Tightness for the cover time of the two dimensional sphere

Abstract

Let ${\mathcal {C}}^*_{\epsilon ,\mathbf{S}^2}$ denote the cover time of the two dimensional sphere by a Wiener sausage of radius $\epsilon$. We prove that

$$\begin{aligned} \sqrt{{\mathcal {C}}^{*}_{\epsilon ,\mathbf{S}^2} } -\sqrt{\frac{2A_{\mathbf{S}^2}}{\pi }}\left( \log \epsilon ^{-1}-\frac{1}{4}\log \log \epsilon ^{-1}\right) \end{aligned}$$

is tight, where $A_{\mathbf{S}^2}=4\pi$ denotes the Riemannian area of $\mathbf{S}^2$.

Appendices

Appendix I: Barrier estimates

Recall that $\alpha (l)=\rho _{L}(L-l) - l^{\gamma }_{L}$, $\gamma <1/2$, see (1.14) and (1.18), where

$$\begin{aligned} \rho _{L}= 2-\frac{\log L}{2L}, \end{aligned}$$

(8.1)

and recall that

$$\begin{aligned} t_{z}=\frac{\left( \rho _{L}L+z\right) ^{ 2}}{2}. \end{aligned}$$

(8.2)

Recall the notation $I_{y }=[y,y+1)$. (Note that in [5], $I_{y }$ is denoted $H_{y }$.)

Lemma 8.1

For any $k<L$ and all $j \ge 0$ and $0\le z\le 100k$ with $t_{z}\in \mathbb {N}$,

$$\begin{aligned} K_{1}\,:= & {} \mathbb {P}\left( \alpha ( l) \le \sqrt{2T_{l}^{0,t_{z}}},\,l=1,\ldots ,k-1;\, \sqrt{2T_{k}^{0, t_{ z }}}\in I_{\alpha (k)+j}\right) \nonumber \\\le & {} ce^{-2k-2z-2k^{\gamma }_{L}+2j} \times \left( 1+z+ k^{ \gamma }_{L}\right) \left( 1+j \right) e^{-\frac{\left( z + k_{L}^{\gamma } - j \right) ^{2}}{4k}}. \end{aligned}$$

(8.3)

Proof of Lemma 8.1

Recall that we denote by $T_{l}, l=0,1,2,\ldots$ the critical Galton–Watson process with geometric distribution, and let $P^{\mathrm{GW}}_n$ denote its law when $T_0=n$. With $v=\rho _{L}(L-k)- k^{ \gamma }_{L}$ we rewrite $K_{1}$ as

$$\begin{aligned} K_{1}= & {} P^{\mathrm{GW}}_{t_{z}} \left( \rho _{L}(L-l)- l^{ \gamma }_{L}\le \sqrt{2T_{l} } \,\, \text{ for } l=1,\ldots , k-1;\right. \nonumber \\&\left. \sqrt{2T_{k} }\in I_{v+ j} \right) . \end{aligned}$$

(8.4)

We show below that for any $0<\gamma <1$, and all $1\le l\le k-1$

$$\begin{aligned} l_{L}^{\gamma }\le k_{L}^{\gamma }+l_{k}^{\gamma }, \end{aligned}$$

(8.5)

from which it follows that

$$\begin{aligned} \rho _{L}\left( L-l\right) - l_{L}^{\gamma }\ge \rho _{L}\left( L-k\right) - k_{L}^{\gamma }+\rho _{L}\left( k-l\right) - l_{k}^{\gamma }. \end{aligned}$$

Hence

$$\begin{aligned} K_{1}\le P^{\mathrm{GW}}_{t_{z}}\left( v+\rho _{L}\left( k-l\right) - l_{k}^{\gamma } \le \sqrt{2T_{l}},l=1,\ldots ,k-1;\sqrt{2T_{k}}\in I_{v+ j} \right) , \end{aligned}$$

and by (8.2), $\sqrt{2t_{z}} =\rho _{L}L+ z=\rho _{L}k+v+ z + k^{\gamma }_{L}$.

