GEORGIAN MATHEMATICAL JOURNAL: Vol. 5, No. 4, 1998, 385-400

BOUNDARY PROPERTIES OF SECOND-ORDER PARTIAL

DERIVATIVES OF THE POISSON INTEGRAL FOR A HALF-SPACE ℝ_{+}^{k+1} (k > 1)

S. TOPURIA

Abstract. The boundary properties of second-order partial derivatives of the Poisson integral are studied for a half-space ℝ_{+}^{k+1}.

In this paper, the theorems generalizing the author’s previous results [1– 5] are proved. It is the continuation of [6] and uses the same notation. Let us recall some of them.

Let M = {1, 2, …, k} (k ≥ 2) and B ⊂ M, B′ = M \ B. For every x ∈ ℝ^{k} and for an arbitrary B ⊂ M the symbol x_{B} denotes a point from ℝ^{k} whose coordinates with indices from the set B coincide with the respective coordinates of the point x, while the 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|)^{\frac{k+1}{2}}} ∈ L(ℝ^{k}); ℝ_{+}^{k+1} = \{(x_{1}, x_{2}, …, x_{k+1}) ∈ ℝ^{k+1}; x_{k+1} >

0\} (half-space); Δ_{x} = Δ_{x_{1}x_{2}···x_{k}} = ∑_{i=1}^{k} \frac{∂^{2}}{∂x_{1}^{2}} ; the Poisson integral of the

function f(x) for the half-space ℝ_{+}^{k+1} is

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}}} .

As in [6], we use the following generalization of dihedral-angular limit introduced by O. P. Dzagnidze: if the point N ∈ ℝ_{+}^{k+1} tends to the point

385 P(x^{0}, 0) by the condition x_{k+1}( ∑_{i∈B}(x_{i} − x_{i}^{0})^{2})^{−1/2} ≥ C > 0, ^{1} then we shall

1991 Mathematics Subject Classification. 31B25.

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

1072-947X/98/0700-0385$15.00/0 \begin{matrix}c \\ ◯ \end{matrix} 1998 Plenum Publishing Corporation

write N(x, x_{k+1}) →^{∧}_{x_{B}} P(x^{0}, 0).

If B = M, then we have an angular limit and 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 tends to P arbitrarily, remaining in the half-space ℝ_{+}^{k+1}.

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

1. 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}) − 2f(x_{B} + x0_{B′})}{u^{2}}

is denoted by:

(a) D_{x_{i}}^{2}f(x^{0}) if B = ∅;

(b) D_{x_{i} (\overline{x}_{B})}^{2}f(x^{0}) if i ∈ B′;

(c) \overline{D}_{x_{i} (\overline{x}_{B})}^{2}f(x^{0}) if i ∈ B.

2. The limit

lim_{\begin{matrix}(u,v)→(0,0) \\ \overline{x}_{B} →\overline{x}_{B}^{0} \end{matrix}} [\frac{f(x_{B} + x0_{B′} + ue_{i} + ve_{j}) − f(x_{B} + x0_{B′} + ue_{i})}{uv} −

−\frac{f(x_{B} + x0_{B′} + ve_{j}) − f(x_{B} + x0_{B′})}{uv}]

is denoted by:

(a) D_{x_{i}x_{j}}f(x_{0}) if B = ∅;

(b) D_{x_{i}x_{j} (\overline{x}_{B})}f(x^{0}) if \{i, j\} ⊂ B′;

(c) D_{[x_{i}x_{j}] (\overline{x}_{B})}f(x^{0}) if \{i, j\} ⊂ B;

(d) D_{[x_{i}]x_{j} (\overline{x}_{B})}f(x^{0}) if i ∈ B, j ∈ B′.

3. The limit

lim_{\begin{matrix}(u,v)→(0,0) \\ \overline{x}_{B} →\overline{x}_{B}^{0} \end{matrix}} [\frac{f(x_{B} + x0_{B′} + ue_{i} + ve_{j}) − f(x_{B} + x0_{B′} + ue_{i} − ve_{j})}{4uv} −

−\frac{f(x_{B} + x0_{B′} − ue_{i} + ve_{j}) − f(x_{B} + x0_{B′} − ue_{i} − ve_{j})}{4uv}]

is denoted by:

(a) D_{x_{i}x_{j}}^{∗}f(x_{0}) if B = ∅;

(b) D_{x_{i}x_{j} (\overline{x}_{B})}^{∗}f(x^{0}) if \{i, j\} ⊂ B′;

(c) D_{[x_{i}x_{j}] (\overline{x}_{B})}^{∗}f(x^{0}) if \{i, j\} ⊂ B;

(d) D_{[x_{i}]x_{j} (\overline{x}_{B})}^{∗}f(x^{0}) if i ∈ B, j ∈ B′.

