Limiting Distributions and Large Deviations for Random Walks in …peterson/research/Thesis... ·...

Thesis Presentation

Limiting Distributions and Large Deviations forRandom Walks in Random Environments

Jonathon Peterson

School of MathematicsUniversity of Minnesota

July 24, 2008

Jonathon Peterson 7/24/2008 1 / 35

Thesis Presentation Model

RWRE in Zd with i.i.d. environment

An environment ω = ω(x , y)x ,y∈Zd , such that∑y∈Zd

ω(x , y) = 1, ∀x ∈ Zd .

ω(x , ·)x∈Zd i.i.d. with distribution P.

Quenched law Pω: fix an environment.Xn a random walk: X0 = 0, and

Pω(Xn+1 = x + y |Xn = x) := ω(x , y).

Annealed law P: average over environments.

P(G) :=

Pω(G)dP(ω)

ω(x , y) = 1, ∀x ∈ Zd .

Pω(Xn+1 = x + y |Xn = x) := ω(x , y).

P(G) :=

Pω(G)dP(ω)

ω(x , y) = 1, ∀x ∈ Zd .

Pω(Xn+1 = x + y |Xn = x) := ω(x , y).

P(G) :=

Pω(G)dP(ω)

Definitions

Nearest neighbor:

ω(x , y) > 0 ⇐⇒ |y | = 1.

Elliptic:P(ω(x , y) ∈ (0,1), ∀x ∈ Zd , ∀|y | = 1

Uniformly Elliptic: ∃κ > 0 such that

P(ω(x , y) ∈ [κ,1− κ], ∀x ∈ Zd , ∀|y | = 1

Thesis Presentation Part I

Part I: Limit Distributions forTransient, One-Dimensional

Thesis Presentation Review of RWRE in Z

RWRE in Z: Recurrence / Transience

A crucial statistic is:ρx :=

ω(x ,−1)

ω(x ,1)

Theorem (Solomon ’75)

Transience or recurrence is determined by EP(log ρ0):

(a) EP(log ρ0) < 0⇒ limn→∞

Xn = +∞, P− a.s.

(b) EP(log ρ0) > 0⇒ limn→∞

Xn = −∞, P− a.s.

(c) EP(log ρ0) = 0⇒ Xn is recurrent, P− a.s.

RWRW in Z: Law of Large Numbers

Assume EP(log ρ) < 0 (transience to the right).Assume EPρ

s = 1 for some s > 0.

Theorem (LLN, Solomon ’75)P− a.s.:

(a) s > 1 (EPρ < 1) ⇒ limn→∞

1− EP(ρ)

1 + EP(ρ)> 0

(b) s ≤ 1 (EPρ ≥ 1) ⇒ limn→∞

Denote limn→∞Xnn =: vP .

RWRE in Z: Annealed Limit Laws

Theorem (Kesten, Kozlov, Spitzer ’75)There exists a constant b such that

(a) s ∈ (0,1)⇒ limn→∞

ns ≤ x)

= 1− Ls,b(x−1/s)

(b) s ∈ (1,2)⇒ limn→∞

Xn − nvP

n1/s ≤ x)

= 1− Ls,b(−x)

(c) s > 2⇒ limn→∞

Xn − nvP

n≤ x

)= Φ(x)

where Ls,b is an s-stable distribution function.

Characteristic Function of Ls,b:

exp−b|t |s

(1− i

tan(πs/2)

)Jonathon Peterson 7/24/2008 7 / 35

Proof: First prove stable limit laws for hitting times

Tn := infk ≥ 0 : Xk = n

(a) s ∈ (0,1)⇒ limn→∞

n1/s ≤ x)

= Ls,b(x)

(b) s ∈ (1,2)⇒ limn→∞

(Tn − nv−1

Pn1/s ≤ x

)= Ls,b(x)

(c) s > 2⇒ limn→∞

(Tn − nv−1

n≤ x

)= Φ(x)

Proof: First prove stable limit laws for hitting times

Tn := infk ≥ 0 : Xk = n

(a) s ∈ (0,1)⇒ limn→∞

n1/s ≤ x)

= Ls,b(x)

(b) s ∈ (1,2)⇒ limn→∞

(Tn − nv−1

Pn1/s ≤ x

)= Ls,b(x)

(c) s > 2⇒ limn→∞

(Tn − nv−1

n≤ x

)= Φ(x)

Thesis Presentation Quenched Central Limit Theorem

Quenched Limit Laws (Gaussian Regime)

Theorem (Goldsheid ’06, P. ’06)If s > 2 then

limn→∞

(Tn − EωTn

n≤ x

)= Φ(x), P − a.s.

where σ2 = EP(VarωT1), and

limn→∞

(Xn − nvP + Zn(ω)

v3/2P σ√

n≤ x

)= Φ(x), P − a.s.

where Zn(ω) depends only on the environment.

