Annales Academiæ Scientiarum Fennicæ Series A. I. Mathematica Volumen 20, 1995, 3–26

ON INTEGRALS OF HARMONIC FUNCTIONS OVER ANNULI

B. Korenblum, P.J. Rippon and K. Samotij

State University of New York, Department of Mathematics, Albany NY12222, USA The Open University, Department of Pure Mathematics, Milton Keynes MK7 6AA, U.K. Technical University, Mathematical Institute, Wrocław, Poland

Abstract. We consider functions u which are harmonic in the unit ball B in R^{n} and satisfy

u(x) ≤ k(|x|), x ∈ B,

where k is a positive, increasing, continuous function on [0, 1) such that

(∗)  ∫_{0}^{1} \sqrt{\frac{k(r)}{1 − r}} dr < ∞.

We show that if η ∈ ∂B, 0 ≤ t_{1} < t_{2} ≤ π,

A_{η}(t_{1}, t_{2}) = {ξ ∈ ∂B : t_{1} < cos^{−1}(ξ.η) < t_{2}},

and σ denotes (n − 1)-dimensional measure on ∂B, then

lim_{r→1^{−}} ∫_{A_{η}(t_{1},t_{2})} u(rξ) dσ(ξ)

exists. Moreover, the growth condition (∗) is best possible. For n = 2, these results were proved by Hayman and Korenblum [2], using distortion theorems for conformal mappings.

1. Introduction

In [2] Hayman and Korenblum considered harmonic functions u in the unit disk {|z| < 1}, which vanish at the origin and satisfy the one-sided condition

u(z) ≤ k(|z|), |z| < 1,

where k is an increasing, positive function on [0, 1). They showed that if

(1.1)  ∫_{0}^{1} \sqrt{\frac{k(r)}{1 − r}} dr < ∞

1991 Mathematics Subject Classification: Primary 31B10, 31B25.

The first author acknowledges support from the National Science Foundation.

then, for each such function u, the following limits exist:

μ(α) = lim_{r→1^{−}} ∫_{0}^{α} u(re^{it}) dt, α ∈ R.

Moreover, the function μ is bounded and u has a generalized Riesz–Herglotz representation

(1.2)  u(z) = ∫_{0}^{2π} \frac{1 − |z|^{2}}{|z − e^{it}|^{2}} dμ(t),  |z| < 1,

where the integral is defined using integration by parts. More recently [5] Samotij showed that the function μ can have discontinuities of the first kind only.

In this paper we extend these results to harmonic functions in the unit ball

B = {x = (x_{1}, x_{2}, …, x_{n}) : |x| < 1}

of R^{n}, n ≥ 3, and we show (as was conjectured in [4]) that condition (1.1) plays exactly the same rôle as in the case n = 2. To be precise, let S = ∂B and consider the spherical annuli

(1.3)  A_{η}(t_{1}, t_{2}) = {ξ ∈ S : t_{1} < θ(ξ, η) < t_{2}},  η ∈ S, 0 ≤ t_{1} < t_{2} ≤ π,

where

(1.4)  θ(x, y) = cos^{−1}(\frac{x.y}{|x| |y|}),  x, y ∈ R^{n}\\{0\}.

To obtain an analogue of the representation (1.2) in B we need to show that the following limits exist:

lim_{r→1^{−}} ∫_{A_{η}(t_{1},t_{2})} u(rξ) dσ(ξ), η ∈ S, 0 < r < 1, 0 ≤ t_{1} < t_{2} ≤ π,

where σ denotes (n − 1)-dimensional measure on S. In [2] the existence of these limits, in the case n = 2, was obtained with the help of conformal mapping, but for n > 2 we are forced to work entirely with the Poisson integral for B. The crucial step in proving that these limits exist is the following one-sided estimate.

Theorem 1. Let k be a positive, increasing, continuous function on [0, 1) which satisfies

(1.5)  J = ∫_{0}^{1} \sqrt{\frac{k(r)}{1 − r}} dr < ∞.

Then there exists a continuous, increasing function κ on [0, π], with κ(0) = 0 and κ(π) ≤ C(n), such that, for each function u which is continuous on \overline{B} and harmonic in B, with u(0) = 0, and for each η ∈ S, the condition

(1.6)  u(x) ≤ \frac{k(|x|)}{sin^{n−2} θ(x, η)},  x ∈ B,

implies that

(1.7)  ∫_{A_{η}(t_{1},t_{2})} u(ξ) dσ(ξ) ≤ J^{2}κ(t_{2} − t_{1}),  0 ≤ t_{1} < t_{2} ≤ π.

A result of this type was obtained earlier by Samotij [5], with (1.1) replaced by the more restrictive assumption (for n ≥ 3):

∫_{0}^{1}(\frac{k(r)}{1 − r} )^{1−(1/n)} dr < ∞.

In [5] it was also shown how the estimate (1.7) leads to a representation like (1.2) (cf. also [3]). We can argue in exactly the same way here to obtain the following result.

Corollary. If k satisfies the assumptions of Theorem 1, u is harmonic in B with u(0) = 0, and

u(x) ≤ k(|x|), x ∈ B,

then

u_{η}(t) = lim_{r→1^{−}} ∫_{A_{η}(0,t)} u(rξ) dσ(ξ),

exists for each η ∈ S and 0 ≤ t ≤ π, and u can be represented in the form

u(x) = \frac{1}{ω_{n}} ∫_{0}^{π} \frac{1 − |x|^{2}}{(1 − 2|x| cos t + |x|^{2})^{n/2}} dux̃(t), x ∈ B,

where x̃ = x/|x|, ω_{n} = σ(S) and the integral is defined using integration by parts. Moreover, each function u_{η}, η ∈ S, has discontinuities of the first kind only and

u_{η}(t) = \frac{1}{2}(u_{η}(t+) + u_{η}(t−)), 0 < t < π.

In [2] it was shown that, for n = 2, condition (1.5) cannot be replaced in the corollary by any weaker condition. This is also true for n ≥ 3.

Theorem 2. If k is positive, increasing and continuous in [0, 1) and

(1.8)  ∫_{0}^{1} \sqrt{\frac{k(r)}{1 − r}} dr = ∞,

then, for each η ∈ S, there is a harmonic function u in B with u(0) = 0 and

(1.9)  u(x) ≤ k(|x|),  x ∈ B,

such that

(1.10)  lim_{r→1^{−}} ∫_{A_{η}(0,π/2)} u(rξ) dσ(ξ) = ∞.

The proofs are arranged as follows. In Section 2 we establish certain properties of the Poisson kernel which are needed in the proof of Theorem 1 and in Section 3 we define an auxiliary harmonic function v_{A} and estimate its behaviour on a certain surface Γ_{A}. The proofs of Theorem 1 and Theorem 2 are then given in Sections 4 and 5, respectively. We assume throughout, as we may, that η = (0, …, 0, 1). Also, we use the notation c(a, b, …) and C(a, b, …) to denote positive constants which depend only on the variables a, b, …, not necessarily the same on each occurrence. Finally, although we assume here that n ≥ 3, similar arguments apply if n = 2 and considerable simplification is possible in this case.

Acknowledgement. The authors thank the referee for very careful reading of this paper, and for many helpful suggestions.