Thus using [5, Theorem 1.1], with $a=\rho _{L}k+v, x=\sqrt{2t_{z}}$ and $b=v, y=v+ j$,

$$\begin{aligned}&K_{1}\le K_{2}=c\frac{\left( 1+ z+ k_{L}^{\gamma }\right) \left( 1+ j\right) }{k} e^{-\frac{\left( \rho _{L}k+ z + k_{L}^{\gamma } - j \right) ^{2}}{2k}}, \end{aligned}$$

We write out

$$\begin{aligned}&\left( \rho _{L}k+ z + k_{L}^{\gamma } - j \right) ^{2}/2k \nonumber \\&\quad = \left( 2k+ (z + k_{L}^{\gamma } - j) -k\log (L)/2L\right) ^{2}/2k \nonumber \\&\quad =2k+ 2(z + k_{L}^{\gamma } - j) -k\log (L)/L + \left( (z + k_{L}^{\gamma } - j) -k\log (L)/2L\right) ^{2}/2k. \nonumber \\ \end{aligned}$$

(8.6)

Using the concavity of the logarithm, we have $k\log (L)/L \le \log k$, and using $(r-s)^{2}\ge r^{2}/2-s^{2}$ we see that

$$\begin{aligned} \left( (z + k_{L}^{\gamma } - j) -k\log (L)/2L\right) ^{2}/2k\ge (z + k_{L}^{\gamma } - j)^{2} /4k-o_{L}(1) \end{aligned}$$

(8.7)

It follows that

$$\begin{aligned} K_{2}\le c \left( 1+ z+ k_{L}^{\gamma }\right) \left( 1+j\right) e^{-2k-2z-2k_{L}^{\gamma }+2j} e^{-\frac{\left( z + k_{L}^{\gamma } - j \right) ^{2}}{4k}}. \end{aligned}$$

We now prove (8.5). This certainly holds if $l\le k/2$, since then $l_{L}^{\gamma }=l_{k}^{\gamma }=l^{\gamma }$. If $L/2\le l\le k$, then (8.5) says that

$$\begin{aligned} (L-l)^{\gamma }\le (L-k)^{\gamma }+ (k-l)^{\gamma }, \end{aligned}$$

(8.8)

which follows from concavity.

Equation (8.5) holds if $k\le L/2$, since then $l_{L}^{\gamma }=l^{\gamma }\le k^{\gamma }= k_{L}^{\gamma }$. Finally, if $k/2\le l\le L/2<k$, then (8.5) says that

$$\begin{aligned} l^{\gamma }\le (L-k)^{\gamma }+ (k-l)^{\gamma }. \end{aligned}$$

(8.9)

Since for $l\le L/2$ we have $l^{\gamma }\le (L-l)^{\gamma }$, (8.9) follows from (8.8). $\square$

Recall that

$$\begin{aligned} \gamma \left( l\right) =\gamma \left( l,L\right) = \rho _{L}(L-l)+ l^{ 1/4}_{L}, \end{aligned}$$

(8.10)

see (4.14).

Lemma 8.2

For all L sufficiently large, and all $0\le z\le 10 L$ with $t_{z}\in \mathbb {N}$,

$$\begin{aligned}&\mathbb {P}\left( \gamma (l)\le \sqrt{2T_{l}^{0,t_{z}}}\,\, \text{ for } l=1,\ldots ,L-1;\,T_{L}^{0,t_{z}}=0 \right] \nonumber \\&\quad \asymp (1+z)e^{ -2L-2z}e^{-z^{2}/4L}. \end{aligned}$$

(8.11)

Similar estimates hold if we delete the barrier condition on some fixed finite interval.

Proof of Lemma 8.2

The left hand side of (8.11) can be written in terms of the critical Galton–Watson process $T_{l}, l=0,1,2,\ldots$ as

$$\begin{aligned}&P^{\mathrm{GW}}_{t_{z}}\left( \rho _{L}(L-l)+ l^{ 1/4}_{L}\le \sqrt{2T_{l} }, \,\, \text{ for } l=1,\ldots , L-1;\,T_{L} =0 \right) . \end{aligned}$$

Equation (8.11) follows immediately from [5, Theorem 1.1], with $a=\rho _{L}L,$$x=\rho _{L}L+z$, $b=y=0$, since $\{ T_{L} =0\}=\{ T_{L} \in [0,1] \}$.

For the last statement, we simply note that following the proof of [5, Lemma 2.3] we can show that the analogue of [5, Theorem 1.1] holds where we skip some fixed finite interval. $\square$

Lemma 8.3

If $(L-k) \log L/L=o_{L}(1)$ then for all L sufficiently large, and all $0\le z \le \log L$ with $t_{z}\in \mathbb {N}$,

$$\begin{aligned}&\mathbb {P}\left( \rho _{L}(L-l)\le \sqrt{2T_{l}^{0,t_{z}}}\,\, \text{ for } l=1,\ldots , k;\,T_{L}^{0,t_{z}}=0 \right) \nonumber \\&\quad \le c (1+z)(L-k)^{ 1/2}e^{ -2L-2z-z^{2}/4L}. \end{aligned}$$

(8.12)

Proof of Lemma 8.3

We rewrite this in terms of the critical Galton–Watson process with geometric offspring distribution, $T_{l}, l=0,1,2,\ldots$, as

$$\begin{aligned} P^{\mathrm{GW}}_{t_{z}}\left( \rho _{L}(L-l)\le \sqrt{2T_{l} } \,\, \text{ for } l=1,\ldots , k;\,T_{L}=0 \right) . \end{aligned}$$