Main results of thesis are for s < 2.Do we get quenched stable laws?

Quenched Limit Laws (Gaussian Regime)

Theorem (Goldsheid ’06, P. ’06)If s > 2 then

limn→∞

(Tn − EωTn

n≤ x

)= Φ(x), P − a.s.

where σ2 = EP(VarωT1), and

limn→∞

(Xn − nvP + Zn(ω)

v3/2P σ√

n≤ x

)= Φ(x), P − a.s.

where Zn(ω) depends only on the environment.

Main results of thesis are for s < 2.Do we get quenched stable laws?

Sketch of Proof

Ti − Ti−1∞i=1 are independent under Pω. Lindberg-Feller⇒

limn→∞

(Tn − EωTn

n≤ x

)= Φ(x), P − a.s.

Define X ∗t := maxXn : n ≤ t. Then,

limn→∞

(X ∗n − nvP + Rn(ω,X ∗n )

v3/2P σ√

n≤ x

)= Φ(x), P − a.s.

Difficulty is to replace Rn(ω,X ∗n ) by Zn(ω), which only depends on theenvironment.

Sketch of Proof

limn→∞

(Tn − EωTn

n≤ x

)= Φ(x), P − a.s.

limn→∞

v3/2P σ√

n≤ x

)= Φ(x), P − a.s.

Sketch of Proof

limn→∞

(Tn − EωTn

n≤ x

)= Φ(x), P − a.s.

limn→∞

v3/2P σ√

n≤ x

)= Φ(x), P − a.s.

Thesis Presentation Quenched Limits: s < 2

Define the potential of the environmnet

V (i) :=

∑i−1

k=0 log ρk , i > 00, i = 0∑−1

k=i − log ρk , i < 0

Trap: An atypical section of environment where the potential isincreasing.

Time to cross a trap is exponential in the height of the uphill.

Largest uphill of V (·) in [0,n] is ∼ 1s log n (Erdos & Renyi ’70).

⇒ scaling of n1/s in limit laws of Tn.

Define the potential of the environmnet

V (i) :=

∑i−1

k=0 log ρk , i > 00, i = 0∑−1

k=i − log ρk , i < 0

Trap: An atypical section of environment where the potential isincreasing.

Time to cross a trap is exponential in the height of the uphill.

Largest uphill of V (·) in [0,n] is ∼ 1s log n (Erdos & Renyi ’70).

⇒ scaling of n1/s in limit laws of Tn.

Blocks of the environment

Ladder locations νn defined by ν0 = 0,

νn := infi > νn−1 : V (i) < V (νn−1)ν−n := supj < ν−n+1 : V (k) > V (j) ∀k < j

##HH\\cc

XX JJcc@@HH

XXJJHHcc

ν1 ν2 ν3 ν4 ν5ν−1ν−3

ν−4ν−5ν−6

ν−2ν0

Define a new measure on environments

Q(·) = P ( · |V (i) > 0,∀i < 0)

Under Q, the environment is stationary under shifts of the νi .Jonathon Peterson 7/24/2008 12 / 35

Blocks of the environment

Ladder locations νn defined by ν0 = 0,

νn := infi > νn−1 : V (i) < V (νn−1)ν−n := supj < ν−n+1 : V (k) > V (j) ∀k < j

##HH\\cc

XX JJcc@@HH

XXJJHHcc

ν1 ν2 ν3 ν4 ν5ν−1ν−3

ν−4ν−5ν−6

ν−2ν0

Define a new measure on environments

Q(·) = P ( · |V (i) > 0,∀i < 0)

Under Q, the environment is stationary under shifts of the νi .Jonathon Peterson 7/24/2008 12 / 35

Heuristics of Quenched Limit Laws

Tνn =n∑

(Tνi − Tνi−1)Law≈

n∑i=1

exp(µi,ω)

where µi,ω = Eω(Tνi − Tνi−1) ≈√

Varω(Tνi − Tνi−1).