2. Averaging the Poisson kernel

We write the Poisson kernel for B in the form

(2.1)  P (x, ξ) = \frac{1 − |x|^{2}}{|x − ξ|^{n}} = \frac{1 − |x|^{2}}{(1 − 2|x| cos θ(x, ξ) + |x|^{2})^{n/2}} ,

where x ∈ B and ξ ∈ S (see [1] for potential theory in R^{n}). With η = (0, … 0, 1) and

S(r, t) = {x : |x| = r, θ(x, η) = t}, 0 ≤ r < 1, 0 ≤ t ≤ π,

we introduce the averaged Poisson kernel

(2.2)  μ(r, s, t) = ∫_{S(1,t)} \frac{1 − r^{2}}{|x − ξ|^{n}} dσ̂(ξ),

where |x| = r, θ(x, η) = s and dσ̂ denotes normalized (n−2)-dimensional measure on S(1, t). Note that μ(r, s, t) also equals the average of P (·, ξ), where ξ = x/|x|, over S(r, t); see Figure 1.

We shall need the following estimates for μ(r, s, t).

Figure 1.

Lemma 1. If 0 ≤ r < 1 and 0 ≤ s, t ≤ π, then

\frac{c(n)(1 − r)}{d^{2}(d^{n−2} + sin^{n−2} s)} ≤ μ(r, s, t) ≤ \frac{C(n)(1 − r)}{d^{2}(d^{n−2} + sin^{n−2} s)},

where d = d(r, s, t) = (1 − 2r cos(s − t) + r^{2})^{1/2} denotes the distance from S(r, s) to S(1, t).

Proof. Without loss of generality we may take x = (r sin s, 0, …, 0, r cos s) in (2.2) and write ξ ∈ S(1, t) in the form

ξ = (ζ_{1} sin t, …, ζ_{n−1} sin t, cos t),

where ζ = (ζ_{1}, …, ζ_{n−1}, 0) ∈ S(1, π/2). Now

|x − ξ|^{2} = (r sin s − ζ_{1} sin t)^{2} + ζ_{2}^{2} sin^{2} t + ⋯ + ζ_{n−1}^{2} sin^{2} t + (r cos s − cos t)^{2}
= 1 − 2r(ζ_{1} sin s sin t + cos s cos t) + r^{2}
= d^{2} + 2r(1 − ζ_{1}) sin s sin t,

where d = (1 − 2r cos(s − t) + r^{2})^{1/2} . Thus

(2.3)  μ(r, s, t) = ∫_{S(1,π/2)} \frac{(1 − r^{2}) dσ̂(ζ)}{(d^{2} + 2r(1 − ζ_{1}) sin s sin t)^{n/2}} ,

where dσ̂ denotes normalized (n − 2)-dimensional measure on S(1, π/2), and hence

(2.4)

μ(r, s, t) = \frac{ω_{n−2}}{ω_{n−1}} ∫_{0}^{π} \frac{(1 − r^{2}) sin^{n−3} τ dτ}{(d^{2} + 2r sin s sin t(1 − cos τ))^{n/2}}

= \frac{ω_{n−2}}{ω_{n−1}} ∫_{0}^{π} \frac{(1 − r^{2}) sin^{n−3} τ dτ}{(d^{2} + 4r sin s sin t sin^{2}(τ/2))^{n/2}} .

Here

ω_{k} = \frac{2π^{k/2}}{Γ(k/2)}

denotes the (k − 1)-dimensional measure of the unit sphere in R^{k}.

To proceed further we need the estimates

(2.5) \frac{c(n)b^{n−2}}{d^{2}(d^{n−2} + a^{n−2}b^{n−2})} ≤ ∫_{0}^{b} \frac{τ^{n−3} dτ}{(d^{2} + a^{2}τ^{2})^{n/2}} ≤ \frac{C(n)b^{n−2}}{d^{2}(d^{n−2} + a^{n−2}b^{n−2})}.

Since

∫_{0}^{b} \frac{τ^{n−3} dτ}{(d^{2} + a^{2}τ^{2})^{n/2}} = \frac{1}{d^{2}a^{n−2}} ∫_{0}^{ab/d} \frac{θ^{n−3} dθ}{(1 + θ^{2})^{n/2}} ,

it is sufficient to prove that, for λ > 0,

\frac{c(n)λ^{n−2}}{1 + λ^{n−2}} ≤ ∫_{0}^{λ} \frac{θ^{n−3} dθ}{(1 + θ^{2})^{n/2}} ≤ \frac{C(n)λ^{n−2}}{1 + λ^{n−2}} ,

which is easily established by, for example, considering the cases 0 < λ ≤ 1 and λ > 1 separately.

Applying the right-hand side of (2.5) to (2.4), we obtain

μ(r, s, t) ≤ \frac{ω_{n−2}}{ω_{n−1}} ∫_{0}^{π} \frac{(1 − r^{2})τ^{n−3} dτ}{(d^{2} + ((2/π)^{2}r sin s sin t)τ^{2})^{n/2}}

≤ \frac{C(n)(1 − r)}{d^{2}(d^{n−2} + (r sin s sin t)^{(n−2)/2})}.

Now we suppose (as we may) that r ≥ \frac{1}{2} . Since

d^{2} = 1 − 2r cos(s − t) + r^{2} = (1 − r)^{2} + 4r sin^{2}((s − t)/2),

we deduce that

(2.6)  d^{2} ≥ \frac{4}{π^{2}} ((1 − r)^{2} + (s − t)^{2}),

and hence that

d^{n−2} + (r sin s sin t)^{(n−2)/2} ≥ c(n)((1 − r)^{n−2} + |s − t|^{n−2} + (sin s sin t)^{(n−2)/2}).

Now

|s − t| ≥ \frac{1}{2} sin s, if |s − t| ≥ \frac{1}{2} min\{s, π − s\},

and

sin t ≥ \frac{1}{2} sin s, if |s − t| ≤ \frac{1}{2} min\{s, π − s\},

so that

d^{n−2} + (r sin s sin t)^{(n−2)/2} ≥ c(n)(d^{n−2} + sin^{n−2} s).

This completes the proof of the upper estimate for μ(r, s, t).

The proof of the lower estimate is similar. On applying the left-hand side of (2.5) to (2.4), we obtain

μ(r, s, t) ≥ c(n) ∫_{0}^{π/2} \frac{(1 − r^{2})τ^{n−3} dτ}{(d^{2} + (sin s sin t)τ^{2})^{n/2}}

≥ \frac{c(n)(1 − r)}{d^{2}(d^{n−2} + (sin s sin t)^{(n−2)/2})}

≥ \frac{c(n)(1 − r)}{d^{2}(d^{n−2} + sin^{n−2} s + sin^{n−2} t)}.

The lower estimate for μ(r, s, t) now follows from the fact that

sin t ≤ sin s + |s − t| ≤ sin s + (π/2)d,

by (2.6). This completes the proof of Lemma 1.

We shall also need a certain monotonicity property of μ(r, s, t). To obtain this, we shall make use of a related property of the function

p(x, y) = \frac{2(x_{n} − 1)}{|x − y|^{n}}, x_{n} > 1, y_{n} = 1,

which is the Poisson kernel of the half-space {x_{n} > 1}. The average of p(·, y) over the set

\overline{S}(a, ϱ) = \{(\overline{x}, x_{n}) : |\overline{x}| = ϱ, x_{n} = a\}, a > 1, ϱ > 0,

where \overline{x} = (x_{1}, …, x_{n−1}), depends only on a, ϱ and σ = |\overline{y}|; we denote this average by \overline{μ}(a, σ, ϱ).

