One-dimensional Discrete Anderson Model in a Decaying Random Potential: from A.C. Spectrum to Dynamical Localization

Abstract

We consider a one-dimensional Anderson model where the potential decays in average like n ^−α, α > 0. This simple model is known to display a rich phase diagram with different kinds of spectrum arising as the decay rate α varies.

We review an article of Kiselev, Last and Simon where the authors show a.c. spectrum in the super-critical case $\alpha >\frac 12$, a transition from singular continuous to pure point spectrum in the critical case $\alpha =\frac 12$, and dense pure point spectrum in the sub-critical case $\alpha <\frac 12$. We present complete proofs of the cases $\alpha \ge \frac 12$ and simplify some arguments along the way. We complement the above result by discussing the dynamical aspects of the model. We give a simple argument showing that, despite of the spectral transition, transport occurs for all energies for $\alpha =\frac 12$. Finally, we discuss a theorem of Simon on dynamical localization in the sub-critical region $\alpha <\frac 12$. This implies, in particular, that the spectrum is pure point in this regime.

Appendix A: Some Technical Lemmas

1.1 A.1 Two Results on Unimodular Matrices

We say that a matrix is unimodular if it has determinant equal to 1. The following lemma is used to compare the asymptotics of the transfer matrix with those of the sequence (R _n)_n.

Lemma A.1

Let A be an unimodular matrix and let $\hat \theta = (\cos \theta ,\, \sin \theta )$. Then, for all pair of angles $|\theta _1-\theta _2|\leq \frac {\pi }{2}$,

$$\displaystyle \begin{aligned} \begin{array}{rcl} \| A \| \leq \sin \left( \tfrac{|\theta_1-\theta_2|}{2}\right)^{-1} \max \{ \| A \hat \theta_1\|,\, \| A\hat \theta_2\| \}. \end{array} \end{aligned} $$

(A.1)

Proof

First, there exists angles θ ₀ and σ ₀ such that $A^* \hat \sigma _0 = \| A \| \hat \theta _0$. Then, for any angle θ,

$$\displaystyle \begin{aligned} \begin{array}{rcl} | \cos{}(\theta-\theta_0) | = |\langle \hat \theta, \hat \theta_0 \rangle | = \frac{1}{\| A \|} | \langle\hat \theta, A^*\hat \sigma_0 \rangle | = \frac{1}{\| A \|} | \langle A \hat \theta, \hat \sigma_0 \rangle | \leq \frac{\| A \hat \theta \|}{\| A \|}. \end{array} \end{aligned} $$

This way, for any pair of angles, we obtain

$$\displaystyle \begin{aligned} \begin{array}{rcl} \| A \| \max\{ | \cos{}(\theta_1-\theta_0) |,\, | \cos{}(\theta_2-\theta_0) | \} \leq \max\{ \| A\hat \theta_1\|,\, \| A\hat \theta_2\|\}. \end{array} \end{aligned} $$

To conclude, just note that for all |γ|≤ π∕2, the minimum of the function $x\mapsto \max \{|\cos {}(x)|,$ $|\cos {}(x+\gamma )|\}$ is attained at π∕2 − γ∕2 and is equal to $|\cos {}(\pi /2-\gamma /2)|=|\sin {}(\gamma /2)|$. □

The following lemma is used to find eigenfunctions with the proper decay.

Lemma A.2

For a unimodular matrix with ∥A∥ > 1, define 𝜗 = 𝜗(A) as the unique angle $\vartheta \in (-\frac {\pi }{2},\frac {\pi }{2}]$ such that $\| A \hat \vartheta \| = \| A \|{ }^{-1}$. We also define r(A) = ∥A(1, 0)^T∥∕∥A(0, 1)^T∥.

Let (A _n)_n be a sequence of unimodular matrices with ∥A _n∥ > 1 and write 𝜗 _n = 𝜗(A ₎ and r _n = r(A _n). Assume that

• (i) lim_n→∞∥A _n∥ = ∞,

• (ii) $\displaystyle \lim _{n\to \infty } \frac {\| A_{n+1}A_n^{-1}\|}{\| A_n \| \, \| A_{n+1}\|}=0$.