Quenched CLT? Only if

limn→∞

maxi≤n

µ2i,ω

VarωTνn

= 0, P − a.s.

Exponential limit if

limn→∞

maxi≤n

µ2i,ω

VarωTνn

= 1, P − a.s.

Tνn =n∑

n∑i=1

exp(µi,ω)

limn→∞

maxi≤n

µ2i,ω

VarωTνn

= 0, P − a.s.

limn→∞

maxi≤n

µ2i,ω

VarωTνn

= 1, P − a.s.

Tνn =n∑

n∑i=1

exp(µi,ω)

limn→∞

maxi≤n

µ2i,ω

VarωTνn

= 0, P − a.s.

limn→∞

maxi≤n

µ2i,ω

VarωTνn

= 1, P − a.s.

Tνn =n∑

n∑i=1

exp(µi,ω)

limn→∞

maxi≤n

µ2i,ω

VarωTνn

= 0, P − a.s.

limn→∞

maxi≤n

µ2i,ω

VarωTνn

= 1, P − a.s.

Theorem (P. ’07)Assume s < 2. Then ∃b > 0 s.t.

limn→∞

VarωTνn

n2/s ≤ x)

= L s2 ,b

α-stable process with α < 1 has jumps.This hints that when s < 2

lim infn→∞

(maxi≤n

µ2i,ω

VarωTνn

lim infn→∞

(maxi≤n

µ2i,ω

VarωTνn

> 1− δ

Theorem (P. ’07)Assume s < 2. Then ∃b > 0 s.t.

limn→∞

VarωTνn

n2/s ≤ x)

= L s2 ,b

α-stable process with α < 1 has jumps.This hints that when s < 2

lim infn→∞

(maxi≤n

µ2i,ω

VarωTνn

lim infn→∞

(maxi≤n

µ2i,ω

VarωTνn

> 1− δ

Quenched Limit Laws (sub-gaussian regime)

Theorem (P.’07)

If s < 2 then P − a.s. there exist random subsequences nk = nk (ω),and mk = mk (ω) such that

(a) limk→∞

(Tnk − EωTnk√

VarωTnk

)= Φ(x)

(b) limk→∞

(Tmk − EωTmk√

VarωTmk

0 if x < −11− e−x−1 if x ≥ −1

Contrast with the annealed results:

s ∈ (0,1)⇒ limn→∞

n1/s ≤ x)

= Ls,b(x)

s ∈ (1,2)⇒ limn→∞

(Tn − nv−1

Pn1/s ≤ x

)= Ls,b(x)

Quenched Limit Laws (sub-gaussian regime)

Theorem (P.’07)

If s < 2 then P − a.s. there exist random subsequences nk = nk (ω),and mk = mk (ω) such that

(a) limk→∞

(Tnk − EωTnk√

VarωTnk

)= Φ(x)

(b) limk→∞

(Tmk − EωTmk√

VarωTmk

0 if x < −11− e−x−1 if x ≥ −1

Contrast with the annealed results:

s ∈ (0,1)⇒ limn→∞

n1/s ≤ x)

= Ls,b(x)

s ∈ (1,2)⇒ limn→∞

(Tn − nv−1

Pn1/s ≤ x

)= Ls,b(x)

Quenched Limit Laws (ballistic, sub-gaussian regime)

Theorem (P.’07)

If s ∈ (1,2) then P − a.s. there exist random subsequences nk = nk (ω)and mk = mk (ω) such that

(a) limk→∞

(Xtk − nk

VarωTnk

)= Φ(x)

(b) limk→∞

(Xt ′k−mk

VarωTmk

ex−1 if x < 11 if x ≥ 1

where tk = EωTnk and t ′k = EωTmk .

Contrast with

limn→∞

Xn − nvP

n1/s ≤ x)

= 1− Ls,b(−x)

Quenched Limit Laws (ballistic, sub-gaussian regime)

Theorem (P.’07)

If s ∈ (1,2) then P − a.s. there exist random subsequences nk = nk (ω)and mk = mk (ω) such that

(a) limk→∞

(Xtk − nk

VarωTnk

)= Φ(x)

(b) limk→∞

(Xt ′k−mk

VarωTmk

ex−1 if x < 11 if x ≥ 1

where tk = EωTnk and t ′k = EωTmk .