(8.13)

We condition on $T_{k}$ as follows: let $v=\rho _{L}\left( L-k\right)$. Then

$$\begin{aligned} J_{1}:= & {} P^{\mathrm{GW}}_{t_{z}}\left( \rho _{L}\left( L-l\right) \le \sqrt{2T_{l}}, l=1,\ldots ,k,T_{L}=0\right) \\\le & {} \sum _{j=0}^{\infty }P^{\mathrm{GW}}_{t_{z}}\left( \rho _{L}\left( L-l\right) \le \sqrt{2T_{l}},l=1,\ldots ,k,\sqrt{2T_{k}}\in I_{v+j}\right) \\&\times \sup _{u \in I_{j}}P^{\mathrm{GW}}_{( v+u)^{2}/2}\left( T_{L-k}=0\right) \\= & {} \sum _{j=0}^{\infty }P^{\mathrm{GW}}_{t_{z}}\left( \rho _{L}\left( L-l\right) \le \sqrt{2T_{l}},l=1, \ldots ,k,\sqrt{2T_{k}}\in I_{v+j}\right) \\&\times \sup _{u\in I_{j}}\left( 1-\frac{1}{L-k}\right) ^{(v+u)^{2}/2}. \end{aligned}$$

Since, by (3.21),

$$\begin{aligned} P^{\mathrm{GW}}_{t_{z}}\left( \sqrt{2T_{k}}\ge 100L\right) \le ce^{-\left( \rho _{L}L+z-100L\right) ^{2}/2L}\le e^{-100L}, \end{aligned}$$

we obtain

$$\begin{aligned} J_1\le & {} \sum _{j=0}^{100L}P^{\mathrm{GW}}_{t_{z}}\left( \rho _{L}\left( L-l\right) \le \sqrt{2T_{l}},l=1,\ldots ,k,\sqrt{2T_{k}}\in I_{v+j}\right) \\&\times e^{-\frac{( v+j)^{2}}{2(L-k)}}+e^{-100L}=: J_{1,1}+e^{-100L}. \end{aligned}$$

By [5, Theorem 1.1], with $a=\rho _{L}L$, $x=\rho _{L}L+z$ recall (8.2), and $b=v, y=v+j$,

$$\begin{aligned} J_{1,1}\le & {} \sum _{j=0}^{100L}c\frac{(1+z)(1+j) }{k} \sqrt{\frac{\rho _{L}L}{k( v+j)}}\,\, e^{-\frac{\left( \rho _{L}L+z-v-j\right) ^{2}}{2k}}e^{-\frac{( v+j)^{2}}{2(L-k)}}\\\le & {} c\sum _{j=0}^{100L}\frac{(1+z)( 1+j) }{k}\sqrt{\frac{L}{k(v+j)}}\,\, e^{-\frac{\left( \rho _{L}k+z-j\right) ^{2}}{2k}}e^{-\frac{\left( \rho _{L}\left( L-k\right) +j\right) ^{2}}{2(L-k)}}\\\le & {} c(1+z)e^{-\frac{ \rho _{L}^{2}L }{2}-2z}\sum _{j=0}^{100L} \frac{(1+j) }{k}\sqrt{\frac{L}{k( v+j)}}\,\,e^{-\frac{(j-z)^{2}}{2k}-\frac{j^{2}}{2(L-k)}} \\\le & {} c(1+z)e^{-2L-2 z-z^{2}/2k}\sum _{j=0}^{100L}(1+j) \frac{1}{\sqrt{ L-k}}\,\,e^{-\frac{j^{2}}{2(L-k)}}e^{jz/k}. \end{aligned}$$

But

$$\begin{aligned} e^{ -z^{2}/4k}e^{-\frac{j^{2}}{4(L-k)}}e^{jz/k}\le 1 \end{aligned}$$

(8.14)

which follows by considering separately $j\ge 4(L-k)z/k$ and $j< 4(L-k)z/k$, since in this case $j<z/4$ because our condition on k implies that $(L-k)/k$ is tiny. It then follows that

$$\begin{aligned} J_{1,1}\le & {} c(1+z)e^{-2L-2 z-z^{2}/4L} \sum _{j=0}^{100L} ( 1+j)\frac{1}{\sqrt{ L-k}}\,\,e^{-\frac{j^{2}}{4(L-k)}}\\\le & {} c(1+z) e^{-2L-2 z-z^{2}/4L}\sqrt{L-k}. \end{aligned}$$

