Quantitative boundedness

Now we prove Proposition 5.1. By rescaling the variance profileV we may take σmax= 1. By Claim 5.1 we may assumes ∈ (0, 2]. As in the proof of Proposition 5.2 we may assumet ≤ 1. We may also assumenis sufficiently large depending ons, σ₀, δandκ. In the remainder of the section we make use of asymptotic notationO( ), ., &^{, allowing} implied constants to depend on the parameterss, σ0, δandκ, but not onnandt.

As in the proof of Proposition 5.2 we assume (5.10) holds for someK > 0and aim to derive a contradiction forKsufficiently large depending ons, σ0, δandκ. The argument follows the same general outline as the proof in the previous subsection. We will reuse Lemmas 5.3 and 5.4 as stated, but we will need versions of Lemmas 5.2 and 5.5 with constants independent ofn.

5.3.1 Lower boundingre_i

The following is an analogue of Lemma 5.2.

Lemma 5.6.There are positive constantsC0(s, σ0, δ, κ), β0(s, σ0, δ, κ)such that for all β ≤ β0, ifTβis non-empty thenTC₀β = [n].

Proof. Letβ > 0 to be taken sufficiently small and assume Tβ is non-empty. Fix an element i0 ∈ Tβ. We will grow the set Tβ in stages by enlarging β by appropriate constant factors. We do this by iterative application of the following:

Claim 5.7.Let β, ε0 ∈ (0, 1/2], and assume 0 < |Tβ| ≤ (1 − ε0)n. There exists C = C(σ0, δ, ε0) > 0such that ifnis sufficiently large depending onκandε0then|TCβ\ Tβ| ≥ (δ ε₀/2)n.

Proof. By the assumption thatA(σ0)is(δ, κ)-robustly irreducible we have

|N₋^(δ)(T_β^c) ∩ Tβ| ≥ min(κ|T_β^c|, |Tβ|) ≥ min(κ ε0n, 1) ≥ 1

ifnis sufficiently large. Fix an elementi ∈ N₋^(δ)(T_β^c) ∩ Tβ. By definition we have

|N+(i) ∩ T_β^c| ≥ δ|T_β^c| ≥ δ ε0n.

Next, we claim that for anyC > 0we have

|N₊(i) ∩ T_Cβ^c | ≤ 2s²

Cσ₀²n. (5.32)

Indeed, sincei ∈ T_β, by (5.8) and (5.4) we have β

kϕke _∞ >rei≥ 1

2min ϕi

s², 1 ϕei

 .

Sinceβ ≤ 1/2it follows that the minimum is attained by the first argument. Thus β

kϕke _∞ > ϕi

2s² > 1 2s²

1 n

n

X

j=1

σ_ij²erj ≥ σ₀² 2s²

1 n

X

j∈N+(i)

erj.

From Markov’s inequality it follows that for anyC > 0, er_j < Cβ/kϕke _∞ for all but at most(2s²/Cσ₀²)nvalues ofj ∈ N₊(i), which gives (5.32). Combining these estimates and takingC = 8s²/(σ²₀δ ε0)we have

|TCβ\ Tβ| ≥ |N+(i) ∩ T_β^c| − |N+(i) ∩ T_Cβ^c | ≥



δ ε0−2s² Cσ₀²



n ≥ (δ ε0/2)n

as desired.

Applying the above claim iteratively withε0= s²/8we obtainC⁰(σ0, δ, s) < ∞such that ifβis sufficiently small depending onσ0, δ, sandnis sufficiently large depending onκ, s, then

|TC⁰β| ≥ (1 − s²/8)n. (5.33) Now letC0> 0to be chosen later, and towards a contradiction supposeTC₀β 6= [n]. Then there exists i ∈ [n] such that rei ≥ C0β/kϕke _∞. From (5.4) we have the upper bounderi ≤ kϕk∞/s², so we concludekϕk∞≥ C0s²β/kϕke ∞. Now from our assumption K ≤ _n¹Pn

j=1rj = Pn

j=1erj, if K is sufficiently large depending on s then the same argument as in the proof of Lemma 5.3 shows thater_j ≥ kϕk∞/4 for at least(s²/4)n values of j ∈ [n]. Thus, er_j ≥ C₀s²β/(4kϕke _∞) for at least (s²/4)n values of j ∈ [n], i.e.|T_C0s2β/4| < (1 −^s₄²)n. Taking C₀ = 4C⁰/s² we contradict (5.33), and we conclude TC₀β= [n].

Now by the same lines as in (5.13)–(5.17) we conclude

eri≥ β0/kϕke _∞ ∀i ∈ [n] (5.34) for someβ0(s, σ0, δ, κ) > 0. Note we are now free to use the estimates in Lemma 5.4 with this value ofβ0.