Then,

• 1. (𝜗 _n)_n has a limit 𝜗 _∞∈ (−π∕2, π∕2) if and only if (r _n)_n has a limit r _∞∈ [0, ∞). If 𝜗 _n →±π∕2, then r _n →∞ but, if r _n →∞, we can only conclude that |𝜗 _n|→ π∕2.

• 2. Suppose (𝜗 _n)_n has a limit $\vartheta _{\infty }\neq 0,\, \frac {\pi }{2}$. Then,

  $$\displaystyle \begin{aligned} \begin{array}{rcl}{} \lim_{n\to\infty} \frac{\log \| A_n \hat\vartheta_{\infty}\|}{\log \| A_n \|} = -1 \quad \mathit{\text{if and only if}} \quad \limsup_n \frac{\log |r_n - r_{\infty}|}{\log \| A_n \|} \leq -2.\qquad \quad \end{array} \end{aligned} $$

  (A.2)

Proof

Recall 𝜗 _n denotes the unique angle in (−π∕2, π∕2] such that $\| A_n \hat \vartheta _n\| = \| A_n \|{ }^{-1}$. Let $\vartheta _n^{\prime } \in (\vartheta _n-\pi /2, \vartheta _n+\pi /2]$ be the unique angle such that $\| A_n \hat \vartheta _n^{\prime }\| = \| A_n \|$ and let $\vartheta _n^\perp$ be either 𝜗 _n − π∕2 or 𝜗 _n + π∕2 and such that $\vartheta _n^{\prime }$ lies between 𝜗 _n and $\vartheta _n^{\perp }$.

Let $\displaystyle v_0 = \begin {pmatrix} 1 \\0 \end {pmatrix}$, $\displaystyle w_0 = \begin {pmatrix} 0\\1 \end {pmatrix}$, v _n = A _nv ₀ and w _n = A _nw ₀. Then,

$$\displaystyle \begin{aligned} \begin{array}{rcl} \hat\vartheta_n = \cos\vartheta_n v_0 + \sin\vartheta_nw_0, \end{array} \end{aligned} $$

(A.3)

and applying A _n on both sides,

$$\displaystyle \begin{aligned} \begin{array}{rcl} \pm\frac{1}{\|A_n\|} \hat\vartheta_n= A_n \hat\vartheta_n = \cos \vartheta_n v_n + \sin \vartheta_n w_n \simeq 0, \end{array} \end{aligned} $$

(A.4)

as ∥A _n∥→∞. Hence,

$$\displaystyle \begin{aligned} \begin{array}{rcl} r_n = r(A_n) = \frac{\|v_n\|}{\|w_n\|} \simeq |\tan \vartheta_n|. \end{array} \end{aligned} $$

(A.5)

This shows that, if 𝜗 _n → 𝜗 _∞, then $r_n\to |\tan \vartheta _\infty |\in [0,\infty ]$. On the other hand, if r _n → r _∞∈ [0, ∞], then $|\vartheta _n| \to \arctan (\rho _{\infty })$, with the convention $\arctan (\infty )=\pi /2$. If ρ _∞ = ∞, we only get that |𝜗 _n|→ π∕2. Otherwise, it is enough to prove that 𝜗 _n − 𝜗 _n−1 → 0 to show the convergence of 𝜗 _n. For this, we use the decomposition

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} \hat\vartheta = \cos{}(\vartheta-\vartheta_n) \hat\vartheta_n + \sin{}(\vartheta-\vartheta_n) \hat\vartheta_n^\perp. \end{array} \end{aligned} $$

(A.6)

Applied to 𝜗 _n+1, this yields

$$\displaystyle \begin{aligned} \begin{array}{rcl} \hat\vartheta_{n+1} = \cos{}(\vartheta_{n+1}-\vartheta_n) \hat\vartheta_n + \sin{}(\vartheta_{n+1}-\vartheta_n) \hat\vartheta_n^\perp, \end{array} \end{aligned} $$