$\square$

Lemma 8.4

If $k\log L/L=o_{L}(1)$ and $m=\rho _{L}(L-k)+j$, then for all L sufficiently large, with $m^{ 2}/2\in \mathbb {N}$,

$$\begin{aligned}&\mathbb {P}\left( \rho _{L}(L-l)\le \sqrt{2T_{l}^{0,k,m^{ 2}/2}}\,\, \text{ for } l=k+1,\ldots , L-1;\,T_{L}^{0,k,m^{ 2}/2}=0 \right) \nonumber \\&\quad \le c \left( 1+j\right) e^{-2\left( L-k\right) -2j -\frac{j^{2}}{4(L-k)}}, \end{aligned}$$

(8.15)

and if $j\le \eta L$ and we skip the barrier from $k+1$ to $k+s$, with $s\le \eta ' \log L$, for some $\eta , \eta '<\infty$, then

$$\begin{aligned}&\mathbb {P}\left( \rho _{L}(L-l)\le \sqrt{2T_{l}^{0,k,m^{ 2}/2}}\,\, \text{ for } l=k+s,\ldots , L-1;\,T_{L}^{0,k,m^{ 2}/2}=0 \right) \nonumber \\&\quad \le c\,( 1+j+\sqrt{s}) e^{-2\left( L-k\right) -2j -\frac{j^{2}}{4(L-k)}}. \end{aligned}$$

(8.16)

Proof

We first turn to the probability in (8.15). By the Markov property, the probability in question can be rewritten in terms of the critical Galton–Watson process $T_{l}, l=0,1,2,\ldots$ as

$$\begin{aligned} M_{1}:=P^{\mathrm{GW}}_{m^{2}/2}\left( \rho _{L}\left( (L-k)-l\right) \le \sqrt{2T_{l}}, l=1,\ldots ,L-k-1;\, T_{L-k}=0\right) . \end{aligned}$$

By [5, Theorem 1.1], with $a=\rho _{L}\left( L-k\right) , x=m$ and $b=y=0$,

$$\begin{aligned} M_{1}\le c\frac{\left( 1+j\right) }{L-k} \,\, e^{-\frac{\left( \rho _{L}\left( L-k\right) + j\right) ^{2}}{2\left( L-k\right) }}. \end{aligned}$$

(8.17)

We have

$$\begin{aligned}&e^{-\frac{\left( \rho _{L}\left( L-k\right) + j\right) ^{2}}{2\left( L-k\right) }} =e^{-\frac{\left( \rho _{L}\left( L-k\right) \right) ^{2}}{2(L-k)}- j\rho _{L}-\frac{j^{2}}{2(L-k)}}\\&\le ce^{-\frac{\left( 2\left( L-k\right) -\frac{\log L}{2}\frac{L-k}{L} \right) ^{2}}{2\left( L-k\right) }-2j-\frac{j^{2}}{ 3(L-k)}}\le Ce^{\log L \frac{L-k}{L}}e^{-2\left( L-k\right) -2j -\frac{j^{2}}{3(L-k)}}. \end{aligned}$$

Using the concavity of the logarithm, we get that

$$\begin{aligned} M_{1}\le c \left( 1+j\right) e^{-2\left( L-k\right) -2j -\frac{j^{2}}{3(L-k)}}, \end{aligned}$$

which gives (8.15).

For (8.16) with $k'=k+ s$, $v= \rho _{L}(L-k')$ we bound

$$\begin{aligned}&\mathbb {P}\left( \rho _{L}(L-l)\le \sqrt{2T_{l}^{0,k,m^{ 2}/2}}\,\, \text{ for } l=k+s,\ldots , L-1;\,T_{L}^{0,k,m^{ 2}/2}=0 \right) \nonumber \\&\quad \le \sum _{j'=0}^{\infty }P^{\mathrm{GW}}_{m^{2}/2}\left( \sqrt{2T_{s}}\in I_{v+j'}\right) M'_{1} \end{aligned}$$

(8.18)

where

$$\begin{aligned}&M'_{1}=\sup _{u\in I_{v+j'}}P^{\mathrm{GW}}_{u^{2}/2}\left( \rho _{L}\left( (L-k')-l\right) \le \sqrt{2T_{l}},l=1, \ldots ,L-k'-1;\, T_{L-k'}=0\right) . \end{aligned}$$

By [5, Proposition 1.4, Remark 2.2]

$$\begin{aligned}&P^{\mathrm{GW}}_{m^{2}/2}\left( \sqrt{2T_{s}}\in I_{v+j'}\right) \le c\sqrt{\frac{m}{(v+j')s}}\,\,e^{-\left( m-(v+j')\right) ^{2}/2s}\nonumber \\&\quad \le c\sqrt{\frac{1}{s}}\,\,e^{-\left( \rho _{L}s+(j-j')\right) ^{2}/2s}\le c\sqrt{\frac{1}{s}}\,\,e^{-2s-2(j-j')- (j-j')^{2}/3s} \end{aligned}$$