Lemma 2. For 0 < a − 1 < (σ − ϱ)/\sqrt{n − 1}, the average \overline{μ}(a, σ, ϱ) is an increasing function of both a and ϱ.

Proof. First note that

\frac{∂}{∂x_{n}} p(x, y) = 2\frac{|\overline{x} − \overline{y}|^{2} − (n − 1)(x_{n} − 1)^{2}}{(|\overline{x} − \overline{y}|^{2} + (x_{n} − 1)^{2})^{n/2+1}}

is positive and decreasing with x_{n}, for 0 < x_{n} − 1 ≤ |\overline{x} − \overline{y}|/\sqrt{n − 1}, so that p is increasing and concave with respect to x_{n}, for such values. Thus \overline{μ}(a, σ, ϱ) is increasing with respect to a, for 0 < a − 1 < (σ − ϱ)/\sqrt{n − 1}. Further, since p is harmonic in x, the concaveness of p with respect to x_{n} implies that, for each x_{n} > 1, p is subharmonic with respect to \overline{x} in

\{\overline{x} : 0 < x_{n} − 1 < |\overline{x} − \overline{y}|/\sqrt{n − 1} \}.

Hence \overline{μ}(a, σ, ϱ) is increasing with respect to ϱ if 0 < a − 1 < (σ − ϱ)/\sqrt{n − 1} and so Lemma 2 follows.

10 B. Korenblum, P.J. Rippon and K. Samotij

Figure 2.

We now relate p to P by means of an inversion in the sphere of radius 2 centred at −η = (0, …, 0, −1), which is pictured in Figure 2.

To do this, put

x^{∗} = −η + \frac{4}{|x + η|^{2}} (x + η).

Then

x_{n}^{∗} − 1 = \frac{2(1 − |x|^{2})}{|x + η|^{2}},

and, by similar triangles (or directly),

\frac{|x^{∗} − ξ^{∗}|}{|x − ξ|} = \frac{|x^{∗} + η|}{|ξ + η|} = \frac{4}{|ξ + η∥x + η|}.

Thus

(2.7)  \begin{aligned}p(x^{∗}, ξ^{∗}) & = \frac{2(x_{n}^{∗} − 1)}{|x^{∗} − ξ^{∗}|^{n}} = \frac{4(1 − |x|^{2})|ξ + η|^{n}|x + η|^{n}}{|x + η|^{2}|x − ξ|^{n}4^{n}} \\ & = \frac{|ξ + η|^{n}|x + η|^{n−2}}{4^{n−1}} P (x, ξ). \end{aligned}

It is easy to check that if S(r, t) maps to \overline{S}(a, ϱ) under this inversion, then

(2.8)  ϱ = \frac{4r sin t}{1 + 2r cos t + r^{2}} and a = \frac{4(1 + r cos t)}{1 + 2r cos t + r^{2}} − 1.

We shall need the following facts about a and ϱ.

Lemma 3. If a and ϱ are given by (2.8) then, for each r, 0 < r < 1 :

(i) a is an increasing function of t, for 0 ≤ t < π; (ii) ϱ is an increasing function of t, for 0 ≤ t ≤ π − \frac{1}{2}π(1 − r).

Proof. Part (i) follows immediately from

\frac{∂a}{∂t} = \frac{4r(1 − r^{2}) sin t}{(1 + 2r cos t + r^{2})^{2}} ≥ 0, 0 ≤ t < π.

To prove part (ii), we show that

\frac{∂ϱ}{∂t} = \frac{4r((1 + r^{2}) cos t + 2r)}{(1 + 2r cos t + r^{2})^{2}} ≥ 0, 0 ≤ t ≤ π − \frac{1}{2}π(1 − r).

If 0 < t ≤ π − \frac{1}{2}π(1 − r), then

(2.9)  cos t ≥ − cos(\frac{1}{2}π(1 − r)) = − sin(\frac{1}{2}πr) ≥ −\frac{2r}{1 + r^{2}} .

To prove the final inequality in (2.9), write r = tan(φ/2), where 0 ≤ φ < \frac{1}{2}π, and note that \frac{1}{2}πr ≤ φ, since

\frac{2r}{φ} = \frac{tan(φ/2)}{(φ/2)} ≤ \frac{tan(π/4)}{(π/4)} = \frac{4}{π} .

This completes the proof of Lemma 3.

The required monotonicity property of μ(r, s, t) is the following.

Lemma 4. If

(2.10)  r ∈ [1 − \frac{1}{4}π/\sqrt{n}, 1) and s ∈ [2\sqrt{n} (1 − r), π − 2\sqrt{n} (1 − r)],

then μ(r, s, t) is an increasing function of t on [0, s−2\sqrt{n}(1−r)] and a decreasing function of t on [s + 2\sqrt{n} (1 − r), π].

Proof. First note that, since μ(r, s, t) = μ(r, π − s, π − t), it is enough to show that μ(r, s, t) is an increasing function of t on [0, s − 2\sqrt{n} (1 − r)].

By (2.7) and the observation following (2.2),

μ(r, s, t) = \frac{4^{n−1}\overline{μ}(a, σ, ϱ)}{(2 + 2 cos s)^{n/2}(1 + 2r cos t + r^{2})^{(n−2)/2}} ,

where σ = |\overline{ξ^{∗}}| = 2 tan(\frac{1}{2}s) and ϱ, a are given by (2.8). Since 1 + 2r cos t + r^{2} decreases as t increases, it follows from Lemma 2 and Lemma 3 that μ(r, s, t) increases as t increases, provided that

0 < r < 1, 0 ≤ t ≤ π − \frac{1}{2}π(1 − r) and 0 < a − 1 < (σ − ϱ)/\sqrt{n − 1}.

Since 2\sqrt{n} ≥ \frac{1}{2}π, for n ≥ 3, the proof will be complete once we prove that if r and s satisfy (2.10) and 0 ≤ t ≤ s − 2\sqrt{n} (1 − r), then

0 < a − 1 < (σ − ϱ)/\sqrt{n − 1}.

But, by (2.8),

a − 1 = \frac{2(1 − r^{2})}{1 + 2r cos t + r^{2}}

and

σ − ϱ = 2 tan(\frac{1}{2}s) − \frac{4r sin t}{1 + 2r cos t + r^{2}} ,

so that, since \frac{1}{2} ≤ r < 1 and 0 ≤ t ≤ s − 2\sqrt{n} (1 − r), we have

σ − ϱ ≥ 2 tan(\frac{1}{2}s) − 2 tan(\frac{1}{2}t) ≥ sec^{2}(\frac{1}{2}t)(s − t)

= \frac{sec^{2}(\frac{1}{2}t)(s − t)((1 − r)^{2} + 4r cos^{2}(\frac{1}{2}t))}{1 + 2r cos t + r^{2}} ≥ \frac{4r(s − t)}{1 + 2r cos t + r^{2}}

≥ \frac{4\sqrt{n} (1 − r)}{1 + 2r cos t + r^{2}} ≥ \frac{2\sqrt{n} (1 − r^{2})}{1 + 2r cos t + r^{2}}

= \sqrt{n} (a − 1) > \sqrt{n − 1}(a − 1),

as required. This completes the proof of Lemma 4.

3. An auxiliary harmonic function

In proving Theorem 1, there is no loss of generality in supposing that the integral

J = ∫_{0}^{1} \sqrt{\frac{k(r)}{1 − r}} dr

takes a particular value, since the general case can be deduced by considering a suitable multiple of u (see the end of Section 4). It is convenient to assume in this section that

(3.1)  J ≤ \frac{1}{πn}.