(A.7)

and

$$\displaystyle \begin{aligned} \begin{array}{rcl} A_n\hat\vartheta_{n+1} &\displaystyle =&\displaystyle \pm \| A_n \|{}^{-1}\cos{}(\vartheta_{n+1}-\vartheta_n) \hat\vartheta_n + \sin{}(\vartheta_{n+1}-\vartheta_n) A_n\hat\vartheta_n^\perp\\ &\displaystyle \simeq&\displaystyle \sin{}(\vartheta_{n+1}-\vartheta_n) A_n\hat\vartheta_n^\perp. \end{array} \end{aligned} $$

(A.8)

Hence,

$$\displaystyle \begin{aligned} \begin{array}{rcl} \|A_n\hat\vartheta_n^\perp\| |\sin{}(\vartheta_{n+1}-\vartheta_n)| &\displaystyle \simeq&\displaystyle \|A_n\hat\vartheta_{n+1}\|\\ &\displaystyle \leq&\displaystyle \| A_n A_{n+1}^{-1}\| \| A_{n+1} \hat\vartheta_{n+1}\| = \frac{\| A_{n+1}^{-1}A_n \|}{\| A_{n+1} \|}. \end{array} \end{aligned} $$

(A.9)

In the last step, we used that the A _n’s are unimodular to switch from $\| A_n A_{n+1}^{-1}\|$ to $\| A_{n+1}^{-1} A_n \|$. By condition (ii), it is enough to show that $\|A_n\hat \vartheta _n^\perp \| \gtrsim \|A_n\|$. For this, decompose

$$\displaystyle \begin{aligned} \begin{array}{rcl} \vartheta_n^\perp = \alpha_n \hat\vartheta_n^{\prime} + \beta_n \hat\vartheta_n, \end{array} \end{aligned} $$

(A.10)

for some coefficients such that α ² + β ² = 1. Note that, by construction, |α _n| > |β _n| and hence |α _n|≥ 1∕2. Therefore,

$$\displaystyle \begin{aligned} \begin{array}{rcl} \| A_n\hat\vartheta_n^\perp \|{}^2 &\displaystyle =&\displaystyle \alpha^2 \| A_n \|{}^2 + \beta^2 \|A_n\|{}^{-2} \geq \frac 14 \|A_n\|{}^2. \end{array} \end{aligned} $$

This finishes the proof Part 1.

To prove Part 2, assume 𝜗 _n → 𝜗 _∞∈ (0, π∕2) and apply (A.6) to 𝜗 = 𝜗 _∞ to obtain

$$\displaystyle \begin{aligned} \begin{array}{rcl} A_n\vartheta_{\infty} = \pm \| A_n \|{}^{-1} \cos{}(\vartheta_{\infty}-\vartheta_n) \hat\vartheta_n + \sin{}(\vartheta_{\infty}-\vartheta_n) A_n \hat\vartheta_n^{\perp}. \end{array} \end{aligned} $$

(A.11)

Recalling that ∥A _n∥→∞, 𝜗 _n − 𝜗 _∞→ 0 and $\|A_n \hat \vartheta _n^\perp \| \asymp \|A_n\|$,

$$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{\log \|A_n \vartheta_{\infty}\|}{\log \|A_n\|} \simeq \max \left\{ -1, \frac{\log |\sin{}(\vartheta_{\infty}-\vartheta_{n})|}{\log\|A_n\|}+1\right\}. \end{array} \end{aligned} $$

(A.12)

Hence, the left condition in (A.2) is satisfied if and only if

$$\displaystyle \begin{aligned} \begin{array}{rcl} \limsup_n \frac{\log |\sin{}(\vartheta_{\infty}-\vartheta_{n})|}{\log\|A_n\|} \leq -2. \end{array} \end{aligned} $$

(A.13)

But, disregarding multiplicative constants which will disappear in the limit,

$$\displaystyle \begin{aligned} \begin{array}{rcl} \sin{}(\vartheta_{\infty}-\vartheta_n) \asymp \tan{}(\vartheta_{\infty}-\vartheta_n) \asymp \tan\vartheta_{\infty}-\tan\vartheta_n = r_{\infty} - r_n. \end{array} \end{aligned} $$

(A.14)