(8.19)

and as in the first part of this proof

$$\begin{aligned} M'_{1}\le c \left( 1+j'\right) e^{-2\left( L-k'\right) -2j' -\frac{j'^{2}}{3(L-k')}}. \end{aligned}$$

(8.20)

Thus

$$\begin{aligned}&\sum _{j'=0}^{\infty }P^{\mathrm{GW}}_{m^{2}/2}\left( \sqrt{2T_{s}}\in I_{v+j'}\right) M'_{1}\nonumber \\&\quad \le c e^{-2\left( L-k\right) -2j }\sqrt{\frac{1}{s}}\,\,\sum _{j'=0}^{\infty } \left( 1+j'\right) e^{ -\frac{j'^{2}}{3(L-k')}} e^{ -(j'-j)^{2}/3s}. \end{aligned}$$

(8.21)

Considering separately the cases where $j'\le 1.1 j$ and $j'> 1.1 j$ we see that for L sufficiently large,

$$\begin{aligned} e^{ -\frac{j'^{2}}{3(L-k')}} e^{ -(j'-j)^{2}/3s}\le ce^{ -\frac{j^{2}}{4(L-k)}} e^{ -(j'-j)^{2}/4s}, \end{aligned}$$

(8.22)

and (8.16) follows. $\square$

Lemma 8.5

If $v=\rho _{L}(L-k)+u$ then for L large with $v^{2}/2\in \mathbb {N}$, and $1\le k,{\widetilde{k}}, u,j\le L^{3/4}$,

$$\begin{aligned}&\mathbb {P}\left( \rho _{L}(L-l)\le \sqrt{2T^{0,k, v^{2}/2}_{l}},l=k+1,\ldots ,k+{\widetilde{k}};\sqrt{2T^{0,k, v^{2}/2}_{k+{\widetilde{k}}}}\in I_{\rho _{L}(L-k-{\widetilde{k}})+j} \right) \nonumber \\&\quad \le C\frac{\left( 1+u\right) \left( 1+j\right) }{{\widetilde{k}}^{3/2}}e^{-2{\widetilde{k}} -2\left( u-j\right) -\frac{(u-j)^{2}}{4{\widetilde{k}}}}. \end{aligned}$$

(8.23)

In addition, if we skip the barrier from $k+1$ to $k+3$,

$$\begin{aligned}&\mathbb {P}\left( \rho _{L}(L-l)\le \sqrt{2T^{0,k, v^{2}/2}_{l}},l=k+3,\ldots ,k+{\widetilde{k}};\sqrt{2T^{0,k, v^{2}/2}_{k+{\widetilde{k}}}}\in I_{\rho _{L}(L-k-{\widetilde{k}})+j} \right) \nonumber \\&\quad \le C\frac{\left( 1+u \right) \left( 1+j\right) }{{\widetilde{k}}^{3/2}}e^{-2{\widetilde{k}}-2\left( u-j\right) -\frac{(u-j)^{2}}{4{\widetilde{k}}}}. \end{aligned}$$

(8.24)

Proof

By the Markov property, the probability in question can be rewritten in terms of the critical Galton–Watson process $T_{l}, l=0,1,2,\ldots$ as

$$\begin{aligned}&V_{1}:=P^{\mathrm{GW}}_{ v^{2}/2}\left( \rho _{L}\left( L-k-l\right) \le \sqrt{2T_{l}},l= 1,\ldots , {\widetilde{k}};\right. \nonumber \\&\quad \left. \sqrt{2T_{ {\widetilde{k}}}}\in I_{\rho _{L}\left( L-k-{\widetilde{k}}\right) + j } \right) . \end{aligned}$$

(8.25)

This is a linear barrier of length ${\widetilde{k}}$. At the start of the barrier it is at distance u from the starting point, and at the end it is at distance j from the end point. Therefore by [5, Theorem 1.1] we have that the probability is at most

$$\begin{aligned}&c\frac{\left( 1+ u\right) \left( 1+ j\right) }{{\widetilde{k}}^{3/2}}\sqrt{ \frac{\rho _{L}\left( L-k\right) + u}{\rho _{L}\left( L-k-{\widetilde{k}}\right) + j}} \\&\quad e^{-\frac{\left( \rho _{L}\left( L-k\right) + u-\left( \rho _{L}\left( L-k-{\widetilde{k}}\right) + j\right) \right) ^{2}}{2{\widetilde{k}}}}\\&\quad \le c\frac{\left( 1+u\right) \left( 1+j\right) }{{\widetilde{k}}^{3/2}}e^{-\frac{\left( \rho _{L}{\widetilde{k}}+ u- j \right) ^{2}}{2{\widetilde{k}}}}, \end{aligned}$$