Now define

f(h) = \sqrt{\frac{h}{k(1 − h)}}, 0 < h ≤ 1.

Since k is increasing, we have

(3.2) \frac{h}{f(h)} = \sqrt{hk(1 − h)} ≤ \frac{1}{2} ∫_{0}^{h} \sqrt{\frac{k(1 − s)}{s}} ds ≤ \frac{J}{2} ≤ \frac{1}{2πn},  0 < h ≤ 1,

by (3.1), and so f(h) ≥ 2πnh. Thus the inverse function

φ(t) = \{ \begin{matrix}f^{−1}(t), & 0 < t ≤ 2πn, \\ 0, & t = 0, \end{matrix}

satisfies

(3.3)  φ(t) ≤ \frac{t}{2πn},  0 ≤ t ≤ 2πn.

By definition

(3.4)  \frac{φ(t)}{t^{2}} = k(1 − φ(t)),  0 < t ≤ 2πn,

and, on integrating by parts, we obtain, for 0 < ε ≤ 2πn,

(3.5)  \begin{aligned}∫_{ε}^{2πn} \frac{φ(t)}{t^{2}} dt & = − ∫_{ε}^{2πn} φ(t) d(1/t) = −[\frac{φ(t)}{t}]_{ε}^{2πn} + ∫_{φ(ε)}^{φ(2πn)} \frac{1}{t} dφ \\ & = \frac{φ(ε)}{ε} − \frac{φ(2πn)}{2πn} + ∫_{φ(ε)}^{φ(2πn)} \sqrt{\frac{k(1 − h)}{h}} dh ≤ J, \end{aligned}

since φ(ε)/ε ⟶ 0 as ε ⟶ 0, by (3.2) and (3.3).

For a given annulus A = A_{η}(t_{1}, t_{2}), we now put

(3.6)  θ_{A}(x) = min{θ(x, η) − t_{1}, t_{2} − θ(x, η)},  x/|x| ∈ A,

and then

v_{A}(ξ) = \{ \begin{matrix}\frac{k(1 − φ(θ_{A}(ξ)))}{sin^{n−2} θ(ξ, η)}, & ξ ∈ A, \\ 0, & ξ ∈ S\A. \end{matrix}

The function v_{A} is integrable on S since, by (3.4) and (3.5),

(3.7)  ∫_{S} v_{A}(ξ) dσ(ξ) = 2ω_{n−1} ∫_{0}^{(t_{2}−t_{1})/2} \frac{φ(t)}{t^{2}} dt ≤ 2ω_{n−1}J.

Hence v_{A} can be extended, via the Poisson integral formula, to a harmonic function in B, which we also call v_{A}. The following lemma allows us to compare the values of v_{A}(x) and μ(|x|, θ(x, η), t_{i}), when x lies on the surface

(3.8)  Γ_{A} = {x : x/|x| ∈ A, |x| = 1 − φ(θ_{A}(x))},

which is pictured in Figure 3.

Figure 3.

Lemma 5. There are constants C_{1} = C_{1}(n) and C_{2} = C_{2}(n) such that

μ(|x|, θ(x, η), t_{i}) ≤ \frac{C_{1}k(|x|)}{sin^{n−2} θ(x, η)}, x ∈ Γ_{A}, i = 1, 2,

and

\frac{k(|x|)}{sin^{n−2} θ(x, η)} ≤ C_{2}v_{A}(x), x ∈ Γ_{A}.

Proof. The first inequality follows from Lemma 1 and (2.6), since if x ∈ Γ_{A} then 1 − |x| = φ(θ_{A}(x)) and so, for i = 1, 2,

μ(|x|, θ(x, η), t_{i}) ≤ \frac{C_{1}(1 − |x|)}{(θ(x, η) − t_{i})^{2} sin^{n−2} θ(x, η)}

≤ \frac{C_{1}(1 − |x|)}{θ_{A}(x)^{2} sin^{n−2} θ(x, η)} = \frac{C_{1}k(|x|)}{sin^{n−2} θ(x, η)},

by (3.4).

To prove the second inequality, we define the spherical cap

A_{x} = {ξ ∈ S : θ(ξ, x/|x|) < 1 − |x|}, x ∈ Γ_{A}.

By (3.3), (3.6) and (3.8),

(3.9)  1 − |x| = φ(θ_{A}(x)) ≤ \frac{θ_{A}(x)}{2πn},  x ∈ Γ_{A},

and so A_{x} ⊆ A. Thus

(3.10)  θ_{A}(ξ) ≤ 2θ_{A}(x),  ξ ∈ A_{x},

and also

(3.11)  sin θ(ξ, η) ≤ 2 sin θ(x, η),  ξ ∈ A_{x}.

Since k(1 − φ(t)) decreases as t increases, we deduce from (3.4), (3.8) and (3.10) that, for ξ ∈ A_{x},

k(1 − φ(θ_{A}(ξ))) ≥ k(1 − φ(2θ_{A}(x))) = \frac{φ(2θ_{A}(x))}{(2θ_{A}(x))^{2}} ≥ \frac{φ(θ_{A}(x))}{4θ_{A}(x)^{2}} = \frac{1}{4}k(|x|).

Thus, for ξ ∈ A_{x},

v_{A}(ξ) = \frac{k(1 − φ(θ_{A}(ξ)))}{sin^{n−2} θ(ξ, η)} ≥ \frac{k(|x|)}{2^{n} sin^{n−2} θ(x, η)},

by (3.11). Hence, for x ∈ Γ_{A},

v_{A}(x) ≥ \frac{1}{ω_{n}} ∫_{A_{x}} v_{A}(ξ)P (x, ξ) dσ(ξ)

≥ (\frac{k(|x|)}{ω_{n}2^{n} sin^{n−2} θ(x, η)})(\frac{1 − |x|^{2}}{2^{n}(1 − |x|)^{n}} )ω_{n−1} ∫_{0}^{1−|x|} sin^{n−2} τ dτ

≥ \frac{k(|x|)}{C_{2} sin^{n−2} θ(x, η)},

in view of (2.1) and the fact that, for ξ ∈ A_{x},

|x − ξ| ≤ |x − x/|x|| + |x/|x| − ξ| ≤ 1 − |x| + θ(x, ξ) ≤ 2(1 − |x|).

This completes the proof of Lemma 5.

4. Proof of Theorem 1

We shall first consider the special case when

(4.1)  J = J_{0} = min\{\frac{ω_{n}}{32C_{1}C_{2}ω_{n−1}} , \frac{1}{πn}\},

which satisfies (3.1). We may assume that u is constant on each set S(r, t), since otherwise we replace u by the average û of the harmonic functions u(T (.)), where T ranges over all orthogonal transformations of R^{n} which keep each point of the x_{n} -axis fixed. The harmonic function û satisfies the assumptions of Theorem 1 and

∫_{A} û dσ = ∫_{A} u dσ,

for each annulus A on S centred at η. Thus we can define, for 0 ≤ t ≤ π,

ũ(t) = u(ξ), where |ξ| = 1, θ(ξ, η) = t.

Now put

(4.2)  K = sup_{A} \frac{1}{ω_{n}} ∫_{A} u(ξ) dσ(ξ),

where the supremum is taken over all annuli on S centred at η. Since u(0) = 0, we have

(4.3)  inf_{A} \frac{1}{ω_{n}} ∫_{A} u(ξ) dσ(ξ) ≥ −2K.

