# Corrections to and Improvements on Results from “The Laplacian Spectrum of Large Graphs Sampled from Graphons”

Federica Garin

Paolo Frasca

Renato Vizuete

July 22, 2024

## Abstract

In this note we correct the proof of Proposition 4 in our paper “The Laplacian Spectrum of Large Graphs Sampled from Graphons” and we improve several results therein. To this end, we prove a new concentration lemma about degrees and Laplacian eigenvalues. This lemma allows us to improve several bounds and to dispense from assuming that the graphon is bounded away from zero in several results. This extension leads, in particular, to correct the proof of Proposition 4. Additionally, we extend Proposition 4 to graphs that are sampled from graphons by using deterministic latent variables.

## 1 Introduction

In this note, we assume all notation and definitions from [1]. Arabic numbering for lemmas, propositions, theorems, remarks, definitions and equations refers to results in [1], while letters for equations and roman numbering for lemmas, propositions etc. refer to this note.

## 2 Typos

We begin by noticing that the following typos that should be corrected in [1]:

* In Definition 2.2 (Large enough $N$), Eq.s (4a)-(4b) should be replaced by the following:

$$2 b_N < \min_{k \in \{1, \dots, K+1\}} (\alpha_k - \alpha_{k-1}),$$

<footnote>F. Garin and P. Frasca are with Univ. Grenoble Alpes, Inria, CNRS, Grenoble INP, GIPSA-Lab, F-38000 Grenoble, France. R. Vizuete is with ICTEAM institute, UCLouvain, Louvain-la-Neuve, Belgium. Emails: federica.garin@inria.fr, palo.frasca@gipsa-lab.fr, renato.vizueteharo@uclouvain.be</footnote>
$$ \frac{1}{N} \log \left( \frac{2N}{\nu} \right) + b_N (2K + 3L) < \max_x d_W(x), $$

where

$$ b_N = \frac{1}{N} + \sqrt{\frac{8 \log(N/\nu)}{N + 1}}. $$

Indeed, we had mistakenly used $b_N = \frac{1}{N}$, which is correct in the case of deterministic sampling, i.e., $X_i = i/N$, while we are considering stochastic sampling.

* Eq. (22) should be replaced by the following

$$ \sqrt{\frac{\log(2N/\nu)}{N}} < \frac{\eta_W^2}{1 + 2\eta_W}. $$

Indeed, this is the right inequality that ensures that $\eta_W - \gamma_N > 0$, so that the denominators in the upper bound of $|R_N^{\text{ave}} - R_{W,N}^{\text{ave}}|$ in Theorem 2 are all positive.

We will see in Sect. 3.2 that, when using tighter upper bounds proposed in this note, Eq. (22) can be further replaced by a simpler expression.

# 3 Dispensing from W being bounded away from zero

## 3.1 Improved Lemma 3

The key improvement is replacing Lemma 3 in [1] with the following result.

**Lemma I (Lemma 3, improved).** Given a graphon W, with probability at least $1 - \nu$ the normalized degrees of the graphs $G_N$ and $\bar{G}_N$ sampled from W satisfy:

$$ \max_{i=1,\dots,N} |\delta_{(i)} - \bar{\delta}_{(i)}| \le \sqrt{\frac{\log(2N/\nu)}{N}} =: \gamma(N), \quad (a) $$

and moreover, if $\bar{d}_{(N)} > \frac{4}{9} \log(2N/\nu)$, then with probability at least $1 - 2\nu$ the normalized eigenvalues of their Laplacian matrices $L_N$ and $\bar{L}_N$ satisfy:

$$ \max_{i=1,\dots,N} |\mu_i - \bar{\mu}_i| \le 3 \sqrt{\frac{\log(2N/\nu)}{N}} =: \varphi(N). \quad (b) $$

*Proof.* The proof follows the same main lines of the proof of Lemma 3, with some modifications. The whole proof is given below, for better readability.