where we have bounded the square root using the fact that $k,{\widetilde{k}}, u,j<L^{3/4}$ so that the ratio inside the square root is $\asymp 2L/2L=1$. Using $k,{\widetilde{k}}, u,j<L^{3/4}$ again we see that

$$\begin{aligned} e^{-\frac{\left( \rho _{L}{\widetilde{k}}+ u- j \right) ^{2}}{2{\widetilde{k}}}}\le & {} ce^{-\frac{\left( \rho _{L}{\widetilde{k}}\right) ^{2}}{2{\widetilde{k}}}}e^{ -2\left( u- j\right) }e^{-\frac{\left( u- j \right) ^{2}}{2{\widetilde{k}}}}\\= & {} ce^{-\frac{\left( 2{\widetilde{k}}-\frac{\log L}{2}\frac{{\widetilde{k}}}{L} \right) ^{2}}{2{\widetilde{k}}}}e^{ -2\left( u- j \right) }e^{-\frac{\left( u- j \right) ^{2}}{2{\widetilde{k}}}}\\\le & {} Ce^{- 2{\widetilde{k}} }e^{ -2\left( u- j \right) }e^{-\frac{\left( u- j \right) ^{2}}{2{\widetilde{k}}}}. \end{aligned}$$

This gives (8.23).

Equation (8.24) follows as in the proof of the previous Lemma. $\square$

Appendix II: Conditional excursion probabilities

The following is a modification of [16] adapted to our situation. Fix $k'>k>2$ and let ${{\mathcal {G}}}_k^y$ to be the $\sigma$-algebra generated by the excursions from $\partial B_{d}(y,h_{k-1})$ to $\partial B_{d}(y,h_{k})$ (and if we start outside $\partial B_{d}(y,h_{k-1})$ we include the initial excursion to $\partial B_{d}(y,h_{k-1})$). Note that ${{\mathcal {G}}}_k^y$ includes the information on the end points of the excursions from $\partial B_{d}(y,h_{k})$ to $\partial B_{d}(y,h_{k-1})$.

Recall that $T_{y,l-1\rightarrow l}^{y,k\overset{n}{\rightarrow }k-1}$ is the number of excursions from $\partial B_{d}(y,h_{l-1})\rightarrow \partial B_{d}(y,h_{l})$ during the first n excursions from $\partial B_{d}(y,h_{k })\rightarrow \partial B_{d}(y,h_{k-1})$, see (2.11).

Lemma 9.1

For any $L-2k\ge k '>k\ge 2$ and $n>1$, let $\mathcal {A}_{k'}$ denote an event, measurable on the excursions of the Brownian motion from $\partial B_{d}(y,h_{k'})$ to $\partial B_{d}(y,h_{k'-1})$ and on the collection $\{T_{y,l-1\rightarrow l}^{y,k\overset{n}{\rightarrow }k-1} \}_{l=k',\ldots ,L}$ . Then,

$$\begin{aligned} \mathbb {P}(\mathcal {A}_{k'} \, | {{\mathcal {G}}}_k^y) \le \left( 1+10(k'-k)\frac{h_{k'-1}}{h_{k}}\right) ^n \mathbb {P}(\mathcal {A}_{k'}) \,. \end{aligned}$$

(9.1)

In particular, for all $m_l ;l=k',\ldots L$, and all $y \in \mathbf{S}^2$,

$$\begin{aligned}&\mathbb {P}(T_{y,l-1\rightarrow l}^{y,k\overset{n}{\rightarrow }k-1} = m_l ;l=k',\ldots L \, | {{\mathcal {G}}}_k^y) \nonumber \\&\quad \le \left( 1+10(k'-k)\frac{h_{k'-1}}{h_{k}}\right) ^n \mathbb {P}(T_{y,l-1\rightarrow l}^{y,k\overset{n}{\rightarrow }k-1} = m_l ;l=k',\ldots L \,) \,. \end{aligned}$$

(9.2)

The key to the proof of Lemma 9.1 is the following Lemma.