Contrast with

limn→∞

Xn − nvP

n1/s ≤ x)

= 1− Ls,b(−x)

Quenched Limit Laws (Zero-Speed Regime)

Theorem (P., Zeitouni ’07)

If s ∈ (0,1), then P − a.s. there exist random subsequencesnk = nk (ω), mk = mk (ω), tk = tk (ω), and uk = uk (ω) s.t.

(a) limk→∞

mk≤ x

0 x ≤ 012 0 < x <∞

and limk→∞

log mk

log nk= s

(b) limk→∞

(Xtk − uk

log2 tk∈ [−δ, δ]

)= 1, ∀δ > 0.

Contrast with

limn→∞

ns ≤ x)

= 1− Ls,b(x−1/s)

Quenched Limit Laws (Zero-Speed Regime)

If s ∈ (0,1), then P − a.s. there exist random subsequencesnk = nk (ω), mk = mk (ω), tk = tk (ω), and uk = uk (ω) s.t.

(a) limk→∞

mk≤ x

0 x ≤ 012 0 < x <∞

and limk→∞

log mk

log nk= s

(b) limk→∞

(Xtk − uk

log2 tk∈ [−δ, δ]

)= 1, ∀δ > 0.

Contrast with

limn→∞

ns ≤ x)

= 1− Ls,b(x−1/s)

Thesis Presentation Part II

Part II: Annealed LargeDeviations for Multidimensional

Thesis Presentation Large Deviations: Background

Large Deviations: Definitions

Rate function: A lower semi-continuous function h : Rd → [0,∞].

Good rate function: x : |h(x)| ≤ C compact ∀C <∞.

ξn ∈ Rd satisfy a large deviation principle (LDP) if:

− infx∈Γ

h(x) ≤ lim infn→∞

log P (ξn ∈ Γ)

≤ lim supn→∞

log P (ξn ∈ Γ) ≤ − infx∈Γ

where h is a good rate function.That is

P(ξn ≈ x) ≈ e−nh(x).

− infx∈Γ

log P (ξn ∈ Γ)

≤ lim supn→∞

− infx∈Γ

log P (ξn ∈ Γ)

≤ lim supn→∞

LLN for multidimensional RWRE?

No known LLN in general.(In fact no 0-1 law for transience in a given direction).

However, the random variable V := limn→∞Xnn exists, P− a.s.

(Due to results of Sznitman and Zerner)

Moreover, either1 V =: vP is P− a.s. constant.2 supp(V ) = v−, v+, with v− = cv+ for some c ≤ 0.

There are known conditions such that a V = vP is constant, P− a.s.

Thesis Presentation Annealed Large Deviations

Annealed Large Deviations

Theorem (Varadhan ’03)

Let Xn be a uniformly elliptic, nearst neighbor RWRE on Zd . Then,there exists a convex good rate function H(v) such that Xn

n satisfies anannealed LDP with rate function H(v).

This implies

limδ→0

lim infn→∞

log P(‖Xn − nv‖ < δ) = H(v).

limδ→0

lim supn→∞

log P(‖Xn − nv‖ < δ) = H(v).

Annealed Large Deviations

Let Xn be a uniformly elliptic, nearst neighbor RWRE on Zd . Then,there exists a convex good rate function H(v) such that Xn

n satisfies anannealed LDP with rate function H(v).

This implies

limδ→0

lim infn→∞

log P(‖Xn − nv‖ < δ) = H(v).

limδ→0

lim supn→∞

log P(‖Xn − nv‖ < δ) = H(v).

Zero Set of the Rate Function

Drift at the origin: d(ω) := EωX1.Possible drifts: K := conv (supp (d(ω))).Nestling: 0 ∈ K.Non-nestling: 0 /∈ K.

The set Z := v : H(v) = 0 is either a single point or an intervalcontaining the origin.Non-nestling ⇒ Z = vP.Nestling, supp(V ) = vP ⇒ Z = [0, vP ].Nestling, supp(V ) = v−, v+ ⇒ Z = [v−, v+].

Varadhan’s proof

Xn is not a Markov chain (long term memory).Study the comets of the random walk:

Wn := (−Xn,−Xn + X1, . . . ,−Xn + Xn−1,0)

Wn is a Markov chain (on a horrible state space W ).Obtain a LDP for the empirical distribution process

Rn :=1n

n∑j=1

with rate function J (µ).Contract for LDP for Xn

n : H(v) = infm(µ)=v J (µ).

Properties of the Annealed Rate Function H(v)

Assume the law P is non-nestling. Then, H(v) is analytic in aneighborhood of vP .