For a given fixed annulus A = A_{η}(t_{1}, t_{2}), 0 ≤ t_{1} < t_{2} ≤ π, we define, for x ∈ B,

u_{1}(x) = \frac{1}{ω_{n}} ∫_{A} u(ξ)P (x, ξ) dσ(ξ)

and

u_{2}(x) = u(x) − u_{1}(x) = \frac{ω_{n−1}}{ω_{n}} (∫_{0}^{t_{1}} + ∫_{t_{2}}^{π})ũ(t)μ(|x|, θ(x, η), t) sin^{n−2} t dt.

Suppose that x ∈ Γ_{A}. Then, by (3.6), (3.8), (3.9), and the fact that θ_{A}(x) ≤ π/2,

|x| ≥ 1 − 1/(4n) ≥ 1 − \frac{1}{4}π/\sqrt{n}

and

θ_{A}(x) ≥ 2πn(1 − |x|) ≥ 2\sqrt{n} (1 − |x|).

It follows, by (3.6) and Lemma 4, with r = |x| and s = θ(x, η), that the function μ(|x|, θ(x, η), t) is an increasing function of t on [0, θ(x, η) − 2\sqrt{n} (1 − |x|)] and hence on [0, t_{1}]. Similarly μ(|x|, θ(x, η), t) is a decreasing function of t on [t_{2}, π]. Thus, by the second mean value theorem for integrals,

u_{2}(x) = \frac{ω_{n−1}}{ω_{n}} [μ(|x|, θ(x, η), 0) ∫_{0}^{τ_{1}} ũ(t) sin^{n−2} t dt
+ μ(|x|, θ(x, η), t_{1}) ∫_{τ_{1}}^{t_{1}} ũ(t) sin^{n−2} t dt
+ μ(|x|, θ(x, η), t_{2}) ∫_{t_{2}}^{τ_{2}} ũ(t) sin^{n−2} t dt
+ μ(|x|, θ(x, η), π) ∫_{τ_{2}}^{π} ũ(t) sin^{n−2} t dt],

for some τ_{1} ∈ (0, t_{1}) and τ_{2} ∈ (t_{2}, π). Hence, by (4.3) and Lemma 4,

u_{2}(x) ≥ −4K(μ(|x|, θ(x, η), t_{1}) + μ(|x|, θ(x, η), t_{2})).

Applying (1.6) and Lemma 5, we deduce that

u_{1}(x) = u(x) − u_{2}(x) ≤ \frac{(1 + 8KC_{1})k(|x|)}{sin^{n−2} θ(x, η)} ≤ C_{2}(1 + 8KC_{1})v_{A}(x).

Thus the function u_{1} − C_{2}(1 + 8KC_{1})v_{A} is non-positive on Γ_{A}. In addition, it is continuous on \overline{B}\(S(1, t_{1}) ∪ S(1, t_{2})), harmonic and bounded above in B and vanishes on S\A. Since S(1, t_{1}) ∪ S(1, t_{2}) is a polar set, we can apply the generalized maximum principle in the domain bounded by S\A and Γ_{A} to obtain

(4.4)  u_{1}(0) ≤ C_{2}(1 + 8KC_{1})v_{A}(0).

Hence, by (3.7),

\frac{1}{ω_{n}} ∫_{A} u(ξ) dσ(ξ) ≤ C_{2}(1 + 8KC_{1})(\frac{2ω_{n−1}J_{0}}{ω_{n}} ).

Since A was an arbitrary annulus centred at η,

K ≤ \frac{2ω_{n−1}C_{2}(1 + 8KC_{1})J_{0}}{ω_{n}} ,

and it follows from (4.1) that K ≤ (8C_{1})^{−1} . Thus, by (3.7) and (4.4),

∫_{A_{η}(t_{1},t_{2})} u(ξ) dσ(ξ) ≤ 2C_{2}ω_{n}v_{A}(0) = 4C_{2}ω_{n−1} ∫_{0}^{(t_{2}−t_{1})/2} \frac{φ(t)}{t^{2}} dt = J_{0}^{2}κ(t_{2} − t_{1}),

where we have put

(4.5)  κ(t) = \frac{4C_{2}ω_{n−1}}{J_{0}^{2}} ∫_{0}^{t/2} \frac{φ(τ)}{τ^{2}} dτ,  0 ≤ t ≤ π.

By (3.5) and (4.5),

κ(π) ≤ \frac{4C_{2}ω_{n−1}}{J_{0}} ,

which depends only on n. This completes the proof of Theorem 1 when J = J_{0} . For an arbitrary J, we put u_{0} = (J_{0}/J)^{2}u and k_{0} = (J_{0}/J)^{2}k. Then

∫_{0}^{1} \sqrt{\frac{k_{0}(r)}{1 − r}} dr = J_{0},

and so we can apply the special case of Theorem 1 to k_{0} and u_{0} , to obtain

∫_{A_{η}(t_{1},t_{2})} u_{0}(ξ) dσ(ξ) ≤ J_{0}^{2}κ(t_{2} − t_{1}), 0 ≤ t_{1} < t_{2} ≤ π,

where κ is given by (4.5), with φ constructed in terms of k_{0} . Multiplying both sides of this inequality by (J/J_{0})^{2} gives (1.7). This completes the proof of Theorem 1.

5. Proof of Theorem 2

Let k be the function described in Theorem 2 and assume, as we may, that k(0) = 1. As in Section 3, put

f(h) = \sqrt{\frac{h}{k(1 − h)}} , 0 < h ≤ 1,

and then

φ(t) = \{ \begin{matrix}f^{−1}(t), & 0 < t ≤ 1, \\ 0, & t = 0. \end{matrix}

By definition

(5.1)  \frac{φ(t)}{t^{2}} = k(1 − φ(t)),  0 < t ≤ 1,

and integration by parts (see (3.5)) gives

∫_{0}^{1} \frac{φ(t)}{t^{2}} dt = ∞.

By using [4, Lemma 4], we may furthermore assume that (1 − r)k(r) ≤ 1, since we may replace k(r) by min{k(r), 1/(1 − r)} without affecting (1.8). This yields

(5.2)  φ(t) ≤ t,  0 ≤ t ≤ 1.

We then put

ψ(t) = φ(\frac{1}{2}t), 0 ≤ t ≤ 1,

so that

(5.3)  ∫_{0}^{1} \frac{ψ(t)}{t^{2}} dt = ∞.

Now put, for 0 ≤ r < 1, 0 ≤ s ≤ π and 0 ≤ t < π/2,

(5.4)  K(r, s, t) = μ(r, s, t) − μ(r, s, π/2),

and then

(5.5)  u(x) = ∫_{π/2−1}^{π/2} \frac{ψ(π/2 − t)}{(π/2 − t)^{2}}K(|x|, θ(x, η), t) dt,  x ∈ B.

The kernel function K(|x|, θ(x, η), t), x ∈ B, is harmonic in x for each fixed t (since μ(|x|, θ(x, η), t) is the average of P (x, T (ξ)), where θ(ξ, η) = t and T ranges over all orthogonal transformations of R^{n} which keep each point of the x_{n} -axis fixed) and so, therefore, is u if the integral in (5.5) is locally uniformly convergent. That this is the case follows from

\frac{ψ(π/2 − t)}{(π/2 − t)^{2}} ≤ \frac{1/2}{π/2 − t}, π/2 − 1 ≤ t < π/2,

by (5.2), and the fact that if x remains in any compact subset of B then

(5.6)  K(|x|, θ(x, η), t) = O(π/2 − t)  as t → π/2^{−},

uniformly in x. To prove this estimate, recall from (2.3) that

(5.7)  μ(r, s, t) = ∫_{S(1,π/2)} \frac{(1 − r^{2}) dσ̂(ζ)}{d(r, s, t, ζ_{1})^{n}}.

