Spectral gap for the Cauchy process on convex, symmetric domains

Spectral gap for the Cauchy process on convex, symmetric domains

Rodrigo Ba˜nuelos^∗ Department of Mathematics

Purdue University West Lafayette, IN 47906 banuelos@math.purdue.edu

Tadeusz Kulczycki^† Institute of Mathematics Wroc law University of Technology

50-370 Wroc law, Poland tkulczyc@im.pwr.wroc.pl August 28, 2005

Abstract

Let D ⊂ R² be a bounded convex domain which is symmetric relative to both coordinate axes. Assume that [−a, a] × [−b, b], a ≥ b > 0 is the smallest rectangle (with sides parallel to the coordinate axes) containing D. Let {λn}^∞_n=1be the eigenvalues corresponding to the semigroup of the Cauchy process killed upon exiting D. We obtain the following estimate on the spectral gap:

λ₂− λ1≥ Cb a²,

where C is an absolute constant. The estimate is obtained by proving new weighted Poincar´e inequalities and appealing to the connection between the eigenvalue problem for the Cauchy process and a mixed boundary value problem for the Laplacian in one dimension higher known as the mixed Steklov problem established in [5].

Contents

§1. Introduction and Statement of Results

§2. Geometric and Analytic Properties of ϕ1

§3. Weighted Poincar´e inequalities

§4. The Spectral Gap Estimate, Proof of Theorem 1.1

∗Supported in part by NSF Grant # 9700585-DMS

†Supported by KBN grant 1 P03A 020 28 and RTN Harmonic Analysis and Related Problems, contract HPRN-CT-2001-00273-HARP

1 Introduction and Statement of Results

Spectral gap estimates for eigenvalues of the Laplacian with Dirichlet bound- ary conditions, henceforth referred to as the Dirichlet Laplacian, have at- tracted considerable attention for many years as evident by the many papers written on this subject. See for example, [1], [2], [3], [10], [28], [29], [32], [31], [34]. The Dirichlet Laplacian is the infinitesimal generator of the semigroup of Brownian motion killed upon leaving a domain. Therefore questions con- cerning eigenvalues of this operator have been studied both by analytic and probabilistic methods. The question of precise lower bounds for the spectral gap for the Dirichlet Laplacian (the difference between the first two eigenval- ues) and for Schr¨odinger operators with non–negative convex potentials, was raised by M. van den Berg [10] (see also [1], [2], and problem #44 in [33]).

The question was motivated by problems in mathematical physics related to the behavior of free Boson gases. The van den Berg conjecture asserts that for any convex bounded domain of diameter d, the spectral gap is bounded below by 3π²/d² and that this should hold not only for the Dirichlet Lapla- cian but also for Schr¨odinger operators with convex non–negative potentials.

The full conjecture is only known in dimension one, see [28]. For zero po- tentials it has been proved in [9] and [20] for certain planar domains with symmetry. From the probabilistic point of view, the spectral gap determines the rate to equilibrium for Brownian motion conditioned to remain forever in D, the Doob h–process corresponding to the ground state eigenfunction.

The natural question arises as to whether spectral gap bounds for the Laplacian, such as those discussed above, can be extended to non–local, pseudo-differential operators. The class of such operators which are most closely related to the Laplacian ∆ from the point of view of Brownian motion are −(−∆)^α/2, α ∈ (0, 2). These are the infinitesimal generators of the symmetric L´evy α-stable processes. These processes do not have continuous paths; a fact related to the non-locality of the operator −(−∆)^α/2. As in the case of Brownian motion, nevertheless, we can consider the semigroup of these processes killed upon exiting domains and we can consider the eigenvalues of such semigroups. Here again, the spectral gap determines the asymptotic exponential rate of convergence to equilibrium for the process conditioned to remain forever in the domain. By abuse of language, instead of speaking of the eigenvalue gap for the operator −(−∆)^α/2 we will often refer to it as the eigenvalue gap for the corresponding process.

The purpose of this paper is to obtain spectral gap estimates for the Cauchy process. This is the symmetric α-stable process of order α = 1 corresponding to the square root of the Laplacian. We do this by by proving

new weighted Poincar´e inequalities and exploring the connection between the eigenvalue problem for the Cauchy process and the mixed Steklov problem established in [5]. These results raise many natural questions concerning spectral gap bounds for other symmetric α-stable processes and for more general L´evy processes. We believe that as with the results in [5] which have already motivated subsequent work by many others (see [21], [22] [18]), the current results will also be of interest and will open doors to further explorations.

Before we state our results we make some of the above notions more precise and introduce some notation. Let Xt be a symmetric α-stable pro- cess in R^d, α ∈ (0, 2]. This is a process with independent and stationary increments and characteristic function E⁰e^iξXt = e^−t|ξ|α, ξ ∈ R^d, t > 0. We will use Ex, Px to denote the expectation and probability of this process starting at x, respectively. By p^(α)(t, x, y) = p^(α)_t (x − y) we will denote the transition density of this process. That is,

Px(Xt∈ B) = Z

B

p^(α)(t, x, y) dy.

When α = 2 the process Xt is just the Brownian motion in R^d running at twice the speed. That is, if α = 2 then

(1.1) p⁽²⁾(t, x, y) = 1

(4πt)^d/2e^{−|x−y|24t} , t > 0, x, y ∈ R^d.

When α = 1, the process Xt is the Cauchy process in R^d whose transition densities are given by

(1.2) p⁽¹⁾(t, x, y) = c_dt

(t²+ |x − y|²)^(d+1)/2, t > 0, x, y ∈ R^d, where

c_d= Γ((d + 1)/2)/π^(d+1)/2.