For any $i$ such that $\bar{d}_i > 0$, we can use Chernoff bound, as in [2, Proof of Theorem 2], thanks to the remark that $\bar{d}_i$ is the expectation of $d_i$, conditioned on $X_1, \dots, X_N$, and that $d_i$ is the $i$th row-sum of $A_N$. The Chernoff bound gives us

$$ \Pr[|d_i - \bar{d}_i| > b \bar{d}_i] \le \frac{\nu}{N} \quad \text{if} \quad b \ge \sqrt{\frac{\log(2N/\nu)}{\bar{d}_i}} $$
and in particular taking $b = \sqrt{\frac{\log(2N/\nu)}{\bar{d}_i}}$ we get

$$ \mathrm{Pr}\left[|d_i - \bar{d}_i| > \sqrt{\log(2N/\nu)}\sqrt{\bar{d}_i}\right] \le \frac{\nu}{N}. $$

Moreover, since $\bar{d}_i \le \bar{d}_{(N)}$, this further implies that

$$ \mathrm{Pr}\left[|d_i - \bar{d}_i| > \sqrt{\log(2N/\nu)}\sqrt{\bar{d}_{(N)}}\right] \le \frac{\nu}{N}. \quad (c) $$

For any $i$ such that $\bar{\delta}_i = 0$, instead, we recall that $\bar{d}_i = \sum_j W(X_{(i)}, X_{(j)})$, so that $\bar{\delta}_i = 0$ implies $W(X_{(i)}, X_{(j)}) = 0$ for all $j = 1, \dots, N$. This also ensures that the edges at $i$ have probability zero, i.e.,

$$ \mathrm{Pr}[d_i > 0] = 0. $$

This trivially implies that (c) holds true also when $\bar{d}_i = 0$.

Hence, for each $i$ (whether or not $\bar{d}_i = 0$), with probability at most $\nu/N$ we have

$$ |d_i - \bar{d}_i| > \sqrt{\log(2N/\nu)}\sqrt{\bar{d}_{(N)}}. $$

Hence, with probability at least $1 - \nu$ we have

$$ |d_i - \bar{d}_i| \le \sqrt{\log(2N/\nu)}\sqrt{\bar{d}_{(N)}} \quad \text{for all } i = 1, \dots, N. $$

Since $D_N - \bar{D}_N$ is diagonal,

$$ \|D_N - \bar{D}_N\|_2 = \max_{i=1,\dots,N} |d_i - \bar{d}_i| \le \sqrt{\log(2N/\nu)}\sqrt{\bar{d}_{(N)}} \quad (d) $$

with probability at least $1 - \nu$.

From Weyl's Theorem, $\max_i |d_{(i)} - \bar{d}_{(i)}| \le \|D_N - \bar{D}_N\|_2$, which ends the proof of (a), recalling that $\bar{d}_{(N)} \le N$, $d_{(i)} = N\delta_{(i)}$ and $\bar{d}_{(i)} = N\bar{\delta}_{(i)}$.

For the second part of the lemma, we have:

$$ \begin{aligned} \|L_N - \bar{L}_N\|_2 &= \|D_N - A_N - \bar{D}_N + \bar{A}_N\|_2 \\ &\le \|D_N - \bar{D}_N\|_2 + \|\bar{A}_N - A_N\|_2. \end{aligned} $$

From [2, Theorem 1], thanks to the assumption $\bar{d}_{(N)} > \frac{4}{9} \ln(2N/\nu)$, we have that with probability at least $1 - \nu$:

$$ \|A_N - \bar{A}_N\|_2 \le \sqrt{4\bar{d}_{(N)} \log(2N/\nu)}. \quad (e) $$

By combining (d) and (e) we have with probability at least $1 - 2\nu$:

$$ \|L_N - \bar{L}_N\|_2 \le \sqrt{\bar{d}_{(N)} \log(2N/\nu)} + 2\sqrt{\bar{d}_{(N)} \log(2N/\nu)} = 3\sqrt{\bar{d}_{(N)} \log(2N/\nu)}. $$

By using Weyl's Theorem and considering the normalized eigenvalues we get:

$$ \max_i |\mu_i - \bar{\mu}_i| \le 3 \frac{1}{N} \sqrt{\bar{d}_{(N)} \log(2N/\nu)}. $$

Finally, since $\bar{d}_{(N)} \le N$, we get the desired result. $\square$
**Remark I** (On the assumption $\bar{d}_{(N)} > \frac{4}{9} \ln(2N/\nu)$).

* If $W$ has infimum $\eta_W > 0$, then the assumption $\bar{d}_{(N)} > \frac{4}{9} \ln(2N/\nu)$ is surely satisfied for all sufficiently large $N$, irrespective of the value of the latent variables $X_1, \dots, X_N$. Indeed, all degrees grow linearly with $N$, since for any $i$

$$ \bar{d}_i = \sum_{j=1}^{N} W(X_i, X_j) \geq \eta_W N. $$

For this reason, the assumption $\bar{d}_{(N)} > \frac{4}{9} \ln(2N/\nu)$ was not mentioned in the statement of Lemma 3, since it was implied by the assumptions $\eta_W > 0$ and large enough $N$ considered therein.

