This article has appeared in: Proceedings of the Royal Society of Edinburgh A 132 (2002), 91–120.

Sharp estimates for iterated Green functions

H.-Ch. Grunau ^{1} and G. Sweers ^{2}

Abstract Optimal pointwise estimates from above and below are obtained for iterated (poly)harmonic Green functions corresponding to zero Dirichlet boundary conditions. For second order elliptic operators these estimates hold true on bounded C^{1,1} domains. For higher order elliptic operators we have to restrict ourselves to the polyharmonic operator on balls. We will also consider applications to noncooperatively coupled elliptic systems and to the lifetime of conditioned Brownian motion.

1 Introduction and main results

Roughly spoken the pointwise deflection of solutions of elliptic partial differential equations in bounded smooth domains Ω ⊂ ℝ^{n} with Dirichlet boundary conditions is submitted to two opposite influences. A positive source terms locally ‘bends’ the solution and hence urges it to increase globally while the zero boundary condition(s) pulls in the opposite direction. As is well known in second order elliptic equations such as −Δ the positive source term will dominate the sign of the solution everywhere in the domain. Indeed, this result is a consequence of the maximum principle. For higher order elliptic operators such a sign preserving property in general does not hold except when considering e.g. the polyharmonic operator (−Δ)^{m} with zero Dirichlet boundary conditions on a ball. These positivity preserving results are reflected in the fact that the kernel of the solution operator is of fixed sign. Sharp pointwise estimates of this Green function will show the balance between the two effects just mentioned.

For studying perturbations of those elliptic operators, both for second and for higher order operators, it also becomes crucial to find optimal pointwise estimates for the corresponding Green functions. And indeed, in recent years such estimates have been developed. Motivated by Schrödinger operators Zhao ([21], see also [4]) was in 1986 the first to prove a sharp estimate from below for the Green function of the Laplace operator. The estimate from above for the Laplacian had been proven earlier in 1967 by Widman ([20], see also [7, Theorem 4.6.11]). In fact, due to a result of Ancona [1], such estimates hold for quite general second order elliptic operators. For the explicit formula of those estimates, both for n = 2 and n > 2, we refer to [18].

The need of having pointwise estimates for iterated Green functions of polyharmonic operators became first obvious to the authors when studying in [11] positivity questions for perturbations of the polyharmonic Dirichlet problem on the (unit-) ball B in ℝ^{n} :

1Fakultät für Mathematik, Otto-von-Guericke-Universität Magdeburg, Postfach 4120, D-39016 Magdeburg, Germany

2Applied Math. Analysis, I.T.S., Delft University of Technology, PObox 5031, NL-2600 GA Delft, The Netherlands

\{ \begin{matrix}(−Δ)^{m} u = f & in B, \\ u = \frac{∂}{∂n}u = ⋯ = \frac{∂^{m−1}}{∂n^{m−1}} u & on ∂B. \end{matrix}  (1.1)

In that paper sharp estimates from above and below for the Green function for (1.1) were proven.

The main difficulty in [11] was to find the appropriate balance between the singularity in the Green function G (x, y) when x → y (a point mass source term) and G (x, y) → 0 when x or y → ∂Ω (the zero boundary condition). For more general operators, but without the effect of the boundary condition, estimates from above for the elliptic kernel are obtained by Krasovskiĭ in [15]. The higher order ‘heat’-kernel K(t, x, y) on ℝ^{n} has been considered e.g. by Davies and Barbatis in [2] and [9]. There for K(t, x, x), i.e. on the diagonal of ℝ^{n} × ℝ^{n}, they give even bounds from below. For a survey on recent results on higher order elliptic equations with emphasis on spectral theory we refer to [8].

1.1 Green function estimates

The starting point in [11] is an explicit expression for the Green function G_{m,n} of (1.1), which was discovered by Boggio [3, p. 126] already at the beginning of the twentieth century:

G_{m,n} (x, y) = k_{m,n} |x − y|^{2m−n} ∫_{1}^{Φ(x,y)} \frac{(v^{2} − 1)^{m−1}}{v^{n−1}} dv (1.2)

where k_{m,n} is a positive constant and

Φ (x, y) = \frac{|x |y| − \frac{y}{|y|} |}{|x − y|}.

Since |x |y| − \frac{y}{|y|} |^{2} − |x − y|^{2} = (1 − |x|^{2}) (1 − |y|^{2}) > 0 one directly finds that

Φ (x, y) > 1 in B^{2}. This result is singular in so far as for general higher order elliptic boundary value problems the Green function is not positive in an arbitrary domain. For a survey on positivity preserving properties of higher order elliptic equations we refer to [12].

In order to develop a perturbation theory of positivity for Dirichlet problems like (1.1) one uses Neumann series, the first term of which is the unperturbed Green function for (1.1) itself. This term is the only one which one is sure to be positive and which consequently has to dominate the other terms of the Neumann series.

As a first step in [11] we derived from Boggio’s formula the following two-sided estimate. There exists C_{m,n} > 0 such that

C_{m,n}^{−1}H_{m,n} (x, y) ≤ G_{m,n} (x, y) ≤ C_{m,n}H_{m,n} (x, y) for all x, y ∈ B, (1.3) where

H_{m,n} (x, y) = \{ \begin{matrix}(|x − y|^{2})^{m− \frac{1}{2} n} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m} & for m − \frac{1}{2}n < 0, \\ log (1 + (\frac{d(x)d(y)}{|x−y|^{2}})^{m}) & for m − \frac{1}{2}n = 0, \\ (d (x) d (y))^{m− \frac{1}{2} n} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{\frac{1}{2} n} & for m − \frac{1}{2}n > 0, \end{matrix} (1.4)

with a ∧ b = min {a, b} and d denoting the distance to the boundary ∂B :

d (x) = 1 − |x| . (1.5)

We also use the notation a ∨ b = max{a, b}.

As we mentioned, these estimates are appropriate for the polyharmonic Dirichlet problem on a ball. For second order elliptic equations, that is m = 1, one does not have this restriction. On arbitrary bounded domains the maximum principle implies a positivity preserving property and even the estimate in (1.3) holds in all bounded C^{1,1}-domains when d (x) is defined by

d (x) = d (x, ∂Ω) = inf {|x − x^{∗}|; x^{∗} ∈ ∂Ω} . (1.6)

Obviously one should replace C_{m,n} in (1.3) by some constant C_{m,Ω}. We just remark that if n = 2 then for any polyharmonic operator also smooth domains may be considered, which are sufficiently close to the disk in a suitable sense (depending on the order of the operator), see [10].

Since our proofs do not distinguish between B (for m ≥ 2) or a C^{1,1}-domain (for m = 1) we consider H_{m,n} defined on Ω^{2} with d as in (1.6).

1.2 Boundary and internal behavior

The estimate (1.4) can be understood with help of distinguishing the cases |x − y| ≤ \frac{1}{2} (d (x) ∨ d (y)) and |x − y| ≥ \frac{1}{2} (d (x) ∨ d (y)) . For later use we will formulate it as a lemma.

Lemma 1.1 Let d ( .) be as in (1.5) or (1.6). If |x − y| ≤ \frac{1}{2} (d (x) ∨ d (y)) then

\frac{1}{2}d (x) ≤ d (y) ≤ 2d (x)  and  1 ≤ \frac{d (x) d (y)}{|x − y|^{2}}.  (1.7)

If |x − y| ≥ \frac{1}{2} (d (x) ∨ d (y)) then

\frac{d (x)}{|x − y|} ≤ 2,  \frac{d (y)}{|x − y|} ≤ 2  and  \frac{d (x) d (y)}{|x − y|^{2}} ≤ 4.  (1.8)

Proof. Assume that |x − y| ≤ \frac{1}{2} (d (x) ∨ d (y)) and suppose d (y) ≤ d (x) . Let y^{∗} ∈ ∂Ω such that d (y) = |y − y^{∗}| . Then d (x) ≤ |x − y^{∗}| ≤ |x − y| + d (y) ≤ \frac{1}{2}d (x) + d (y) implying that d (x) ≤ 2d (y) . The second case is immediate. □

In the first case both x and y are not as close to the boundary as to each other, and the term, which describes the boundary behavior, may be neglected; the singularity |x − y|^{2m−n} (if n > 2m) is dominating. In the second case both x and y are closer to the boundary than to each other, and the boundary behavior is described by the term (d (x) d (y))^{m} |x − y|^{−n}.

1.3 Results

The second step in [11] in order to obtain perturbation results for positivity consists in estimating the modulus of higher terms in the Neumann series mentioned above by the first term. That means that we have to estimate iterated Green functions from above by the Green function itself. The estimates as needed for this purpose were derived in [11] without taking care e.g. of the fact that the singular character of iterated Green functions becomes weaker and weaker when the number of iterations increases.

A first goal of this paper is to derive precise estimates from above and below by the same simple function for iterated Green functions. Below, after presenting the main result we will also comment on a number of further applications.

For the iterated polyharmonic Dirichlet problem

\{ \begin{matrix}(−Δ)^{mk} u = f  in B, \\ u = \frac{∂}{∂n}u = ⋯ = \frac{∂^{m−1}}{∂n^{m−1}} u  on ∂B (−Δ)^{m} u = \frac{∂}{∂n} (−Δ)^{m} u = ⋯ = \frac{∂^{m−1}}{∂n^{m−1}} (−Δ)^{m} u ⋮ (−Δ)^{(k−1)m} u = \frac{∂}{∂n} (−Δ)^{(k−1)m} u = ⋯ = \frac{∂^{m−1}}{∂n^{m−1}} (−Δ)^{(k−1)m} u & \} \end{matrix} (1.9)

the Green function G_{m,n}^{(k)} satisfies

G_{m,n}^{(k)} (x, y) =∫_{B} … ∫_{B} G_{m,n} (x, z_{1}) G_{m,n} (z_{1}, z_{2}) … G_{m,n} (z_{k−1}, y) dz_{1} … dz_{k−1}.

