GEORGIAN MATHEMATICAL JOURNAL: Vol. 4, No. 6, 1997, 585-600

BOUNDARY PROPERTIES OF FIRST-ORDER PARTIAL DERIVATIVES OF THE POISSON INTEGRAL FOR THE

HALF-SPACE ℝ_{k+1}^{+} (k > 1)

S. TOPURIA

Abstract. Boundary properties of first-order partial derivatives of the Poisson integral are studied in the half-space ℝ_{k+1}^{+} (k > 1).

The boundary properties of the Poisson integral for a circle were thoroughly studied by Fatou [1]. In particular, he showed that the following theorems are valid:

Theorem A. If there exists a finite f′(x_{0}), then

lim_{re^{ix}→^{∧}e^{ix_{0}}} \frac{∂u(f; r, x)}{∂x} = f′(x_{0}),

where u(f; r, x) is the Poisson integral for a circle, and the symbol re^{ix} →^{∧} e^{ix_{0}} means that the point re^{ix} tends to e^{ix_{0}} along the paths which are nontangential to the circumference (see [2], p. 100, and [3], p. 156).

Theorem B. If there exists a finite or infinite D_{1}f(x_{0}) which is a first symmetric derivative of f at the point x_{0} (see [2], p. 99 − 100), i.e.,

D_{1}f(x_{0}) = lim_{h→0} \frac{f(x_{0} + h) − f(x_{0} − h)}{2h},

then

lim_{r→1−} \frac{∂u(f; r, x_{0})}{∂x} = D_{1}f(x_{0}).

585

1991 Mathematics Subject Classification. 31B25.

Key words and phrases. The Poisson integral for the half-space, angular limit, derivatives of the integral, differential properties of the density.

1072-947X/97/1100-0585$12.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1998 Plenum Publishing Corporation

In [4] a continuous 2π-periodic function f(x) is constructed such that D_{1}f(x_{0}) = 0, but

lim_{re^{ix}→^{∧}e^{ix_{0}}} \frac{∂u(f; r, x)}{∂x}

does not exist. Thus it is shown that Theorem B cannot be strengthened in the sense of the existence of an angular limit.

An analogue of Theorem A for a half-plane ℝ_{+}^{2} is proved in [5, Theorem 4], while an analogue of Theorem B given in [6, Theorem 1] shows that this theorem cannot be strengthened in the sense of the existence of an angular limit.

The question as to the validity of Fatou’s theorem for a bicylinder was considered in [7], where it is proved that in the neighborhood of some point the density of the Poisson integral can have no smoothness that would ensure the existence of a boundary value of partial derivatives of the Poisson integral at the considered point. Furthermore, in this paper sufficient conditions are found for the convergence of first- and second- order partial derivatives of the Poisson integral for a bicylinder, and it is shown that the results obtained cannot be strengthened (in the definite sense).

The boundary properties of the integral D_{k}u(f; r, ϑ_{1}, ϑ_{2}, …, ϑ_{k−2}, φ) were studied in [8] (see also [9], p. 118), where u(f; r, ϑ_{1}, ϑ_{2}, …, ϑ_{k−2}φ) is the Poisson integral for the unit sphere in ℝ^{k} (k > 2), and D_{k} is the Laplace operator on the sphere, i.e., the angular part of the Laplace operator written in terms of spherical coordinates (see [9], p. 14). The boundary properties of first- and second- order partial derivatives of the Poisson integral for the unit sphere in ℝ^{3} are given a detailed consideration in [10, 11, 12], but for the half-space ℝ_{+}^{3} in [13], [14], [15]. In [14] it is shown that there exists a continuous function of two variables f(x, y) ∈ L(ℝ^{2}) which, at the point (x_{0}, y_{0}), has the partial derivatives f′_{x}(x_{0}, y_{0}) and f′_{y}(x_{0}, y_{0}), but the integrals \frac{∂u(f;x,y,z)}{∂x} and \frac{∂u(f;x,y,z)}{∂y} (u(f; x, y, z) is the Poisson integral for ℝ_{+}^{3}) of this function have no values at the point (x_{0}, y_{0}) even along the normal.

Hence the question arises how to generalize the notion of derivatives of a function of many variables so that a Fatou type theorem would hold for the integral u(f; x, x_{k+1}) (u(f; x, x_{k+1}) is the Poisson integral for ℝ_{+}^{k+1} (k > 1)).

In this paper, the notion of a generalized partial derivative is introduced for a function of many variables and Fatou type theorems are proved on boundary properties of first-order partial derivatives of the Poisson integral for a half-space. These results complement and generalize the author’s studies in [13], [14], [15]. In particular, in this paper it is shown that the boundary properties of derivatives of the Poisson integral for a half-space essentially depend on the sense in which the integral density is differentiable. Examples are constructed testifying to the fact that the results obtained are unimprovable (in the definite sense).

1. Notation, Definitions, and Auxiliary Propositions

The following notation is used in this paper:

ℝ^{k} is a k-dimensional Euclidean space (ℝ = ℝ^{1});

x = (x_{1}, x_{2}, …, x_{k}), t = (t_{1}, t_{2}, …, t_{k}), x^{0} = (x_{1}^{0}, x_{2}^{0}, …, x_{k}^{0}) are the points (vectors) of the space ℝ^{k};

(x, t) = ∑_{i=1}^{k} x_{i}t_{i} is the scalar product;

|x| = \sqrt{(x, x)}; x + t = (x_{1} + t_{1}, x_{2} + t_{2}, …, x_{k} + t_{k});

e_{i} (i = 1, 2, …, k) is the coordinate vector.