* If $W$ is piecewise Lipschitz and satisfies the corrected version of Eq.s (4a)-(4b) in the definition of 'large enough $N$' in Definition 2.2, as discussed in Sect. 1, i.e., satisfies

(i) $2b_N < \min_{k \in \{1, \dots, K+1\}} (\alpha_k - \alpha_{k-1})$,

(ii) $\frac{1}{N} \log(\frac{2N}{\nu}) + b_N(2K + 3L) < \max_x d_W(x)$

with $b_N = \frac{1}{N} + \sqrt{\frac{8 \log(N/\nu)}{N+1}}$, then [3, Lemma 5] ensures that $\bar{d}_{(N)} > \frac{4}{9} \ln(2N/\nu)$ with probability at least $1-\nu$; this holds for any $\nu \in (Ne^{-N/5}, e^{-1})$. Notice that (ii) implies that $W$ is not identically zero.

## 3.2 Improvements

Notice that Lemma I improves upon Lemma 3 in two ways: we have removed the assumption that $W$ is bounded away from zero (that is, $W$ has infimum $\eta_W > 0$), which replaced by a milder assumption, and tighter upper bounds are obtained, which do not involve $\eta_W$.

The latter is easy to see by comparing the new definitions of $\gamma(N)$ and $\varphi(N)$ in (a)-(b) with the following ones, that appeared as Eq.s (13)-(14) in Lemma 3:

$$ \gamma(N) = \sqrt{\frac{1}{\eta_W}} \sqrt{\frac{\log(2N/\nu)}{N}}, \quad \varphi(N) = \left(\frac{1}{\eta_W} + 2\right) \sqrt{\frac{\log(2N/\nu)}{N}}, $$

recalling that $0 < \eta_W \le 1$.

**Remark II** (Improvement in the constants). The new definitions of $\gamma(N)$ and $\varphi(N)$ given in (a)-(b) can replace the old definitions in all results of the paper in which $\gamma(N)$ and $\varphi(N)$ appear. When using the expressions (a)-(b) in Theorem 2, we can modify the assumption stated in there as (22) (or rather the corrected version of (22) discussed in Sect. 1), which is used to ensure that $\gamma(N) < \eta_W$ and $\varphi(N) < \eta_W$ so that the terms in the denominator are positive. Hence (22) can be replaced by the following simpler assumption:

$$ \frac{\log(2N/\nu)}{N} < \frac{\eta_W^2}{9}. $$
**Remark III (Dispensing from W bounded away from zero).** Thanks to Lemma I and Remark I, the assumption 'W with infimum $\eta_W > 0$' (and the equivalent assumption 'W with minimum $\eta_W > 0$', as it is sometimes phrased because the regularity of W ensures existence of the minimum) can be removed from the following results:

* Theorem 1 (and hence also Remark 1), and

* Lemma 4.

Indeed, in such results, this assumption was only used in order to apply Lemma 3, which can be replaced by Lemma I; the assumptions W piecewise Lipschitz and N large enough, that are already present in these results, ensure that the assumption on $d_N$ is satisfied, as discussed in Remark I.

On the other hand, the following results still require the assumption 'W with infimum $\eta_W > 0$: Proposition 3, Remark 3, Theorem 2, and Remark 4. Indeed, these results rely on the bound $\bar{\mu} \ge \eta_W > 0$ obtained from Lemma 5 and the assumption $\eta_W > 0$.

Remark 2 and Proposition 4 deserve some more detailed discussion, see Sect. 3.3

## 3.3 Remark 2 and Proposition 4

Remark 2 is about the asymptotic behavior for $N \to \infty$ of the explicit bounds given in Lemma 3. For this reason, it holds under the same assumptions as Lemma 3. When replacing Lemma 3 with Lemma I, Remark 2 clearly holds under the same assumptions as Lemma I. However, it is useful to replace the assumption $\bar{d}_{(N)} > \frac{4}{9} \log(2N/\nu)$ by some assumption that implies it (at least for large N), and that is easier to verify; in particular, it is useful to consider an assumption that depends on the graphon W only. From Remark I, we know that we can choose the assumption that W has infimum $\eta_W > 0$, or the assumption that W is piecewise Lipschitz. Another useful alternative, that will be crucial to obtain a correct proof of Proposition 4, is to assume that W is continuous; this can be done thanks to the following technical lemma.