Here σ̂ is normalized (n−2)-dimensional measure on S(1, π/2), ζ = (ζ_{1}, …, ζ_{n−1}, 0) ∈ S(1, π/2), and

(5.8)

d(r, s, t, ζ_{1}) = |(r sin s, 0, ⋯, 0, r cos s) − (ζ_{1} sin t, …, ζ_{n−1} sin t, cos t)|
= (1 − 2r(ζ_{1} sin s sin t + cos s cos t) + r^{2})^{1/2}
= (d^{2} + 2r(1 − ζ_{1}) sin s sin t)^{1/2},

where

d = d(r, s, t) = (1 − 2r cos(s − t) + r^{2})^{1/2}.

Now

(5.9)  \frac{1}{d(r, s, t, ζ_{1})^{n}} − \frac{1}{d(r, s, π/2, ζ_{1})^{n}} = \frac{1 − (d(r, s, t, ζ_{1})/d(r, s, π/2, ζ_{1}))^{n}}{d(r, s, t, ζ_{1})^{n}}

and, for 0 ≤ t < π/2, by the triangle inequality,

(5.10)

|d(r, s, π/2, ζ_{1}) − d(r, s, t, ζ_{1})| ≤ |(ζ_{1}, …, ζ_{n−1}, 0)
− (ζ_{1} sin t, ⋯, ζ_{n−1} sin t, cos t)|
= ((1 − sin t)^{2} + cos^{2} t)^{1/2} = 2 sin(\frac{1}{2}(π/2 − t))
≤ π/2 − t.

Thus, by applying the inequality

|1 − x^{n}| ≤ 2n|x − 1|, x ∈ R, |x − 1| ≤ 1/(2n),

to (5.9), we obtain

|\frac{1}{d(r, s, t, ζ_{1})^{n}} − \frac{1}{d(r, s, π/2, ζ_{1})^{n}} | ≤ \frac{2n(π/2 − t)}{d(r, s, t, ζ_{1})^{n}d(r, s, π/2, ζ_{1})},

provided that

|d(r, s, t, ζ_{1}) − d(r, s, π/2, ζ_{1})| ≤ \frac{1}{2n}d(r, s, π/2, ζ_{1}).

Hence, if 0 ≤ r ≤ r_{0} < 1, say, and π/2 − t ≤ (1 − r_{0})/(2n), then

|\frac{1}{d(r, s, t, ζ_{1})^{n}} − \frac{1}{d(r, s, π/2, ζ_{1})^{n}} | ≤ (\frac{2n}{1 − r_{0}} )\frac{(π/2 − t)}{d(r, s, t, ζ_{1})^{n}} ,

by (5.10). It follows that, for 0 ≤ r ≤ r_{0} and π/2 − t ≤ (1 − r_{0})/(2n),

|K(r, s, t)| ≤ (\frac{2n}{1 − r_{0}} )(π/2 − t)μ(r, s, t) ≤ \frac{C(n)}{(1 − r_{0})^{n+1}} (π/2 − t),

by (5.4), (5.7) and Lemma 1, (since d ≥ 1 − r_{0} ). This proves (5.6), so that u is harmonic in B. Note that u(0) = 0, in particular.

We now claim that for some positive constant C, depending on n and the function k,

(5.11)  u(x) ≤ Ck(|x|),  x ∈ B ∩ {x_{n} > 0},

so that (1.9) can be satisfied in B ∩ {x_{n} > 0} by a scaling. To carry out the proof we need two upper estimates for K, the first being

(5.12)  \begin{aligned}K(r, s, t) ≤ \frac{C(n)(1 − r)(π/2 − t)}{d(r, s, t)^{2}d(r, s, π/2)}, & 0 ≤ r < 1, 0 ≤ s ≤ π/2, \\ & π/2 − 1 ≤ t < π/2. \end{aligned}

We prove (5.12) by applying the inequality

1 − x^{n} ≤ n|1 − x|, x > 0,

to (5.9), and using (5.8) and (5.10), to obtain

\begin{aligned}\frac{1}{d(r, s, t, ζ_{1})^{n}} & − \frac{1}{d(r, s, π/2, ζ_{1})^{n}} ≤ \frac{n|1 − d(r, s, t, ζ_{1})/d(r, s, π/2, ζ_{1})|}{d(r, s, t, ζ_{1})^{n}} \\ & = \frac{n|d(r, s, π/2, ζ_{1}) − d(r, s, t, ζ_{1})|}{d(r, s, t, ζ_{1})^{n}d(r, s, π/2, ζ_{1})} ≤ \frac{n(π/2 − t)}{d(r, s, t, ζ_{1})^{n}d(r, s, π/2)}. \end{aligned}

Thus, by Lemma 1,

K(r, s, t) ≤ \frac{C(n)(1 − r)(π/2 − t)}{d^{2}(d^{n−2} + sin^{n−2} s)d(r, s, π/2)},

where d = d(r, s, t), so that (5.12) follows from the fact that if π/2 − 1 ≤ t < π/2, then max\{d(r, s, t), sin s\} ≥ sin(\frac{1}{2}(π/2 − 1)).

The other upper estimate for K is

(5.13)  \begin{aligned}K(r, s, t) ≤ \frac{−c(n)(1 − r)}{d(r, s, π/2)^{2}}, & 0 ≤ r < 1, 0 ≤ s ≤ π/2, \\ & π/2 − 1 ≤ t ≤ π/2 − C_{3}d(r, s, π/2), \end{aligned}

where the constant C_{3} = C_{3}(n) is suitably large. To prove (5.13), note that Lemma 1 gives

(5.14)  \begin{aligned}K(r, s, t) & ≤ \frac{C(n)(1 − r)}{d(r, s, t)^{2}(d(r, s, t)^{n−2} + sin^{n−2} s)} \\ & − \frac{c(n)(1 − r)}{d(r, s, π/2)^{2}(d(r, s, π/2)^{n−2} + sin^{n−2} s)} \\ & ≤ \frac{C(n)(1 − r)}{d(r, s, t)^{2}} − \frac{c(n)(1 − r)}{d(r, s, π/2)^{2}} , \end{aligned}

because \sqrt{2} ≥ max\{d(r, s, t), sin s\} ≥ sin(\frac{1}{2}(π/2 − 1)). Now, if π/2 − 1 ≤ t ≤ π/2 − C_{3}d(r, s, π/2), then, by the triangle inequality,

d(r, s, t) ≥ 2 sin \frac{1}{2}(π/2 − t) − d(r, s, π/2) ≥ (2C_{3}/π − 1)d(r, s, π/2), and so (5.13) follows from (5.14), provided that C_{3} = C_{3}(n) is chosen large enough.