Our main concern in this paper are the eigenvalues of the semigroup of the process X_t killed upon leaving a domain. Let D ⊂ R^d be a bounded connected domain and τ_D = inf{t ≥ 0 : X_t∈ D} be the first exit time of D./ By {T_t^D}_t≥0 we denote the semigroup on L²(D) of Xt killed upon exiting D. That is,

T_t^Df (x) = Ex(f (Xt), τD > t), x ∈ D, t > 0, f ∈ L²(D).

The semigroup has transition densities pD(t, x, y) satisfying T_t^Df (x) =

Z

D

p_D(t, x, y)f (y) dy.

The kernel pD(t, x, y) is strictly positive symmetric and

pD(t, x, y) ≤ p^(α)(t, x, y) ≤ c_α,dt^−d/α, x, y ∈ D, t > 0.

The fact that D is bounded implies that for any t > 0 the operator T_t^D maps L²(D) into L^∞(D). From the general theory of semigroups (see [19]) it follows that there exists an orthonormal basis of eigenfunctions {ϕn}^∞_n=1 for L²(D) and corresponding eigenvalues {λn}^∞_n=1 satisfying

0 < λ₁< λ₂ ≤ λ₃≤ . . .

with λ_n→ ∞ as n → ∞. That is, the pair {ϕ_n, λ_n} satisfies (1.3) T_t^Dϕ_n(x) = e^−λntϕ_n(x), x ∈ D, t > 0.

The eigenfunctions ϕ_n are continuous and bounded on D. In addition, λ₁ is simple and the corresponding eigenfunction ϕ1, often called the ground state eigenfunction, is strictly positive on D. For more general properties of the semigroups {T_t^D}_t≥0, see [24], [12], [16].

It is well known (see [4], [16], [17], [27]) that if D is a bounded connected Lipschitz domain and α = 2, or that if D is a bounded connected domain for 0 < α < 2, then {T_t^D}_t≥0 is intrinsically ultracontractive. Intrinsic ultracontractivity is a remarkable property with many consequences. It implies, in particular, that

t→∞lim

e^λ1tpD(t, x, y) ϕ₁(x)ϕ₁(y) = 1,

uniformly in both variables x, y ∈ D. In addition, the rate of convergence is given by the spectral gap λ₂− λ₁. That is, for any t ≥ 1 we have

(1.4) e^{−(λ2−λ1)t}≤ sup

x,y∈D

e^λ1tpD(t, x, y) ϕ₁(x)ϕ₁(y) − 1

≤ C(D, α)e^{−(λ2−λ1)t}. The proof of this for α = 2 may be found in [32]. The proof in our setting is exactly the same.

In the Brownian motion case the properties of eigenfunctions and eigen- values have been extensively studied for many years, both analytically and

probabilistically. It is well known that geometric information on D, such as convexity, symmetry, volume growth, smoothness of its boundary, etc., provide information not only on the ground state eigenfunction ϕ₁ and the ground state eigenvalue λ1, but also on the spectral gap λ2− λ₁, and on the geometry of the nodal domains of ϕ₂.

In the case of stable processes of index 0 < α < 2 surprisingly very little seems to be known. We refer the reader to [6] where some of the known results are reviewed and for a discussion of the many questions that remain open. In the case of the one dimensional interval the spectral gap for the Cauchy process has been estimated in [5] (see Theorem 5.3 and Corollary 2.2). Recently, Z.-Q. Chen and R. Song in [18] (see (5.1)) obtained some estimates of eigenvalues for all α-stable symmetric processes which can be used to obtain estimates of the spectral gap for α–stable processes in dimension one, when α > 1. However, except for the one-dimensional case, lower bound estimates on λ₂−λ₁ remain completely open. In [6], we studied domains with one axis of symmetry and obtained estimates for λ∗− λ₁ in the Cauchy process case where λ∗ is the eigenvalue corresponding to the

“first” antisymmetric eigenfunction for D. However, due to the fact that we do not know the location of the nodal lines for the second eigenfunction for symmetric domains (an interesting question on its own right), the results from [6] do not give estimates on the spectral gap λ₂− λ₁. In this paper we restrict ourselves to bounded convex planar domains which are symmetric with respect to both coordinate axes. For these domains we are able to obtain estimates on the spectral gap for the Cauchy process. Here is the main result of this paper.

Theorem 1.1. Let D ⊂ R²be a bounded convex domain which is symmetric relative to both coordinate axes. Assume that [−L, L] × [−1, 1], L ≥ 1, is the smallest rectangle (with sides parallel to the coordinate axes) containing D.

Let {T_t^D}_t≥0 be the semigroup of the Cauchy process killed upon exiting D with eigenvalues {λn}^∞_n=1. Then we have

(1.5) λ2− λ₁ ≥ C

L², where C = 10⁻⁷ is an absolute constant.

The eigenvalues λ_n satisfies the scaling property λ_n(kD) = λ_n(D)/k, k > 0. This leads to the following easy corollary.

Corollary 1.1. Let D ⊂ R² satisfy the assumptions of Theorem 1.1 except that now the smallest rectangle containing D is [−a, a] × [−b, b], a ≥ b > 0.

Then we have

(1.6) λ2− λ₁ ≥ Cb

a², where C is the same constant as in Theorem 1.1.

Remark 1.1. We remark here that for any bounded convex domain in R^d, d ≥ 2, of inradius r_D (the radius of the largest ball contained in D), we have

(1.7) λ₂− λ₁ ≤

√µ₂−¹₂√ µ₁ r_D

where µ1 and µ2 are, respectively, the first and second eigenvalues for the Laplacian for the unit ball, B(0, 1), in R^d. In fact, µ₂ = j_d/2,1² and µ₁ = j_d/2−1,1² where jp,k denotes the k^th positive zero of the Bessel function Jp(x).