Let (see [16], p. 174) M = {1, 2, …, k} (k ∈ N, k ≥ 2), B be an arbitrary subset from M and B′ = M \ B. For any x ∈ ℝ^{k} and an arbitrary set B ⊂ M, the symbol x_{B} denotes a point from ℝ^{k} whose coordinates with indices from the set B coincide with the corresponding coordinates of the point x, while coordinates with indices from the set B′ are zeros (x_{M} = x, B \ i = B \ {i}); if B = {i_{1}, i_{2}, …, i_{s}}, 1 ≤ s ≤ k (i_{l} < i_{r} for l < r), then \overline{x}_{B} = (x_{i_{1}}, x_{i_{2}}, …, x_{i_{s}}) ∈ ℝ^{s}; m(B) is the number of elements of the set B;

L̃(ℝ^{k}) is the set of functions f(x) = f(x_{1}, x_{2}, …, x_{k}) such that

\frac{f(x)}{(1 + |x|^{2})^{\frac{k+1}{2}}} ∈ L(ℝ^{k});

ℝ_{+}^{k+1} = {(x, x_{k+1}) ∈ ℝ^{k+1}; x_{k+1} > 0};

u(f; x, x_{k+1}) is the Poisson integral of the function f(x) for the half-space ℝ_{+}^{k+1}, i.e.,

u(f; x, x_{k+1}) = \frac{x_{k+1}Γ(\frac{k+1}{2})}{π^{\frac{k+1}{2}}} ∫_{ℝ^{k}} \frac{f(t) dt}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+1}{2}}} .

In investigating the boundary properties of the partial derivatives \frac{∂}{∂ϑ}u_{f}(r, ϑ, φ) and \frac{∂}{∂φ}u_{f}(r, ϑ, φ) of the spherical Poisson integral u_{f}(r, ϑ, φ) for the summable function f(ϑ, φ) on the rectangle [0, π]×[0, 2π], Dzagnidze introduced the notion of a dihedral-angular limit [10] which is applicable to ℝ_{+}^{k+1} in the manner as follows: if the point N ∈ ℝ_{+}^{k+1} converges to the point P(x^{0}, 0) under the condition x_{k+1}( ∑_{i∈B}(x_{i} − x_{i}^{0})^{2})^{−1/2} ≥ C > 0,^{1}then

we shall write N(x, x_{k+1}) →^{∧}_{x_{B}} P(x^{0}, 0). When B = M, we have an angular

convergence and thus we write N(x, x_{k+1}) →^{∧} P(x^{0}, 0). Finally, the notation N(x, x_{k+1}) → P(x^{0}, 0) means that the point N(x, x_{k+1}) remaining in ℝ_{+}^{k+1} converges to P(x^{0}, 0) without any restrictions.

1Here and further C denotes absolute positive constants which, generally speaking, may be different in different relations.

It is known that \frac{∂}{∂ϑ}u_{f}(r, ϑ, φ) and \frac{∂}{∂φ}u_{f}(r, ϑ, φ) have dihedral-angular limits if partial derivatives of the function f(ϑ, φ) exist in a strong sense [10], [12]. This notion admits various generalizations when the function depends on three and more variables and we shall also discuss them below.

Let u ∈ ℝ. We shall consider the following derivatives of the function f(x):

1. Denote the limit

lim_{(u,\overline{x}_{B})→(0,\overline{x}_{B}^{0} )} \frac{f(x_{B} + x0_{B′} + ue_{i}) − f(x_{B} + x0_{B′})}{u}

by:

(a) f′_{x_{i}}(x^{0}) for B ≠ ∅;

(b) D_{x_{i}(\overline{x}_{B})}f(x^{0}) for i ∈ B′;

(c) \overline{D}_{x_{i}(\overline{x}_{B})}f(x^{0}) for i ∈ B.

2. Denote the limit

lim_{(u,\overline{x}_{B})→(0,\overline{x}_{B}^{0})} \frac{f(x_{B} + x0_{B′} + ue_{i}) − f(x_{B} + x0_{B′} − ue_{i})}{2u}

by:

(a) D_{x_{i}}^{∗}(x^{0}) for B ≠ ∅;

(b) D_{x_{i}(\overline{x}_{B})}^{∗}f(x^{0}) for i ∈ B′;

(c) \overline{D}_{x_{i}(\overline{x}_{B})}f(x^{0}) for i ∈ B. The following propositions are valid:

(1) If B_{2} ⊂ B_{1}, the existence of D_{x_{i}(\overline{x}_{B_{1}} )}f(x^{0}) implies the existence of D_{x_{i}(\overline{x}_{B_{2}} )}f(x^{0}) and D_{x_{i}(\overline{x}_{B_{1}} )}f(x^{0}) = f′_{x_{i}}(x^{0}). The converse does not hold.

(2) The existence of \overline{D}_{x_{i}(\overline{x}_{B_{1}} )}f(x^{0}) implies the existence of \overline{D}_{x_{i}(\overline{x}_{B_{2}} )}f(x^{0}) and their equality.

(3) The existence of \overline{D}_{x_{i}(\overline{x}_{B})}f(x^{0}) implies the existence of \overline{D}_{x_{i}(\overline{x}_{B_{1}} )}f(x^{0}) and \overline{D}_{x_{i}(\overline{x}_{B})}f(x^{0}) = \overline{D}_{x_{i}(\overline{x}_{B\i})}f(x^{0}) = f′_{x_{i}}(x^{0}).

(4) If f′_{x_{i}}(x) is a continuous function at x^{0}, then for any B ⊂ M all derivatives \overline{D}_{x_{i}(\overline{x}_{B})}f(x^{0}) exist and

\overline{D}_{x_{i}(\overline{x}_{B})}f(x^{0}) = f′_{x_{i}}(x^{0}).