Idea:1 Define a new function J(v), which is analytic near vP .2 Show H(v) = J(v) near vP .

Properties of the Annealed Rate Function H(v)

Assume the law P is non-nestling. Then, H(v) is analytic in aneighborhood of vP .

Idea:1 Define a new function J(v), which is analytic near vP .2 Show H(v) = J(v) near vP .

Regeneration Times

Let ` ∈ Rd with ‖`‖2 = 1.Regeneration times (in direction `):

τ4τ3

Xt · `Xτ1 · ` Xτ2 · ` Xτ3 · ` Xτ4 · `

Regeneration Times

Assume that P(limn→∞ Xn · ` = +∞) = 1.Define P(·) := P (·|Xn · ` ≥ 0, ∀n) .

(Xτ1 , τ1), (Xτ2 − Xτ1 , τ2 − τ1), (Xτ3 − Xτ2 , τ3 − τ2), . . .

independent sequence under Pi.i.d. under P

Moreover,

vP := limn→∞

Regeneration Times

Moreover,

vP := limn→∞

Regeneration Times

Moreover,

vP := limn→∞

The function IDefine for λ ∈ Rd+1

Λ(λ) := log Eeλ·(Xτ1 ,τ1),

andI(x , t) := sup

λ∈Rd+1λ · (x , t)− Λ(λ).

Cramer’s Theorem:(

Xτkk , τk

)∈ Rd+1 satisfies a LDP under P with rate

function I.

I(x , t) is convex.I(Eτ1vP ,Eτ1) = 0.Λ(λ) is analytic in the interior of its domain and I(x , t) is convexand analytic in a neighborhood of (Eτ1vP ,Eτ1).

The function IDefine for λ ∈ Rd+1

Λ(λ) := log Eeλ·(Xτ1 ,τ1),

andI(x , t) := sup

λ∈Rd+1λ · (x , t)− Λ(λ).

Cramer’s Theorem:(

Xτkk , τk

)∈ Rd+1 satisfies a LDP under P with rate

function I.

I(x , t) is convex.I(Eτ1vP ,Eτ1) = 0.Λ(λ) is analytic in the interior of its domain and I(x , t) is convexand analytic in a neighborhood of (Eτ1vP ,Eτ1).

The function J

J(v) := infr∈(0,1]

J(v) is convex.J(v) is analytic in a neighborhood of vP .

We want to show

limδ→∞

lim supn→∞

log P(‖Xn

n− v‖ < δ

)≤ −J(v),

limδ→∞

lim infn→∞

log P(‖Xn

n− v‖ < δ

)≥ −J(v).

For convenience we’ll work with P instead of P.

The function J

J(v) := infr∈(0,1]

J(v) is convex.J(v) is analytic in a neighborhood of vP .

We want to show

limδ→∞

lim supn→∞

log P(‖Xn

n− v‖ < δ

)≤ −J(v),

limδ→∞

lim infn→∞

log P(‖Xn

n− v‖ < δ

)≥ −J(v).

For convenience we’ll work with P instead of P.

Thesis Presentation Sketch of Proof

Sketch of the proof

P(Xn ≈ nv) ≈ P(Xτk ≈ nv , τk ≈ n, k = rn)

≈ e−nrI( vr ,

Lower bound:Force Xτk ≈ nv and τk ≈ n for some k = rn.Choose optimal r .

Upper bound:Harder. Need to show that above strategy is optimal.That is, rule out long regeneration times.

Sketch of the proof

≈ e−nrI( vr ,

Sketch of the proof

≈ e−nrI( vr ,

Lower bound

Fix r ∈ (0,1], and let k = rn.

log P(‖Xn − nv‖ < 2δn)

≥ 1n

log P(‖Xτk − nv‖ < δn, |τk − n| < δn)

log P(‖

k− v

r‖ < δ

r, |τk

k− 1

r| < δ

)Limit as k →∞ and then δ → 0 : r I

( vr ,

This lower bound is true for all r ∈ (0,1] and so

limδ→∞

lim infn→∞

log P(‖Xn

n− v‖ < δ

)≥ − inf

r∈(0,1]r I(

Note: Lower bound holds for any v · ` > 0.