This estimate, and its version for all α ∈ (0, 2), is proved in [6]. From this point on we will concentrate ourselves on the lower bound for the gap and upper bounds will not be mentioned any further. (We refer the reader to [6]

for further questions related to the upper bound including a version of the Payne–P´olya–Weinberger Conjecture for stable processes.)

2 Geometric and Analytic Properties of ϕ

We start with the following theorem which as its proof shows, holds for any stable processes of order 0 < α ≤ 2 and other L´evy processes obtained from Brownian motion by subordination.

Theorem 2.1. Let D ⊂ R² satisfy the assumptions of Theorem 1.1. Let ϕ1

be the first eigenfunction of the Cauchy semigroup {T_t^D}_t≥0. Then we have (i) ϕ1 is continuous and strictly positive in D.

(ii) ϕ₁ is symmetric in D with respect to both coordinate axes. That is, ϕ1(x1, −x2) = ϕ1(x1, x2) and ϕ1(−x1, x2) = ϕ1(x1, x2).

(iii) ϕ₁ is unimodal in D with respect to both coordinate axes. That is, if we take any a2 ∈ (−1, 1) and p(a₂) > 0 such that (p(a2), a2) ∈ ∂D, then the function v(x₁) = ϕ₁(x₁, a₂) defined on (−p(a₂), p(a₂)) is non–

decreasing on (−p(a₂), 0) and non–increasing on (0, p(a₂)). Similarly, if we take any a1 ∈ (−L, L) and r(a₁) > 0 such that (a1, r(a1)) ∈ ∂D, then the function u(x₂) = ϕ₁(a₁, x₂) defined on (−r(a₁), r(a₁)) is non–

decreasing on (−r(a₁), 0) and non–increasing on (0, r(a₁)).

The fact that ϕ1 is continuous has already been proved in [5]. In fact, as shown there (Theorem 4.1) the eigenfunctions ϕ_n are real analytic in D.

This is true not only for the Cauchy process but also for all symmetric stable processes of order α ∈ (0, 2). Also, the fact that ϕ1 is strictly positive is well known. It follows from the general theory of semigroups and from the Harnack inequality proved below. Thus we only need to prove (ii) and (iii) above. These will follow from subordination and the following lemma.

Lemma 2.2. Let B_t be Brownian motion in R² and let D ⊂ R² satisfy the assumptions of Theorem 1.1. For x = (x1, x2) ∈ D and 0 < t1 < t2 < · · · <

t_n< ∞, set

(2.1) F (x) = Px{B_t1 ∈ D, B_t2 ∈ D, . . . , B_tn ∈ D}.

Then properties (ii) and (iii) of Theorem 2.1 hold for F . That is, F is symmetric and unimodal in D with respect to both coordinate axes.

Let us assume the lemma for the moment. We recall that for 0 < α < 2 the symmetric stable process X_tin R^d has the representation

Xt= B2σt,

where σt is a stable subordinator of index α/2 independent of Bt (see [11]).

Thus

(2.2) p^(α)_t (x − y) = Z ∞

0

p⁽²⁾_s (x − y)g_α/2(t, s)ds,

where g_α/2(t, s) is the transition density of σtand p⁽²⁾s (x − y) is the Gaussian density with t replaced by 2t as in (1.1). Now, let D and t₁, t₂, . . . , t_n be as in the statement of Lemma 2.2. Set z0 = x and t0 = 0. Using the Markov property of the stable process Xt, the subordination formula (2.2), Fubini’s theorem, and the Markov property of the Brownian motion, in this order,

we obtain that,

H(x) = Px{X_t1 ∈ D, . . . , X_tn∈ D}

= Z

D

· · · Z

D n

Y

i=1

p^(α)_ti−ti−1(zi−1− z_i) dz1. . . dzn

= Z ∞

0

. . . Z ∞

0

Z

D

· · · Z

D n

Y

i=1

p⁽²⁾_si (zi−1− z_i) dz1. . . dzn

!

×

n

Y

i=1

g_α/2(t_i− t_i−1, s_i) ds₁. . . ds_n

= Z ∞

0

. . . Z ∞

0

P_x{B_2s1 ∈ D, B_2(s1+s2) ∈ D, . . . , B_{2(s1+s2+···+sn)}∈ D}

×

n

Y

i=1

g_α/2(t_i− t_i−1, s_i) ds₁. . . ds_n.

From this and Lemma 2.2 we see that the function H(x) = Px{X_t1 ∈ D, . . . , X_tn ∈ D}

also satisfies the assertions (ii) and (iii) of Theorem 2.1.

Let us now denote by τ_D^α the first exit time of the symmetric stable process from D. D is intrinsically ultracontractive (see [27]) and we have that

(2.3) ϕ^α₁(x) = c(α) lim

t→∞e^λα1tP_x{τ_D^α > t},

where λ^α₁ is the first eigenvalue for the semigroup of the stable process killed upon leaving D and ϕ^α₁(x) is its corresponding eigenfunction. Furthermore, the convergence in (2.3) is uniform for x ∈ D. Thus to prove properties (ii) and (iii) of Theorem 2.1 for ϕ^α₁(x), it is enough to prove them for P_x{τ_D^α > t}.

By the right continuity of the sample paths and the fact that for our domains Px{X_τ^α

D ∈ ∂D} = 0 for x ∈ D we have

P_x{ τ_D^α > t } = P_z{ X_s∈ D, 0 ≤ s ≤ t }

= lim

n→∞P_x{ Xit n

∈ D, i = 1, . . . , n }.

(2.4)

Thus Px{ τ_D^α > t } satisfies properties (ii) and (iii) of Theorem 2.1 and hence so does ϕ^α₁(x).