Indeed, by virtue of the Lagrange theorem

\frac{f(x_{B} + x0_{B′} + ue_{i}) − f(x_{B} + x0_{B′})}{u} = \frac{f′_{x_{i}}[x_{B} + x0_{B′} + θ(x)ue_{i}]u}{u} =

= f′_{x_{i}}[x_{B} + x0_{B′} + θ(x)ue_{i}], 0 < θ < 1.

Hence we conclude that statement (4) is valid.

(5) There exists a function f(x) for which \overline{D}_{x_{i}(x)}f(x^{0}) exist, but on an everywhere dense set in the neighborhood of the point x^{0} there are no f′_{x_{i}}(x).

(6) If the function f(x) has finite derivatives

D_{x_{1}(x_{2},...,x_{k})}f(x^{0}), D_{x_{2}(x_{3},...,x_{k})}f(x^{0}), …, D_{x_{k−1}(x_{k})}f(x^{0})

at the point x^{0}, then its continuity at x^{0} with respect to the argument x_{k} is a necessary and sufficient condition for f(x) to be continuous at x^{0} (see [12], p.15).

(7) The existence of the derivatives D_{x_{1}(x_{2},...,x_{k})}f(x^{0}), D_{x_{2}(x_{3},...,x_{k})}f(x^{0}), …, D_{x_{1}(x_{2},...,x_{k})}f(x^{0}) and f′_{x_{k}}(x^{0}) implies the existence of the differential df(x^{0}) (see [12], p. 16).

In what follows it will be assumed that f ∈ L̃(ℝ^{k}).

Lemma. The equalities

I_{1} = ∫_{ℝ^{k}} \frac{(t_{i} − x_{i})f(t − t_{i}e_{i} + x_{i}e_{i})dt}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} = 0,

I_{2} = \frac{(k + 1)x_{k+1}Γ(\frac{k+1}{2})}{π^{\frac{k+1}{2}}} ∫_{ℝ^{k}} \frac{(t_{i} − x_{i})^{2}dt}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} = 1

hold for any (x, x_{k+1}).

Proof. We have

I_{1} = ∫_{ℝ^{k−1}} f(x + t − t_{i}e_{i}) dS(\overline{t}_{M\i}) ∫_{−∞}^{∞} \frac{t_{i} dt_{i}}{(|t|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} = 0.

Further, if we use the spherical coordinates, ρ, θ_{1}, …, θ_{k−2}, φ, we shall have

I_{2} = \frac{(k + 1)x_{k+1}Γ(\frac{k+1}{2})}{π^{\frac{k+1}{2}}} ∫_{ℝ^{k}} \frac{t_{i} dt_{i}}{(|t|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} =

= \frac{(k+1)x_{k+1}Γ(\frac{k+1}{2})}{π^{\frac{k+1}{2}}}∫_{ℝ^{k}} \frac{ρ^{2} sin^{2} ϑ_{1} sin^{2} ϑ_{2}⋯sin^{2} ϑ_{i−1} cos^{2} ϑ_{i}}{(ρ^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} ×

×ρ^{k−1} sin^{k−2} ϑ_{1} ⋯ sin^{k−i} ϑ_{i−1} ⋯ sin ϑ_{k−2}dρdϑ_{1} ⋯ dϑ_{k−2}dφ =

= \frac{2(k+1)Γ(\frac{k+1}{2})}{π^{\frac{k−1}{2}}}∫_{0}^{∞}\frac{ρ^{k+1}dρ}{(1 + ρ^{2})^{\frac{k+3}{2}}} \overbrace{∫_{0}^{π}⋯∫_{0}^{π}}^{k−2} sin^{k} ϑ_{1} sin^{k−1} ϑ_{2}⋯sin^{k−i+2} ϑ_{i−1}×

× cos^{2} ϑ_{i} sin^{k−i−1} ϑ_{i} ⋯ sin ϑ_{k−2}dϑ_{1}dϑ_{2} ⋯ dϑ_{k−2} =

= \frac{2(k + 1)Γ(\frac{k+1}{2})}{π^{\frac{k−1}{2}}} · \frac{kΓ(\frac{k}{2} )\sqrt{π}}{2(k + 1)Γ(\frac{k+1}{2})} · \frac{π^{\frac{k−2}{2}}}{kΓ(\frac{k}{2} )} = 1.

2. Boundary Properties of the Integral \frac{∂u(f;x,x_{k+1})}{∂x_{i}}

The following theorem is valid.

Theorem 1.

(a) If a finite derivative \overline{D}_{x_{i}(x)}f(x^{0}) exists at the point x^{0}, then

lim_{(x,x_{k+1})→(x^{0},0)} \frac{∂u(f; x, x_{k+1})}{∂x_{i}} = \frac{∂f(x^{0})}{∂x_{i}}.  (1)

(b) There is a continuous function f ∈ L(ℝ^{k}) such that for any B ⊂ M, m(B) < k, at the point x^{0} = (0, 0, …, 0) = 0 all derivatives \overline{D}_{x_{i}(\overline{x}_{B})}f(0) = 0, i = \overline{1, k}, but the limits

lim_{x_{k+1}→0+} \frac{∂u(f; 0, x_{k+1})}{∂x_{i}}, i = \overline{1, k},

do not exist.

Proof.

(a) Let x^{0} = 0, C_{k} = \frac{(k+1)Γ( \frac{k+1}{2} )}{π^{\frac{k+1}{2}}}. It is easy to check that

\frac{∂u(f; x, x_{k+1})}{∂x_{i}} = C_{k}x_{k+1} ∫_{ℝ^{k}} \frac{(t_{i} − x_{i})f(t) dt}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} .