□

We apply this with A _n = T _ω,n to prove Proposition 5.5. Define

$$\displaystyle \begin{aligned} \begin{array}{rcl} \begin{pmatrix} x^{(1)}_{n+1} \\ x^{(1)}_n \end{pmatrix} = {\mathbf{T}}_{\omega,n} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad {\text and} \quad \begin{pmatrix} x^{(2)}_{n+1} \\ x^{(2)}_n1 \end{pmatrix} = {\mathbf{T}}_{\omega,n} \begin{pmatrix} 0 \\ 1 \end{pmatrix} \end{array} \end{aligned} $$

(A.15)

and let $R^{(i)}_n,\, \theta ^{(i)}_n,\, n\geq 1$, i = 1, 2 be the corresponding Prüfer radii and phases. We let $r_n=R^{(1)}_n/R^{(2)}_n$ and 𝜗 _n be as in Lemma A.2. Then it follows from (3.5) and some elementary trigonometry that

$$\displaystyle \begin{aligned} \begin{array}{rcl} x^{(1)}_{n+1}x^{(2)}_n-x^{(1)}_nx^{(2)}_{n+1}=R^{(1)}_n R^{(2)}_n \sin k \sin{}(\theta^{(2)}_n-\theta^{(1)}_n), \end{array} \end{aligned} $$

(A.16)

where $\bar {\theta }^{(i)}_n = nk -\theta ^{(i)}_n$. On the other hand,

$$\displaystyle \begin{aligned} \begin{array}{rcl} x^{(1)}_{n+1}x^{(2)}_n-x^{(1)}_nx^{(2)}_{n+1} &\displaystyle =&\displaystyle \det\left( {\mathbf{T}}_{\omega,n} \begin{pmatrix} 1 &\displaystyle 0 \\ 0 &\displaystyle 1 \end{pmatrix} \right) =1. \end{array} \end{aligned} $$

(A.17)

This, together with the convergence

$$\displaystyle \begin{aligned} \begin{array}{rcl} \lim_{n\to\infty} \frac{R^{(i)}_n}{\log n} = \beta,\quad i=1,2, \end{array} \end{aligned} $$

(A.18)

gives

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} \lim_{n\to\infty}\frac{\log |\sin{}(\theta^{(2)}_n-\theta^{(1)}_n)|}{\log n} =\lim_{n\to\infty}\frac{\log |\sin{}(\bar\theta^{(2)}_n-\bar\theta^{(1)}_n)|}{\log n} = -2\beta. \end{array} \end{aligned} $$

(A.19)

Remember the decomposition (5.10). We have to estimate the difference of the expansions for $\log R^{(1)}_n$ and $\log R^{(2)}_n$. By (A.19), $|\sin {}(\bar \theta ^{(2)}_n-\bar \theta ^{(1)}_n)| \lesssim n^{-\beta +\epsilon }$, for any 𝜖 > 0. Hence, there exists random sequences and such that $\bar \theta ^{(1)}_n-\bar \theta ^{(2)}_n=m_n\pi + \Delta _n$ and $|\Delta _n|\lesssim n^{-\beta +\epsilon }$. Therefore,

$$\displaystyle \begin{aligned} \sin{}(2\bar\theta^{(2)}_n) = \sin{}(2\bar\theta^{(1)}_n+2\Delta_n) \simeq \sin{}(2\bar\theta^{(1)}_n)+2\cos{}(2\bar\theta^{(1)}_n)\Delta_n. \end{aligned} $$

(A.20)

This shows that

$$\displaystyle \begin{aligned} \begin{array}{rcl} \left|V_j \left(\sin{}(2\bar{\theta}^{(1)}_j) - \sin{}(2\bar{\theta}^{(2)}_j)\right)\right| &\displaystyle \lesssim&\displaystyle j^{-\frac{1}{2}-2\beta + \epsilon},\\ \text{and} \quad \left|V_j^2 \left(\cos^2(\bar{\theta}^{(1)}_j) - \cos^2(\bar{\theta}^{(2)}_j)\right)\right| &\displaystyle \lesssim&\displaystyle j^{-1-2\beta + \epsilon}, \end{array} \end{aligned} $$

by a similar argument. Hence,

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} \log r_n &\displaystyle =&\displaystyle -\sum^n_{j=1} \frac{V_{\omega,j}}{\sin k} \left(\sin{}(2\bar{\theta}^{(1)}_j) - \sin{}(2\bar{\theta}^{(2)}_j) \right) + \sum^n_{j=1} A_j \end{array} \end{aligned} $$

(A.21)

where the first sum is a convergent martingale by Lemma 5.3 with $\gamma =\frac 12 + 2\beta -\epsilon$ and the second one is absolutely convergent as A _j = O(j ^{−1−2β+𝜖}). This shows that r _n → r _∞∈ (0, ∞) almost surely which implies that 𝜗 _n has a limit 𝜗 _∞≠ 0, π∕2 by the first part of Lemma A.2.