Proof of Lemma 2.2. By the symmetry properties of D and the the reflec- tion properties of Brownian motion, F is symmetric relative to both coordi- nate axes. That is, F satisfies (ii) of Theorem 2.1. With z₀ = x = (x₁, x₂) and t0 = 0 we have as above that

(2.5) F (x) = Z

D

· · · Z

D n

Y

i=1

p^R_ti²_−ti−1(z_i−1− z_i) dz₁. . . dz_n, where

p^R_t²(z, w) = 1

2πte^{−|z−w|22t}

is the Gaussian density in R². Setting z = (u₁, v₁) and w = (u₂, v₂) we see that

p^R_t²(z − w) = p^R_t(u₁− u₂)p^R_t(v₁− v₂), where

p^R_t(r) = 1

√

2πte^−r22t

is the standard one dimensional Gaussian distribution. Let us now write the domain as D = {(u, v) ∈ R² : v ∈ (−1, 1), −f (v) < u < f (v) } for some function f so that L = sup_(u,0)∈D|u|. With t₀ = 0 and x = (x₁, x₂) = (u0, v0), (2.5) gives

(2.6) F (x) = Z 1

−1

· · · Z 1

−1 n

Y

i=1

p^R_ti−ti−1(vi− v_i−1)Ψn(x1; v1, . . . , vn)dv1· · · dv_n, where

Ψn(x1; v1, . . . , vn) = Z f (v1)

−f (v1)

· · · Z f (vn)

−f (vn) n

Y

i=1

p^R_ti−ti−1(ui− u_i−1) du1· · · du_n Fix v1, v2, . . . , vn and set a1= f (v1), . . . , an= f (vn). Let

(2.7) Φn(x1) = Z a1

−a1

· · · Z an

−an

n

Y

i=1

p^R_ti−ti−1(ui− u_i−1) du1· · · du_n, where as above x₁= u₀. Let us observe that

Φ_n(x₁) = Z a1

−a1

· · · Z an

−an

p^R_t1(u₁− x₁)

n

Y

i=2

p^R_ti−ti−1(u_i− u_i−1) du₁. . . du_n

= Z a1

−a₁

p^R_t1(u₁− x₁) Z a2

−a₂

· · · Z an

−a_n n

Y

i=2

p^R_ti−ti−1(u_i− u_i−1) du₂. . . du_n

! du₁

= Z a1

−a₁

p^R_t1(u₁− x₁)Φ_n−1(u₁)du₁

where

Φn−1(u1) = Z a2

−a₂

· · · Z an

−an

n

Y

i=2

p^R_ti−ti−1(ui− u_i−1) du2. . . dun.

With this, we have

(2.8) Φn(x1) = 1

√2πt1

Z a1

−a1

e^−(u1−x1)

2

2t1 Φn−1(u1) du1.

Using the same arguments as in the proof of Lemma 2.1 in [7] one can show that Φn(x1) is decreasing on (0, ∞) and increasing on (−∞, 0), for all n ≥ 1. Thus it follows that _∂x^∂F

1(x₁, x₂) ≥ 0, for x₁ < 0 and _∂x^∂F

1(x₁, x₂) ≤ 0, for x₁ > 0. This completes the proof of Lemma 2.2.

Next, we prove the Harnack inequality. While this is well known for all symmetric stable processes, our purpose here is to give a proof in the case of the Cauchy process which will give an explicit (and reasonable) constant.

The available proofs (see e.g. Theorem 6.1 in [14]) do not seem to provide such a constant.

Theorem 2.3. Let D ⊂ R² satisfy the assumptions of Theorem 1.1. Let ϕ₁ be the first eigenfunction of the Cauchy semigroup {T_t^D}_t≥0. If B(x0, 3/8) ⊂ D then on B(x0, 1/8) ϕ1 satisfies the Harnack inequality with the constant C = 70.5. That is, for any x, y ∈ B(x₀, 1/8) we have ϕ₁(x) ≤ Cϕ₁(y) where C = 70.5.

Corollary 2.1. Let D ⊂ R² satisfy the assumptions of Theorem 1.1. Let ϕ₁ be the first eigenfunction of the Cauchy semigroup {T_t^D}_t≥0. If B(x0, 3/8) ⊂ D then on B(x₀, 1/8) ϕ²₁ satisfies the Harnack inequality with the constant C = 5 · 10³. That is, for any x, y ∈ B(x₀, 1/8) we have ϕ²₁(x) ≤ Cϕ²₁(y) with C = 5 · 10³

Proof of Theorem 2.3. Let B ⊂ D be any ball (B 6= D). G_B(x, y) = R∞

0 pB(t, x, y) dt is the Green function for B, where pB(t, x, y) is the density of the Cauchy process killed upon exiting B, x, y ∈ B. We have

(2.9) p_B(t, x, y) =

∞

X

n=1

e^−λn(B)tϕ_n,B(x)ϕ_n,B(y),

where λn(B) and ϕn,B are the eigenvalues and eigenfunctions for the semi- group {T_t^B}_t≥0.

We will use the fact that the first eigenfunction is q-harmonic in B ⊂ D according to the 1-stable Schr¨odinger operator.

Let ϕ₁, λ₁ = λ₁(D) be the first eigenfunction and eigenvalue for the semigroup {T_t^D}_t≥0. Let A be the infinitesimal generator of this semigroup.

For x ∈ D we have Aϕ₁(x) = lim

t→0⁺

T_t^Dϕ₁− ϕ₁(x)

t = e^−λ1(D)tϕ₁(x) − ϕ₁(x)

t = −λ₁(D)ϕ₁(x).

This gives that (A + λ1(D))ϕ1= 0 on D. Let B ⊂ D be any ball (B 6= D).