For x ∈ B ∩ {x_{n} > 0} and d(r, s, π/2) ≤ C_{3}^{−1} , where r = |x| and s = θ(x, η), we put

\begin{aligned}E_{1} & = \{t : |t − s| ≤ \frac{1}{2}(π/2 − s)\}, \\ E_{2} & = \{t : |t − s| ≥ \frac{1}{2}(π/2 − s), π/2 − C_{3}d(r, s, π/2) ≤ t < π/2\}, \\ E_{3} & = \{t : |t − s| ≥ \frac{1}{2}(π/2 − s), π/2 − 1 ≤ t ≤ π/2 − C_{3}d(r, s, π/2)\}, \end{aligned}

and then

u_{i}(x) = ∫_{E_{i}} \frac{ψ(π/2 − t)}{(π/2 − t)^{2}}K(r, s, t) dt, i = 1, 2, 3,

so that u(x) = u_{1}(x) + u_{2}(x) + u_{3}(x), by (5.5). We now estimate u_{1}(x), u_{2}(x) and u_{3}(x) from above.

First note that

\frac{1}{2}(π/2 − s) ≤ π/2 − t ≤ 2(π/2 − s), for t ∈ E_{1},

and ψ(π/2 − t) = φ(\frac{1}{2}(π/2 − t)), so that (5.12) gives

(5.15)  \begin{aligned}u_{1}(x) & ≤ \frac{C(n)(1 − r)}{d(r, s, π/2)} ∫_{E_{1}} \frac{ψ(π/2 − t)}{(π/2 − t)d(r, s, t)^{2}} dt \\ & ≤ \frac{C(n)φ(π/2 − s)}{d(r, s, π/2)(π/2 − s)} ∫_{E_{1}} \frac{1 − r^{2}}{d(r, s, t)^{2}} dt \\ & ≤ \frac{C(n)φ(π/2 − s)}{d(r, s, π/2)(π/2 − s)}, \end{aligned}

because (1 − r^{2})/d(r, s, t)^{2} is the two-dimensional Poisson kernel.

22 B. Korenblum, P.J. Rippon and K. Samotij

Next note that, for t ∈ E_{2} , d(r, s, t) ≥ d(r, s, s+ \frac{1}{2}(π/2−s)) ≥ d(r, s, π/2)− \frac{1}{2}(π/2−s) ≥ d(r, s, π/2)(1−π/4), the last step of which follows from

(5.16)  d(r, s, π/2) ≥ sin(π/2 − s) ≥ \frac{2}{π} (π/2 − s).

Thus (5.12) gives

(5.17)  \begin{aligned}u_{2}(x) & ≤ \frac{C(n)(1 − r)}{d(r, s, π/2)} ∫_{E_{2}} \frac{ψ(π/2 − t)}{(π/2 − t)d(r, s, t)^{2}} dt \\ & ≤ \frac{C(n)(1 − r)}{d(r, s, π/2)^{3}} ∫_{E_{2}} \frac{ψ(π/2 − t)}{(π/2 − t)} dt ≤ \frac{C(n)(1 − r)}{d(r, s, π/2)^{2}}, \end{aligned}

since ψ(π/2 − t) ≤ π/2 − t, by (5.2), and E_{2} ⊆ [π/2 − C_{3}d(r, s, π/2), π/2]. Finally, we deduce from (5.13) that

(5.18)  u_{3}(x) ≤ −\frac{c(n)(1 − r)}{d(r, s, π/2)^{2}} ∫_{C_{3}d(r,s,π/2)}^{1} \frac{ψ(τ)}{τ^{2}} dτ.

We are now able to verify (5.11). First suppose that 1 − r ≤ φ(π/2 − s). Then, by (5.1), (5.15), (5.16), (5.17) and (5.18),

(5.19)  u(x) ≤ \frac{C(n)φ(π/2 − s)}{(π/2 − s)^{2}} = C(n)k(1 − φ(π/2 − s)) ≤ C(n)k(r).

On the other hand, if 1 − r ≥ φ(π/2 − s), then

\frac{φ(π/2 − s)}{π/2 − s} ≤ \frac{2(1 − r)}{d(r, s, π/2)},

in view of (5.2) and the estimate

d(r, s, π/2) ≤ (1 − r) + (π/2 − s) ≤ 2 max{1 − r, π/2 − s}.

Thus, in this case, (5.15), (5.17) and (5.18) give

(5.20)  u(x) ≤ \frac{1 − r}{d(r, s, π/2)^{2}} (C(n) − c(n) ∫_{C_{3}d(r,s,π/2)}^{1} \frac{ψ(τ)}{τ^{2}} dτ) ≤ 0,

provided that d(r, s, π/2) ≤ c_{1} ≤ C_{3}^{−1} , for some positive c_{1} which depends on n and the function k.

To complete the proof of (5.11), we observe that if x ∈ B ∩ {x_{n} > 0} and d(r, s, π/2) ≥ c_{1} , where r = |x| and s = θ(x, η), then, by (5.12),

(5.21)  u(x) ≤ \frac{C(n)}{d(r, s, π/2)} ∫_{π/2−1}^{π/2}(\frac{ψ(π/2 − t)}{π/2 − t})(\frac{1 − r^{2}}{d(r, s, t)^{2}} ) dt ≤ \frac{C(n)}{c_{1}},

since ψ(π/2−t) ≤ π/2−t, by (5.2), and (1−r^{2})/d(r, s, t)^{2} is the two-dimensional Poisson kernel. On combining (5.19), (5.20) and (5.21), we obtain (5.11), since k(0) = 1.

To prove that (1.9) is satisfied in B ∩ {x_{n} ≤ 0}, it is sufficient to show that

(5.22)  K(r, s, t) ≤ 0,  0 ≤ r < 1, 0 ≤ t < π/2 ≤ s ≤ π.

We do this by proving that, if 0 ≤ r < 1, π/2 ≤ s ≤ π and 0 ≤ ζ_{1} ≤ 1, then the function

λ(t) = \frac{1}{d(r, s, t, ζ_{1})^{n}} + \frac{1}{d(r, s, t, −ζ_{1})^{n}}

= \frac{1}{(1 − 2r(ζ_{1} sin s sin t + cos s cos t) + r^{2})^{n/2}}

+ \frac{1}{(1 − 2r(−ζ_{1} sin s sin t + cos s cos t) + r^{2})^{n/2}}

(see (5.8)) increases as t increases, for 0 ≤ t ≤ π/2. Indeed

λ′(t) = nrζ_{1} sin s cos t[\frac{1}{d(r, s, t, ζ_{1})^{n+2}} − \frac{1}{d(r, s, t, −ζ_{1})^{n+2}} ]

− nr cos s sin t[\frac{1}{d(r, s, t, ζ_{1})^{n+2}} + \frac{1}{d(r, s, t, −ζ_{1})^{n+2}} ] ≥ 0,

since 0 < d(r, s, t, ζ_{1}) ≤ d(r, s, t, −ζ_{1}), cos t ≥ 0, sin t ≥ 0, sin s ≥ 0 and cos s ≤ 0. Thus, on integration over S(1, π/2) ∩ {ζ_{1} ≥ 0}, we deduce from (5.7) that

μ(r, s, t) ≤ μ(r, s, π/2), 0 ≤ r < 1, 0 ≤ t < π/2 ≤ s ≤ π,

which is (5.22).

Finally we prove that u satisfies (1.10). To do this we need the following facts:

(5.23)  \frac{∂u}{∂x_{n}} ≥ 0,  x ∈ B ∩ \{x_{n} = 0\},

and

(5.24)  I = ∫_{B∩\{x_{n}=0\}}(1 − |x|)\frac{∂u}{∂x_{n}} dσ(x) = ∞,

where σ denotes (n − 1)-dimensional measure.