**Lemma II.** Assume that there exists $\eta > 0$ and intervals $J_1, J_2 \subseteq [0, 1]$ with length $0 < \ell \le 1$ such that $W(x, y) \ge \eta$ for all $(x, y) \in J_1 \times J_2$. Then,

$$ \mathrm{Pr}[\bar{d}_{(N)} \ge \eta N \ell/4] \ge (1 - \exp(-N \ell^2/4)) (1 - (1 - \ell)^{\lfloor N/2 \rfloor}) $$

and for $N \to \infty$

$$ \bar{d}_{(N)} = \Theta(N) \text{ a.s.} $$

Notice that the existence of $\eta$, $J_1$ and $J_2$ as in the lemma is ensured for example if W is continuous and not identically zero.

*Proof.* We will use the notation $[N] := \{1, \dots, N\}$, $[N]_{\text{odd}} := \{n \in [N] : n \text{ odd}\}$ and $[N]_{\text{evn}} := \{n \in [N] : n \text{ even}\}$.

Consider the following events:
* $E_1 := \{\exists i \in [N] \text{ s.t. } X_i \in J_1\}$, and

* $E_{2,c} := \{\exists S \subseteq [N] \text{ with } |S| \ge cN, \text{ s.t. } X_j \in J_2 \forall j \in S\}$, where $c > 0$ will be chosen later.

Recalling that $X_1, \dots, X_N$ are obtained as a reordering of $U_1, \dots, U_N$, with $U_i$ i.i.d. uniform random variables, it is obvious that

* $E_1 := \{\exists i \in [N] \text{ s.t. } U_i \in J_1\}$, and

* $E_{2,c} := \{\exists S \subseteq [N] \text{ with } |S| \ge cN \text{ s.t. } U_j \in J_2 \forall j \in S\}$.

Now consider the following events:

* $\tilde{E}_1 := \{\exists i \in [N]_{\text{evn}} \text{ s.t. } U_i \in J_1\}$, and

* $\tilde{E}_{2,c} := \{\exists S \subseteq [N]_{\text{odd}} \text{ with } |S| \ge cN \text{ s.t. } U_j \in J_2 \forall j \in S\}$.

Notice that $E_1 \supseteq \tilde{E}_1$ and $E_{2,c} \supseteq \tilde{E}_{2,c}$. Moreover, $\tilde{E}_1$ and $\tilde{E}_{2,c}$ are independent events. Hence,

$$ \mathrm{Pr}[E_1 \cap E_{2,c}] \ge \mathrm{Pr}[\tilde{E}_1 \cap \tilde{E}_{2,c}] = \mathrm{Pr}[\tilde{E}_1] \mathrm{Pr}[\tilde{E}_{2,c}]. $$

We are interested in the events $E_1$ and $E_{2,c}$ because of the following simple remark: if $X_i \in J_1$ and $X_j \in J_2$ for all $j \in S$, then $\bar{d}_i \ge \eta|S|$, and this fact implies the inclusion

$$ E_1 \cap E_{2,c} \subseteq \{d_{(N)} \ge \eta cN\}, $$

which gives

$$ \mathrm{Pr}[d_{(N)} \ge \eta cN] \ge \mathrm{Pr}[E_1 \cap E_{2,c}] \ge \mathrm{Pr}[\tilde{E}_1] \mathrm{Pr}[\tilde{E}_{2,c}]. \quad (f) $$

We start by studying $\mathrm{Pr}[\tilde{E}_1]$. Since $|[N]_{\text{evn}}| = \lfloor N/2 \rfloor$,

$$ \mathrm{Pr}[\tilde{E}_1] = 1 - \mathrm{Pr}[X_i \notin J_1 \forall i \in [N]_{\text{evn}}] = 1 - (1 - \ell)^{\lceil N/2 \rceil}. $$

To study $\mathrm{Pr}[\tilde{E}_{2,c}]$, we define random variables $Y_1, \dots, Y_N$, where $Y_j$ is equal to one if $U_j \in J_2$ and is equal to zero otherwise. Then, we define the random variable $Z := \sum_{j \in [N]_{\text{odd}}} Y_j$, i.e., $Z$ is the number of indexes $j \in [N]_{\text{odd}}$ such that $U_j \in J_2$. Hence, $\tilde{E}_{2,c} = \{Z \ge cN\}$.