Theorem 1.2 Let G_{m,n}^{(k)} be the Green function above. Then there exists C_{k,m,n} > 0 such that for x, y ∈ B

C_{k,m,n}^{−1} H_{m} (mk − \frac{1}{2}n; x, y) ≤ G_{m,n}^{(k)} (x, y) ≤ C_{k,m,n} H_{m} (mk − \frac{1}{2}n; x, y) (1.10)

where

H_{m} (θ; x, y) =\{ \begin{matrix}|x − y|^{2θ} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m} & for θ < 0, \\ log (1 + (\frac{d(x)d(y)}{|x−y|^{2}})^{m}) & for θ = 0, \\ (d (x) d (y) )^{θ} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m−θ} & for 0 < θ < m, \\ (d (x) d (y) )^{m} log (2 + \frac{1}{|x−y|^{2}+d(x)d(y)}) & for θ = m, \\ (d (x) d (y) )^{m} & for θ > m. \end{matrix}

If m = 1 these estimates hold on bounded C^{1,1}-domains Ω with the obvious replacements. There exists a positive constant C_{k,1,Ω} such that for x, y ∈ Ω

C_{k,1,Ω}^{−1} H_{1} (k − \frac{1}{2}n; x, y) ≤ G_{1,Ω}^{(k)} (x, y) ≤ C_{k,1,Ω} H_{1} (k − \frac{1}{2}n; x, y) . (1.11)

This theorem will be proved in Section 3.

As far as only estimates from above for the singular behaviour of iterated Green functions are concerned, this partial result is well known and goes back to Hadamard (see e.g. [17, p. 39]). Here the crucial point is to give twosided estimates in particular for the boundary behaviour, resp. the interplay of singular and boundary behaviour.

Defining the operator G by

G(f)(x) = ∫_{Ω} G_{m,n}(x, y)f(y)dy, (1.12)

and hence G^{k}(f)(x) = ∫_{Ω} G_{m,n}^{(k)}(x, y)f(y)dy, we find the following scheme.

[figure: |x − y|^{2θ} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}} )^{m}; log (1 + ( \frac{d(x)d(y)}{|x−y|^{2}} )^{m}); (d (x) d (y))^{θ} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}} )^{m−θ}; (d (x) d (y))^{m} log (2 + \frac{1}{|x−y|^{2}+d(x)d(y)} ); (d (x) d (y))^{m}; 0; m; θ; G^{5}; G^{6}; G^{7}; G with 10m < n < 12m]

Remark 1.3 The dependence of C_{k,m,n} (resp. C_{k,m,Ω}) on k in Theorem 1.2 can be made more explicit. Let k_{0} be the smallest number such that mk_{0} − \frac{1}{2}n > m. Since there exist constants 0 < c_{φ_{1,m}} < C_{φ_{1,m}} (see [5]) such that

c_{φ_{1,m}}φ_{1,m} (x) ≤ d (x)^{m} ≤ C_{φ_{1,m}}φ_{1,m} (x) ,

where φ_{1,m} is a suitably normalized positive first eigenfunction of (−Δ)^{m} under Dirichlet boundary conditions, one concludes for k ≥ k_{0} from

G_{m,n}^{(k)} (x, y) = \underbrace{∫_{B} … ∫_{B}}_{(k − k_{0})-times}G_{m,n} (x, z_{1}) G_{m,n} (z_{1}, z_{2}) … G_{m,n} (z_{k−k_{0}−1}, z_{k−k_{0}}) ×
× G_{m,n}^{(k_{0})} (z_{k−k_{0}}, y) dz_{1} … dz_{k−k_{0}}

that

C̃_{m,n}^{−1} λ_{1,m}^{k−k_{0}} (d (x) d (y) )^{m} ≤ G_{m,n}^{(k)} (x, y) ≤ C̃_{m,n} λ_{1,m}^{k−k_{0}} (d (x) d (y) )^{m}

with C̃_{m,n} = C_{k_{0},m,n} \frac{C_{φ_{1,m}}}{c_{φ_{1,m}}} .

Remark 1.4 Let H_{m} (Θ; x, y) be as in Theorem 1.2. If Θ ≥ θ then there exists a constant c_{θ}^{Θ} ∈ ℝ^{+} such that

H_{m} (Θ; x, y) ≤ c_{θ}^{Θ} H_{m} (θ; x, y) for all x, y ∈ Ω. (1.13)

Moreover, if (1.13) holds for some constant c_{θ}^{Θ} ∈ ℝ^{+}, then θ ≤ Θ or m < Θ.

As a first immediate consequence of Theorem 1.2 and the preceding remark we have the so called ‘3-G-theorems’ from [11]:

Corollary 1.5 There exists a constant Ĉ_{m,n} > 0, such that for x, z ∈ B

∫_{B} G_{m,n}(x, y) G_{m,n}(y, z) dy ≤ Ĉ_{m,n}G_{m,n}(x, z).

In the remainder we will use the following symbols for the sake of notation.

Notation 1.6 Let f, g be nonnegative functions on Ω (resp. Ω^{2}, Ω^{3}). We say f ⪯ g on Ω (resp. Ω^{2}, Ω^{3}) if there exists a constant c > 0, such that

f (x) ≤ c g (x) for all x ∈ Ω (resp. Ω^{2}, Ω^{3}).

We say f ∼ g on Ω if f ⪯ g and g ⪯ f on Ω (resp. Ω^{2}, Ω^{3}).

In what follows the constants in the estimates, which appear implicitly by the use of the symbols ∼ and ⪯, depend on the space dimension n (resp. the domain Ω), the order m of the differential operator and on the number k of iterations.

2 Applications

Another important motivation to find optimal estimates is the use of these estimates in necessary and sufficient conditions for uniform anti-maximum-principles to hold. The idea to use Green function estimates for anti-maximum-principles goes back to a paper by Takáč [19].

Anti-maximum-principles concern the resolvent for boundary value problems like (1.9), when the resolvent parameter λ is beyond the first eigenvalue. Usually for sufficiently regular fixed right hand side f there exists a small λ-interval where f ≥ 0 implies u ≤ 0. If this interval does not depend on f the anti-maximum-principle is called uniform.

The estimates obtained here allow us to solve an open problem from [5]. This application will be treated in a separate paper [13].

2.1 Application to coupled elliptic systems

As studied in [16], noncooperatively coupled elliptic systems may still satisfy some positivity preserving property. For example for the system

−Δu = f − ε^{2}v in Ω,
−Δv = u  in Ω,
u = v = 0  on ∂Ω,

it holds true that f ≥ 0 implies (u, v) ≥ 0 when the operator G and G − εG^{2} are positive and ∑_{k=0}^{∞} (εG)^{4k} converges. The result follows from v = Gu and

u = (I + εG) ∑_{k=0}^{∞} (εG)^{4k} (G − εG^{2}) f. (2.1)

Here G is as in (1.12) with m = 1. In more involved systems, see [16, Theorem 6.3], it was necessary for proving this sign preserving property to have for k large that

G^{(k)} (x, y) ∼ G^{(k+1)} (x, y) , (2.2)

where we denote G^{(k)}(x, y) = G_{1,Ω}^{(k)}(x, y). Then, instead of considering an infinite series in the solution operator for u, it is possible to restrict oneself to a finite number of terms. Indeed in [16, page 272] it is proven that (2.2) holds for k ≥ k_{n} = [\frac{n+2}{2}] + 1 with [ ·] the entier function. In fact it is shown that on Ω × Ω the following ‘ordering’ exists:

G ⪰ G^{(2)} ⪰ G^{(3)} ⪰ ⋯ ⪰ G^{(k_{n}−1)} ⪰ G^{(k_{n})} ∼ G^{(k_{n}+1)} ∼ G^{(k_{n}+2)} ∼ ….

It was conjectured in that paper that k_{n} is optimal in the sense that G^{(k_{n}−1)} ≁ G^{(k_{n})}.

This conjecture holds true since from Theorem 1.2 and Remark 1.4 it follows that (2.2) holds if and only if k− \frac{1}{2}n > 1. Notice that for k ∈ ℕ this bound can be rewritten as k ≥ [\frac{n+2}{2}] + 1.

2.2 Application to Brownian motion

We assume that n ≥ 2. Let y ∈ Ω and let {X (t)}_{t≥0} denote a G (·, y)–conditioned Brownian motion in Ω\ \{y\} , normalized for Δ instead of \frac{1}{2}Δ. See [4, Chapter 5]. This Brownian motion is conditioned to be killed at exiting Ω\ {y} . Since x ↦ G (x, y) is harmonic in Ω\ {y} with G (x, y) = 0 for x ∈ ∂Ω it will be killed almost surely at y. The lifetime is denoted by τ_{Ω\{y}}. Denoting by ℙ^{x} and 𝔼^{x} the probability respectively expectation for Brownian motion starting at x ∈ Ω, that is ℙ^{x} [X (0) = x] = 1, the expected lifetime 𝔼_{y}^{x} [τ_{Ω\{y}}] of the G (·, y)-conditioned Brownian motion equals

𝔼_{y}^{x} [τ_{Ω\{y}}] : = 𝔼_{G(·,y)}^{x} [τ_{Ω\{y}}] = − ∫_{t=0}^{∞} t dℙ_{y}^{x} [τ_{Ω\{y}} > t]
= ∫_{t=0}^{∞} ℙ_{y}^{x} [τ_{Ω\\{y\}} > t] dt = ∫_{t=0}^{∞} ∫_{z∈Ω} \frac{G (z, y)}{G (x, y)}p (t, x, z) dz dt
= ∫_{z∈Ω} \frac{G (z, y)}{G (x, y)} ∫_{t=0}^{∞} p (t, x, z) dt dz
= ∫_{z∈Ω} \frac{G (z, y)}{G (x, y)} G (x, z) dz = \frac{G^{(2)} (x, y)}{G (x, y)}.

Here p (t, x, z) is the kernel for the diffusion equation on Ω with Dirichlet boundary conditions.

A similar expression holds for the expectation of the moments of τ_{Ω\{y}}.

Lemma 2.1 Denoting G^{(k)} (x, y) = G_{1,Ω}^{(k)} (x, y) with Ω as above one has:

𝔼_{y}^{x} [τ_{Ω\\{y\}}^{k}] = k!\frac{G^{(k+1)} (x, y)}{G (x, y)}.

Remark 2.2 𝔼_{y}^{x} [τ_{Ω\{y}}^{k}] is the expectation of the kth moment of the lifetime of G( . , y)–conditioned Brownian motion that starts in x and stays inside Ω until it is killed when reaching y.