The equivalence (A.2) in our context corresponds to

$$\displaystyle \begin{aligned} \begin{array}{rcl} \lim_{n\to\infty}\frac{\log R_n( \vartheta_{\infty})}{\log n}=-\beta \quad \text{if and only if} \quad \limsup_n \frac{\log |r_n-r_{\infty}|}{\log n} \leq -2\beta.\qquad \end{array} \end{aligned} $$

(A.22)

Let us denote by $\log r_n = M_n + S_n$ the decomposition (A.21) and $\log r_{\infty }=M_{\infty }+S_{\infty }$ where M _∞ and S _∞ are the almost sure limits of M _n and S _n respectively. Then,

$$\displaystyle \begin{aligned} \begin{array}{rcl} |r_{\infty}-r_n| &\displaystyle =&\displaystyle e^{M_{\infty}+S_{\infty}}\left| 1-\mathrm{e}^{M_n-M_{\infty}+S_n-S_{\infty}}\right| \simeq \mathrm{e}^{M_{\infty}+S_{\infty}} \end{array} \end{aligned} $$

(A.23)

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \left| M_n-M_{\infty}+S_n-S_{\infty}\right| \lesssim \mathrm{e}^{M_{\infty}+S_{\infty}} n^{-2\beta+2\epsilon}, \end{array} \end{aligned} $$

(A.24)

by the last statement of Lemma 5.3 with $\gamma = \frac 12 + 2\beta - \epsilon$. This finishes the proof of Proposition 5.5.

1.2 A.2 Analysis of the Prüfer Phases

We begin with a simple observation: from (3.6),

$$\displaystyle \begin{aligned} \begin{array}{rcl} \left| e^{i(\theta_{n+1}-\theta_n)}-1\right| = \left| \frac{\rho_{n+1}}{\rho_n}-1\right|\leq \frac{| V_{\omega,a}|}{|\sin k|}\leq 1, \end{array} \end{aligned} $$

(A.25)

for n large enough. Hence, for n large enough, |θ _n+1 − θ _n| < π∕2 and

$$\displaystyle \begin{aligned} \begin{array}{rcl} |\theta_{n+1}-\theta_n| \leq \frac{\pi}{2} |\sin{}(\theta_{n+1}-\theta_n)| \leq \frac{\pi}{2} \left|e^{i(\theta_{n+1}-\theta_n)}-1\right| \lesssim n^{-\alpha}. \end{array} \end{aligned} $$

(A.26)

This can be written in the equivalent form

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} |\bar{\theta}_{n+1}-\bar{\theta}_n-k|\leq c_0 n^{-\alpha}, \end{array} \end{aligned} $$

(A.27)

for some c ₀ > 0, which will be more suitable for our purposes.

The next lemma provides the control of the Prüfer phases needed to complete the proof of Proposition 5.2.

Lemma A.3

Let $0<\alpha \leq \frac 12$. Let E ∈ (−2, 2) corresponding to values of k different from $\frac {\pi }{4}$, $\frac {\pi }{2}$ and $\frac {3\pi }{4}$. Then,

(A.28)

Proof

The key to the proof is [31, Lemma 8.5] which states: suppose that is not in . Then, there exists a sequence of integers q _l →∞ such that

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} \left|\sum^{q_l}_{j=1} \cos \theta_j \right| \leq 1 + \sum^{q_l}_{j=1} \left| \theta_j-\theta_0 - jy \right|, \end{array} \end{aligned} $$

(A.29)

for all .