By virtue of the lemma we have

\frac{∂u(f; x, x_{k+1})}{∂x_{i}} − \overline{D}_{x_{i}(x)}f(0) =

= C_{k}x_{k+1} ∫_{ℝ^{k}} \frac{t_{i}^{2}}{(|t|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} [\frac{f(x + t) − f(x + t − t_{i}e_{i})}{t_{i}} − \overline{D}_{x_{i}(x)}f(0)]dt =

= I_{1} + I_{2},

where

I_{1} = C_{k}x_{k+1} ∫_{V_{δ}} , I_{2} = C_{k}x_{k+1} ∫_{CV_{δ}} ,

V_{δ} is the ball with center at 0 and radius δ. Let ε > 0. Choose δ > 0 such that

|\frac{f(x + t) − f(x + t − t_{i}e_{i})}{t_{i}} − \overline{D}_{x_{i}(x)}f(0)| < ε

for |x| < δ, |t| < 2dl.

Hence

|I_{1}| < C_{k}x_{k+1}ε ∫_{V_{δ}} \frac{t_{i}^{2}dt}{(|t|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} <

< C_{k}x_{k+1}ε ∫_{ℝ^{k}} \frac{t_{i}^{2}dt}{(|t|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} = ε.  (2)

It is likewise easy to show that

lim_{(x,x_{k+1})→(x^{0},0)} I_{2} = 0.  (3)

Equalities (2) and (3) imply the validity of (1).

(b) Let D = (0 ≤ t_{1} < ∞; 0 ≤ t_{2} < ∞; …, 0 ≤ t_{k} < ∞). Define the function f as follows:

f(t) = \begin{cases}\sqrt[k+1]{t_{1}t_{2} ⋯ t_{k}} & if (t_{1}, t_{2}, …, t_{k}) ∈ D, \\ 0 & if (t_{1}, t_{2}, …, t_{k}) ∈ CD. \end{cases}

Clearly, f(t) is continuous in ℝ^{k} and \overline{D}_{x_{i}(\overline{x}_{B})}f(0) = 0, i = \overline{1, k}, for any B when m(B) < k.

If in the integral

\frac{∂u(f; x, x_{k+1})}{∂x_{i}} = C_{k}x_{k+1} ∫_{ℝ^{k}} \frac{(t_{i} − x_{i})f(t) dt}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}}

we use spherical coordinates, then for the considered function we shall have

\frac{∂u(f; 0, x_{k+1})}{∂x_{i}} = C_{k}x_{k+1} ∫_{ℝ^{k}} \frac{t_{i}f(t) dt}{(|t|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} =

= Cx_{k+1} ∫_{0}^{∞} \frac{ρ \sqrt[k−1]{ρ^{k}}}{(ρ^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} ρ^{k−1}dρ =

= Cx_{k+1} ∫_{0}^{∞} \frac{ρ^{k+ \frac{k}{k+1}} dρ}{(ρ^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} > Cx_{k+1} ∫_{0}^{x_{k+1}} \frac{ρ^{k+ \frac{k}{k+1}} dρ}{(ρ^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} ρ^{k−1}dρ >

> Cx_{k+1} ∫_{0}^{x_{k+1}} \frac{ρ^{k+ \frac{k}{k+1}} dρ}{x_{k+1}^{k+3}} = \frac{C}{\sqrt[k+1]{x_{k+1}}} .

Hence

lim_{x_{k+1}→0+} \frac{∂u(f; 0, x_{k+1})}{∂x} = +∞.

Corollary 1. If finite derivatives \overline{D}_{x_{i}(x)}f(x^{0}), i = \overline{1, k}, exist at the point x^{0}, then

lim_{(x,x_{k+1})→(x^{0},0)} d_{x}u(f; x, x_{k+1}) = df(x^{0}).

Corollary 2. If f has a continuous partial derivative at the point x^{0}, then

lim_{(x,x_{k+1})→(x^{0},0)} \frac{∂u(f; x, x_{k+1})}{∂x_{i}} = f′_{x_{i}}(x^{0}).

Corollary 3.

(a) If f is a continuously differentiable function at the point x^{0}, then

lim_{(x,x_{k+1})→(x^{0},0)} d_{x}u(f; x, x_{k+1}) = df(x^{0}).

(b) There exists a differentiable function f(t_{1}, t_{2}) at the point (0, 0) such that df(0, 0) = 0 but the limits

lim_{(x_{1},x_{2},x_{3})→(0,0,0)} \frac{∂u(f; x_{1}, x_{2}, x_{3})}{∂x_{1}}, lim_{(x_{1},x_{2},x_{3})→(0,0,0)} \frac{∂u(f; x_{1}, x_{2}, x_{3})}{∂x_{2}}

do not exist.

Proof of assertion (b) of Corollary 3. We set D = [0, 1; 0, 1]. Let

f(t_{1}, t_{2}) = \begin{cases}\sqrt[5]{t_{1}^{3}t_{2}^{3}} & for (t_{1}, t_{2}) ∈ D, \\ 0 & for (t_{1}, t_{2}) ∈ [−∞, 0; 0, ∞[∪] − ∞, ∞; −∞, 0[, \end{cases}

and continue f onto the set ]0, ∞; 0, ∞[\D so that f ∈ L(ℝ^{2}). It is easy to check that f(t_{1}, t_{2}) is differentiable at the point (0, 0) and

f′_{t_{1}}(0, 0) = f′_{t_{2}}(0, 0) = 0.