Lemma 9.2

For $k'>k \ge 2$ and a Brownian path $X_\cdot$ starting at $z\in \partial B_{d}(y,h_{k'-1})$, let $Z_l=T_{y,l-1\rightarrow l}^{y,k'-1\overset{1}{\rightarrow }k}$, $l=k',\ldots L$, denote the number of excursions of the path from $\partial B_{d}(y,h_{l-1})\rightarrow \partial B_{d}(y,h_{l})$, prior to $\bar{\tau }=\inf \{t>0\, :\,X_t\in \partial B_{d}(y,h_{k})\}$, and let F denote an event measurable with respect to the path of the Brownian motion inside $B_d(y,h_{k'})$, prior to $\bar{\tau }$. Then, for some $c<\infty$ and all $\{ m_l : l=k',\ldots L \}$, uniformly in $v\in \partial B_{d}(y,h_{k})$ and y,

$$\begin{aligned} \left| \frac{\mathbb {E}^z (F ;Z_l = m_l ,\, l=k',\ldots L \,\big |\,w_{\bar{\tau }}=v)}{\mathbb {E}^z (F; Z_l = m_l ,\, l=k',\ldots L)}-1\right| \le 10 (k'-k)\frac{h_{k'-1}}{h_k}. \end{aligned}$$

(9.3)

In words, conditioning by the endpoint of the excursion at level k has only minor influence on the probability of events involving pieces of excursions at levels larger than $k'$.

Proof of Lemma 9.2

Without loss of generality we can take $y=0$. Fixing $k \ge 2$ and $z \in \partial B_{d}(0, h_{k'-1})$ it suffices to consider $\{m_l ,\, l=k',\ldots L \}$ for which $\mathbb {P}^z\left( Z_{l} =m_l,\, l=k',\ldots L\right) >0$. Fix such $\{m_l ,\, l=k',\ldots L \}$ and a positive continuous function g on $\partial B_{d}(0,h_{k})$. Let $\bar{\tau } = \inf \{ t:\, w_t \in \partial B_{d}(0,h_{k})\}$, $\tau _0=0$ and for $i=0,1,\ldots$ define

$$\begin{aligned} \tau _{2i+1}= & {} \inf \{ t \ge \tau _{2i} :\; X_t \in \partial B_{d}(0,h_{k'}) \cup \partial B_{d}(0,h_{k}) \} \\ \tau _{2i+2}= & {} \inf \{ t \ge \tau _{2i+1} :\; X_t \in \partial B_{d}(0,h_{k'-1})\}\,. \end{aligned}$$

Set $j=m_{k'}$ and let $Z^j_l$, $l = k',\ldots L$ be the corresponding number of excursions by the Brownian path prior to time $\tau _{2j}$. Then, by the strong Markov property at $\tau _{2j}$,

$$\begin{aligned}&{\mathbb {E}}^z [ g(X_{\bar{\tau }}) ; F,Z_l = m_l,\, l = k',\ldots L ] \\&\quad = \mathbb {E}^z \left[ {\mathbb {E}}^{X_{\tau _{2j}}} (g(X_{\bar{\tau }}); Z_{k'}=0) ;F, Z^j_l =m_l,\, l = k',\ldots L, {\bar{\tau }} \ge \tau _{2j} \right] \,. \end{aligned}$$

and, substituting $g=1$,

$$\begin{aligned}&\mathbb {P}^z \left( F,Z_l=m_l,\, l = k',\ldots L \right) \\&\quad = \mathbb {E}^z \left[ \mathbb {P}^{X_{\tau _{2j}}} ( Z_{k'}=0) ;F, Z^j_l =m_l,\, l = k',\ldots L, {\bar{\tau }} \ge \tau _{2j} \right] \;. \end{aligned}$$

Consequently,

$$\begin{aligned}&\mathbb {P}^z \left( F, Z_l =m_l,\, l = k',\ldots L \right) \inf _{x\in \partial B_{d}(0,h_{k'-1})} \frac{{\mathbb {E}}^{x} \left( g(X_{\bar{\tau }});\, Z_{k'}=0 \right) }{\mathbb {P}^x \left( Z_{k'}=0\right) }\\&\quad \le \mathbb {E}^z [ g(X_{\bar{\tau }}); F, Z_l =m_l,\, l = k',\ldots L ]\\&\quad \le \mathbb {P}^z \left( F, Z_l =m_l,\, l = k',\ldots L \right) \sup _{x\in \partial B_{d}(0,h_{k'-1})} \frac{{\mathbb {E}}^{x} \left( g(X_{\bar{\tau }});\, Z_{k'}=0 \right) }{\mathbb {P}^x \left( Z_{k'}=0\right) } \,, \end{aligned}$$

and, using again the strong Markov property at time $\tau _{2}$,

$$\begin{aligned} {\mathbb {E}}^{x} \left( g(X_{\bar{\tau }});\, Z_{k'}=0 \right)= & {} {\mathbb {E}}^{x} \left( g(X_{\bar{\tau }}) \right) - {\mathbb {E}}^{x} \left( {\mathbb {E}}^{X_{\tau _2}} (g(X_{\bar{\tau }}));\, Z_{k'} \ge 1 \right) \\\le & {} {\mathbb {E}}^{x} \left( g(X_{\bar{\tau }}) \right) - \mathbb {P}^{x} (Z_{k'} \ge 1) \inf _{y\in \partial B_{d}(0,h_{k'-1})} {\mathbb {E}}^{y} (g(X_{\bar{\tau }})), \end{aligned}$$