Proof. For k = 1 it is just the result mentioned before. A repeated integration by parts shows

𝔼_{y}^{x} [τ_{Ω\{y}}^{k}] = − ∫_{t=0}^{∞} t^{k} dℙ_{y}^{x} [τ_{Ω\{y}} > t]

= k! ∫_{t_{1}=0}^{∞} ∫_{t_{2}=t_{1}}^{∞} … ∫_{t_{k}=t_{k−1}}^{∞} ℙ_{y}^{x} [τ_{Ω\{y}} > t_{k}] dt_{1}dt_{2} … dt_{k}

= k! ∫_{t_{1}=0}^{∞} ∫_{t_{2}=t_{1}}^{∞} … ∫_{t_{k}=t_{k−1}}^{∞} ∫_{z_{1}∈Ω} \frac{G (z_{1}, y)}{G (x, y)}p (t_{k}, x, z_{1}) dz_{1} dt_{1}dt_{2} … dt_{k}

= \frac{k!}{G (x, y)} ∫_{t_{1}=0}^{∞} ∫_{t_{2}=0}^{∞} … ∫_{t_{k}=0}^{∞} ∫_{z_{1}∈Ω} p (t_{1} + t_{2} + ⋯ + t_{k}, x, z_{1}) ×

× G (z_{1}, y) dz_{1} dt_{1}dt_{2} … dt_{k}. (2.3)

By Chapman-Kolmogorov one obtains

p (t_{1} + t_{2} + ⋯ + t_{k}, x, z_{1})

= ∫_{z_{k}∈Ω} … ∫_{z_{2}∈Ω} p (t_{k}, x, z_{k}) p (t_{k−1}, z_{k}, z_{k−1}) … p (t_{1}, z_{2}, z_{1}) dz_{2} … dz_{k} and hence with Fubini-Tonelli

(2.3) = \frac{k!}{G (x, y)} ∫_{z_{k}∈Ω} … ∫_{z_{1}∈Ω} G (x, z_{k}) G (z_{k}, z_{k−1}) ⋯ ×
× G (z_{2}, z_{1}) G (z_{1}, y) dz_{1}dz_{2} … dz_{k}
= k!\frac{G^{(k+1)} (x, y)}{G (x, y)}.
□

As a consequence of Theorem 1.2 we obtain the following estimates for 𝔼_{y}^{x} [τ_{Ω\{y}}^{k}]. This result improves and optimizes related ones in [18]. Proposition 2.3 Let Ω be a bounded C^{1,1}-domain in ℝ^{n} and let 𝔼_{y}^{x} [τ_{Ω\{y}}^{k}] be as above with k ∈ ℕ^{+}.

1. If n ≥ 3 then 𝔼_{y}^{x} [τ_{Ω\{y}}^{k}] ∼ …

for x, y ∈ Ω :  for x ∈ Ω, y ∈ ∂Ω :
for k < \frac{n−2}{2} : ⋯ ∼ |x − y|^{2k}  ∼ |x − y|^{2k} ,
for k = \frac{n−2}{2} : ⋯ ∼ |x − y|^{n−2} log (2 + \frac{d(x)d(y)}{|x−y|^{2}}) ∼ |x − y|^{n−2} ,
for k = \frac{n−1}{2} : ⋯ ∼ |x − y|^{n−1} (1 + \frac{d(x)d(y)}{|x−y|^{2}})^{\frac{1}{2}}  ∼ |x − y|^{n−1} ,
for k = \frac{n}{2} :  ⋯ ∼ |x − y|^{n} (1 + \frac{d(x)d(y)}{|x−y|^{2}}) log (2 + \frac{1}{|x−y|^{2}+d(x)d(y)})
∼ |x − y|^{n} log (2 + \frac{1}{|x−y|^{2}} ) ,
for k > \frac{n}{2} : ⋯ ∼ |x − y|^{n} (1 + \frac{d(x)d(y)}{|x−y|^{2}}) ∼ |x − y|^{n} .

2. If n = 2 then 𝔼_{y}^{x} [τ_{Ω\{y}}^{k}] ∼ …

for x, y ∈ Ω :  for x ∈ Ω, y ∈ ∂Ω :
for k = 1 :  ⋯ ∼ (|x − y|^{2} + d (x) d (y)) \begin{matrix}log & (2 + \frac{1}{|x−y|^{2}+d(x)d(y)}) \\ log & (2 + \frac{d(x)d(y)}{|x−y|^{2}}) \end{matrix}
∼ |x − y|^{2} log (2 + \frac{1}{|x−y|^{2}} ) ,
for k > 1 :  ⋯ ∼ (|x − y|^{2} + d (x) d (y)) \frac{1}{log (2 + \frac{d(x)d(y)}{|x−y|^{2}})}
∼ |x − y|^{2} .

The constants in the two-sided estimates denoted by ∼ depend on Ω and k. Remark 2.4 For y ∈ ∂Ω one recovers the Brownian motion in Ω that is conditioned to exit Ω at y.

Remark 2.5 The results in this proposition give local estimates for the lifetime with the explicit dependence on d (x) , d (y) and |x − y| . Global estimates have been studied intensively. See the book by Chung and Zhao [4]. For bounded Lipschitz domains there exists C_{Ω} < ∞ such that for all x ∈ Ω and y ∈ Ω̄ the estimate 𝔼_{y}^{x} [τ_{Ω\{y}}] ≤ C_{Ω} holds. For n = 2 Cranston and McConnell [6] even proved that C_{Ω} = C · m (Ω) with C an absolute constant and m (Ω) the Lebesgue measure of Ω.

Proof of Proposition 2.3. The proof is rather straightforward by using the expression in Lemma 2.1, Lemma 3.4 below and the estimates for G^{(k)} (x, y) of Theorem 1.2. Consider for example the case n ≥ 3 and k = \frac{n−1}{2}. Then

\frac{G^{(k+1)} (x, y)}{G (x, y)} ∼ \begin{matrix}(d (x) d (y) ) & \frac{1}{2} & (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}}) & \frac{1}{2} \\ |x − y| & 2−n & (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}}) \end{matrix}

∼ |x − y|^{n−1} (\frac{d(x)d(y)}{|x−y|^{2}})^{\frac{1}{2}} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{− \frac{1}{2}}
∼ |x − y|^{n−1} (1 + \frac{d(x)d(y)}{|x−y|^{2}})^{\frac{1}{2}} ,

where in the last step the cases \frac{d(x)d(y)}{|x−y|^{2}} ≥ 1 and \frac{d(x)d(y)}{|x−y|^{2}} ≤ 1 were distinguished. □

3 Proofs

The main estimates that we need will be stated in the following lemmata. We will split the estimates according to three types for G_{m,n} = G_{m,n}^{(1)} over three lemmata which cover respectively the cases n > 2m, n = 2m and n < 2m. The principal terms contain logarithms when \frac{n}{2m} = k and \frac{n}{2m} = k − 1, hence in dimensions n that are a multiple of 2m.

Lemma 3.1 Let n, m ∈ ℕ^{+} with 2m < n. Suppose that α, δ ∈ ℕ^{+} with 1 ≤ α ≤ n − 2m and 1 ≤ δ < 2m. Then on Ω^{2} holds:

1. ∫_{y∈Ω} |x − y|^{−α} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m} |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ∼

∼ \{ \begin{matrix}|x − z|^{2m−α} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m} & if & α > 2m, \\ log (1 + (\frac{d(x)d(z)}{|x−z|^{2}})^{m}) & if & α = 2m, \\ (d (x) d (z) )^{m− \frac{1}{2} α} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{\frac{1}{2} α} & if & α < 2m. \end{matrix}  (3.1)

2.

∫_{y∈Ω} (d (x) d (y) )^{\frac{1}{2} δ} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m− \frac{1}{2} δ} |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ∼
∼ (d (x) d (z) )^{m} .  (3.2)

3.

∫_{y∈Ω} log (1 + (\frac{d(x)d(y)}{|x−y|^{2}})^{m}) |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ∼
∼ (d (x) d (z) )^{m} log (2 + \frac{1}{|x−z|^{2}+d(x)d(z)}) .  (3.3)

4. ∫_{y∈Ω} (d (x) d (y) )^{m} log (2 + \frac{1}{|x−y|^{2}+d(x)d(y)}) |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ∼

∼ (d (x) d (z) )^{m} .  (3.4)

Lemma 3.2 Let n = 2m ∈ ℕ^{+}. Then:

1.

∫_{y∈Ω} log (1 + (\frac{d(x)d(y)}{|x−y|^{2}})^{m}) log (1 + (\frac{d(y)d(z)}{|y−z|^{2}})^{m}) dy ∼
∼ (d (x) d (z) )^{m} log (2 + \frac{1}{|x−z|^{2}+d(x)d(z)}) .  (3.5)

2.

∫_{y∈Ω} (d (x) d (y) )^{m} log (2 + \frac{1}{|x−y|^{2}+d(x)d(y)}) log (1 + (\frac{d(y)d(z)}{|y−z|^{2}})^{m}) dy ∼
∼ (d (x) d (z) )^{m} .  (3.6)

Lemma 3.3 Let m, n ∈ ℕ^{+} with 2m > n. Suppose that δ ∈ ℕ^{+} with 1 ≤ δ < 2m. Then

∫_{y∈Ω} (d (x) d (y) )^{\frac{1}{2} δ} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m− \frac{1}{2} δ} (d (y) d (z) )^{m− \frac{1}{2} n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{\frac{1}{2} n} dy ∼

∼ (d (x) d (z) )^{m} .  (3.7)

Proof of Theorem 1.2. The results stated in Theorem 1.2 immediately follow from the estimates in the lemmata above. Indeed if k = 1 the estimates follow for m, n ∈ ℕ^{+} from (1.4) since H_{m} (m − \frac{1}{2}n; x, y) = H_{m,n} (x, y) . For k > 1 one should note that the results of the previous three lemmata can be condensed to

∫_{Ω} H_{m} (θ; x, y) H_{m} (m − \frac{1}{2}n; y, z) dy ∼ H_{m} (θ + m; x, z)

for all bounded θ of the form θ = ℓm − \frac{1}{2}n, ℓ ∈ ℕ^{+}. Repeating this step (k − 1) times the statement of Theorem 1.2 follows. □

First we recall and modify some estimates from [11].