Let (x_{1}, x_{2}, x_{3}) → (0, 0, 0) for x_{1} = 0, x_{3} = x_{2}^{2}, x_{2} > 0. Then for the considered function

\frac{∂u(f; 0, x_{2}, x_{3})}{∂x_{1}} = \frac{3x_{3}}{2π} ∫_{0}^{∞}∫_{0}^{∞} \frac{t_{1}f(t_{1}, t_{2}) dt_{1}dt_{2}}{[t_{1}^{2} + (t_{2} − x_{2})^{2} + x_{2}^{4}]^{5/2}} =

= \frac{3x_{2}^{2}}{2π} ∫_{0}^{∞} ∫_{−x_{2}}^{∞} \frac{t_{1}f(t_{1}, t_{2} + x_{2}) dt_{1}dt_{2}}{(t_{1}^{2} + t_{2}^{2} + x_{2}^{4})^{5/2}} =

= \frac{3x_{2}^{2}}{2π} ∫_{0}^{∞} ∫_{−x_{2}}^{∞} \frac{t_{1} \sqrt[5]{t_{1}^{3}(t_{2} + x_{2})^{3}} dt_{1}dt_{2}}{(t_{1}^{2} + t_{2}^{2} + x_{2}^{4})^{5/2}} >

> \frac{3x_{2}^{2}}{2π} ∫_{x_{2}^{2}}^{2x_{2}^{2}}∫_{x_{2}^{2}}^{2x_{2}^{2}} \frac{t_{1} \sqrt[5]{t_{1}^{3}} \sqrt[5]{x_{2}^{3}} dt_{1}dt_{2}}{(t_{1}^{2} + t_{2}^{2} + x_{2}^{4})^{5/2}} >

> \frac{3x_{2}^{2}x_{2}^{3/5}}{2π} ∫_{x_{2}^{2}}^{2x_{2}^{2}}∫_{x_{2}^{2}}^{2x_{2}^{2}} \frac{x_{2}^{2} · x_{2}^{6/5} dt_{1}dt_{2}}{(4x)2^{4} + 4x_{2}^{4} + x_{2}^{4})^{5/2}} =

= \frac{1}{162π\sqrt[5]{x_{2}}} → ∞ for x_{2} → 0 + .

Theorem 2. If a finite derivative D_{x_{i}(\overline{x}_{M\i})}f(x^{0}) exists at the point x^{0}, then

lim_{(x,x_{k+1})→^{∧}_{x_{i}}(x^{0},0)} \frac{∂u(f; x, x_{k+1})}{∂x_{i}} = \frac{∂f(x^{0})}{∂x_{i}}.

Proof. Let x^{0} = 0. By virtue of the lemma we have the equality

\frac{∂u(f; x, x_{k+1})}{∂x_{i}} − D_{x_{i}(\overline{x}_{M\i})}f(0) =

= C_{k}x_{k+1} ∫_{ℝ^{k}} \frac{t_{i}(t_{i} − x_{i})}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} [\frac{f(t) − f(t − t_{i}e_{i})}{t_{i}} − D_{x_{i}(\overline{x}_{M\i})}f(0)]dt =

= I_{1} + I_{2},

where I_{1} = C_{k}x_{k+1} ∫_{V_{δ}}, I_{2} = C_{k}x_{k+1} ∫_{CV_{δ}}.

Let ε > 0 and choose δ > 0 such that

|\frac{f(t) − f(t − t_{i}e_{i})}{t_{i}} − D_{x_{i}(\overline{x}_{M\i})}f(0)| < ε for |r| < δ.

Now,

|I_{1}| < C_{k}x_{k+1}ε ∫_{ℝ^{k}} \frac{|t_{i}(t_{i} − x_{i})| dt}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} <

< C_{k}x_{k+1}ε ∫_{ℝ^{k}} \frac{t_{i}^{2} dt}{(|t|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} +

+ C_{k}x_{k+1}ε|x_{i}| ∫_{ℝ^{k}} \frac{|t_{i}| dt}{(|t|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} =

= ε + Cx_{k+1}ε|x_{i}| ∫_{0}^{∞} \frac{ρ^{k} dρ}{(ρ^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} =

= ε + \frac{Cx_{k+1}^{k+2}|x_{i}|ε}{x_{k+1}^{k+3}} ∫_{0}^{∞} \frac{ρ^{k} dρ}{(1 + ρ^{2})^{\frac{k+3}{2}}} = (1 + \frac{C|x_{i}|}{x_{k+1}})ε.

594 S. TOPURIA

Hence we obtain

lim_{(x,x_{k+1})→^{∧}_{x_{i}}(x^{0},0)} I_{1} = 0.

In a similar manner we prove that

lim_{(x,x_{k+1})→^{∧}_{x_{i}}(x^{0},0)} I_{2} = 0.

Theorem 3. If at the point x^{0} there exist finite derivatives D_{x_{i}(\overline{x}_{M\i})}f(x^{0}) and D_{x_{j}(\overline{x}_{B})}f(x^{0}), i ≠ j, B = M \ \{i, j\}, then

lim_{(x,x_{k+1})→^{∧}_{x_{i}}(x^{0},0)} \frac{∂u(f; x, x_{k+1})}{∂x_{i}} = \frac{∂f(x^{0})}{∂x_{i}},

lim_{(x,x_{k+1})→^{∧}_{x_{i}}(x^{0},0)} \frac{∂u(f; x, x_{k+1})}{∂x_{j}} = \frac{∂f(x^{0})}{∂x_{j}}.