1Here and in what follows C denotes absolute positive constants which, generally speaking, can be different in different relations.

POISSON INTEGRAL FOR A HALF-SPACE R_{+}^{k+1} (k > 1) 387

4. The limit

lim_{\begin{matrix}ρ→0 \\ x_{B}+x_{B′}^{0} →x^{0} \end{matrix}} \frac{\frac{1}{|S_{ρ}|} ∫_{S_{ρ}(x_{B} +x_{B′}^{0} )} f(t) dS(t) − f(x_{B} + x0_{B′})}{ρ^{2}/2k},

where S_{ρ}(x) is the sphere in ℝ^{k} with center at x, and radius ρ, and the (k − 1)-dimensional surface area |S_{ρ}(x)| is denoted by:

(a) \overline{Δ}f(x^{0}) if B = ∅;

(b) \overline{Δ}_{x}f(x^{0}) if B = M;

(c) \overline{Δ}_{x_{B}}f(x^{0}) if B ≠ ∅ and B ≠ M.

Remark. The λ-derivatives of a function of two variables have been studied by the author in [1, 2] and hence are not considered here.

The following propositions are valid:

(1) For any B ⊂ M, the existence of \overline{D}_{x_{i} (\overline{x}_{B})}^{2}f(x^{0}) implies that there exists D_{x_{i} (\overline{x}_{B\i})}f(x^{0}) and

\overline{D}_{x_{i} (\overline{x}_{B})}f(x^{0}) = D_{x_{i} (\overline{x}_{B\i})}^{2}f(x^{0}) = D_{x_{i}}^{2}f(x^{0}).

If there exists f′′_{x_{i}^{2}}(x^{0}), then D_{x_{i}}^{2}f(x^{0}) exists too, and they have the same value.

(2) If in the neighborhood of the point x^{0} there exists a partial derivative f′′_{x_{i}^{2}}(x) which is continuous at x^{0}, then \overline{D}_{x_{i} (x)}^{2}f(x^{0}) exists too and

\overline{D}_{x_{i} (x)}^{2}f(x^{0}) = f′′_{x_{i}^{2}}(x^{0}).

Indeed, if we apply the Cauchy formula for the function f(x + ue_{i}) + f(x − ue_{i}) − 2f(x^{0}) and for u^{2} with respect to u, then we have

\frac{f(x + ue_{i}) + f(x − ue_{i}) − 2f(x^{0})}{u^{2}} =

= \frac{f′_{x_{i}}(x + θ(x)ue_{i}) − f′_{x_{i}}(x − θ(x)ue_{i})}{2θ(x)u}, 0 < θ < 1.

Now applying the Lagrange formula, we obtain

\frac{f(x + ue_{i}) + f(x − ue_{i}) − 2f(x^{0})}{u^{2}} =

= \frac{2θuf′′_{x_{i}^{2}}(x − θue_{i} + 2θ_{1}θue_{i})}{2θu} = f′′_{x_{i}^{2}}(x − θue_{i} + 2θ_{1}θue_{i}), 0 < θ < 1,

which by the continuity of f′′_{x_{i}^{2}}(x) implies that Proposition (2) is valid. Note that the continuity of the partial derivative f′′_{x_{i}^{2}}(x) at x^{0} is only a

sufficient condition for the existence of the derivative \overline{D}_{x_{i} (\overline{x}_{B})}f(x^{0}) for any B ⊂ M.

(3) If in the neighborhood of the point x there exists a derivative f′′_{x_{i}x_{j}}(x) which is continuous at x^{0}, then there exists \overline{D}_{[x_{i}x_{j}] (x)}f(x^{0}) and \overline{D}_{[x_{i}x_{j}] (x)}f(x^{0}) = f′′_{x_{i}x_{j}}f(x^{0}).