Lemma 3.4 Let γ ∈ ℕ^{+}. For all x, y, z ∈ Ω the following holds:

i.  (1 ∧ \frac{d(x)}{|x−y|}) (1 ∧ \frac{d(y)}{|x−y|}) ∼ (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}}) ,
ii.  (1 ∧ \frac{d(x)}{|x−y|}) ∼ (1 ∧ \frac{d(x)}{d(y)} ∧ \frac{d(x)}{|x−y|}) ,
iii. (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}}) (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}}) ⪯ (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}}) ,
iv.  log (1 + (\frac{d(x)d(y)}{|x−y|^{2}})^{γ}) ∼ (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{γ−1} log (1 + \frac{d(x)d(y)}{|x−y|^{2}}) ,
v.  log (1 + (\frac{d(x)d(y)}{|x−y|^{2}})^{γ}) ∼ (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{γ} log (2 + \frac{d(x)}{|x−y|}) ,
vi.  log (2 + (\frac{d(x)d(y)}{|x−y|^{2}})^{γ}) ∼ log (2 + \frac{d(x)}{|x−y|}) .

Remark 3.5 We will also use the obvious estimate (1 ∧ s) ⪯ log (1 + s) for s ∈ [0, ∞) .

Proof. For i and ii, respectively iii see Lemma 3.2 and Lemma 4.3 of [11]. For the last three we distinguish two cases according to Lemma 1.1. Assume that |x − y| ≤ \frac{1}{2} (d (x) ∨ d (y)) and by using (1.7) and log (1 + s) ∼ log (2 + s) ∼ log (1 + s^{γ}) ∼ log (2 + s^{γ}) for s ∈ [1, ∞) the estimates iv, v and vi follow. Next assume that |x − y| ≥ \frac{1}{2} (d (x) ∨ d (y)) . Using (1.8) and log (1 + s) ∼ s respectively log (2 + s) ∼ 1 for s ∈ [0, 4^{γ}] we obtain the estimates iv, v and vi also in the latter case. □

The proofs of Lemmata 3.1, 3.2 and 3.3 will be split into two parts, the estimates from above, respectively those from below.

3.1 Estimates from above

The lemmata will be proven in several steps. For the estimates from above we will split the domain of integration Ω ∋ y in three parts, Ω = O_{x} ∪ O_{z} ∪ R, which are defined by

O_{x} = B_{\frac{2}{3}}_{|x−z|} (x) ∩ Ω, O_{z} = B_{\frac{2}{3}}_{|x−z|} (z) ∩ Ω and R = Ω\ (O_{x} ∪ O_{z}) .

[figure: Ω; z; x]

A similar distinction was made in [17, Lemma 4.1.1].

Iterated Green Functions 13 3.1.1 On O_{x} Note that for y ∈ O_{x} we have

|z − y| ∼ |z − x| . (3.8)

• Lemma 3.1-1. Using Lemma 3.4.iii and (3.8) we find ∫_{y∈O_{x}} |x − y|^{−α} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m} |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ⪯

⪯ (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m} ∫_{y∈O_{x}} |x − y|^{−α} |y − z|^{2m−n} dy
∼ |z − x|^{2m−n} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m} ∫_{y∈O_{x}} |x − y|^{−α} dy
∼ |x − z|^{2m−α} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m}

⪯ \{ \begin{matrix}|x − z|^{2m−α} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m} & if α > 2m; \\ (d (x) d (z) )^{m− \frac{1}{2} α} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{\frac{1}{2} α} & if α ≤ 2m. \end{matrix}

• Lemma 3.1-2. Using Lemma 3.4.iii, i, ii and (3.8) we obtain ∫_{y∈O_{x}} (d (x) d (y) )^{\frac{1}{2} δ} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m− \frac{1}{2} δ} |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ⪯

⪯ d (x)^{\frac{1}{2} δ} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m− \frac{1}{2} δ} |x − z|^{2m−n} ∫_{y∈O_{x}} d (y)^{\frac{1}{2} δ} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{\frac{1}{2} δ} dy
⪯ d (x)^{\frac{1}{2} δ} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m− \frac{1}{2} δ} |x − z|^{2m−n} ∫_{y∈O_{x}} d (y)^{\frac{1}{2} δ} (1 ∧ \frac{d(z)}{|y−z|})^{\frac{1}{2} δ} dy
⪯ d (x)^{\frac{1}{2} δ} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m− \frac{1}{2} δ} |x − z|^{2m−n} ∫_{y∈O_{x}} d (y)^{\frac{1}{2} δ} (\frac{d(z)}{d(y)})^{\frac{1}{2} δ} dy
⪯ d (x)^{\frac{1}{2} δ} d (z)^{\frac{1}{2} δ} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m− \frac{1}{2} δ} |x − z|^{2m}
⪯ (d (x) d (z) )^{m} .

• Lemma 3.1-3. First we need the following auxiliary result:

∫_{y∈O_{x}} |x − y|^{β} log (2 + \frac{s}{|x−y|}) dy ⪯ |x − z|^{β+n} log (2 + \frac{s}{|x−z|}) , (3.9)

14 H.-Ch. Grunau, G. Sweers for any s > 0 and β > −n. The constant in (3.9) only depends on β + n > 0. Indeed one has with τ > 0 :

∫_{r=0}^{τ} log (2 + \frac{s}{r}) r^{β+n−1}dr = [\frac{r^{β+n}}{β + n} log (2 + \frac{s}{r})]_{0}^{τ} + ∫_{r=0}^{τ} \frac{r^{β+n−1}}{β + n} \frac{s}{2r + s}dr
≤ \frac{1}{β+n} (τ^{β+n} log (2 + \frac{s}{τ} ) + \frac{1}{β+n}τ^{β+n})
≤ (\frac{1}{β+n} + \frac{2}{(β+n)^{2}} ) τ^{β+n} log (2 + \frac{s}{τ} ) .  (3.10)

Next we employ Lemma 3.4.v, iii, vi and (3.8)

∫_{y∈O_{x}} log (1 + (\frac{d(x)d(y)}{|x−y|^{2}})^{m}) |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ⪯

⪯ ∫_{y∈O_{x}} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m} log (2 + \frac{d(x)}{|x−y|}) |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy
⪯ (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m} |x − z|^{2m−n} ∫_{y∈O_{x}} log (2 + \frac{d(x)}{|x−y|}) dy
⪯ (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m} |x − z|^{2m} log (2 + \frac{d(x)}{|x−z|})
⪯ (|x − z|^{2} ∧ d (x) d (z) )^{m} log (2 + (\frac{d(x)d(z)}{|x−z|^{2}})^{m})
⪯ (d (x) d (z) )^{m} .

In the last step we distinguished the cases as in Lemma 1.1.

• Lemma 3.1-4. By Lemma 3.4.i, d (y) ⪯ 1, again (3.8) and (3.9):

∫_{y∈O_{x}} (d (x) d (y) )^{m} log (2 + \frac{1}{|x−y|^{2}+d(x)d(y)}) |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ⪯

⪯ (d (x) d (z) )^{m} |x − z|^{m−n} ∫_{y∈O_{x}} log (2 + \frac{1}{|x−y|}) dy
⪯ (d (x) d (z) )^{m} |x − z|^{m−n} |x − z|^{n} log (2 + \frac{1}{|x−z|})
⪯ (d (x) d (z) )^{m} .

• Lemma 3.2-1. We have n = 2m. Suppose |x − z| ≤ \frac{1}{2} (d (x) ∨ d (z) ) . Then d (x) ∼ d (z) and with (3.8) and (3.9)

∫_{y∈O_{x}} log (1 + (\frac{d(x)d(y)}{|x−y|^{2}})^{m}) log (1 + (\frac{d(y)d(z)}{|y−z|^{2}})^{m}) dy ⪯

⪯ ∫_{y∈O_{x}} (1 ∧ \frac{d(x)}{|x−y|})^{m} log (2 + \frac{d(x)}{|x−y|}) (1 ∧ \frac{d(z)}{|y−z|})^{m} log (2 + \frac{d(z)}{|y−z|}) dy
⪯ log (2 + \frac{d(z)}{|x−z|}) ∫_{y∈O_{x}} log (2 + \frac{d(x)}{|x−y|}) dy
⪯ log (2 + \frac{d(z)}{|x−z|}) |x − z|^{n} log (2 + \frac{d(x)}{|x−z|})
⪯ \frac{d(z)}{|x−z|} |x − z|^{2m} \frac{d(x)}{|x−z|}
⪯ (d (x) d (z) )^{m} .

If |x − z| ≥ \frac{1}{2} (d (x) ∨ d (z) ) then

∫_{y∈O_{x}} log (1 + (\frac{d(x)d(y)}{|x−y|^{2}})^{m}) log (1 + (\frac{d(y)d(z)}{|y−z|^{2}})^{m}) dy ⪯

⪯ ∫_{y∈O_{x}} (1 ∧ \frac{d(x)}{|x−y|})^{m} log (2 + \frac{d(x)}{|x−y|}) (1 ∧ \frac{d(z)}{|y−z|})^{m} log (2 + \frac{d(z)}{|y−z|}) dy
⪯ ∫_{y∈O_{x}} (1 ∧ \frac{d(x)}{|x−y|})^{m} log (2 + \frac{d(x)}{|x−y|}) (1 ∧ \frac{d(z)}{|x−z|})^{m} log (2 + \frac{d(z)}{|x−z|}) dy
⪯ (\frac{d(z)}{|x−z|})^{m} ∫_{y∈O_{x}} (1 ∧ \frac{d(x)}{|x−y|})^{m} log (2 + \frac{d(x)}{|x−y|}) dy

⪯ (\frac{d(z)}{|x−z|})^{m} (∫_{0}^{\frac{1}{4} d(x)} log (2 + \frac{d(x)}{r}) r^{n−1}dr +

+ d (x)^{m} ∫_{\frac{1}{4} d(x)}^{|x−z|} r^{−m}r^{n−1}dr) , (3.11)

and using (3.10) we get

(3.11) ⪯ (\frac{d(z)}{|x−z|})^{m} (d (x)^{n} log 6 + d (x)^{m} |x − z|^{m} ) ⪯ (d (x) d (z) )^{m} .