5.3.2 Upper boundingr_i

Here our task is essentially to modify the proof of Lemma 5.5 to show we can take α⁰ sufficiently small andindependent ofnsuch that|Sα⁰\ Sα| & n, rather than merely nonempty. We can then conclude the proof by iterating this fact a bounded number of times.

Let us summarize the key new ideas. In the proof of Lemma 5.5 we used the irreducibility assumption to find an element i^∗ ∈ Sα such that the average of the componentsrk overS_α^c was of order&α,nri^∗≥ αkϕke _∞ (see (5.30)). In a similar spirit, Lemma 5.8 below controls the average ofrkoverk ∈ S_α^c from below by the average ofri

overi ∈ U0, for a setU0⊂ Sαthat isdensely connected toS_α^c. By averaging over a large setU₀ we are able to use the full strength of the bounds in Lemma 5.4 and avoid any dependence of the constants onn.

Proceeding naïvely, one can then use Lemma 5.8 to deduce|S_c0α2\ S_α| ≥ c₀α|S_α^c|for a sufficiently small constantc0= c0(s, σ0, δ, κ) > 0. However, when iterating this bound over a sequence of valuesαk+1= c0α²_k, the setsSα_kgrow by an exponentially decreasing proportion ofn, so this is not enough to find a value ofαfor which|Sα|is close ton.

Instead, in Lemma 5.9 we are able to growSαby a constant factor using a nested iteration argument, which we now describe. We would like to find some value of α⁰ ∈ (0, α)for which

|Sα⁰ \ Sα| ≥ c|Sα⁰| (5.35) where c > 0 is small constant. Suppose that (5.35) fails. By the expansion assump-tion, we know thatSα⁰ contains a fairly large set U = N₋^(δ)(S_α^c0) ∩ Sα⁰ (of size at least

min(|S_α⁰|, κ|S_α^c0|)) that is densely connected toS_α^c0. In particular, ifcis sufficiently small depending onκ, thenU must have large overlap withS_α. Denoting the overlap byU₀, Lemma 5.8 can now be applied to deduce

|Sc₀αα⁰\ Sα⁰| ≥ c0α|S^c_α0|

for some c₀ = c₀(s, σ₀, δ, κ) > 0 sufficiently small. The key is that the constant of proportionality on the right hand side is independent ofα⁰. Hence, for fixedα, as long as (5.35) fails we can iteratively lowerα⁰ to increase|Sα⁰\ Sα|by an amount& α|Sα^c0|, until eventually (5.35) holds. This whole procedure can then be iterated a bounded number of times to obtainα⁰⁰such|Sα⁰⁰|is close ton.

Having motivated the key ideas, we turn now to the proofs.

Lemma 5.8.Letα ∈ (0, 1) and suppose that0 < |Sα| ≤ (1 − δ/2)n. IfKis sufficiently

Proof. First we prove the comparison X

where in the second bound we applied the robust irreducibility assumption and our assumed bounds onU₀andS_α, and in the bound we used that bothS_αand its complement are of linear size inn. From our assumption (5.10), (5.4) and the above it follows that

K ≤ 1

From (5.20) and Lemma 5.4 we have

X

where in the bound we have applied (5.21) and (5.22). Applying (5.37) and rearranging yields

Now sinceU0 ⊂ N₋^(δ)(S^c_α), for anyi ∈ U0 we have|N+(i) ∩ S^c_α| ≥ δ|S_α^c|. Together with (5.21) and (5.23) this implies

δ|S_α^c|X

i∈U0

W_ij^Sα ≤ X

i∈Sα

W_ij^Sα|N+(i) ∩ S_α^c| ≤ nkϕke _∞ β0σ₀²

 rej. Rearranging we obtain a bound onP

i∈U0W_ij^Sα, which we substitute in (5.39) to obtain X

i∈U0

ri ≤ 2kϕke ∞

β0σ₀²δ|S_α^c| X

j∈Sα

X

k∈S_α^c

σ²_kjerjrk ≤ 2|Sα| s²αβ0σ²₀δ|S_α^c|

X

k∈S_α^c

rk,

where in the second inequality we applied (5.5) to bounderj≤ 1/s²αkϕke ∞for allj ∈ Sα. The result now follows by rearranging.

Lemma 5.9.For anyα ∈ (0, 1)there existsα⁰ = α⁰(α, s, σ0, δ, κ) > 0such that either

|S_α⁰| ≥ (1 − δ/2)n (5.40)

or

|Sα⁰\ Sα| ≥1

2min(|Sα⁰|, κ|S_α^c0|) (5.41) (or both).

Proof. Forα⁰∈ (0, α)denote byP (α⁰)the statement that at least one of (5.40) and (5.41) holds. We will show that whileP (α⁰)fails, we can lowerα⁰ by a controlled amount to increase the size ofSα⁰\ Sαby a little bit. We can then iterate this untilP (α⁰)holds.