The continuity of f′′_{x_{i}x_{j}}(x) at the point x^{0} is only a sufficient condition for the existence of D_{[x_{i}x_{j}] (x)}f(x^{0}).

(4) If for the function f(x) at x^{0} there exists D_{x_{i}x_{j}}f(x^{0}), then at the same point there exists D_{x_{i}x_{j}}^{∗}f(x^{0}), and their values coincide.

(5) If the function f(x) at the point x^{0} has continuous partial derivatives up to second order inclusive, then at the same point there exists \overline{Δ}f(x^{0}), and \overline{Δ}f(x^{0}) = Δf(x^{0}) (see [7], p. 18).

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

The next lemma is proved analogously to the lemma from [6].

Lemma 1. For every (x_{1}, x_{1}, …, x_{k}) the following equalities are valid:

∫_{∞}^{∞} \frac{(k + 3)(t_{i} − x_{i})^{2} − |t − x|^{2} − x_{k+1}^{2}}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}} dt_{i} = 0,

∫_{ℝ^{k}} \frac{(k + 3)(t_{i} − x_{i})^{2} − |t − x|^{2} − x_{k+1}^{2}}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}} f(t − t_{i}e_{i} + x_{i}e_{i}) dt = 0,

\frac{(k + 1)x_{k+1}Γ(\frac{k+1}{2})}{2π^{\frac{k+1}{2}}} ∫_{ℝ^{k}} \frac{(k + 3)(t_{i} − x_{i})^{2} − |t − x|^{2} − x_{k+1}^{2}}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}} t_{i}^{2} dt = 1.

Theorem 1.

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

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

(b) There exists a continuous function f ∈ L(ℝ^{k}) such that for any B ⊂ M, m(B) < k all derivatives \overline{D}_{x_{i} (\overline{x}_{B})}^{2}f(0) = 0, i = \overline{1, k}, but the limits

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

do not exist.

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

\frac{∂^{2}u(f; x, x_{k+1})}{∂x_{i}^{2}} = C_{k}x_{k+1} ∫_{ℝ^{k}} \frac{(k + 3)(t_{i} − x_{i})^{2} − |t − x|^{2} − x_{k+1}^{2}}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}} f(t) dt =

POISSON INTEGRAL FOR A HALF-SPACE R_{+}^{k+1} (k > 1) 389

= C_{k}x_{k+1} ∫_{ℝ^{k}} \frac{(k + 3)t_{i}^{2} − |t|^{2} − x_{k+1}^{2}}{(|t|^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}} f(x + t) dt =

= C_{k}x_{k+1} ∫_{ℝ^{k}} \frac{(k + 3)t_{i}^{2} − |t|^{2} − x_{k+1}^{2}}{(|t|^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}} f(x + t − 2t_{i}e_{i}) dt.

By Lemma 1 this gives

\frac{∂^{2}u(f; x, x_{k+1})}{∂x_{i}^{2}} − \overline{D}_{x_{i} (x)}^{2}f(0) = \frac{1}{2} c_{k}x_{k+1} ∫_{ℝ^{k}} \frac{[(k + 3)t_{i}^{2} − |t|^{2} − x_{k+1}^{2}]t_{i}^{2}}{(|t|^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}} ×

×[\frac{f(x + t) + f(x + t − 2t_{i}e_{i}) − 2f(x + t − t_{i}e_{i})}{t_{i}^{2}} − \overline{D}_{x_{i} (x)}^{2}f(0)] dt,

which implies that equality (1) is valid. Proof of part (b). Assume D = [0 ≤ t_{1} < ∞; 0 ≤ t_{2} < ∞; 0 ≤ t_{3} < ∞]. Let

f(t) = \begin{cases}\sqrt[3]{t_{1}t_{2}t_{3}} & if (t_{1}, t_{2}, t_{3}) ∈ D, \\ 0 & if (t_{1}, t_{2}, t_{3}) ∈ CD. \end{cases}

We can easily find that \overline{D}_{x_{i} (x_{j})}^{2}f(0) = 0, i, j = 1, 2, 3, i ≠ j. However, for the given function we have