• Lemma 3.2-2. By Lemma 3.4.v, d (y) ⪯ 1, (3.8), (3.9) and |x − z| ⪯ 1 :

∫_{y∈O_{x}} (d (x) d (y) )^{m} log (2 + \frac{1}{|x−y|^{2}+d(x)d(y)}) log (1 + (\frac{d(y)d(z)}{|y−z|^{2}})^{m}) dy ⪯

⪯ d (x)^{m} ∫_{y∈O_{x}} log (2 + \frac{1}{|x−y|^{2}} ) (\frac{d(z)}{|x−z|})^{m} log (2 + \frac{1}{|x−z|^{2}} ) dy
⪯ (d (x) d (z) )^{m} log (2 + \frac{1}{|x−z|^{2}} ) |x − z|^{−m} ∫_{y∈O_{x}} log (2 + \frac{1}{|x−y|^{2}} ) dy
⪯ (d (x) d (z) )^{m} (log (2 + \frac{1}{|x−z|^{2}} ))^{2} |x − z|^{n−m}
⪯ (d (x) d (z) )^{m} ,

16 H.-Ch. Grunau, G. Sweers where we used that n − m = m > 0.

• Lemma 3.3. We have to estimate

∫_{y∈O_{x}} (d (x) d (y) )^{\frac{1}{2} δ} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m− \frac{1}{2} δ} (d (y) d (z) )^{m− \frac{1}{2} n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{\frac{1}{2} n} dy

= d (x)^{\frac{1}{2} δ} d (z)^{m− \frac{1}{2} n} ×
× ∫_{y∈O_{x}} d (y)^{m− \frac{1}{2} (n−δ)} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m− \frac{1}{2} δ} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{\frac{1}{2} n} dy. (3.12)

If δ ≥ n then Lemma 3.4.ii implies

(3.12) ⪯ d (x)^{\frac{1}{2} δ} d (z)^{m− \frac{1}{2} n} ∫_{y∈O_{x}} d (y)^{m+ \frac{1}{2} (δ−n)} (\frac{d(x)}{d(y)})^{m− \frac{1}{2} δ} (\frac{d(z)}{d(y)})^{\frac{1}{2} n} dy
∼ (d (x) d (z) )^{m} ∫_{y∈O_{x}} d (y)^{δ−n} dy
⪯ (d (x) d (z) )^{m} |x − z|^{n} ⪯ (d (x) d (z) )^{m} .

If δ < n then

(3.12) ⪯ d (x)^{\frac{1}{2} δ} d (z)^{m− \frac{1}{2} n} ∫_{y∈O_{x}} d (y)^{m− \frac{1}{2} (n−δ)} (\frac{d(x)}{d(y)})^{m− \frac{1}{2} n} ×
× (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{^{\frac{1}{2}} (n−δ)} (\frac{d(z)}{d(y)})^{\frac{1}{2} δ} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{^{\frac{1}{2}} (n−δ)} dy
⪯ (d (x) d (z) )^{m− \frac{1}{2} (n−δ)} ∫_{y∈O_{x}} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{^{\frac{1}{2}} (n−δ)} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{^{\frac{1}{2}} (n−δ)} dy
⪯ (d (x) d (z) )^{m− \frac{1}{2} (n−δ)} ∫_{y∈O_{x}} (\frac{d(x)}{|x−y|})^{^{\frac{1}{2}} (n−δ)} (\frac{d(z)}{|y−z|})^{^{\frac{1}{2}} (n−δ)} dy
⪯ (d (x) d (z) )^{m} |x − z|^{− \frac{1}{2} (n−δ)} ∫_{0}^{|x−z|} r^{\frac{1}{2} (δ−n)+n−1}dr
⪯ (d (x) d (z) )^{m} |x − z|^{δ} ⪯ (d (x) d (z) )^{m} .

3.1.2 On O_{z} Notice that y ∈ O_{z} implies

|x − y| ∼ |x − z| . (3.13)

• Lemma 3.1-1. The proof is similar as on O_{x}. Using Lemma 3.4.iii we find

∫_{y∈O_{z}} |x − y|^{−α} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m} |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ⪯  (3.14)

⪯ (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m} ∫_{y∈O_{z}} |x − y|^{−α} |y − z|^{2m−n} dy
∼ |z − x|^{−α} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m} ∫_{y∈O_{z}} |y − z|^{2m−n} dy
∼ |x − z|^{2m−α} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m}

⪯ \{ \begin{matrix}|x − z|^{2m−α} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m} & if α > 2m; \\ (d (x) d (z) )^{m− \frac{1}{2} α} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{\frac{1}{2} α} & if α ≤ 2m. \end{matrix}

• Lemma 3.1-2. Using Lemma 3.4.iii, i and ii we conclude in an almost similar way as for O_{x} :

∫_{y∈O_{z}} (d (x) d (y) )^{\frac{1}{2} δ} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m− \frac{1}{2} δ} |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ⪯

⪯ d (x)^{\frac{1}{2} δ} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m− \frac{1}{2} δ} ∫_{y∈O_{z}} d (y)^{\frac{1}{2} δ} |y − z|^{2m−n} (\frac{d(z)}{d(y)})^{\frac{1}{2} δ} dy
⪯ (d (x) d (z) )^{\frac{1}{2} δ} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m− \frac{1}{2} δ} |x − z|^{2m}
⪯ (d (x) d (z) )^{m} .

• Lemma 3.1-3. Using Lemma 3.4.v and iii we obtain similarly as on O_{x} using here (3.13).

∫_{y∈O_{z}} log (1 + (\frac{d(x)d(y)}{|x−y|^{2}})^{m}) |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ⪯

⪯ (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m} log (2 + \frac{d(x)}{|x−z|}) ∫_{y∈O_{z}} |y − z|^{2m−n} dy
⪯ (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m} log (2 + (\frac{d(x)d(z)}{|x−z|^{2}})^{m}) |x − z|^{2m}
⪯ (d (x) d (z) )^{m} .

• Lemma 3.1-4. With (3.13) and d (y) ⪯ 1

∫_{y∈O_{z}} (d (x) d (y) )^{m} log (2 + \frac{1}{|x−y|^{2}+d(x)d(y)}) |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ⪯

⪯ (d (x) d (z) )^{m} log (2 + \frac{1}{|x−z|^{2}} ) ∫_{y∈O_{z}} |y − z|^{m−n} dy
⪯ (d (x) d (z) )^{m} log (2 + \frac{1}{|x−z|^{2}} ) |x − z|^{m}
⪯ (d (x) d (z) )^{m} .

• Lemma 3.2-1. This statement is symmetric in x and z and hence the same result as on O_{x} holds.

• Lemma 3.2-2. Again we observe (3.13), (3.9) and |x − z| ⪯ 1:

∫_{y∈O_{z}} (d (x) d (y) )^{m} log (2 + \frac{1}{|x−y|^{2}+d(x)d(y)}) log (1 + (\frac{d(y)d(z)}{|y−z|^{2}})^{m}) dy ⪯

⪯ (d (x) d (z) )^{m} log (2 + \frac{1}{|x−z|^{2}} ) ∫_{y∈O_{z}} |y − z|^{−m} log (2 + \frac{1}{|y−z|}) dy
⪯ (d (x) d (z) )^{m} (log (2 + \frac{1}{|x−z|^{2}} ))^{2} |x − z|^{m}
⪯ (d (x) d (z) )^{m} .

• Lemma 3.3. Similarly as for O_{x} we get

∫_{y∈O_{z}} (d (x) d (y) )^{\frac{1}{2} δ} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m− \frac{1}{2} δ} ×
× (d (y) d (z) )^{m− \frac{1}{2} n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{\frac{1}{2} n} dy ⪯  (3.15)

If δ ≥ n then Lemma 3.4.ii implies

(3.15) ⪯ d (x)^{\frac{1}{2} δ} d (z)^{m− \frac{1}{2} n} ∫_{y∈O_{z}} d (y)^{m+ \frac{1}{2} (δ−n)} (\frac{d(x)}{d(y)})^{m− \frac{1}{2} δ} (\frac{d(z)}{d(y)})^{\frac{1}{2} n} dy
⪯ (d (x) d (z) )^{m} ∫_{y∈O_{z}} d (y)^{δ−n} dy
⪯ (d (x) d (z) )^{m} |x − z|^{n} ⪯ (d (x) d (z) )^{m} .

If δ < n then by (3.13)

(3.15) ⪯ d (x)^{\frac{1}{2} δ} d (z)^{m− \frac{1}{2} n} ∫_{y∈O_{z}} d (y)^{m− \frac{1}{2} (n−δ)} (\frac{d(x)}{d(y)})^{m− \frac{1}{2} n} (\frac{d(x)}{|x−y|})^{^{\frac{1}{2}} (n−δ)} ×
× (\frac{d(z)}{d(y)})^{\frac{1}{2} δ} (\frac{d(z)}{|y−z|})^{^{\frac{1}{2}} (n−δ)} dy
⪯ (d (x) d (z) )^{m} ∫_{y∈O_{z}} |x − y|^{− \frac{1}{2} (n−δ)} |y − z|^{− \frac{1}{2} (n−δ)} dy

⪯ (d (x) d (z) )^{m} |x − z|^{− \frac{1}{2} (n−δ)} ∫_{0}^{|x−z|} r^{\frac{1}{2} (δ−n)+n−1}dr
⪯ (d (x) d (z) )^{m} |x − z|^{δ} ⪯ (d (x) d (z) )^{m} .

3.1.3 On R

Here we have |y − z| ≥ \frac{2}{3} |x − z| implying that |x − y| ≤ |x − z| + |z − y| ≤ \frac{5}{2} |y − z| . By symmetry |y − x| ≥ \frac{2}{3} |x − z| implies |y − z| ≤ \frac{5}{2} |x − y| and hence

|x − y| ∼ |y − z| . (3.16)

Iterated Green Functions 19 Let us denote

D = 2 diam (Ω) . (3.17)

• Lemma 3.1-1. We have by (3.16)

∫_{y∈R} |x − y|^{−α} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m} |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ⪯

⪯ ∫_{y∈R} |x − y|^{2m−n−α} (1 ∧ \frac{d(x)}{|x−y|})^{m} (1 ∧ \frac{d(z)}{|x−y|})^{m} dy
⪯ ∫_{|x−z|}^{D} r^{2m−α−1} (1 ∧ \frac{d(x)}{r})^{m} (1 ∧ \frac{d(z)}{r})^{m} dr.  (3.18)

Notice that we may replace \frac{2}{3} |x − z| by |x − z|, as D = 2 diam (Ω) .

