{5}/{2}} = o\Big(\|\boldsymbol\mu\|_2^{{9}/{2}}\Big), \end{align&#x002A;}

and thus (45) holds for $\ell=3$ and 4.

In the following, we verify (45) for $A_1$ , $A_2$ , and $A_5$ under either (8) or (9). We first assume (4) and (8).

Consider $A_1$ . Noting that $\|\boldsymbol\mu\|_{3/2}^{3/4} = O\big(\|\boldsymbol\mu\|_2^3\big)$ under (8) and $\|\boldsymbol\mu\|_{5/2}^{5/4} = o\big(\|\boldsymbol\mu\|_2\big)$ under (4), we obtain

\begin{equation&#x002A;} A_1=\|\boldsymbol\mu\|_{{3}/{2}}^{{3}/{4}}\|\boldsymbol\mu\|_{{5}/{2}}^{{5}/{4}} = o\big(\|\boldsymbol\mu\|_{2}^4\big) = o\Big(\|\boldsymbol\mu\|_{2}^{{9}/{2}}\Big), \end{equation&#x002A;}

and thus (45) holds for $\ell=1$ .

Similarly, using the facts $\|\boldsymbol\mu\|_{7/2}^{7/4}=o(\|\boldsymbol\mu\|_{3}^{3/2})$ and $\|\boldsymbol\mu\|_{4}^{2}=o(\|\boldsymbol\mu\|_{3}^{3/2})$ , for $A_2$ we have

\begin{equation&#x002A;} A_2 = \|\boldsymbol\mu\|_{{3}/{2}}^{{3}/{4}}\|\boldsymbol\mu\|_{2}^{1/2}\|\boldsymbol\mu\|_{{7}/{2}}^{{7}/{4}}\|\boldsymbol\mu\|_{4}^{2} = o\Big(\|\boldsymbol\mu\|_{2}^{{7}/{2}}\|\boldsymbol\mu\|_{3}^{3}\Big) = o\Big(\Big(\|\boldsymbol\mu\|_2^{{5}/{2}}\|\boldsymbol\mu\|_3^6\Big)^{1/2}\Big(\|\boldsymbol\mu\|_2^{{9}/{2}}\Big)^{1/2}\Big). \end{equation&#x002A;}

Then it follows from the basic inequality $\sqrt{xy}\le (x+y)/2$ for $x,y>0$ that (45) holds for $\ell=2$ .

For $A_5$ , Lemma 3 with $s=\frac32$ and $t=2$ yields $\|\boldsymbol\mu\|_{{7}/{4}}^{{7}/{2}} \le \|\boldsymbol\mu\|_{{3}/{2}}^{{3}/{2}}\|\boldsymbol\mu\|_2^2$ , which implies that

(46) \begin{equation} A_5 = \|\boldsymbol\mu\|_{{7}/{4}}^{{7}/{4}}\|\boldsymbol\mu\|_{{7}/{2}}^{{7}/{4}} \le \|\boldsymbol\mu\|_{{3}/{2}}^{{3}/{4}}\|\boldsymbol\mu\|_2\|\boldsymbol\mu\|_{{7}/{2}}^{{7}/{4}}. \end{equation}

Thus, by (8) and the fact that $\|\boldsymbol\mu\|_{7/2}^{7/4}=o(\|\boldsymbol\mu\|_{3}^{3/2})$ , we have

(47) \begin{equation} A_5 = o\Big(\|\boldsymbol\mu\|_{2}^{4}\|\boldsymbol\mu\|_3^{{3}/{2}}\Big) = o\Big(\Big(\|\boldsymbol\mu\|_2^{{5}/{2}}\|\boldsymbol\mu\|_3^6\Big)^{1/4}\Big(\|\boldsymbol\mu\|_2^{{9}/{2}}\Big)^{3/4}\Big). \end{equation}

Then it follows from the inequality $(xy^3)^{1/4}\le (x+3y)/4$ for $x,y>0$ that (45) holds for $\ell=5$ .

Finally, under (4) and (9), we invoke (44) to obtain

(48) \begin{equation} \|\boldsymbol\mu\|_s^{s/2}\|\boldsymbol\mu\|_t^{t/2} = O\Big(\|\boldsymbol\mu\|_{2}^{{{3(s+t)}/{8}}}\|\boldsymbol\mu\|_{({s+t})/{2}}^{({s+t})/{2}}\Big). \end{equation}

We also consider $A_1$ , $A_2$ , and $A_5$ in turn.

For $s = \frac{3}{2}$ and $t = \frac{5}{2}$ , (48) gives

(49) \begin{equation} A_1 = \|\boldsymbol\mu\|_{{3}/{2}}^{{3}/{4}}\|\boldsymbol\mu\|_{{5}/{2}}^{{5}/{4}} = O\Big(\|\boldsymbol\mu\|_{2}^{7/2}\Big) = o\Big(\|\boldsymbol\mu\|_{2}^{{9}/{2}}\Big), \end{equation}

and thus (45) holds for $\ell=1$ .

For $A_2$ , by the second equality in (49) and the simple fact that $\|\boldsymbol\mu\|_{4}^{2}=o\big(\|\boldsymbol\mu\|_{3}^{3/2}\big)$ , we have

\begin{align&#x002A;} A_2 = \|\boldsymbol\mu\|_{{3}/{2}}^{{3}/{4}}\|\boldsymbol\mu\|_{2}^{1/2}\|\boldsymbol\mu\|_{{7}/{2}}^{{7}/{4}}\|\boldsymbol\mu\|_{4}^{2} = o\Big(\|\boldsymbol\mu\|_{{3}/{2}}^{{3}/{4}}\|\boldsymbol\mu\|_{2}^{1/2}\|\boldsymbol\mu\|_{{5}/{2}}^{{5}/{4}}\|\boldsymbol\mu\|_{4}^{2}\Big) = o\Big(\|\boldsymbol\mu\|_{2}^{4}\|\boldsymbol\mu\|_{3}^{3/2}\Big), \end{align&#x002A;}

which matches (47), and thus (45) holds for $\ell=2$ .

Consider $A_5$ . Note that (46) holds under (4), and Lemma 3 with $s=2$ and $t=3$ gives $\|\boldsymbol\mu\|_{5/2}^{5/2}\le \|\boldsymbol\mu\|_2\|\boldsymbol\mu\|_3^{3/2}$ . Applying (48) with $s = \frac{3}{2}$ and $t = \frac{7}{2}$ to (46), we thus have

\begin{align&#x002A;} A_5 = o\Big(\|\boldsymbol\mu\|_{2}^{{23}/{8}}\|\boldsymbol\mu\|_{{5}/{2}}^{{5}/{2}}\Big) = o\Big(\|\boldsymbol\mu\|_{2}^3\|\boldsymbol\mu\|_{{5}/{2}}^{{5}/{2}}\Big) = o\Big(\|\boldsymbol\mu\|_{2}^{4}\|\boldsymbol\mu\|_3^{{3}/{2}}\Big), \end{align&#x002A;}

which also matches (47). This proves (45) for $\ell=5$ , and completes the proof of Theorem 2.