\frac{∂^{2}u(f; 0, x_{4})}{∂x_{1}^{2}} = \frac{4x_{4}}{π^{2}} ∫_{D} \frac{6t_{1}^{2} − |t|^{2} − x_{4}^{2}}{(|t|^{2} + x_{4}^{2})^{4}} \sqrt[3]{t_{1}t_{2}t_{3}} dt =

= \frac{4x_{4}}{π^{2}} ∫_{0}^{∞}∫_{0}^{\frac{π}{2}} ∫_{0}^{\frac{π}{2}} \frac{6ρ^{2} sin^{2} ϑ cos^{2} φ − ρ^{2} − x_{4}^{2}}{(ρ^{2} + x_{4}^{2})^{4}} ×

× \sqrt[3]{ρ^{3} sin^{2} ϑ cos ϑ sin φ cos φ} ρ^{2} sin ϑ dρ dϑ dφ =

= \frac{C}{x_{4}} (4 ∫_{0}^{\frac{π}{2}} \sqrt[3]{sin^{2} ϑ cos ϑ} sin^{3} ϑ dϑ ∫_{0}^{\frac{π}{2}} \sqrt[3]{sin 2φ} sin^{2} φ dφ −

− ∫_{0}^{\frac{π}{2}} \sqrt[3]{sin^{2} ϑ cos ϑ} sin ϑ dϑ ∫_{0}^{\frac{π}{2}} \sqrt[3]{sin 2φ} dφ) =

= \frac{4C}{9x_{4}} Γ(\frac{1}{3})Γ(\frac{2}{3}) ∫_{0}^{\frac{π}{2}} ( sin^{2} φ − \frac{3}{8})\sqrt[3]{sin 2φ} dφ =

= \frac{C}{x_{4}} ( \begin{matrix}arcsin & \sqrt{\frac{3}{8}} \\ ∫ \\ 0 \end{matrix}+ \frac{π}{2} −arcsin \sqrt{\frac{3}{8}} ; ∫ ; \sqrt[arcsin]{\frac{3}{8}}+ ∫_{\frac{π}{2} −arcsin \sqrt{\frac{3}{8}}}^{\frac{π}{2}} ) =

= \frac{C}{x_{4}} (I_{1} + I_{2} + I_{3}).

Clearly, I_{1} < 0, I_{2} > 0, I_{3} > 0. Further,

I_{3} = \begin{matrix}arcsin & \sqrt{\frac{3}{8}} \\ ∫ \\ 0 \end{matrix}( cos^{2} φ − \frac{3}{8})\sqrt[3]{sin 2φ} dφ.

Since I_{1} + I_{3} > 0, we have I_{1} + I_{2} + I_{3} > 0. Finally, we obtain

\frac{∂^{2}u(f; 0, x_{4})}{∂x_{1}^{2}} → +∞ as x_{4} → 0 + .

Corollary. If the function f has a continuous partial derivative \frac{∂^{2}f(x)}{∂x_{i}^{2}} at the point x^{0}, then equality (1) is fulfilled.

Theorem 2.

(a) If at the point x^{0} there exists a finite derivative D_{x_{i} (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{∂^{2}u(f; x − x_{i}e_{i} + x_{i}^{0}e_{i}, x_{k+1})}{∂x_{i}^{2}} = D_{x_{i}}^{2}f(x^{0}).

(b) There exists a function f ∈ L(ℝ^{k}) such that D_{x_{i} (x_{M\i})}f(x^{0}) = 0, but the limit

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

does not exist. Proof of part (a). Let x^{0} = 0. By Lemma 1 we easily obtain

\frac{∂^{2}u(f; x − x_{i}e_{i}, x_{k+1})}{∂x_{i}^{2}} − D_{x_{i} (x_{M\i})}^{2}f(0) =

= \frac{1}{2} c_{k}x_{k+1} ∫_{ℝ^{k}} \frac{[(k + 3)t_{i}^{2} − |t|^{2} − x_{k+1}^{2}]t_{i}^{2}}{(|t|^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}} ×

×[\frac{f(t + x − x_{i}e_{i}) + f(t − 2t_{i}e_{i} + x − x_{i}e_{i}) − 2f(t − t_{i}e_{i} + x − x_{i}e_{i})}{t_{i}^{2}} −

−D_{x_{i} (x_{M\i})}^{2}f(0)] dt.