Clearly $Y_1, \dots, Y_N$ are i.i.d. Bernoulli random variables with mean $\ell$. Since $|[N]_{\text{odd}}| = \lfloor N/2 \rfloor$, $Z$ is a Binomial random variable $B(\lfloor N/2 \rfloor, \ell)$. We can now use Hoeffding inequality: for any $k < EZ$, $\mathrm{Pr}[Z \le k] \le \exp(-2(EZ - k)^2/\lfloor N/2 \rfloor)$. Choosing $k = \frac{1}{2}EZ = \frac{\ell}{2} \lceil \frac{N}{2} \rceil$, we get

$$ \mathrm{Pr}[Z \le \frac{\ell}{4}N] \le \mathrm{Pr}[Z \le \frac{\ell}{2} \lceil \frac{N}{2} \rceil] \le \exp(-\frac{\ell^2}{2} \lceil \frac{N}{2} \rceil) \le \exp(-\ell^2 N/4), $$

which ends the proof of the first claim, by considering $c = \frac{\ell}{4}$ in $\{Z \ge cN\} = \tilde{E}_{2,c}$ and in (f).

Then, for $N \to \infty$, by Borel-Cantelli Lemma we obtain $\bar{d}_{(N)} = \Omega(N)$ a.s., and then it is also $\Theta(N)$ a.s. due to the trivial bound $\bar{d}_{(N)} \le N$. $\square$
**Remark IV (Remark 2, improved).** If W has infimum $\eta_W > 0$, or if W is piecewise Lipschitz, or if W is continuous, then

$$|\mu_2 - \bar{\mu}_2| = O((\log(N)/N)^{1/2}) \text{ a.s.}$$

Indeed, under either of the first two assumptions, the assumption of Lemma I is satisfied for large N with probability at least $1 - \nu$ (by Remark I) and the above asymptotic behavior is obtained by applying Borel-Cantelli Lemma, same as in the original Remark 2.

We will now consider W continuous. Consider $\nu = N^{-\alpha}$ for some $\alpha > 1$. With this $\nu$, by Lemma II, there exists $\ell > 0$ and $\eta > 0$ such that, for all sufficiently large N,

$$ \begin{aligned} \Pr[\bar{d}_{(N)} \ge \frac{4}{9} \log(2N/\nu)] &\ge \Pr[\bar{d}_{(N)} \ge \frac{4}{9}(1+\alpha) \log(2N)] \\ &\ge \Pr[\bar{d}_{(N)} \ge \eta N \ell/4] \\ &\ge (1 - \exp(-N\ell^2/4)) (1 - (1-\ell)^{\lfloor N/2 \rfloor}). \end{aligned} $$

Moreover, by Lemma I with the same $\nu = N^{-\alpha}$,

$$\Pr\left[\max_{i=1,\dots,N} |\mu_i - \bar{\mu}_i| \le \varphi(N) \middle| \bar{d}_{(N)} \ge \frac{4}{9} \log(2N/\nu)\right] \ge 1 - 2\nu.$$

Hence,

$$ \begin{aligned} \Pr\left[\max_{i=1,\dots,N} |\mu_i - \bar{\mu}_i| \le \varphi(N)\right] &\ge \Pr\left[\left\{\max_{i=1,\dots,N} |\mu_i - \bar{\mu}_i| \le \varphi(N)\right\} \cap \left\{\bar{d}_{(N)} \ge \frac{4}{9} \log(2N/\nu)\right\}\right] \\ &\ge (1 - 2\nu) (1 - \exp(-N\ell^2/4)) (1 - (1-\ell)^{\lfloor N/2 \rfloor}). \end{aligned} $$

Recalling that $\nu = N^{-\alpha}$ with $\alpha > 1$, by Borel-Cantelli Lemma, for $N \to \infty$

$$\max_{i=1,\dots,N} |\mu_i - \bar{\mu}_i| = O(\varphi(N)) = O\left(\sqrt{\log N/N}\right) \text{ a.s.}.$$

To be precise, the above proof holds if W is not identically zero. However, in case W is identically zero, then the statement is trivially true, since in this case $\mu_2 = 0$ and $\bar{\mu}_2 = 0$.