It follows that ϕ₁ is q-harmonic on B according to the 1-stable Schr¨odinger operator A + q with q ≡ λ₁(D). Formally this follows from Proposition 3.17, Theorem 5.5, Definition 5.1 from [14] and the fact that (B, λ1(D)) is gaugable because B is a proper open subset of D and λ₁(B) > λ₁(D).

Let V_B(x, y) =R∞

0 e^λ1(D)tp_B(t, x, y) dt. Here, V_Bis the q-Green function, for q ≡ λ1(D), see page 58 in [14]. The q-harmonicity of ϕ1 (Definition 5.1 in [14]), Theorem 4.10 in [14] (formula (4.15)) and formula (2.17) in [14]

(page 61) give that for z ∈ B,

ϕ1(z) = E^z[e_λ1(D)(τB)ϕ1(X(τB))]

= Z

B

V_B(z, y) Z

D\B

(2π)⁻¹|y − w|⁻³ϕ₁(w) dw dy, (2.10)

where e_λ1(D)(τ_B) = exp(λ₁(D)τ_B). Of course (2.10) is a standard fact in the theory of q-harmonic functions for the α-stable Schr¨odinger operators.

For us this will be a key formula for proving the Harnack inequality for ϕ1. By the well known formula for the distribution of the harmonic measure [26] we have

(2.11) E^zϕ₁(X(τ_B)) = Z

B

G_B(z, y) Z

D\B

(2π)⁻¹|y − w|⁻³ϕ₁(w) dw dy.

To obtain our Harnack inequality for ϕ₁ we will first compare (2.10) and (2.11) and then we will use the formula for E^zϕ1(X(τB)). In order to compare (2.10) and (2.11) we need to compare VB(z, y) and GB(z, y). This will be done in a sequence of lemmas.

Lemma 2.4. Let B ( D be any ball. Then for any z, y ∈ B and t0 > 0 we have

V_B(z, y) ≤ e^λ1(D)t0 Z t0

0

p_B(t, z, y) dt + 1 2π(1 − β)²t₀, where β = λ₁(D)/λ₁(B).

Proof. We have

(2.12) V_B(z, y) ≤ e^λ1(D)t0 Z t0

0

p_B(t, z, y) dt + Z ∞

t0

e^λ1(D)tp_B(t, z, y) dt.

By (2.9) we obtain pB(t, z, y) =

∞

X

n=1

e^−λn(B)tϕn,B(z)ϕn,B(y) ≤ 1 2

∞

X

n=1

e^−λn(B)t(ϕ²_n,B(z)+ϕ²_n,B(y)).

It follows that the second integral in (2.12) is bounded above by (2.13) 1

2 Z ∞

t0

∞

X

n=1

e^{(λ1(D)−βλn(B))t}e^{−λn(B)(1−β)t}(ϕ²_n,B(z) + ϕ²_n,B(y)) dt, where β = λ1(D)/λ1(B). Since λ1(D) < λ1(B) we have β ∈ (0, 1). Note also that e^{λ1(D)−βλn(B)}≤ e^{λ1(D)−βλ1(B)} = e⁰= 1.

For any w ∈ B (w = z or w = y) we have Z ∞

t0

∞

X

n=0

e^{−λn(B)(1−β)t}ϕ²_n,B(w) dt = Z ∞

t0

pB((1 − β)t, w, w) dt

≤ Z ∞

t0

p((1 − β)t, 0, 0) dt = Z ∞

t0

1

2π(1 − β)²t² dt = 1 2π(1 − β)²t₀. We have used the formula for p(t, 0, 0) (see (1.2)). It follows that (2.13) is bounded from above by 1/(2π(1 − β)²t₀).

Lemma 2.5. Let B = B(w, r), r > 0, w ∈ R². For any y ∈ B and z ∈ B(w, r/3) we have

E^y(τ_B) ≤ (32√

2r²/3)G_B(z, y).

Proof. We may and do assume that w = 0. It is well known ([15] formula (2.10)) that E^y(τB) = (2/π)(r²− |y|²)^1/2. It is also well known (see [13]) that

G_B(0,1)(z, y) = 1 2π²|z − y|

Z w(z,y) 0

dr

r^1/2(r + 1), z, y ∈ B(0, 1), where w(z, y) = (1 − |z|²)(1 − |y|²)/|z − y|².

Note that for t > 0 (by substituting s = r^1/2) Z t

0

dr

r^1/2(r + 1) = 2 arctan(t^1/2).

Now, for t ∈ (0, 1] we have 2 arctan(t^1/2) ≥ 2t^1/2arctan(1) = t^1/2π/2, so for any t > 0 we have 2 arctan t^1/2 ≥ (π/2)(t^1/2∧ 1) where a ∧ b denotes min(a, b). Hence G_B(0,1)(z, y) ≥ (4π|z − y|)⁻¹(1 ∧ (w(z, y))^1/2). By scaling it follows that for any z, y ∈ B = B(0, r),

GB(z, y) = r⁻¹G_B(0,1)(z r,y

r)

≥ 1

4πr ^z_r −^y_r





1 ∧

 1 −

^z_r  

21/2 1 −

^y_r  

21/2

 ^z_r −^y_r







= 1

4πr|z − y| r ∧ (r²− |z|²)^1/2(r²− |y|²)^1/2

|z − y|

! . (2.14)

But for z ∈ B(0, r/3) and y ∈ B(0, r) we have |z − y| ≤ 4r/3 and (r² −

|z|²)^1/2 ≥ (8/9)^1/2r. Hence (r²− |z|²)^1/2|z − y|⁻¹ ≥ 1/√

2. It follows that for z ∈ B(0, r/3) and y ∈ B(0, r), (2.14) is bounded below by

3(r²− |y|²)^1/2 16√

2πr² = 3E^y(τ_B) 32√

2r² , and this completes the proof.