POISSON INTEGRAL FOR A HALF-SPACE R_{+}^{k+1} (k > 1) 391

Proof of part (b). This will be given for the cases i = 1 and k = 2. We have

\frac{∂^{2}u(f; x_{1}, x_{2}, x_{3})}{∂x_{1}^{2}} = \frac{3x_{3}}{2π} ∫_{ℝ^{2}} \frac{4(t_{1} − x_{1})^{2} − (t_{2} − x_{2})^{2} − x_{3}^{2}}{(|t − x|^{2} + x_{3}^{2})^{7/2}} f(t) dt =

= \frac{3x_{3}}{2π} ∫_{0}^{∞}∫_{0}^{2π}\frac{4(ρ cos φ−x_{1})^{2}−(ρ sin φ−x_{2})−x_{3}^{2}}{[(ρ cos φ−x_{1})^{2}+(ρ sin φ−x_{2})^{2}+x_{3}^{2}]^{7/2}} f(ρ cos φ, ρ sin φ)ρ dρ dφ.

Let N(x_{1}, x_{2}, x_{3}) → (0, 0, 0) so that x_{2} = 0 and x_{3} = x_{1}. Then

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

= \frac{3x_{3}}{2π} ∫_{0}^{∞}∫_{0}^{2π}\frac{4ρ^{2} cos^{2} φ−8ρx_{3} cos φ−ρ^{2} sin^{2} φ+3x_{3}^{2}}{(ρ^{2}−2ρx_{3} cos φ+2x_{3})^{7/2}} f(ρ cos φ, ρ sin φ)ρ dρ dφ.

Using Lemma 1, for any x_{1}, x_{2}, and x_{3} we have

∫_{ℝ^{2}} \frac{4(t_{1} − x_{1})^{2} − (t_{2} − x_{2})^{2} − x_{3}^{2}}{(|t − x|^{2} + x + 3^{2})^{7/2}} dt_{1} dt_{2} = 0,

which, in particular, yields

∫_{0}^{∞}∫_{0}^{2π} \frac{4ρ^{2} cos^{2} φ − 8ρx_{3} cos φ − ρ^{2} sin^{2} φ + 3x_{3}^{2}}{(ρ^{2} − 2ρx_{3} cos φ + 2x_{3})^{7/2}} ρ dρ dφ =

=2∫_{0}^{∞}∫_{0}^{π}\frac{4ρ^{2} cos^{2} φ−8ρx_{3} cos φ−ρ^{2} sin^{2} φ+3x_{3}^{2}}{(ρ^{2} − 2ρx_{3} cos φ + 2x_{3})^{7/2}} f(ρ cos φ, ρ sin φ)ρdρdφ=0.

Therefore

∫_{0}^{∞}∫_{0}^{π} \frac{4ρ^{2} cos^{2} φ − 8ρx_{3} cos φ − ρ^{2} sin^{2} φ + 3x_{3}^{2}}{(ρ^{2} − 2ρx_{3} cos φ + 2x_{3})^{7/2}} ρ dρ dφ = 0.  (2)

In the interval \frac{π}{2} ≤ φ ≤ π we have

x_{3} ∫_{0}^{∞}∫_{\frac{π}{2}}^{π} \frac{4ρ^{2} cos^{2} φ − 8ρx_{3} cos φ − ρ^{2} sin^{2} φ + 3x_{3}^{2}}{(ρ^{2} − 2ρx_{3} cos φ + 2x_{3})^{7/2}} ρ dρ dφ >

> x_{3} ∫_{0}^{∞}∫_{\frac{π}{2}}^{π}\frac{5ρ^{2} cos^{2} φ − ρ^{2}}{(ρ^{2} + 2ρx_{3} + 4x_{3})^{7/2}} ρ dρ dφ=x_{3} ∫_{0}^{∞}∫_{\frac{π}{2}}^{π}\frac{5ρ^{2} cos^{2} φ − ρ^{2}}{(ρ + 2x_{3})^{7}} ρ dρ dφ =

392 S. TOPURIA

= \frac{3π}{4} x_{3} ∫_{0}^{∞} \frac{ρ^{3} dρ}{(ρ + 2x_{3})^{7}} > cx_{3} ∫_{0}^{x_{3}} \frac{ρ^{3} rho}{(ρ + 2x_{3})^{7}} = \frac{c}{x_{3}^{2}} .

