One-dimensional geometric random graphs with nonvanishing densities II: a very strong zero-one law for connectivity

Abstract

We consider a collection of n independent points which are distributed on the unit interval [0,1] according to some probability distribution function F. Two nodes communicate with each other if their distance is less than some given threshold value. When F admits a density f which is strictly positive on [0,1], we give conditions on f under which the property of graph connectivity for the induced geometric random graph obeys a very strong zero–one law when the transmission range is scaled appropriately with n large. The very strong critical threshold is identified. This is done by applying a version of the method of first and second moments.

Appendices

Appendix A: A proof of Proposition 4

We begin with easy facts to be used repeatedly in the proofs. With 0≤x<1, it is a simple matter to check that

$$ \log (1 - x) = - \int_{0}^x \frac{1}{1-t} \,dt = - x - \varPsi (x) $$

(113)

where

$$\varPsi (x) := \int_0^x \frac{t}{1-t}\, dt. $$

The mapping x→Ψ(x) is increasing and convex on the interval [0,1) with

$$ 0 < \varPsi (x) \leq \frac{x^2}{2(1-x)}, \quad 0 < x < 1 . $$

(114)

The standard bound

$$ 1 - x \leq e^{-x}, \quad x \in [0,1], $$

(115)

is now a simple consequence of decomposition (113) and of the nonnegativity of Ψ.

Proof of (56)

Fix n=1,2,… and ρ in (0,1). The bound (115) readily yields

(116)

since

$$ 0 \leq b(x;\rho) \leq 1, \quad x \in [0, 1 - \rho ]. $$

(117)

The assumption lim _n→∞ ρ _n =0 on the scaling ρ:ℕ₀→ℝ₊ implies the existence of a positive integer n _⋆ such that ρ _n <1 for all n≥n _⋆. Using this fact into (116) (with ρ replaced by ρ _n for n≥n ^⋆) gives the desired conclusion (56). □

Proof of (58)

Fix n=1,2,… and ρ in (0,1). Using (117) in expression (53), we get

where the last step used decomposition (113).

Now pick ρ>0 small enough so that both constraints ρ<1 and ρf ^⋆<1 hold. With such a choice, the inequalities

$$0 \leq b(x;\rho) \leq f^\star \rho, \quad x \in [0, 1 - \rho ], $$

hold, and the uniform bound

$$ \sup_{x \in [0,1-\rho ] } \varPsi \bigl(b(x;\rho)\bigr) \leq \frac{ ( f^\star \rho )^2 }{ 2 ( 1 - f^\star \rho ) } $$

(118)

immediately follows from (114).

To conclude, pick a scaling ρ:ℕ₀→ℝ₊ which satisfies (57) (hence also (55)). The latter convergence implies both ρ _n <1 and f ^⋆ ρ _n <1 for large enough n, while the former yields

$$\lim_{n \rightarrow \infty} \sup_{x \in [0,1-\rho_n ] } \bigl( n \varPsi \bigl(b(x; \rho_n)\bigr) \bigr) = 0 $$

upon invoking (118) (where ρ _n is substituted to ρ). By the continuity of the exponential mapping, for each ε in (0,1), there exists a positive integer n ^⋆(ε) such that

$$\inf_{x \in [0,1-\rho_n ] } e^{- n \varPsi (b(x;\rho_n)) } \geq 1 - \varepsilon, \quad n \geq n^\star (\varepsilon), $$

and the bound (58) follows. □

Appendix B: A proof of Lemma 4

Pick ε in the interval (0,a). Under (11), there exists δ(ε)>0 such that

$$-\varepsilon |x-x_\star|^r \leq h(x) \leq \varepsilon |x-x_\star|^r $$

whenever |x−x _⋆|≤δ(ε) in [0,1]. In this range, representation (10) yields the inequalities

$$ f_\star + (a-\varepsilon) |x-x_\star|^r \leq f(x) $$

(119)

and

$$ f(x) \leq f_\star + (a+\varepsilon) |x-x_\star|^r . $$

(120)

The upper bound Whenever |x−x _⋆|≥δ(ε) in [0,1], the finiteness of f ^⋆ implies

(121)

Combining (120) and (121), we get the upper bound f≤f ₊ on the entire interval [0,1] with f ₊ given by (60), where

$$a_+ := \max \biggl\{ \frac{f^\star - f_\star }{\delta (\varepsilon)^r} , a+\varepsilon \biggr\}. $$

The lower bound Assume at first that x _⋆ is an element of the open interval (0,1). For each δ>0, we set

$$f_-(\delta) := \sup \bigl( f(x):\ 0 \leq x < x_\star, \ |x - x_\star| \leq \delta \bigr) $$

and

$$f_+(\delta) := \sup \bigl( f(x):\ x_\star < x \leq 1, \ |x - x_\star| \leq \delta \bigr) . $$

Under Assumption 3, the point x _⋆ is a unique isolated minimum for the density function f which is continuous in a neighborhood of x _⋆. The strict inequalities b _±(δ):=f _±(δ)−f _⋆>0 follow.

Once δ(ε) has been determined, we note that δ can be selected small enough in the interval (0,δ(ε)) such that both inequalities

$$ f_-(\delta) \leq f(x), \quad 0 \leq x < x_\star - \delta $$

(122)

and

$$ f_+(\delta) \leq f(x), \quad x_\star + \delta < x \leq 1 $$

(123)

hold with δ<min(x _⋆,1−x _⋆).

From (122), whenever 0≤x<x _⋆−δ, we get

(124)

while from (123), whenever x _⋆+δ<x≤1, we have

(125)

since |x−x _⋆|≤1 on these ranges. Taken together, the two inequalities (124) and (125) yield

$$f(x) \geq f_\star + \min \bigl( b_+ (\delta), b_-(\delta) \bigr) \cdot | x - x_\star|^r $$

when |x−x _⋆|≥δ.

Finally, we get the desired lower bound f ₋≤f on the entire interval [0,1] with f ₋ given by (60) where a ₋:=min(b ₊(δ),b ₋(δ),a−ε). The boundary cases x _⋆=0,1 are handled mutatis mutandis; details are left out in the interest of brevity.

Appendix C: A technical lemma

Throughout this appendix let p≥3 denote a constant. The proof of Proposition 3 relies on our ability to determine the validity of the inequality

$$ \bigl( 1 - (u+v) \bigr)^p \leq \bigl( (1-u) (1-v) \bigr)^{p+1} $$

(126)

on the range 0≤u,v≤1 under the constraint u+v≤1. The next technical lemma provides a simple characterization of a large region where this inequality holds.

Lemma 13

Fix p≥3. For 0≤u,v≤1 with u+v≤1, inequality (126) holds, provided that

$$ \frac{2}{p} \leq \min (u,v) . $$

(127)

Proof

Fix u,v in the interval [0,1] such that u+v≤1. This pair satisfies (126) if and only if

$$ \frac{1}{ (1-u)(1-v) } \leq \biggl( 1 + \frac{uv}{1 -(u+v)} \biggr)^p . $$

(128)

Since 1−(u+v)≤(1−u)(1−v), we see that (128) holds if we can show that

$$ \frac{1}{1-(u+v)} \leq 1 + p \cdot \frac{uv}{1-(u+v)} $$

(129)

as we make use of the standard inequality (1+t)^p≥1+pt valid for all t≥0. Inequality (129) is equivalent to u+v≤puv. In short, the pair u,v satisfies (126) if

$$\frac{1}{u} + \frac{1}{v} \leq p. $$

This last inequality is clearly satisfied if we select u and v according to (127). □