Lemma 2.6. Assume that B = B(x, 3/8) ( D. Then for any z ∈ B(x, 1/8) and y ∈ B we have G_B(z, y) ≤ V_B(z, y) ≤ 16.6 G_B(z, y).

Proof. The inequality GB(z, y) ≤ VB(z, y) is trivial, it follows from the definition of G_B(z, y) and V_B(z, y).

We will prove the inequality V_B(z, y) ≤ 16.6 G_B(z, y). By Lemma 4.8 in [14] we have

(2.15) V_B(z, y) = G_B(z, y) + λ₁(D) Z

B

V_B(z, u)G_B(u, y) du.

By Lemma 2.4,R

BV_B(z, u)G_B(u, y) du is bounded above by (2.16)

e^λ1(D)t0 Z

B

Z t0

0

pB(t, z, u) dtGB(u, y) du + 1 2π(1 − β)²t0

Z

B

GB(u, y) du, where β = λ₁(D)/λ₁(B). Let us denote the above sum by I + II. We have

Z

B

Z _t0

0

p_B(t, z, u) dtG_B(u, y) du = Z _t0

0

Z ∞ 0

Z

B

p_B(t, z, u)p_B(s, u, y) du ds dt

= Z t0

0

Z ∞ 0

pB(t + s, z, y) ds dt ≤ t0GB(z, y).

It follows that I ≤ t0e^λ1(D)t0GB(z, y).

By applying Lemma 2.5 for r = 3/8 and z ∈ B(x, 1/8) we get II = E^y(τB)

2π(1 − β)²t₀ ≤ 3√

2GB(z, y) 4π(1 − β)²t₀.

Putting the estimates (2.15), (2.16) together with those for I and II gives

(2.17) VB(z, y) ≤ GB(z, y) 1 + λ1(D)t0e^λ1(D)t0 + 3√

2λ1(D) 4π(1 − β)²t0

! . By our geometric assumptions on D, the inradius of D is no smaller than √

2/2. Hence λ1(D) ≤ λ1(B(0,√

2/2)) = √

2λ1(B(0, 1)). By scaling we also have λ1(B) = (8/3)λ1(B(0, 1)) so β = λ1(D)/λ1(B) ≤ 3√

2/8.

Hence, (1 − β)⁻² ≤ (1 − 3√

2/8)⁻². By (2.15) in [5], λ₁(B(0, 1)) ≤ 2π/3 so λ1(D) ≤ 2√

2π/3. By applying this to (2.17) and putting t0 = 1/2 we obtain VB(z, y) ≤ 16.6 GB(z, y), which completes the proof.

We now return to the proof of Theorem 2.3. Let z1, z2 ∈ B(x, 1/8) ⊂ B(x, 3/8) ⊂ D. By (2.10), (2.11) and Lemma 2.6 we obtain

(2.18) ϕ₁(z₂) ≥ E^z2[ϕ₁(X(τ_B(x,3/8)))]

and

(2.19) ϕ1(z1) ≤ 16.6 E^z1[ϕ1(X(τ_B(x,3/8)))].

So to compare ϕ₁(z₂) and ϕ₁(z₁) we have to compare E^z1[ϕ₁(X(τ_B(x,3/8)))]

and E^z2[ϕ1(X(τ_B(x,3/8)))]. Let Pr,x(z, y) be the Poisson kernel for the ball B(x, r) ⊂ R² for the Cauchy process. That is,

P^w(X(τ_B(x,r)) ∈ A) = Z

A

Pr,x(z, y) dy, where w ∈ B(x, r), A ⊂ B^c(x, r). We have [13]

Pr,x(z, y) = 1 π²

(r²− |z − x|²)^1/2 (|y − x|²− r²)^1/2|y − z|², z ∈ B(x, r) and y ∈ int(B^c(x, r)). Put r = 3/8. We have (2.20) E^zi[ϕ1(X(τ_B(x,3/8)))] =

Z

D\B

ϕ1(y)Pr,x(zi, y) dy,

for i = 1, 2. We have reduce to comparing Pr,x(z1, y) and Pr,x(z2, y). Recall that z₁, z₂∈ B(x, 1/8). For y ∈ B^c(x, 3/8) we have

|y − z₂|

|y − z₁| ≤ 2 so |y − z₂|²

|y − z₁|² ≤ 4 and

(r²− |z₁− x|²)^1/2

(r²− |z₂− x|²)^1/2 ≤ (r²)^1/2

(r²− (r/3)²)^1/2 = 3√ 2 4 . It follows that P_r,x(z₁, y)/P_r,x(z₂, y) ≤ 3√

2. Using this, (2.20), (2.19) and (2.18) we obtain that for z1, z2 ∈ B(x, 1/8) we have ϕ₁(z1) ≤ 16.6 · 3√

2 ϕ1(z2) < 70.5 ϕ1(z2).

3 Weighted Poincar´ e inequalities

The aim of this section is to prove Theorem 3.1 which gives a weighted Poincar´e type inequality. This theorem will be a key step in proving our esti- mates for λ₂−λ₁in §4. We will apply our Poincar´e inequality to the function g = ϕ²₁, where ϕ1 is the first eigenfunction for the Cauchy semigroup. Such weighted Poincar´e type inequalities are well known in the literature (see e.g., [32]) under the assumption that g is log–concave. The novelty of Theorem 3.1 is that we do not assume that g is log-concave. Instead, we assume that g is symmetric and unimodal with respect to both coordinate axis, that it is continuous, strictly positive and that it satisfies an appropriate Har- nack inequality. These hypothesis replaces the log–concavity assumption.

Log–concavity for the function ϕ1 remains an interesting open question.

