Monketoo's picture
Add files using upload-large-folder tool
89d0916 verified

Convergence of Eulerian triangulations

If it does not leave this layer, then its length is bounded below by

i=1Nr(2m,lm)Diam(G2m(i,lm)). \sum_{i=1}^{N_r(2m,lm)} \mathrm{Diam}(\mathcal{G}_{2m}(i,lm)).

Then, from Lemma 7.3, and from the definition of f, we get that, for r/m large enough, for some c' > 0 independent of r, l, m,

P({Nr(2m,lm)<c(rm)2}{i{1,,Nr(2m,lm)}Diam(G2m(i,lm))cf(2m)})14, P\left(\{N_r(2m, lm) < c \left(\frac{r}{m}\right)^2\} \cup \{\exists i \in \{1, \dots, N_r(2m, lm)\} \mid \text{Diam}(\mathcal{G}_{2m}(i, lm)) \le c' f(2m)\} \right) \le \frac{1}{4},

so that

P(Diam(Gr)m/3cc(r/m)2f(2m))14, \mathbb{P}(\mathrm{Diam}(\mathcal{G}_r) \leq m/3 \wedge c \cdot c'(r/m)^2 f(2m)) \leq \frac{1}{4},

which implies the desired bound, by the definition of f.

The details of the proof can be adapted from the proof of [14, Proposition 1]. $\square$

Theorem 7.2 is then a purely analytic consequence of Proposition 7.4, and its proof is a straightforward adaptation of that of [14, Theorem 5].

We can now use Theorem 7.2 to obtain the following lower bounds for the distances along the boundary of $\mathcal{L}$:

Proposition 7.5. For every $\varepsilon > 0$, there exists an integer $K > 0$ such that, for every $r \ge 1$,

P(minjKr2dL((0,0),(j,0))r)1ε. \mathbb{P} \left( \min_{|j| \ge K r^2} \vec{d}_{\mathcal{L}}((0,0), (j,0)) \ge r \right) \ge 1 - \varepsilon.

Consequently, for $K' = 9K$, we also have, for every $r \ge 1$,

P(minj2Kr2minKr2iKr2dL((i,0),(j,0))r)12ε. \mathbb{P} \left( \min_{|j| \ge 2K'r^2} \min_{-K'r^2 \le i \le K'r^2} \vec{d}_{\mathcal{L}}((i,0), (j,0)) \ge r \right) \ge 1 - 2\varepsilon.

Proof. Let us start with the first assertion. Let $\epsilon > 0$. Fix $r \ge 1$, and $K \ge 1$. Then, from (5.18), the number $N_{(K,r)}$ of trees that reach height $r$ between $(0,0)$ and $(j,0)$ is bounded below by a binomial variable of parameters $(Kr^2, 3/((r+2)^2-1))$, so that, using Chebyshev's inequality, for any $a > 0$,

P(N(K,r)38Ka)3Ka2. \mathbb{P}\left(N_{(K,r)} \le \frac{3}{8}K - a\right) \le \frac{3K}{a^2}.

(Note that the binomial variable in question has expectation greater than or equal to $3K/8$, with equality when $r=1$, and a variance smaller than $3K$.)

Taking $a = \sqrt{(6K/\epsilon)}$, for $K$ large enough that $a \le (1/8)K + 1$, we get

P(N(K,r)14K+1)ϵ2.(7.1) \mathbb{P}\left(N_{(K,r)} \le \frac{1}{4}K + 1\right) \le \frac{\epsilon}{2}. \tag{7.1}

Now, on the event that $N_{(K,r)} > K/4$, for any $j \ge Kr^2$, we have

dL((0,0),(j,0))i=1K4+1Diam(Gr(i))r, \vec{d}_{\mathcal{L}}((0,0), (j,0)) \geq \sum_{i=1}^{\lfloor \frac{K}{4} \rfloor + 1} \mathrm{Diam}(\mathcal{G}_r(i)) \wedge r,

so that, using Theorem 7.2,

P(dL((0,0),(j,0))<crK4r)12K/4. \mathbb{P}\left(\vec{d}_{\mathcal{L}}((0,0), (j,0)) < c r \frac{K}{4} \wedge r\right) \leq \frac{1}{2^{K/4}}.