Proof. Let x^{0} = 0. By virtue of the lemma we have

\frac{∂u(f; x, x_{k+1})}{∂x_{j}} = C_{k}x_{k+1} ∫_{ℝ^{k}} \frac{(t_{j} − x_{j})f(t) dt}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} =

=C_{k}x_{k+1}∫_{ℝ^{k}} \frac{(t_{j} −x_{j})\{[f(t) − f(t − t_{i}e_{i})]+[f(t−t_{i}e_{i})−f(t−t_{i}e_{i}−t_{j}e_{j})]\}dt}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} =

= C_{k}x_{k+1} ∫_{ℝ^{k}} \frac{(t_{j} − x_{i})[f(t) − f(t − t_{i}e_{i})]}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} dt +

+C_{k}x_{k+1} ∫_{ℝ^{k}} \frac{(t_{j} − x_{i})[f(t − t_{i}e_{i}) − f(t − t_{i}e_{i} − t_{j}e_{j})]}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} dt = I_{1} + I_{2},

where

I_{1} = C_{k}x_{k+1} ∫_{ℝ^{k}} \frac{t_{i}(t_{j} − x_{j})}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} · \frac{f(t) − f(t − t_{i}e_{i})}{t_{i}} dt =

=C_{k}x_{k+1} ∫_{ℝ^{k}} \frac{t_{i}(t_{j} −x_{j})}{(|t−x|^{2}+x_{k+1}^{2})^{\frac{k+3}{2}}} [\frac{f(t)−f(t−t_{i}e_{i})}{t_{i}} −D_{x_{i}(\overline{x}_{M\i})}f(0)]dt+

+ C_{k}x_{k+1}D_{x_{i}(\overline{x}_{M\i})}f(0) ∫_{ℝ^{k}} \frac{t_{i}(t_{j} − x_{j}) dt}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} = I′_{1} + I′′_{1}.

PARTIAL DERIVATIVES OF THE POISSON INTEGRAL 595

It is easy to see that I′′_{1} and

|I′_{1}|<C_{k}x_{k+1} ∫_{ℝ^{k}} \frac{|t_{i}(t_{j} − x_{j})|}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} |\frac{f(t) − f(t − t_{i}e_{i})}{t_{i}} −D_{x_{i}(\overline{x}_{M\i})}f(0)|dt.

Hence we obtain

lim_{(x,x_{k+1})→^{∧}_{x_{i}}0} I′_{1} = lim_{(x,x_{k+1})→^{∧}_{x_{i}}0} I_{1} = 0.

Now we shall show that

lim_{(x,x_{k+1})→^{∧}_{x_{j}}0} I_{2} = D_{x_{j}(\overline{x}_{M\\{i,j\}})}f(0).

Indeed,

I_{2} = C_{k}x_{k+1} ∫_{ℝ^{k}} \frac{|t_{j}(t_{j} − x_{j})|}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} ×

×[\frac{f(t − t_{i}e_{i})−f(t − t_{i}e_{i} − t_{j}e_{j})}{t_{j}} −D_{x_{j}(\overline{x}_{M\\{i,j\}})}f(0)]dt+D_{x_{j}(\overline{x}_{M\\{i,j\}})}f(0).

This readily implies

lim_{(x,x_{k+1})→^{∧}_{x_{j}}0} I_{2} = D_{x_{j}(\overline{x}_{M\\{i,j\}})}f(0) = \frac{∂f(0)}{∂x_{j}}.

Finally, we obtain

lim_{(x,x_{k+1}) →^{∧}_{x_{i}x_{j}}0} \frac{∂u(f; x, x_{k+1})}{∂x_{j}} = \frac{∂f(0)}{∂x_{j}}.

By a similar reasoning we prove Theorem 4. If at the point x^{0} there exist finite derivatives

D_{x_{1}(x_{2},x_{3},...,x_{k})}f(x^{0}), D_{x_{2}(x_{3},...,x_{k})}f(x^{0}), …, D_{x_{k−1}(x_{k})}f(x^{0}), f′_{x_{k}}(x^{0}), then

lim_{(x,x_{k+1})→^{∧}_{x_{1}}(x^{0},0)} \frac{∂u(f; x, x_{k+1})}{∂x_{1}} = \frac{∂f(x^{0})}{∂x_{1}},

lim_{(x,x_{k+1}) →^{∧}_{x_{1}x_{2}}(x^{0},0)} \frac{∂u(f; x, x_{k+1})}{∂x_{2}} = \frac{∂f(x^{0})}{∂x_{2}},

.............................................................

lim_{(x,x_{k+1})→^{∧}(x^{0},0)} \frac{∂u(f; x, x_{k+1})}{∂x_{k}} = \frac{∂f(x^{0})}{∂x_{k}}.

Corollary. If at the point x^{0} there exist finite derivatives D_{x_{1}(\overline{x}_{M\1})}f(x^{0}), D_{x_{2}(\overline{x}_{M\\{1,2\}})}f(x^{0}), …, D_{x_{k−1}(x_{k})}f(x^{0}), f′_{x_{k}}(x^{0}), then

lim_{(x,x_{k+1})→^{∧}(x^{0},0)} d_{x}u(f; x, x_{k+1}) = df(x^{0}).

Theorem 5.

(a) If at the point x^{0} there exists a finite derivative D_{x_{i}(\overline{x}_{M\i})}^{∗}f(x^{0}), then

lim_{(x−x_{i}e_{i}+x_{i}^{0}e_{i},x_{k+1})→(x^{0},0)} \frac{∂u(f; x − x_{i}e_{i} + x_{i}^{0}e_{i}, x_{k+1})}{∂x_{i}} = D_{x_{i}}^{∗}f(x^{0}).