The proof of Theorem 3.1 is quite technical. Generally speaking, this is so because we can only efficiently control the function g in the center of the domain D where the Harnack inequality holds. One may ask whether Theorem 3.1 is true without the assumption that g satisfies the Harnack inequality with a constant C_H. The answer is negative as we shall see later (Example 3.1).

A remark is perhaps in order here concerning the constants that appear in the formulation and in the proof of the Theorem 3.1. Our goal was to obtain a reasonable concrete constant in the conclusion of the Theorem 3.1 and at the same time emphasise our method rather than to strive for best constants which which these techniques do not have a hope of providing.

In the case when the function g is log–concave, the inequality holds with C_P = π²/8, see for example [32] (note that the diameter d_D of D satisfies d_D ≤ 2√

2L).

Theorem 3.1. Let D ⊂ R² satisfy the assumptions of Theorem 1.1. We assume that g : D → R satisfies the following conditions.

(i) g is continuous and strictly positive.

(ii) g is symmetric with respect to both coordinate axes. That is, g satisfies g(x1, −x2) = g(−x1, x2) = g(−x1, −x2) = g(x1, x2).

(iii) g is unimodal with respect to both coordinate axes as defined in part (iii) of Theorem 2.1.

(iv) If B(x₀, 3/8) ⊂ D, then on B(x₀, 1/8) g satisfies the Harnack inequal- ity with a constant CH ≥ 1. That is, for any x, y ∈ B(x₀, 1/8) we have g(x) ≤ C_Hg(y).

Suppose that f ∈ C¹(D) is bounded and symmetric with respect to both coordinate axes and that

Z

D

f (x)g(x) dx = 0.

Then we have (3.1)

Z

D

|∇f (x)|²g(x) dx ≥ CP

L² Z

D

f²(x)g(x) dx, where

(3.2) CP = min 16

10⁵, 2 10²CH

 .

Throughout this section we will use the following notation. The func- tion p : (−1, 1) → (0, L] is defined so that for any a₂ ∈ (−1, 1) we have (p(a₂), a₂) ∈ ∂D. Similarly, r : (−L, L) → (0, 1] is defined so that for any a1 ∈ (−L, L) we have (a₁, r(a1)) ∈ ∂D. Set

D = {(x˜ 1, x2) ∈ D : x1 ≥ 0, x₂≥ 0}.

Since f and g are symmetric with respect to both axis we can consider R

D˜ |∇f |g instead of R

D|∇f |g. Note that in particular we have R

D˜f g = 0.

We also write ˜D = D1∪ D₂∪ D₃∪ D₄ where

D1 = {(x₁, x2) ∈ ˜D : x1 ≤ L/4, x₂ ≤ 1/4}, D2 = {(x₁, x2) ∈ ˜D : x1 > L/4, x2 ≤ 1/4}, D3 = {(x₁, x2) ∈ ˜D : x1 ≤ L/4, x₂ > 1/4}, D₄ = {(x₁, x₂) ∈ ˜D : x₁ > L/4, x₂ > 1/4}.

The proof of Theorem 3.1 will require several auxiliary lemmas. First, for any function f we set

f+(x) =







f (x), when f (x) ≥ 0 0, when f (x) < 0 and

f−(x) =







0, when f (x) ≥ 0

−f (x), when f (x) < 0.

For brevity, we will write f₊⁰(x) for (f₊(x))⁰. Note that if f ∈ C¹[a, b] then f+0

(x0) is well defined except only for those x₀ for which f (x₀) = 0 and such that f (x) does not vanish in some neighborhood of x₀. Therefore there are at most countably many points on (a, b) for which f+0

does not exist. Note also that for any a1, a2 ∈ [a, b], a < b, we have

Z a2

a1

f₊⁰(x) dx = f₊(a₂) − f₊(a₁).

Similar remarks apply to f−.

Lemma 3.2. Let g : [a, b) → R be continuous, non–increasing and strictly positive. Let f ∈ C¹[a, b) and f (a) = 0. Then,

Z b a

(f₊⁰(x))²g(x) dx ≥ 1 (b − a)²

Z b a

f₊²(x)g(x) dx.

The same inequality holds if we replace f₊ by f− or f .

Proof. Let M = sup_x∈[a,b)f₊²(x)g(x). We consider the following 2 cases:

Case 1: There exists x₀ ∈ [a, b) such that f₊²(x₀)g(x₀) = M .

Case 2: There exists {x_n} ⊂ [a, b), x_n→ b, such that f₊²(x_n)g(x_n) → M . First we deal with Case 1. We have

Z b a

(f+0

(x))²g(x) dx ≥ Z x0

a

(f+0

(x))²g(x) dx

≥ g(x₀) Z x0

a

(f₊⁰(x))²dx.

By Schwarz inequality this is larger than or equal to g(x0)

x₀− a

Z x0

a

(f₊⁰(x)) dx

2

= g(x0)

x₀− a(f₊(x₀) − f₊(a))²

= f₊²(x₀)g(x₀) x₀− a

≥ 1

(b − a)² Z b

a

f₊²(x)g(x) dx.

Next, consider Case 2. Arguing as in Case 1 we have Z b

a

(f+0

(x))²g(x) dx ≥ f₊²(x_n)g(x_n) x_n− a . By letting n → ∞ we obtain

Z b a

(f+0

(x))²g(x) dx ≥ M

b − a ≥ 1 (b − a)²

Z b a

f₊²(x)g(x) dx.

The proof for f− or f is similar.

Lemma 3.3. Let f ∈ C¹[a, b) and f (a) = 0. Then, Z b

a

(f+0

(x))²dx ≥ 1 (b − a)²

Z b a

f₊²(x) dx.

The same inequality holds if we replace f+ by f− or by f . Also, the same conclusion is true when we assume that f ∈ C¹(a, b] and f (b) = 0.