with the reversed inequality if the $\inf$ is replaced by $\sup$. Writing $p=\mathbb {P}^{x} (Z_{k'} \ge 1) =1-1/(k'-k)$ whenever $x\in \partial B_{d}(0,h_{k'-1})$, c.f. (2.7), it thus follows that

$$\begin{aligned}&\mathbb {E}^z \left[ g(X_{\bar{\tau }}); F,Z_l =m_l,\, l = k',\ldots L \right] \nonumber \\&\quad \le \mathbb {P}^z\left( F,Z_l =m_l,\, l = k',\ldots L \right) \nonumber \\&\quad \cdot {\mathbb {E}}^z\left( g(w_{\bar{\tau }})\right) (1-p)^{-1} \left\{ {\sup _{x\in \partial B_{d}(0,h_{k'-1})} E^x( g(X_{\bar{\tau }})) \over \inf _{y\in \partial B_{d}(0,h_{k'-1})} E^y( g(X_{\bar{\tau }}))} -p\right\} , \end{aligned}$$

(9.4)

with the reversed inequality if the $\inf$ and $\sup$ are interchanged.

As in (2.8), let $p_{B_{d}( 0, h_{k})}( z,x)$ denote the Poisson kernel for $B_{d}( 0, h_{k})\subseteq \mathbf{S}^{ 2}$, see (2.8).

Recall that

$$\begin{aligned} {\mathbb {E}}^{z'} g (X_{\bar{\tau } }) = \int _{\partial B_{d}(0,h_{k})} p_{B_{d}(0,h_{k})}( z',u) g(u) du. \end{aligned}$$

Therefore, we get the Harnack inequality

$$\begin{aligned} \frac{\sup _{x\in \partial B_{d}(0,h_{k'-1})} {\mathbb {E}}^x( g(X_{\bar{\tau }}))}{\inf _{y\in \partial B_{d}(0,h_{k'-1})} {\mathbb {E}}^y( g(X_{\bar{\tau }}))}\le & {} \frac{\max _{x\in \partial B_{d}(0,h_{k'-1}),u\in \partial B_{d}(0,h_{k })}p_{B_{d}(0,h_{k})}( x,u)}{\min _{y\in \partial B_{d}(0,h_{k'-1}),u\in \partial B_{d}(0,h_{k })} p_{B_{d}(0,h_{k})}( y,u)}\nonumber \\= & {} \frac{\max _{y\in \partial B_{d}(0,h_{k'-1}),u\in \partial B_{d}(0,h_{k })}\sin ^{ 2} ( d( u,y)/2)}{\min _{x\in \partial B_{d}(0,h_{k'-1}),u\in \partial B_{d}(0,h_{k })} \sin ^{ 2} ( d( u,x)/2)} \nonumber \\\le & {} \frac{\sin ^{ 2}((h_k+h_{k'-1})/2) }{\sin ^{ 2}((h_k-h_{k'-1})/2) }. \end{aligned}$$

(9.5)

Writing $\alpha =(h_k-h_{k'-1})/2$, $\beta =h_{k'-1}$ we have

$$\begin{aligned} \frac{\sin ^{ 2}((h_k+h_{k'-1})/2) }{\sin ^{ 2}((h_k-h_{k'-1})/2) }.=\frac{\sin (\alpha +\beta ) }{\sin (\alpha ) }=\cos (\beta )+\frac{\sin ( \beta ) }{\tan (\alpha ) }. \end{aligned}$$

(9.6)

Using the bounds, valid for $|\alpha |,|\beta |<0.1$, $|\sin (\alpha )|\ge |2\alpha /3|$, $|\cos \beta -1|<|\beta |/2$, we obtain that the right side of (9.6) is bounded by $1+4h_{k'-1}/h_k$, and therefore the right side of (9.5) is bounded by $1+10h_{k'-1}/h_k$. Substituting this bound into (9.4) and using the value of p yields the claim. $\square$

Proof of Lemma 9.1

This follows from Lemma 9.2 in the same manner as [16, Lemma 7.3] was derived from [16, Lemma 7.4]: using the strong Markov property, each excursion from $\partial B_{d}(y,h_{k})$ to $\partial B_{d}(y,h_{k-1})$ will contribute a multiplicative factor $(1+10 (k'-k)h_{k'-1}/{h_{k}})$ to the probability. We omit further details. $\square$