Hence

lim_{x_{3}→0+} x_{3} ∫_{0}^{∞}∫_{\frac{π}{2}}^{π} \frac{4ρ^{2} cos^{2} φ − 8ρx_{3} cos φ − ρ^{2} sin^{2} φ + 3x_{3}^{2}}{(ρ^{2} − 2ρx_{3} cos φ + 2x_{3})^{7/2}} ρ dρ dφ = +∞. (3)

Equations (2) and (3) imply

lim_{x_{3}→0+} x_{3} ∫_{0}^{∞}∫_{0}^{\frac{π}{2}} \frac{4ρ^{2} cos^{2} φ − 8ρx_{3} cos φ − ρ^{2} sin^{2} φ + 3x_{3}^{2}}{(ρ^{2} − 2ρx_{3} cos φ + 2x_{3})^{7/2}} ρ dρ dφ = −∞. (4)

Next, we define f(t_{1}, t_{2}) as follows:

f(t_{1}, t_{2}) = \begin{cases}−1 & for t_{1} > 0, t_{2} > 0, \\ 1 & for t_{1} < 0, t_{2} > 0, \\ 0 & for − ∞ < t_{1} < ∞, t_{2} ≤ 0, \\ 0 & for t_{1} = 0, 0 < t_{2} < ∞. \end{cases}

Clearly, for this function

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

However, by (3) and (4), \frac{∂^{2}u(f;x_{1},0,x_{3})}{∂x_{1}^{2}} →+∞, as (x_{1}, x_{2}, x_{3})→(0, 0, 0). Lemma 2. For any (x_{1}, x_{2}, …, x_{k+1}) the following equalities are valid:

∫_{ℝ^{k}} \frac{(t_{i} − x_{i})(t_{j} − x_{j})f(t − t_{i}e_{i} + x_{i}e_{i})}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}} dt = 0,
∫_{ℝ^{k}} \frac{(t_{i} − x_{i})(t_{j} − x_{j})f(t − t_{i}e_{i} − t_{j}e_{j} + x_{i}e_{i} + x_{j}e_{j})}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}} dt = 0,
\frac{(k + 1)(k + 3)x_{k+1}Γ(\frac{k+1}{2})}{2π^{\frac{k+1}{2}}} ∫_{ℝ^{k}} \frac{(t_{i} − x_{i})^{2}(t_{j} − x_{j})^{2}}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}} t_{i}^{2} dt = 1.

Theorem 3.

(a) If at the point x^{0} there exists a finite derivative D_{[x_{i}x_{j}] (x)}f(x^{0}), i ≠ j, then

lim_{(x,x_{k+1})→(x^{0},0)} \frac{∂^{2}u(f; x, x_{k+1})}{∂x_{i}∂x_{j}} = D_{x_{i}x_{j}}f(x^{0}).

(b) There exists a continuous function f ∈ L(ℝ^{k}) such that for any B ⊂ M, m(B) < k all derivatives D_{[x_{i}x_{j}] (x)}f(0) = 0, but the limits

lim_{x_{k+1}→0+} \frac{∂^{2}u(f; 0, x_{k+1})}{∂x_{i}∂x_{j}}

do not exist.

Proof of part (a). Let x^{0} = 0 and B_{k} = \frac{(k+1)(k+3)Γ( \frac{k+1}{2} )}{π^{\frac{k+1}{2}}}. It is easy to

verify that

\frac{∂^{2}u(f; x, x_{k+1})}{∂x_{i}∂x_{j}} = B_{k}x_{k+1} ∫_{ℝ^{k}} \frac{(t_{i} − x_{i})(t_{j} − x_{j})f(t) dt}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}}.

By Lemma 2 we obtain

\frac{∂^{2}u(f; x, x_{k+1})}{∂x_{i}∂x_{j}} − D_{[x_{i}x_{j}] (x)}f(0) = B_{k}x_{k+1} ∫_{ℝ^{k}} \frac{t_{i}^{2}t_{j}^{2}}{(|t|^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}} ×

×[\frac{f(x + t) − f(x + t − t_{i}e_{i}) + f(x + t − t_{j}e_{j}) + f(x + t − t_{i}e_{i} − t_{j}e_{j})}{t_{i}t_{j}} −

−D_{[x_{i}x_{j}] (x)}f(0)] dt,

which implies that part (a) is valid.