We will treat the term with $\cos {}(4\bar {\theta }_j)$ as the other one is similar. We will take y = 4k above. Let n be large enough so that it can be written as n = n ₀ + Kq _l with $n_0\ge q_l^2$ and $4c_0 n_0^{-\alpha }\le q_l^{-2}$ (where c ₀ is the constant from (A.27)). Then,

$$\displaystyle \begin{aligned} \begin{array}{rcl} \left|\sum_{j=n_0+1}^n j^{-2\alpha}\cos{}(4\bar\theta_j)\right| &\displaystyle =&\displaystyle \left|\sum_{m=0}^K \sum_{r=1}^{q_l}(n_0+mq_l+r)^{-2\alpha}\cos{}(4\bar\theta(n_0+mq_l+r))\right| \end{array} \end{aligned} $$

(A.30)

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle \leq&\displaystyle \sum_{m=0}^K(n_0+mq_l)^{-2\alpha} \left|\sum_{r=1}^{q_l}\cos{}(4\bar\theta(n_0+mq_l+r))\right| \end{array} \end{aligned} $$

(A.31)

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle +&\displaystyle \sum_{m=0}^K \sum_{r=1}^{q_l} \left|(n_0+mq_l+r)^{-2\alpha}-(n_0+mq_l)^{-2\alpha}\right| \end{array} \end{aligned} $$

(A.32)

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle =:&\displaystyle A + B. \end{array} \end{aligned} $$

(A.33)

By (A.29),

$$\displaystyle \begin{aligned} \begin{array}{rcl} A &\displaystyle \leq&\displaystyle \sum_{m=0}^{K}(n_0+mq_l)^{-2\alpha}\left(1+4\sum_{r=1}^{q_l}|\bar\theta(n_0+mq_l+r)-\bar\theta(n_0+mq_l)-kr|\right). \end{array} \end{aligned} $$

(A.34)

Now, by (A.27),

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle 4\sum_{r=1}^{q_l}|\bar\theta(n_0+mq_l+r)-\bar\theta(n_0+mq_l)-kr| \leq c_0 \sum_{r=1}^{q_l}\sum_{s=1}^r(n_0+mq_l+r)^{-\alpha} \end{array} \end{aligned} $$

(A.35)

(A.36)

Thus,

$$\displaystyle \begin{aligned} \begin{array}{rcl} A \leq 2 \sum_{m=0}^{K}(n_0+mq_l)^{-2\alpha} \leq 2 q_l^{-2\alpha} \sum_{m=0}^{K}(n_0q_l^{-1}+m)^{-2\alpha} \leq c_1 q_l^{-2\alpha} K^{1-2\alpha} \leq c_1 q_l^{-1} n^{1-2\alpha}, \end{array} \end{aligned} $$

for some finite c ₁ > 0. To estimate B, we use that

$$\displaystyle \begin{aligned} \begin{array}{rcl} \left| (n_0+mq_l+r)^{-2\alpha}-(n_0+mq_l)^{-2\alpha}\right| \leq c_2 (n_0+mq_l)^{-2\alpha-1} r, \end{array} \end{aligned} $$

(A.37)

for some finite c ₂ > 0 which shows that

$$\displaystyle \begin{aligned} \begin{array}{rcl} B &\displaystyle \leq&\displaystyle c_2 \sum_{m=0}^K \sum_{r=1}^{q_l} (n_0+mq_l)^{-2\alpha-1} r \leq c_2 q_l^2 n_0^{-1} \sum_{m=0}^K (1+n_0^{-1}mq_l)^{-1}(n_0+mq_l)^{-2\alpha}\qquad \end{array} \end{aligned} $$

(A.38)

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle \leq&\displaystyle c_2 \sum_{m=0}^K (n_0+mq_l)^{-2\alpha},\qquad \end{array} \end{aligned} $$

(A.39)

where we used $q_l^2 n_0^{-1}\leq 1$. This last sum can be estimated as above. Summarizing,

$$\displaystyle \begin{aligned} \begin{array}{rcl} \left|\sum^{q_l}_{j=1} \cos \theta_j \right| \leq c_3 q_l^{-1} n^{1-2\alpha}, \end{array} \end{aligned} $$

(A.40)

for some finite c ₃ > 0. This finishes the proof. □