Letα⁰∈ (0, α)be arbitrary and assumeP (α⁰)fails. We claim there existsc0(s, σ0, δ, κ) >

0such that

1

|S_α^c0| X

k∈S^c

α0

rk≥ c0αα⁰kϕke ∞. (5.42)

PutU₀= N₋^(δ)(S_α^c0) ∩ S_α.By the robust irreducibility assumption and the fact that (5.41) fails,

|U0| ≥ |N₋^(δ)(S_α^c0) ∩ Sα⁰| − |Sα⁰\ Sα| ≥ 1

2|N₋^(δ)(S_α^c0) ∩ Sα⁰| (5.43)

≥ 1

2min(|Sα⁰|, κ|S_α^c0|). (5.44)

By (5.43) and Lemma 5.8, 1

|S_α^c0| X

k∈S^c

α0

r_k & α⁰

|Sα⁰| X

i∈U₀

r_i ≥ αα⁰kϕke _∞|U₀|

|Sα⁰| & αα⁰kϕke _∞,

where in the last inequality we applied (5.44) and the fact that (5.40) fails. This gives (5.42) as desired.

Now denoting

U⁰=n

k ∈ S^c_α0 : r_k ≥1

2c₀αα⁰kϕke _∞o we have

X

k∈S^c

α0

rk ≤ X

k∈U⁰

rk+ X

k∈S^c

α0\U⁰

rk≤ α⁰kϕke _∞|U⁰| +1

2c0αα⁰kϕke _∞|S_α^c0|,

where we used that by definition,r_k ≤ α⁰kϕk∞for allk ∈ S_α^c0. Combining with (5.42) and rearranging gives

|Sc₀αα⁰/2\ Sα⁰| ≥ |U⁰| ≥ 1

2c0α|S_α^c0|. (5.45) Since (5.45) holds as long asP (α⁰)fails, we can repeatedly lowerα⁰ by a factorc0α/2 to obtainα⁰= α⁰(α, s, σ0, δ, ε)such thatP (α⁰)holds. More explicitly, for eachk ≥ 0put αk = (c0α/2)^kαand abbreviateSk := Sα_k. Then for allk ≥ 1such thatP (αk)fails we have|Sk+1\ Sk| ≥ ¹₂c0α|S_k^c|,so

|Sk+1\ U0| = |Sk+1\ Sk| + · · · + |S1\ U0| ≥1

2c0α(|S^c_k| + · · · + |S₀^c|) ≥ (k + 1)1

2c0α|S_k+1^c |.

Thus, we must have that P (α_k) holds for some k ≤ 2κ/c₀α. (This gives α⁰ of size O(1/α)^−O(1/α).)

Now we conclude the proof of Proposition 5.1. From Lemma 5.3 we have|S1/4| ≥ (s²/4)n. Applying Lemma 5.9O(1)times we obtainα⁰⁰& 1^{such that}

|Sα⁰⁰| ≥ (1 − δ/2)n. (5.46)

Now from (5.5) we have

rej ≤ 1

s²α⁰⁰kϕke _∞ ^(5.47)

for allj ∈ Sα⁰⁰. On the other hand, for anyj ∈ S_α^c00, 1/erj ≥ϕej ≥ (V^Tr)j≥ 1

n X

i∈S_α00

σ²_ijri≥ 1

nσ²₀α⁰⁰kϕke _∞|N₋(j) ∩ Sα⁰⁰|.

From (5.46) and the robust irreducibility assumption (specifically the condition (2.13)),

|N−(j) ∩ Sα⁰⁰| ≥ δn − |S_α^c00| ≥ δn/2.

Combining the previous two displays we obtain

erj ≤ 2 δσ₀²α⁰⁰kϕke _∞ for allj ∈ S_α^c00. Together with (5.47) we have

erj. 1 α⁰⁰kϕke _∞ for allj ∈ [n]. Applying (5.3),

α⁰⁰kϕke ∞n/2 ≤ α⁰⁰kϕke ∞|Sα⁰⁰| ≤

n

X

j=1

rj=

n

X

j=1

erj. n α⁰⁰kϕke _∞ and rearranging giveskϕke _∞. 1/α⁰⁰. 1. Finally, since

K ≤ 1 n

n

X

j=1

rj ≤ kϕke ∞/s²

by (5.4), we obtain a contradiction ifKis sufficiently large depending ons, σ₀, δandκ. It follows that (5.10) fails for sufficiently largeK, which concludes the proof of Proposition 5.1.

Remark 5.1. We note that in the above proof we only applied the expansion bound (2.14) to sets of size at leastδn/10.

6 Proof of Theorem 2.3-(1): tail estimates and asymptotics of the

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