(b) There exists a continuous function f(x) such that D_{x_{i}(\overline{x}_{M\i})}^{∗}f(x^{0}) = 0, but the limit

lim_{(x,x_{k+1})→^{∧}(x^{0},0)} \frac{∂u(f; x, x_{k+1})}{∂x_{i}}

does not exist.

Proof. (a) Let x^{0} = 0. The validity of (a) follows from the equality

\frac{∂u(f; x − x_{i}e_{i} + x_{i}^{0}e_{i}, x_{k+1})}{∂x_{i}} − D_{x_{i}(\overline{x}_{M\i})}^{∗}f(x^{0}) =

= C_{k}x_{k+1} ∫_{ℝ^{k}} \frac{t_{i}^{2}}{(|t|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} [\frac{f(t + x − x_{i}e_{i}) − f(t + x − x_{i}e_{i} − 2t_{i}e_{i})}{2t_{i}} −

−D_{x_{i}(\overline{x}_{M\i})}f(x^{0})]dt.

(b) We set D_{1} = [0, 1; 0, 1], D_{2} = p − 1, 0; 0, 1]. Let

f(t_{1}, t_{2}) = \begin{cases}\sqrt{t_{1}\sqrt{t_{2}}} & for (t_{1}, t_{2}) ∈ D_{1}, \\ \sqrt{−t_{1}\sqrt{t_{2}}} & for (t_{1}, t_{2}) ∈ D_{2}, \\ 0 & for t_{2} ≤ 0 \end{cases}

and continue f(t_{1}, t_{2}) onto the set ℝ_{+}^{2} \ (D_{1} ∪ D_{2}) so that f ∈ L(ℝ^{2}). It is easy to check that D_{t_{1}(t_{2})}^{∗}f(0) = 0. Let x_{1}^{0} = x_{2}^{0} = 0 and (x_{1}, x_{2}, x_{3}) → (0, 0, 0) so that x_{2} = 0 and x_{3} = x_{1}. Then for the constructed function we have

\frac{∂u(f; x_{1}, x_{2}, x_{3})}{∂x_{1}} = \frac{3x_{3}}{2π} ∫_{ℝ^{2}} \frac{(t_{1} − x_{1})f(t_{1}, t_{2}) dt_{1} dt_{2}}{[(t_{1} − x_{1})^{2} + (t_{2} − x_{2})^{2} + x_{3}^{2}]^{5/2}} =

= Cx_{3}\{ ∫_{−1}^{0} ∫_{0}^{1} \frac{(t_{1} − x_{1})\sqrt{−t_{1}\sqrt{t_{2}}} dt_{1} dt_{2}}{[(t_{1} − x_{1})^{2} + t_{2}^{2} + x_{3}^{2}]^{5/2}} +

+ ∫_{0}^{1} ∫_{0}^{1} \frac{(t_{1} − x_{1})\sqrt{t_{1}\sqrt{t_{2}}} dt_{1} dt_{2}}{[(t_{1} − x_{1})^{2} + t_{2}^{2} + x_{3}^{2}]^{5/2}} \} + o(1) =

= Cx_{1}[ ∫_{x_{1}}^{1+x_{1}}∫_{0}^{1} \frac{t_{1}\sqrt{(t_{1} − x_{1})\sqrt{t_{2}}}}{(t_{1}^{2} + t_{2}^{2} + x_{1}^{2})^{5/2}} dt_{1}dt_{2} +

+ ∫_{−x_{1}}^{1−x_{1}}∫_{0}^{1} \frac{t_{1}\sqrt{(t_{1} + x_{1})\sqrt{t_{2}}}}{(t_{1}^{2} + t_{2}^{2} + x_{1}^{2})^{5/2}} dt_{1}dt_{2}] + o(1) =

= Cx_{1}\{ ∫_{−x_{1}}^{x_{1}} ∫_{0}^{1} \frac{t_{1}\sqrt{(t_{1} + x_{1})\sqrt{t_{2}}}}{(t_{1}^{2} + t_{2}^{2} + x_{1}^{2})^{5/2}} dt_{1}dt_{2} +

+ ∫_{x_{1}}^{1−x_{1}}∫_{0}^{1} \frac{t_{1}[\sqrt{(t_{1} + x_{1})\sqrt{t_{2}}} − \sqrt{(t_{1} − x_{1})\sqrt{t_{2}}}]}{(t_{1}^{2} + t_{2}^{2} + x_{1}^{2})^{5/2}} dt_{1}dt_{2}\} =

= Cx_{1}(I_{1} + I_{2}) + o(1),

where

I_{1} = ∫_{0}^{x_{1}} ∫_{0}^{1} \frac{t_{1}[\sqrt{(t_{1} + x_{1})\sqrt{t_{2}}} − \sqrt{(x_{1} − t_{1})\sqrt{t_{2}}}]}{(t_{1}^{2} + t_{2}^{2} + x_{1}^{2})^{5/2}} dt_{1}dt_{2} > 0,

I_{2} = ∫_{x_{1}}^{1−x_{1}}∫_{0}^{1} \frac{t_{1} \sqrt[4]{t_{2}}(\sqrt{t_{1} + x_{1}} − \sqrt{t_{1} − x_{1}})}{(t_{1}^{2} + t_{2}^{2} + x_{1}^{2})^{5/2}} dt_{1}dt_{2} >

> ∫_{x_{1}}^{2x_{1}}∫_{x_{1}}^{2x_{1}} \frac{t_{1} \sqrt[4]{t_{2}}(\sqrt{t_{1} + x_{2}} − \sqrt{t_{1} − x_{1}})}{(t_{1}^{2} + t_{2}^{2} + x_{1}^{2})^{5/2}} dt_{1}dt_{2} >

> ∫_{x_{1}}^{2x_{1}}∫_{x_{1}}^{2x_{1}} \frac{x_{1} \sqrt[4]{x_{1}}(\sqrt{2x_{1}} − \sqrt{x_{1}})}{(9x_{1}^{2})^{5/2}} dt_{1}dt_{2} = \frac{\sqrt{2} − 1}{128} · \frac{1}{\sqrt[4]{x_{1}^{5}}} .