Assuming that |x − z|^{2} ≤ d (x) d (z) we find

(3.18) ⪯ ∫_{|x−z|}^{√\overline{d(x)d(z)}} r^{2m−α−1}dr + (d (x) d (z) )^{m} ∫_{√\overline{d(x)d(z)}}^{D} r^{−α−1}dr
⪯ \{ \begin{matrix}|x − z|^{2m−α} & + & (d (x) d (z) )^{m− \frac{1}{2} α} & if α > 2m; \\ log ^{\frac{\sqrt{d(x)d(z)}}{|x−z|}} & + & 1 & if α = 2m; \\ (d (x) d (z) )^{m− \frac{1}{2} α} & + & (d (x) d (z) )^{m− \frac{1}{2} α} & if α < 2m; \end{matrix}
⪯ \{ \begin{matrix}|x − z|^{2m−α} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m} & if α > 2m; \\ log (1 + (\frac{d(x)d(z)}{|x−z|^{2}})^{m}) & if α = 2m; \\ (d (x) d (z) )^{m− \frac{1}{2} α} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{\frac{1}{2} α} & if α < 2m. \end{matrix}  (3.19)

If |x − z|^{2} ≥ d (x) d (z) we estimate

(3.18) ⪯ (d (x) d (z) )^{m} ∫_{|x−z|}^{D} r^{−α−1}dr
⪯ (d (x) d (z) )^{m} |x − z|^{−α} ⪯ (3.19) .

• Lemma 3.1-2. By Lemma 3.4.i and ii we find, again using (3.16), that ∫_{y∈R} (d (x) d (y) )^{\frac{1}{2} δ} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m− \frac{1}{2} δ} |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ⪯

⪯ d (x)^{\frac{1}{2} δ}∫_{y∈R} d (y)^{\frac{1}{2} δ} (1 ∧ \frac{d(x)}{|x−y|})^{m− \frac{1}{2} δ} |x − y|^{2m−n} (1 ∧ \frac{d(z)}{|x−y|})^{m− \frac{1}{2} δ} (\frac{d(z)}{d(y)})^{\frac{1}{2} δ} dy
⪯ (d (x) d (z) )^{\frac{1}{2} δ} ∫_{\frac{2}{3} |x−z|}^{D} (1 ∧ \frac{d(x)}{r})^{m− \frac{1}{2} δ} r^{2m−1} (1 ∧ \frac{d(z)}{r})^{m− \frac{1}{2} δ} dr
⪯ (d (x) d (z) )^{m} ∫_{\frac{2}{3} |x−z|}^{D} r^{δ−1}dr ⪯ (d (x) d (z) )^{m} .  (3.20)

• Lemma 3.1-3. With Lemma 3.4.i and v and (3.16)

∫_{y∈R} log (1 + (\frac{d(x)d(y)}{|x−y|^{2}})^{m}) |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ⪯

⪯ ∫_{y∈R} (1 ∧ \frac{d(x)}{|x−y|})^{m} log (2 + \frac{d(x)}{|x−y|}) |x − y|^{2m−n} (1 ∧ \frac{d(z)}{|x−y|})^{m} dy
⪯ ∫_{\frac{2}{3} |x−z|}^{D} (1 ∧ \frac{d(x)}{r})^{m} log (2 + \frac{d(x)}{r}) r^{2m−1} (1 ∧ \frac{d(z)}{r})^{m} dr.  (3.21)

Assuming that |x − z| ≤ \frac{1}{2} (d (x) ∨ d (z) ) we may use as in (1.7) that d (x) ∼ d (z) to proceed by

(3.21) ⪯ ∫\begin{matrix}√ & \overline{d(x)d(z)} \\ \frac{2}{3} & |x−z| \end{matrix} log (2 + ^{\frac{\sqrt{d(x)d(z)}}{r}}) r^{2m−1}dr
+ (d (x) d (z) )^{m} ∫_{√\overline{d(x)d(z)}}^{D} log (2 + ^{\frac{\sqrt{d(x)d(z)}}{r}}) r^{−1}dr
⪯ (d (x) d (z) )^{m} (∫_{0}^{1} log (2 + \frac{1}{s}) s^{2m−1}ds + ∫1^{\frac{D}{\sqrt{d(x)d(z)}}} s^{−1}ds)
⪯ (d (x) d (z) )^{m} (1 + log (\frac{D}{\sqrt{d(x)d(z)}}))
⪯ (d (x) d (z) )^{m} log (2 + \frac{1}{|x−z|^{2}+d(x)d(z)}) .

If |x − z| ≥ \frac{1}{2} (d (x) ∨ d (z) ) we have r = |x − y| ≥ \frac{2}{3} |x − z| ≥ \frac{1}{3}d (x) and hence

(3.21) ⪯ (d (x) d (z) )^{m} ∫_{\frac{2}{3} |x−z|}^{D} r^{−1}dr ⪯ (d (x) d (z) )^{m} log (\frac{3}{2} \frac{D}{|x−z|})
⪯ (d (x) d (z) )^{m} log (2 + \frac{1}{|x−z|^{2}+d(x)d(z)}) .

• Lemma 3.1-4. With d (y) ⪯ 1 and (3.16)

∫_{y∈R} (d (x) d (y) )^{m} log (2 + \frac{1}{|x−y|^{2}+d(x)d(y)}) |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ⪯

⪯ (d (x) d (z) )^{m} ∫_{y∈R} log (2 + \frac{1}{|x−y|^{2}} ) |x − y|^{m−n} dy
⪯ (d (x) d (z) )^{m} ∫_{0}^{D} log (2 + \frac{1}{r^{2}} ) r^{m−1}dr
⪯ (d (x) d (z) )^{m} .

• Lemma 3.2-1. We have n = 2m and need to estimate

∫_{y∈R} log (1 + (\frac{d(x)d(y)}{|x−y|^{2}})^{m}) log (1 + (\frac{d(y)d(z)}{|y−z|^{2}})^{m}) dy ⪯

⪯ ∫_{\frac{2}{3} |x−z|}^{D} (1 ∧ \frac{d(x)}{r})^{m} log (2 + \frac{d(x)}{r}) (1 ∧ \frac{d(z)}{r})^{m} log (2 + \frac{d(z)}{r}) r^{n−1}dr. (3.22)

Assuming that |x − z| ≤ \frac{1}{2} (d (x) ∨ d (z) ) we take from (1.7) that d (x) ∼ d (z) in order to find

(3.22) ⪯ ∫\begin{matrix}√ & \overline{d(x)d(z)} \\ \frac{2}{3} & |x−z| \end{matrix} log (2 + ^{\frac{\sqrt{d(x)d(z)}}{r}})^{2} r^{n−1}dr
+ (d (x) d (z) )^{m} ∫_{√\overline{d(x)d(z)}}^{D} log (2 + ^{\frac{\sqrt{d(x)d(z)}}{r}})^{2} r^{−1}dr
⪯ (d (x) d (z) )^{m} (∫_{0}^{1} log (2 + \frac{1}{s})^{2} s^{2m−1}ds + ∫1^{\frac{D}{\sqrt{d(x)d(z)}}} s^{−1}ds)
⪯ (d (x) d (z) )^{m} (1 + log (\frac{D}{\sqrt{d(x)d(z)}}))
⪯ (d (x) d (z) )^{m} log (2 + \frac{1}{|x−z|^{2}+d(x)d(z)}) .

If |x − z| ≥ \frac{1}{2} (d (x) ∨ d (z) ) we have as above |x − y| ≥ \frac{1}{3}d (x) , |x − y| ≥ \frac{1}{3}d (z) and consequently

(3.22) ⪯ (d (x) d (z) )^{m} ∫_{\frac{2}{3} |x−z|}^{D} r^{−1}dr ⪯ (d (x) d (z) )^{m} log (\frac{3}{2} \frac{D}{|x−z|})
⪯ (d (x) d (z) )^{m} log (2 + \frac{1}{|x−z|^{2}+d(x)d(z)}) .

• Lemma 3.2-2. Since 2m = n it follows that

∫_{y∈R} (d (x) d (y) )^{m} log (2 + \frac{1}{|x−y|^{2}+d(x)d(y)}) log (1 + (\frac{d(y)d(z)}{|y−z|^{2}})^{m}) dy ⪯

⪯ (d (x) d (z) )^{m} ∫_{y∈R} log (2 + \frac{1}{|x−y|^{2}} ) |y − z|^{−m} log (2 + \frac{1}{|y−z|^{2}} ) dy
⪯ (d (x) d (z) )^{m} ∫_{\frac{2}{3} |x−z|}^{D} (log (2 + \frac{1}{r^{2}} ))^{2} r^{m−1}dr
⪯ (d (x) d (z) )^{m} .

• Lemma 3.3. As in the first steps of the case on O_{x} we have

∫_{y∈R}(d (x) d (y) )^{\frac{1}{2} δ} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m− \frac{1}{2} δ} (d (y) d (z) )^{m− \frac{1}{2} n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{\frac{1}{2} n}dy ⪯

⪯ d (x)^{\frac{1}{2} δ} d (z)^{m− \frac{1}{2} n} ∫_{y∈R} d (y)^{m− \frac{1}{2} n+ \frac{1}{2} δ} (1 ∧ \frac{d(x)}{|x−y|})^{m− \frac{1}{2} δ} (1 ∧ \frac{d(z)}{|y−z|})^{\frac{1}{2} n} dy. (3.23)

If δ ≥ n then by Lemma 3.4.ii

(3.23) ⪯ d (x)^{\frac{1}{2} δ} d (z)^{m− \frac{1}{2} n} ∫_{y∈R} d (y)^{m− \frac{1}{2} n+ \frac{1}{2} δ} (\frac{d(x)}{d(y)})^{m− \frac{1}{2} δ} (\frac{d(z)}{d(y)})^{\frac{1}{2} n} dy
⪯ (d (x) d (z) )^{m} ∫_{y∈R} d (y)^{δ−n} dy ⪯ (d (x) d (z) )^{m} .