We can now turn our attention to Proposition 4. In [1], Proposition 4 was stated without the assumption 'W with infimum $\eta_W > 0'$, but its proof relied on Lemma 3 (more precisely, on Remark 2, which was based on Lemma 3) and on results from [4]; Lemma 3 has the assumption $\eta_W > 0$, and the paper [4] states that W is bounded away from zero among the general assumptions to be considered throughout the paper. However, we can now confirm that Proposition 4 is correct, i.e., it holds true without the assumption $\eta_W > 0$. Concerning Remark 2, we can now use the new version, see Remark IV. Concerning [4], a careful look at all the proofs confirms that the assumption that W is bounded away from zero is only used for the case of normalized Laplacian, while it is not needed for the un-normalized Laplacian, which is of interest here.
# 4 Sampling with deterministic latent variables

Finally, we also have another small extension, concerning deterministic sampling. Section 3.4 in the paper already discusses how the results easily extend to the case of deterministic sampling, with the exception of Proposition 4. The reason is that Proposition 4 is built upon results in [4] that are obtained there for the stochastic sampling only. However, it is easy to extend such results to deterministic sampling. Indeed, in the results that we use from [4] and their proofs, the only claim that concerns the choice of stochastic $X_i$'s is [4, Proposition 11] (Glivenko-Cantelli classes); in the case of deterministic latent variables $X_i = i/N$, [4, Proposition 11] can be replaced by the following result, which is analogous in the statement but much simpler to prove.

**Proposition I.** W a continuous graphon, $f: [0, 1] \to \mathbb{R}$ continuous. Then

$$ \lim_{N \to \infty} \sup_{x \in [0,1]} \left| \frac{1}{N} \sum_{j=1}^{N} W(x, X_j) - \int_{0}^{1} W(x, y) dy \right| = 0 $$

and

$$ \lim_{N \to \infty} \sup_{x \in [0,1]} \left| \frac{1}{N} \sum_{j=1}^{N} W(x, X_j) f(X_j) - \int_{0}^{1} W(x, y) f(y) dy \right| = 0. $$

*Proof.* W being continuous on a compact, it is uniformly continuous: for any $\varepsilon > 0$, there exists $\delta > 0$ such that for all $(x_1, y_1)$ and $(x_2, y_2)$ with $\|(x_1, y_1) - (x_2, y_2)\| < \delta$ we have $|W(x_1, y_1) - W(x_2, y_2)| < \varepsilon$.

We consider the following simple upper bound:

$$ \left| \frac{1}{N} \sum_{j=1}^{N} W(x, X_j) - \int_{0}^{1} W(x, y) dy \right| \le \frac{1}{N} \sum_{j=1}^{N} (u_j(x) - l_j(x)), $$

where $u_j(x) := \max_{y \in I_j} W(x, y)$ and $l_j(x) := \min_{y \in I_j} W(x, y)$, with $I_j = [\frac{j-1}{N}, \frac{j}{N}]$. For all $N > 1/\delta$, we have that $\|(x, y_1) - (x, y_2)\| < \delta$ for any $x$ and any $(y_1, y_2) \in I_j$ (any $I_j$), which ensures $(u_j(x) - l_j(x)) < \varepsilon$ for any $x$ and any $j$.

We have obtained that for any $\varepsilon > 0$, there exists $\delta > 0$ such that for all $N > \frac{1}{\delta}$,

$$ \sup_{x \in [0,1]} \left| \frac{1}{N} \sum_{j=1}^{N} W(x, X_j) - \int_{0}^{1} W(x, y) dy \right| \le \varepsilon, $$

which concludes the proof of the first claim.

The proof of the second claim is the same, just considering the function $W(x, y)f(y)$ instead of $W(x, y)$. $\square$

To ensure that Proposition 4 holds true in the case of deterministic latent variables, we use the above-mentioned modification of the results from [4], together with the following simple adaptation of Lemma II, from which it follows that Remark IV remains true also when $X_i = i/N$.
**Lemma III.** Consider $\bar{G}_N$ the weighted graph obtained with deterministic sampling, i.e., with adjacency matrix $[\bar{A}_N]_{ij} = W(X_i, X_j)$, $X_i = i/N$. Assume that there exists $\eta > 0$ and closed intervals $J_1, J_2 \subseteq [0, 1]$ with length $0 < \ell \le 1$ such that $W(x, y) \ge \eta$ for all $(x, y) \in J_1 \times J_2$. Then, for all $N \ge 1/\ell$,

$$ \bar{d}_{(N)} \ge \eta \lfloor \ell N \rfloor $$

and for $N \to \infty$

$$ \bar{d}_{(N)} = \Theta(N). $$

*Proof.* Simply notice that in a closed interval of length $\ell$ there are at least $\lfloor \ell N \rfloor$ points of the form $i/N$. Hence, if $\lfloor \ell N \rfloor \ge 1$, there is at least one point $X_i \in J_1$ and there are $\lfloor \ell N \rfloor$ points in $J_2$, so that $\bar{d}_i \ge \eta \lfloor \ell N \rfloor$. $\square$

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