Thus, by the chosen path, we obtain

\frac{∂u(f; x_{1}, 0, x_{1})}{∂x_{1}} > \frac{C}{\sqrt[4]{x_{1}}} ,

which yields \frac{∂u(f;x_{1},0,x_{1})}{∂x_{1}} → +∞ when (x_{1}, x_{2}, x_{3}) → (0, 0, 0) by the chosen path.

By a similar reasoning we prove

Theorem 6.

(a) If at the point x^{0} there exists a finite derivative \overline{D}_{x_{i}(x)}^{∗}f(x^{0}), i = \overline{1, k}, then

lim_{(x,x_{k+1})→(x^{0},0)} \frac{∂u(f; x, x_{k+1})}{∂x_{i}} = D_{x_{i}}^{∗}f(x^{0}).

(b) There exists a continuous function f(x) such that for any B ⊂ M, m(B) < k, all derivatives \overline{D}_{x_{i}(\overline{x}_{B})}^{∗}f(0) = 0, i = \overline{1, k}, but the limits

lim_{x_{k+1}→0+} \frac{∂u(f; 0, x_{k+1})}{∂x_{i}}

do not exist.

Statement (a) of Theorem 1 is a corollary of statement (a) of Theorem 6.

The validity of (b) follows from statement (b) of Theorem 1.

Theorem 7.

(a) If f has a total differential df(x^{0}) at the point x^{0}, then

lim_{(x,x_{k+1})→^{∧}(x^{0},0)} d_{x}u(f; x, x_{k+1}) = df(x^{0}).  (4)

(b) there exists a continuous function f which has partial derivatives of any order, but the limits

lim_{x_{k+1}→0+} \frac{∂u(f; x^{0}, x_{k+1})}{∂x_{i}}

do not exist.

Proof.

(a) By virtue of the lemma we have (x^{0} = 0)

\frac{∂u(f; x, x_{k+1})}{∂x_{i}} − \frac{∂f(0)}{∂x_{i}} =

= C_{k}x_{k+1} ∫_{ℝ^{k}} \frac{(t_{i} − x_{i}) ∑_{ν=1}^{k} |t_{ν}|}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+3}{2}}} · \frac{f(t) − f(0) − ∑_{ν=1}^{k} \frac{∂f(0)}{∂x_{i}} t_{i}}{∑_{ν=1}^{k} |t_{ν}|} dt.

This equality implies

lim_{(x,x_{k+1})→^{∧}0} \frac{∂u(f; x, x_{k+1})}{∂x_{i}} = \frac{∂f(0)}{∂x_{i}}, i = \overline{1, k}.

Thus equality (4) is valid.

(b) Consider the function

f(t_{1}, t_{2}) = \begin{cases}\sqrt[4]{(2t_{1} − t_{2})(t_{2} − \frac{1}{2}t_{1})} & for (t_{1}, t_{2}) ∈ D = \{(t_{1}, t_{2}) : \\ & 0 ≤ t_{1} < ∞; \frac{1}{2}t_{1} ≤ t_{2} ≤ 2t_{1}\}, \\ 0 & for (t_{1}, t_{2}) ∈ CD. \end{cases}

This function is continuous in ℝ^{2}, has partial derivatives of any order at the point (0, 0) which are equal to zero, but

\frac{∂u(f; 0, 0, x_{3})}{∂x_{1}} = \frac{3x_{3}}{2π} ∫_{0}^{∞}dt_{1} ∫_{\frac{1}{2} t_{1}}^{2t_{1}} \frac{t_{1} \sqrt[4]{(2t_{1} − t_{2})(t_{2} − \frac{1}{2}t_{1})}}{(t_{1}^{2} + t_{2}^{2} + x_{3}^{2})^{5/2}} dt_{2} >

> Cx_{3} ∫_{x_{3}}^{2x_{3}}t_{1} dt_{1} ∫_{t_{1}}^{\frac{3}{2} t_{1}} \frac{\sqrt[4]{(2t_{1} − t_{2})(t_{2} − \frac{1}{2}t_{1})}}{(t_{1}^{2} + t_{2}^{2} + x_{3}^{2})^{5/2}} dt_{2} >

> Cx_{3} ∫_{x_{3}}^{2x_{3}}t_{1} dt_{1} ∫_{t_{1}}^{\frac{3}{2} t_{1}} \frac{\sqrt[4]{(2t_{1} − \frac{3}{2}t_{1})(t_{1} − \frac{1}{2}t_{1})}}{(\frac{13}{4}t_{1}^{2} + x_{3}^{2})^{5/2}} dt_{2} >

> Cx_{3} ∫_{x_{3}}^{2x_{3}}t_{1} dt_{1} ∫_{t_{1}}^{\frac{3}{2} t_{1}} \frac{\sqrt[4]{t_{1}^{2}} dt_{2}}{x_{3}^{5}} = \frac{C}{x_{3}^{4}} ∫_{x_{3}}^{2x_{3}}t_{1}^{5/2}dt_{1} =

= \frac{C}{\sqrt{x_{3}}} → +∞ for x_{3} → 0 + .

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(Received 3.11.1995; revised 15.02.1996)

Author’s address:

Department of Mathematics (No. 63)

Georgian Technical University

77, M. Kostava St., Tbilisi 380093

Georgia