If δ < n then by Lemma 3.4.ii and (3.16) we also have

(3.23) ⪯ d (x)^{\frac{1}{2} δ} d (z)^{m− \frac{1}{2} n} ∫_{y∈R} d (y)^{m− \frac{1}{2} n+ \frac{1}{2} δ} (\frac{d(x)}{|x−y|})^{^{\frac{1}{2}} (n−δ)} (\frac{d(x)}{d(y)})^{m− \frac{1}{2} n} ×
× (\frac{d(z)}{|y−z|})^{^{\frac{1}{2}} (n−δ)} (\frac{d(z)}{d(y)})^{\frac{1}{2} δ} dy
⪯ (d (x) d (z) )^{m} ∫_{\frac{2}{3} |x−z|}^{D} r^{δ−n+n−1}dr ⪯ (d (x) d (z) )^{m} ,

which completes the proofs for the estimates from above.

3.2 Estimates from below

Let us start with the simplest estimates. Since H_{m} (θ; x, y) ⪰ (d (x) d (y) )^{m} for all θ we immediately find that

∫_{Ω} H_{m} (θ; x, y) H_{m} (m − \frac{1}{2}n; y, z) dy
⪰ (d (x) d (z) )^{m} ∫_{Ω} d (y)^{2m} dy ⪰ (d (x) d (z) )^{m} .

This implies the estimate from below for Lemma 3.1-2 and 4, Lemma 3.2-2 and Lemma 3.3. In the remaining cases, i.e. for θ ≤ 0, we will exploit the following idea. The fact that the H_{m} (θ; x, y) are positive allows us to identify for each of these cases a region R_{i} ⊂ Ω such that

∫_{R_{i}} H_{m} (θ; x, y) H_{m} (m − \frac{1}{2}n; y, z) dy ⪰ H_{m} (m + θ; x, z) . (3.24)

The main difficulty will be to choose R_{i} such that it is large enough to find (3.24) but nice enough to simplify the formula.

3.2.1 Some auxiliary geometrical results

We will start with some geometric properties of C^{1,1}-domains. Let us first define a family of cones for ε ∈ (0, 1) and e ∈ ℝ^{n}\ {0} :

K (ε, e) = {x ∈ ℝ^{n}; x · e > ε |x| |e|} . (3.25)

Without proof we state a lemma related to a cone condition, observe that bounded C^{1,1}-domains satisfy uniform interior sphere conditions.

Lemma 3.6 For every bounded C^{1,1}-domain Ω and ε ∈ (0, 1) there is a constant C = C (ε, Ω) > 0 such that:

1. if x ∈ ∂Ω and n_{x} denotes the inward normal at x, then

x + (K (ε, n_{x}) ∩ B_{C} (0) ) ⊂ Ω;

2. if x ∈ Ω and x^{∗} ∈ ∂Ω is such that d (x) = |x − x^{∗}| , then

x + (K (ε, x − x^{∗}) ∩ B_{C} (0) ) ⊂ Ω.

All C (ε, Ω) satisfy C (ε, Ω) ≤ \frac{1}{2}diam (Ω) .

Let x ≠ z ∈ Ω and fix a closest point x^{∗} ∈ ∂Ω. We take ε = \frac{1}{8}, set R = \frac{1}{3}C (\frac{1}{8}, Ω) and define

R_{x}^{z} = x + (K (\frac{1}{4}, n_{x^{∗}}) \K (\frac{3}{4}, z − x)) ∩ B_{R} (0) . (3.26)

Two crucial estimates that hold true on these subsets are given in the next lemma.

[figure: x^{∗}; x; possible z-positions; 3R; R; Ω; R_{x}^{z}; ^{\sqrt{\frac{63}{64}}} |x − x^{∗}|]

Lemma 3.7 Let R be as above. Then there is a constant C_{1} = C_{1} (Ω, R) such that for all x ≠ z ∈ Ω and y ∈ R_{x}^{z} the following holds:

d (y) ≥ C_{1} (|x − y| + d (x) ) and |z − y| ≥ C_{1} |x − y| .

Proof. We use the following definition of the distance between two sets A, B ⊂ ℝ^{n}:

d (A, B) = inf {|a − b|; a ∈ A, b ∈ B} . (3.27)

Let y ∈ R_{x}^{z}. By Lemma 3.6 O := x + (K (\frac{1}{8}, x − x^{∗}) ∩ B_{2R} (0) ) ⊂ Ω. Since y ∈ O ⊂ Ω ⊂ ℝ^{n} it follows that

d (y) ≥ d (y, ∂O) + d (O, ∂Ω) . (3.28)

The distance between O and ∂Ω satisfies

d (O, ∂Ω) ≥ min (\sqrt{\frac{63}{64}}d (x) , R) ≥ min (\sqrt{\frac{63}{64}}d (x) , R\frac{d (x)}{D}) ≥ \frac{R}{D} d (x) , (3.29)

where we have set

D = 2 diam (Ω) . (3.30)

To estimate d (y, ∂O) let ỹ ∈ ∂O be such that |y − ỹ| = d (∂O, y) . One has either ỹ ∈ ∂B_{2R} (x) or ỹ ∈ ∂ (x + K (\frac{1}{8}, x − x^{∗})) . In the first case

|y − ỹ| ≥ |x − ỹ| − |y − x| ≥ 2R − R = R ≥ |x − y| .

In the second case let y_{p} denote the projection of y on the hyperplane through x perpendicular to x − x^{∗}, that is

y_{p} = y − \frac{(x − x^{∗}) · (y − x)}{|x − x^{∗}|^{2}} (x − x^{∗}) ,

and let ỹ_{p} denote the intersection of ∂ (x + K (\frac{1}{8}, x − x^{∗})) with the line through y and y_{p}. Then

|y − ỹ| = \sqrt{\frac{63}{64}} |y − ỹ_{p}|

and

|y − ỹ_{p}| ≥ |y − y_{p}| − |y_{p} − ỹ_{p}| = |\frac{(x − x^{∗}) · (y − x)}{|x − x^{∗}|}| − \frac{1}{8} |x − ỹ_{p}|
≥ \frac{1}{4} |y − x| − \frac{1}{8} |y − x| = \frac{1}{8} |y − x| .

One concludes that

d (y, ∂O) = |y − ỹ| ≥ min (\frac{\sqrt{63}}{64}, 1) |x − y| . (3.31)

The first claim follows from (3.28), (3.29) and (3.31).

For the second claim notice that y ∉ x + K (\frac{3}{4}, z − x) implies

|x − y|^{2} + |x − z|^{2} = |y − z|^{2} + 2 (x − z) · (x − y)
≤ |y − z|^{2} + \frac{3}{2} |x − z| |x − y|
≤ |y − z|^{2} + |x − z|^{2} + \frac{9}{16} |x − y|^{2}

and hence \frac{7}{16} |x − y|^{2} ≤ |y − z|^{2} . □

The following two sets will be used:

R_{1} = B_{\frac{1}{2}}_{|x−z|} (x) . (3.32) and

R_{2} = R_{x}^{z}\B_{\frac{R}{D}} _{|x−z|} (x) . (3.33) The first will only be considered under the assumption |x − z| ≤ \frac{1}{2} (d (x) ∨ d (z)) . We will list some estimates for later use. Case I Inequalities for y ∈ R_{1} assuming that |x − z| ≤ \frac{1}{2} (d (x) ∨ d (z)) :

i. by assumptions and (1.7):  |x − y| ≤ \frac{1}{2} |x − z| ≤ \frac{1}{4} (d (x) ∨ d (z) ) ≤ \frac{1}{2}d (x) ,
ii. from the previous and (1.7): |y − z| ≤ |x − y| + |x − z| ≤ \frac{3}{2} |x − z| ≤ \frac{3}{2}d (z) ,
iii. since |y − x| ≤ \frac{1}{2}|x − z| :  |y − z| ≥ |x − z| − |x − y| ≥ \frac{1}{2} |x − z| ,
iv. from i:  d (y) ≥ d (x) − |x − y| ≥ |x − y| ,
v. and from i, (1.7) and ii:  d (y) ≥ d (x) − |x − y| ≥ \frac{1}{2}d (x) ≥ \frac{1}{2} |x − z| ≥
≥ \frac{1}{3} |y − z| .

A similar list for R_{2} : Case II Inequalities for y ∈ R_{2} :

i. by Lemma 3.7:  d (y) ≥ C_{1} |x − y| ,
ii. and also:  |y − z| ≥ C_{1} |x − y| ,
iii. since y ∉ B_{\frac{R}{D}} _{|x−z|} (x) : |y − z| ≤ |x − z| + |x − y| ≤ (\frac{D}{R} + 1) |x − y| ,
iv. from i and iii:  d (y) ≥ C_{1} |x − y| ≥ C_{1} \frac{R}{D+R} |y − z| .

3.2.2 Integral estimates when |x − z| ≤ \frac{1}{2} (d (x) ∨ d (z)) The estimates

\frac{1}{2}d (x) ≤ d (z) ≤ 2d (x) (3.34)

by Lemma 1.1 will be used throughout this section. First suppose that d (x) is bounded from below by \frac{1}{2}R. Hence d (z) ≥ \frac{1}{4}R by (3.34) and the proof of the estimates simplifies. The right hand sides of Lemma 3.1-1 for α < 2m, Lemma 3.1-3 and Lemma 3.2-1 reduce to (d (x) d (z))^{m−(α/2)} and (d (x) d (z))^{m} respectively. Left over is the case Lemma 3.1-1 for n − 2m ≥ α ≥ 2m. If α > 2m then we integrate over R_{1}. By the inequalities iv and i of Case I we find (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}}) ∼ 1 and ii, v imply (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}}) ∼ 1. These equivalence relations with ii, iii yield

∫_{y∈R_{1}} |x − y|^{−α} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m} |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ∼

∼ ∫_{y∈R_{1}} |x − y|^{−α} |x − z|^{2m−n} dy

∼ |x − z|^{2m−n} ∫_{0}^{\frac{1}{2} |x−z|} r^{n−α−1}dr ∼ |x − z|^{2m−α}
∼ |x − z|^{2m−α} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m} .  (3.35)

If α = 2m then integrating over R_{2} yields with i, iv, ii and iii of Case II, as d(x), d(z) are here assumed to be bounded from below:

∫_{y∈R_{2}} |x − y|^{−α} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m} |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ∼

∼ ∫_{y∈R_{2}} |x − y|^{−α} (1 ∧ \frac{d(x)}{|x−y|})^{m} |y − z|^{2m−n} (1 ∧ \frac{d(z)}{|y−z|})^{m} dy  (3.36)
∼ ∫_{R_{2}} |x − y|^{−α} |y − z|^{2m−n} dz ∼ ∫_{R_{2}} |x − y|^{−α} |x − y|^{2m−n} dz

∼ ∫_{\frac{R}{D} |x−z|}^{R} r^{−α+2m−1}dr ∼ log (\frac{D}{|x−z|}) ∼ log (1 + (\frac{d(x) d(z)}{|x−z|^{2}})^{m}) .