Lower bound

≥ 1n

log P(‖

k− v

r‖ < δ

r, |τk

k− 1

r| < δ

( vr ,

limδ→∞

lim infn→∞

log P(‖Xn

n− v‖ < δ

)≥ − inf

r∈(0,1]r I(

Lower bound

≥ 1n

log P(‖

k− v

r‖ < δ

r, |τk

k− 1

r| < δ

( vr ,

limδ→∞

lim infn→∞

log P(‖Xn

n− v‖ < δ

)≥ − inf

r∈(0,1]r I(

Upper bound

First, note that by Chebychev’s inequality

P(Xτk = x , τk = t) ≤ e−λ·(x ,t)Eeλ·(Xτk ,τk )

= e−λ·(x ,t)+kΛ(λ) = e−k(λ·( xk ,

tk )−Λ(λ)).

True for any λ ∈ Rd+1 thus

P(Xτk = x , τk = t) ≤ e−kI( xk ,

tk ) = e−t k

t I( xt

tk ) ≤ e−tJ( x

Note: The final bound does not depend on k .

Would like to say that

P(Xn ≈ nv) ≤ C P(∃k : Xτk ≈ nv , τk ≈ n).

Upper bound

= e−λ·(x ,t)+kΛ(λ) = e−k(λ·( xk ,

tk )−Λ(λ)).

tk ) = e−t k

t I( xt

tk ) ≤ e−tJ( x

Upper bound

= e−λ·(x ,t)+kΛ(λ) = e−k(λ·( xk ,

tk )−Λ(λ)).

tk ) = e−t k

t I( xt

tk ) ≤ e−tJ( x

Upper bound

Since P is non-nestling, τ1 has exponential tails:

P(τ1 ≥ εn) ≤ Ce−Cεn.

Fix ε small.Since J(vP) = 0, J(v) < Cε in a neighborhood of vP .

Thus we may assume τk − τk−1 < εn for all k ≤ n.Need an upper bound for

P(∃k : τk ∈ (n − εn,n], ‖Xn − nv‖ < nδ, τk+1 > n).

The event τk ∈ (n − εn,n], ‖Xn − nv‖ < nδ, τk+1 > n impliesτk = (1− s)n for some s ∈ [0, ε)

‖Xτk − nv‖ < n(δ + s)

τk+1 − τk > ns

P(∃k : τk ∈ (n − εn,n], ‖Xn − nv‖ < nδ, τk+1 > n)

≤∑k≤n

∑s∈[0,ε)

∑‖x−v‖<δ+s

P(τk = (1− s)n, Xτk = xn)P(τ1 > ns)

≤ Cnd+2 sups∈[0,ε)

sup‖x−v‖<δ+s

e−n(1−s)J( x1−s )e−Csn

Claim: Since J(v) is quadratic near vP , for v near vP

infs∈[0,ε)

inf‖x−v‖<δ+s

(1− s)J(

x1− s

)+ Cs = inf

‖x−v‖<δJ(x).

‖Xτk − nv‖ < n(δ + s)

τk+1 − τk > ns

≤∑k≤n

∑s∈[0,ε)

∑‖x−v‖<δ+s

sup‖x−v‖<δ+s

infs∈[0,ε)

inf‖x−v‖<δ+s

(1− s)J(

x1− s

)+ Cs = inf

‖x−v‖<δJ(x).

‖Xτk − nv‖ < n(δ + s)

τk+1 − τk > ns

≤∑k≤n

∑s∈[0,ε)

∑‖x−v‖<δ+s

sup‖x−v‖<δ+s

infs∈[0,ε)

inf‖x−v‖<δ+s

(1− s)J(

x1− s

)+ Cs = inf

‖x−v‖<δJ(x).

infs∈[0,ε)

inf‖x−v‖<δ+s

(1− s)J(

x1− s

)+ Cs = inf

‖x−v‖<δJ(x).

eeeeeeee

infs∈[0,ε)

inf‖x−v‖<δ+s

(1− s)J(

x1− s

)+ Cs = inf

‖x−v‖<δJ(x).

eeeeeeee

x x1−s vP

infs∈[0,ε)

inf‖x−v‖<δ+s

(1− s)J(

x1− s

)+ Cs = inf

‖x−v‖<δJ(x).

eeeeeeee

x x1−s vP

(1− s)J(

x1−s

Thesis Presentation Future Work:

Other Results and Future Work:

When d = 1, have shown H(v) = J(v) for all v > 0 (even innestling case).When d ≥ 2, does H(v) = J(v) for all v · ` > 0?Analytic behavior of H(v) for ”speedup” in nestling case?Can anything be done for v · ` < 0?

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