Proof. This follows from Lemma 3.2 with g ≡ 1.

Lemma 3.4. Let a < b, g : [a, b) → R continuous, non–increasing, strictly positive and let f ∈ C¹[a, b). Let η ∈ (0, 1) and A > 0. Assume that f (a) ≤ Aη. Put F = {x ∈ [a, b) : f (x) ≥ A}. Then,

Z b a

(f+0

(x))²g(x) dx ≥ (1 − η)² (b − a)²

Z

F

f₊²(x)g(x) dx.

The same inequality holds if we replace f+ by f .

Proof. We can and do assume that F is not empty. That is, there exists x ∈ (a, b) such that f (x) ≥ A. Let h(x) = f (x) − Aη. We have h(a) ≤ 0.

Put

a₁ = inf{x ∈ [a, b) : f (x) = Aη} = inf{x ∈ [a, b) : h(x) = 0}.

For x ∈ [a, b) such that f (x) ≥ Aη we have f+0(x) = h+0(x) so Z b

a

(f+0(x))²g(x) dx ≥ Z b

a1

(h+0(x))²g(x) dx.

By Lemma 3.2 this is bounded below by

(3.3) 1

(b − a1)² Z b

a1

h²₊(x)g(x) dx.

Note that [a, a₁] ∩ F is empty and that for x ∈ F we have h₊(x) = f₊(x) − Aη ≥ (1 − η)f+(x). It follows that (3.3) is bounded below by

(1 − η)² (b − a)²

Z

F

f₊²(x)g(x) dx, and this completes the proof.

Lemma 3.5. Let a < b, f ∈ C¹(a, b], η ∈ (0, 1) and A > 0. Assume that f (b) ≤ Aη. Put F = {x ∈ (a, b] : f (x) ≥ A}. Then

Z _b

a

(f₊⁰(x))²dx ≥ (1 − η)² (b − a)²

Z

F

f₊²(x) dx.

The same inequality holds if we replace f+ by f . Proof. This follows from Lemma 3.4 with g ≡ 1.

Lemma 3.6. Let 0 < a1 < a2. Suppose g : [0, a2) → R is continuous, non–increasing, strictly positive and let f ∈ C¹[0, a₂). Fix δ₁ ∈ (0, 1). Then at least one of the following inequalities holds

(3.4)

Z a1

0

f₊²(x)g(x) dx ≥ δ1

a1

a2

Z a2

0

f₊²(x)g(x) dx,

(3.5)

Z a2

0

(f+0

(x))²g(x) dx ≥ (1 −√ δ1)² a²₂

Z a2

0

f₊²(x)g(x) dx.

The same conclusion holds if in (3.4) and (3.5) we replace f₊ with f− or f . Proof. Assume that

(3.6)

Z a1

0

f₊²(x)g(x) dx < δ1

a1

a₂ Z a2

0

f₊²(x)g(x) dx.

Let

M = sup

x∈[a1,a2)

f₊²(x)g(x) and x₁ ∈ [0, a₁] be such that

f₊²(x1)g(x1) = min

x∈[0,a1]f₊²(x)g(x).

Let C =Ra2

0 f₊²g. We will show that

(3.7) f₊²(x1)g(x1) ≤ Cδ1

a₂ and

(3.8) M ≥ C

a₂.

Observe that (3.7) follows easily from (3.6). To show (3.8) note that by (3.6) we have

Z a1

0

f₊²g < δ₁Ca₁/a₂. If (3.8) does not hold we would have

Z a2

a1

f₊²g < C(a₂− a₁)/a₂

so Z a2

0

f₊²g < C, which gives contradiction and hence (3.8) holds.

Let ε ∈ (0, 1) and let x0∈ [a₁, a2) be such that f₊²(x0)g(x0) ≥ M (1 − ε).

(3.8) gives that f₊²(x₀)g(x₀) ≥ C(1 − ε)/a₂. Note also that f₊²(x1) ≤ δ1C

a2g(x1) ≤ δ1C

a2g(x0) ≤ f₊²(x0)δ1

1 − ε , so f₊(x₁) ≤ f₊(x₀)√

δ₁/√ 1 − ε.

Now let ε ∈ (0, 1 − δ1), so δ1/(1 − ε) < 1. We have Z a2

0

(f₊⁰(x))²g(x) dx ≥ g(x₀) Z x0

x1

(f₊⁰(x))²dx

≥ g(x₀)

x0− x₁(f₊(x₀) − f₊(x₁))²

≥ f²(x0)g(x0) x₀− x₁

 1 −

√δ1

√1 − ε

2

≥ C(1 − ε) a²₂

 1 −

√δ1

√1 − ε

2

. By letting ε → 0, we obtain (3.5).

Lemma 3.7. Let a < b. Suppose g : [a, b) → (0, ∞) is continuous and non–increasing and let f ∈ C¹[a, b). Let η ∈ (0, 1) and A > 0. Assume that f (a) ≥ Aη². Then at least one of the following inequalities holds

(3.9)

Z b a

f²(x)g(x) dx ≥ A²η⁶ 2

Z b a

g(x) dx,

(3.10)

Z b a

(f⁰(x))²g(x) dx ≥ A²η⁴(1 − η)² 2(b − a)²

Z b a

g(x) dx.

Proof. Put M =Rb

ag(x) dx. If there exists x ∈ [a, b) such that f (x) = Aη³ let x0= min{x ∈ [a, b) : f (x) = Aη³}, in the other case let x₀= b. If

Z x0

a

g(x) dx ≥ M/2 then we have

Z x0

a

f²(x)g(x) dx ≥ A²η⁶M

2 ,

which gives (3.9). If

Z x0

a

g(x) dx < M/2 then

(b − a)g(x₀) ≥ Z b

x0

g(x) dx > M 2 ,