In the remaining part for the case |x − z| ≤ \frac{1}{2} (d (x) ∨ d (z)) we may restrict ourselves to d (x) , d (z) ≤ R.

• Lemma 3.1-1. If α > 2m, then we use again R_{1} = B_{\frac{1}{2}}_{|x−z|} (x) and proceed precisely as in (3.35).

If α ≤ 2m, then we integrate on R_{2}. It follows from the definition of R that \frac{R}{D} < \frac{1}{4}, see (3.30). Beginning as in (3.36) and using (3.34)

∫_{y∈R_{2}} |x − y|^{−α} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m} |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ∼

∼ ∫_{y∈R_{2}} |x − y|^{−α} (1 ∧ \frac{d(x)}{|x−y|})^{m} |y − z|^{2m−n} (1 ∧ \frac{d(z)}{|y−z|})^{m} dy
∼ ∫_{y∈R_{2}} |x − y|^{2m−α−n} (1 ∧ \frac{d(x)}{|x−y|})^{2m} dy

∼ ∫_{\frac{R}{D} |x−z|}^{d(x)} r^{2m−α−1}dr + d (x)^{2m} ∫_{d(x)}^{R} r^{−α−1}dr ⪰ ∫_{\frac{R}{D} |x−z|}^{d(x)} r^{2m−α−1}dr. (3.37)

If α = 2m, then

(3.37) ∼ log (\frac{D}{R} ) + log (\frac{d(x)}{|x−z|}) ∼ log (1 + \frac{d(x)d(z)}{|x−z|^{2}})
∼ log (1 + (\frac{d(x)d(z)}{|x−z|^{2}})^{m}) .

If α < 2m, then we use \frac{R}{D} |x − z| < \frac{1}{4} d(x) and obtain

(3.37) ∼ d (x)^{2m−α} ∼ (d (x) d (z) )^{m− \frac{1}{2} α} ∼ (d (x) d (z) )^{m− \frac{1}{2} α} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{\frac{1}{2} α} .

• Lemma 3.1-3. We will use the same set R_{2} from (3.33). By Lemma 3.4.v one estimates

∫_{y∈R_{2}} log (1 + (\frac{d(x)d(y)}{|x−y|^{2}})^{m}) |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ∼

∼ ∫_{y∈R_{2}} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m} log (2 + \frac{d(y)}{|x−y|}) |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy (3.38)

and continues with the estimates i, iv in Case II, (3.34), and ii, iii again in Case II to find

(3.38) ⪰ ∫_{y∈R_{2}} (1 ∧ \frac{d(x)}{|x−y|})^{m} |y − z|^{2m−n} (1 ∧ \frac{d(z)}{|y−z|})^{m} dy
∼ ∫_{y∈R_{2}} |x − y|^{2m−n} (1 ∧ \frac{d(x)}{|x−y|})^{2m} dy

∼ ∫_{\frac{R}{D} |x−z|}^{d(x)} r^{2m−1}dr + d (x)^{2m} ∫_{d(x)}^{R} r^{−1}dr
∼ d (x)^{2m} (1 + log \frac{R}{d(x)}) ∼ d (x)^{2m} log (2 + \frac{1}{d(x)})
∼ (d (x) d (z) )^{m} log (2 + \frac{1}{|x−z|^{2}+d(x)d(z)}) .

• Lemma 3.2-1. Here we have n = 2m. Again using the set R_{2} from (3.33) we obtain by Lemma 3.4.v that

∫_{y∈R_{2}} log (1 + (\frac{d(x)d(y)}{|x−y|^{2}})^{m}) log (1 + (\frac{d(y)d(z)}{|y−z|^{2}})^{m}) dy ∼

∼ ∫_{y∈R_{2}} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m} log (2 + \frac{d(x)}{|x−y|}) (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} log (2 + \frac{d(z)}{|y−z|}) dy. (3.39)

We again continue with the estimates i, iv in Case II, (3.34), and ii, iii in Case II to see that

(3.39) ⪰ ∫_{y∈R_{2}} (1 ∧ \frac{d(x)}{|x−y|})^{m} log (2 + \frac{d(x)}{|x−y|}) (1 ∧ \frac{d(z)}{|y−z|})^{m} log (2 + \frac{d(z)}{|y−z|}) dy
∼ ∫_{y∈R_{2}} (1 ∧ \frac{d(x)}{|x−y|})^{2m} (log (2 + \frac{d(x)}{|x−y|}))^{2} dy

∼ ∫_{\frac{R}{D} |x−z|}^{d(x)} (log (2 + \frac{d(x)}{r}))^{2} r^{2m−1}dr + d (x)^{2m} ∫_{d(x)}^{R} r^{−1}dr
⪰ d (x)^{2m} + d (x)^{2m} log (\frac{R}{d(x)})
∼ d (x)^{2m} log (2 + \frac{1}{d(x)})
∼ (d (x) d (z) )^{m} log (2 + \frac{1}{|x−z|^{2}+d(x)d(z)}) .

28 H.-Ch. Grunau, G. Sweers 3.2.3 Integral estimates when |x − z| ≥ \frac{1}{2} (d (x) ∨ d (z)) Now the second part of Lemma 1.1 applies and we find

d (x) ≤ 2 |x − z| and d (z) ≤ 2 |x − z| . Again we consider R_{2} = R_{x}^{z}\B_{\frac{R}{D}} _{|x−z|} (x) . In addition to the inequalities in Case II we have that

d (x) ≤ 2 |x − z| ≤ \frac{2D}{R} |x − y| and d (z) ≤ 2 |x − z| ≤ \frac{2D}{R} |x − y| . (3.40)

• Lemma 3.1-1. By the estimates i, iv in Case II, respectively the estimates ii, iii in Case II and finally (3.40) it follows that

∫_{y∈R_{2}} |x − y|^{−α} (1 ∧ \frac{d(x)d(y)}{|x−y|^{2}})^{m} |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ∼

∼ ∫_{y∈R_{2}} |x − y|^{−α} (1 ∧ \frac{d(x)}{|x−y|})^{m} |y − z|^{2m−n} (1 ∧ \frac{d(z)}{|y−z|})^{m} dy
∼ ∫_{y∈R_{2}} |x − y|^{2m−n−α} (1 ∧ \frac{d(x)}{|x−y|})^{m} (1 ∧ \frac{d(z)}{|x−y|})^{m} dy
∼ (d (x) d (z) )^{m} ∫_{y∈R_{2}} |x − y|^{−n−α} dy
∼ (d (x) d (z) )^{m} ∫_{\frac{R}{D} |x−z|}^{R} r^{−1−α}dr
∼ (d (x) d (z) )^{m} |x − z|^{−α} .  (3.41)

Distinguishing the three cases of α we conclude by \frac{d(x)d(z)}{|x−z|^{2}} ⪯ 1 that

(3.41) ∼ \{ \begin{matrix}if α > 2m : & |x − z|^{2m−α} (\frac{d(x)d(z)}{|x−z|^{2}})^{m} ∼ \\ & ∼ |x − z|^{2m−α} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{m}; \\ if α = 2m : & (\frac{d(x)d(z)}{|x−z|^{2}})^{m} ∼ log (1 + (\frac{d(x)d(z)}{|x−z|^{2}})^{m}); \\ if α < 2m : & (d (x) d (z) )^{m− \frac{1}{2} α} (\frac{d(x)d(z)}{|x−z|^{2}})^{\frac{1}{2} α} ∼ \\ & ∼ (d (x) d (z) )^{m− \frac{1}{2} α} (1 ∧ \frac{d(x)d(z)}{|x−z|^{2}})^{\frac{1}{2} α} . \end{matrix}

• Lemma 3.1-3. By similar steps as in the previous item we find ∫_{y∈R_{2}} log (1 + (\frac{d(x)d(y)}{|x−y|^{2}})^{m}) |y − z|^{2m−n} (1 ∧ \frac{d(y)d(z)}{|y−z|^{2}})^{m} dy ⪰

⪰ ∫_{y∈R_{2}} log (1 + (\frac{d(x)}{|x−y|})^{m}) |y − z|^{2m−n} (1 ∧ \frac{d(z)}{|y−z|})^{m} dy
∼ ∫_{y∈R_{2}} log (1 + (\frac{d(x)}{|x−y|})^{m}) |x − y|^{2m−n} (1 ∧ \frac{d(z)}{|x−y|})^{m} dy
∼ ∫_{y∈R_{2}} |x − y|^{2m−n} (\frac{d(x)}{|x−y|})^{m} (\frac{d(z)}{|x−y|})^{m} dy

∼ (d (x) d (z) )^{m} ∫_{\frac{R}{D} |x−z|}^{R} r^{−1}dr
∼ (d (x) d (z) )^{m} log (\frac{D}{|x−z|}) ∼ (d (x) d (z) )^{m} log (2 + \frac{1}{|x−z|^{2}} ) .  (3.42)

Using again d (x) d (z) ⪯ |x − z|^{2} we conclude that

(3.42) ∼ (d (x) d (z) )^{m} log (2 + \frac{1}{|x−z|^{2}+d(x)d(z)}) . (3.43)

• Lemma 3.2-1. Again by similar steps and as n = 2m

∫_{y∈R_{2}} log (1 + (\frac{d(x)d(y)}{|x−y|^{2}})^{m}) log (1 + (\frac{d(y)d(z)}{|y−z|^{2}})^{m}) dy ⪰

⪰ ∫_{y∈R_{2}} log (1 + (\frac{d(x)}{|x−y|})^{m}) log (1 + (\frac{d(z)}{|y−z|})^{m}) dy
∼ ∫_{y∈R_{2}} (\frac{d(x)}{|x−y|})^{m} (\frac{d(z)}{|x−y|})^{m} dy

∼ (d (x) d (z) )^{m} ∫_{\frac{R}{D} |x−z|}^{R} r^{−1}dr
∼ (d (x) d (z) )^{m} log (2 + \frac{1}{|x−z|^{2}+d(x)d(z)}) .
□

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