The inequality (5.23) follows immediately from (5.5) and

\frac{∂}{∂s}K(r, s, t) |_{s=π/2}= \frac{∂}{∂s}μ(r, s, t) |_{s=π/2}≤ 0, 0 ≤ r < 1, 0 ≤ t < π/2,

which in turn follows from

\frac{∂}{∂s}(\frac{1}{d(r, s, t, ζ_{1})^{n}} ) |_{s=π/2}= \frac{−nr cos t}{d(r, π/2, t, ζ_{1})^{n+2}} ≤ 0,

and (5.7). To verify (5.24), note that (5.4), (5.5) and (5.7) give

I = ∫_{B∩\{x_{n}=0\}}(1 − |x|)(∫_{π/2−1}^{π/2} \frac{ψ(π/2 − t)}{(π/2 − t)^{2}}(\frac{−1}{|x|})\frac{∂}{∂s}K(|x|, s, t) |_{s=π/2} dt) dσ(x)

= ∫_{B∩\{x_{n}=0\}}(1 − |x|)(∫_{π/2−1}^{π/2}\frac{ψ(π/2 − t)}{(π/2 − t)^{2}}(∫_{S(1,t)}\frac{n(1 − |x|^{2}) cos t}{|x − ξ|^{n+2}} dσ̂(ξ))dt)dσ(x)

= n ∫_{π/2−1}^{π/2} \frac{ψ(π/2 − t)}{(π/2 − t)^{2}} cos t(∫_{S(1,t)} dσ̂(ξ)(∫_{B∩\{x_{n}=0\}} \frac{(1 − |x|)^{2}}{|x − ξ|^{n+2}} dσ(x))) dt,

where σ̂ denotes normalized (n − 2)-dimensional Lebesgue measure on S(1, t).

Now, if π/2 − 1 ≤ t < π/2 and ξ = (ζ_{1} sin t, …, ζ_{n−1} sin t, cos t) ∈ S(1, t), then we consider the (n − 1)-dimensional ball B_{ξ} in B ∩ {x_{n} = 0} with centre x_{ξ} = (1 − (π/2 − t))(ζ_{1}, …, ζ_{n−1}, 0) and radius \frac{1}{2}(π/2 − t). For x ∈ B_{ξ}, 1 − |x| ≥ \frac{1}{2}(π/2 − t) and

|x − ξ| ≤ |x − x_{ξ}| + |x_{ξ} − ξ| ≤ \frac{1}{2}(π/2 − t) + 2(π/2 − t).

Hence

∫_{B∩\{x_{n}=0\}} \frac{(1 − |x|)^{2}}{|x − ξ|^{n+2}} dσ(x) ≥ ∫_{B_{ξ}} \frac{\frac{1}{4}(π/2 − t)^{2}}{(3(π/2 − t))^{n+2}} dσ(x) = \frac{c(n)}{π/2 − t} ≥ \frac{c(n)}{cos t},

so that

I ≥ c(n) ∫_{π/2−1}^{π/2} \frac{ψ(π/2 − t)}{(π/2 − t)^{2}} dt = ∞,

by (5.3), as required.

We deduce (1.10) from (5.23) and (5.24) by applying Green’s theorem with the functions u and v(x) = |x|^{2−n} − 1 on the half-annulus

H(r) = \{x ∈ R^{n} : \frac{1}{2} ≤ |x| ≤ r, x_{n} ≤ 0\}, \frac{1}{2} < r < 1.

The boundary of H(r) consists of

\begin{aligned}_{∑1}(r) & = ∂H(r) ∩ \{x_{n} = 0\}, \\ _{∑2}(r) & = ∂H(r) ∩ \{|x| = r\}, \\ _{∑3}(r) & = ∂H(r) ∩ \{|x| = \frac{1}{2}\}. \end{aligned}

With ∂/∂n denoting differentiation along the outward normal, we obtain

(5.25)  ∫_{∑_{2}(r)∪∑_{3}(r)}(u\frac{∂v}{∂n} − v \frac{∂u}{∂n}) dσ = ∫_{∑_{1}(r)} v \frac{∂u}{∂n} dσ,

since u and v are harmonic in H(r) and ∂v/∂n vanishes on ∑_{1}(r). Now

v(r) = \frac{1}{r^{n−2}} − 1 ≥ (n − 2)(1 − r), \frac{1}{2} ≤ r < 1,

and ∂u/∂n = ∂u/∂x_{n} on ∑_{1}(r). Hence, by (5.23) and (5.24),

(5.26)  ∫_{∑_{1}(r)} v \frac{∂u}{∂n} dσ → ∞  as r → 1^{−}.

Thus, if

I(r) = ∫_{∑_{2}(1)} u(rξ) dσ(ξ) = \frac{1}{r^{n−1}} ∫_{∑_{2}(r)} u(x) dσ(x),

so that

I′(r) = ∫_{∑_{2}(1)} \frac{∂}{∂r} u(rξ) dσ(ξ) = \frac{1}{r^{n−1}} ∫_{∑_{2}(r)} \frac{∂u}{∂n}(x) dσ(x),

then (5.25) and (5.26) give

r^{n−1}(v′(r)I(r) − v(r)I′(r)) → ∞ as r → 1^{−}.

Hence

v(r)^{2} \frac{d}{dr} (\frac{I(r)}{v(r)}) → −∞ as r → 1^{−}.

Thus, if M > 0, then there exists r_{0} ≥ \frac{1}{2} such that

v(r)^{2} \frac{d}{dr} (\frac{I(r)}{v(r)}) ≤ −M, r_{0} ≤ r < 1,

26 B. Korenblum, P.J. Rippon and K. Samotij

and hence

\frac{I(r)}{v(r)} − \frac{I(r_{0})}{v(r_{0})} ≤ −M ∫_{r_{0}}^{r} \frac{dt}{v(t)^{2}} , r_{0} ≤ r < 1,

so that, by the convexity of v,

\begin{aligned}I(r) & ≤ \frac{v(r)}{v(r_{0})}I(r_{0}) − M \frac{v(r)}{(−v′(\frac{1}{2}))} ∫_{r_{0}}^{r} \frac{(−v′(t))}{v(t)^{2}} dt \\ & = \frac{v(r)}{v(r_{0})}I(r_{0}) − \frac{M}{(n − 2)2^{n−1}} [1 − \frac{v(r)}{v(r_{0})}] ≤ 1 − \frac{M}{(n − 2)2^{n}} , \end{aligned}

for r_{1} < r < 1, say. Hence I(r) → −∞ as r → 1^{−}, and (1.10) follows from the mean value property. This completes the proof of Theorem 2.

References

[1] Hayman, W.K., and P.B. Kennedy: Subharmonic functions, Vol. I. - London Math. Soc. Monographs 9, Academic Press, London–New York, 1976.

[2] Hayman, W.K., and B. Korenblum: An extension of the Riesz–Herglotz formula. - Ann. Acad. Sci. Fenn. Ser A I Math. 2, 1976, 175–201.

[3] Korenblum, B.: An extension of the Nevanlinna theory. - Acta Math. 135, 1975, 187–219.

[4] Rippon, P.J.: A boundary estimate for harmonic functions. - Math. Proc. Camb. Phil. Soc. 91, 1982, 79–90.

[5] Samotij, K.: A representation theorem for harmonic functions in the ball in R^{n}. - Ann. Acad. Sci. Fenn. Ser. A I Math. 11, 1986, 29–38.

Received 18 March 1992