Proof of part (b). Let k = 4 and D = (∑_{i=1}^{4} t_{i}^{2} ≤ 1, t_{i} ≥ 0, i = \overline{1, 4}). We

define f as f(t) = \sqrt[5]{t_{1}t_{2}t_{3}t_{4}} for t ∈ D, and extend it continuously to the set ℝ^{4} \ D so that f ∈ L(ℝ^{4}). It is easy verify that for any B ⊂ M, m(B) < 4 we have D_{[x_{i}x_{j}] (x)}f(0) = 0. But

\frac{∂^{2}u(f; 0, x_{5})}{∂x_{1}∂x_{1}} = cx_{5} ∫_{D} \frac{t_{1}t_{2} \sqrt[5]{t_{1}t_{2}t_{3}t_{4}}}{(|t|^{2} + x_{5}^{2})^{9/2}} dt + o(1) =

= cx_{5} ∫_{0}^{1} \frac{ρ^{29/5}dρ}{(ρ^{2} + x_{5}^{2})^{9/2}} + o(1) → +∞ as x_{5} → 0 + .

Corollary. If the function f at the point x^{0} has a continuous derivative f′′_{x_{i}x_{j}}(x), then

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

The following theorem is valid by the corollaries of Theorems 1 and 3.

Theorem 4. If the function f is twice continuously differentiable at the point x^{0}, then lim_{(x,x_{k+1})→(x^{0},0)} d_{x}^{2}u(f; x, x_{k+1}) = d^{2}f(x^{0}).

Theorem 5.

(a) If at the point x^{0} there exists a finite derivative D_{[x_{i}]x_{j} (x)}f(x^{0}), then

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

(b) There exists a continuous function f ∈ L(ℝ^{k}) such that for every B ⊂ M, m(B) < k − 1 all derivatives D_{[x_{i}]x_{j} (\overline{x}_{B})}f(0) = 0, but the limits

lim_{x_{k+1}→0+} \frac{∂^{2}u(f; 0, x_{k+1})}{∂x_{i}∂x_{j}}

do not exist.

Theorem 6.

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

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

.

(b) There exists a continuous function f ∈ L(ℝ^{k}) such that for every B ⊂ M, m(B) < k − 2 all derivatives D_{x_{i}x_{j} (\overline{x}_{B})}f(0) = 0, but the limits

lim_{x_{k+1}→0+} \frac{∂^{2}u(f; 0, x_{k+1})}{∂x_{i}∂x_{j}}

do not exist.

Theorem 7.

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

lim_{(x,x_{k+1})→^{∧}(x^{0},0)} d_{x}^{2}u(f; x, x_{k+1}) = d^{2}f(x^{0}).

(b) There exists a continuous function f ∈ L(ℝ^{k}) such that it is differentiable at the point x^{0} = (0, 0) and has at this point all partial derivatives of any order; however the limits

lim_{x_{k+1}→0+} \frac{∂^{2}u(f; 0, x_{k+1})}{∂x_{i}∂x_{j}}, i, j = \overline{1, k}

do not exist.

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

\frac{∂^{2}u(f; x, x_{k+1})}{∂x_{i}^{2}} − \frac{∂^{2}f(0)}{∂x_{i}^{2}} =

= c_{k}x_{k+1} ∫_{ℝ^{k}} \frac{[(k + 3)(t_{i} − x_{i})^{2} − |t − x|^{2} − x_{k+1}^{2}]|t|^{2}}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}} ×

×[\frac{f(t) + f(0) − ( ∑_{ν=1}^{k} t_{ν} \frac{∂}{∂t_{ν}})f(0) − \frac{1}{2}( ∑_{ν=1}^{k} t_{ν} \frac{∂}{∂t_{ν}})^{2}f(0)}{|t|^{2}} dt,

\frac{∂^{2}u(f; x, x_{k+1})}{∂x_{i}∂x_{j}} − \frac{∂^{2}f(0)}{∂x_{i}∂x_{j}} =

= B_{k}x_{k+1} ∫_{ℝ^{k}} \frac{(t_{i} − x_{i})(t_{j} − x_{j})|t|^{2}}{−} (|t − x|^{2} + x_{k+1}^{2})^{\frac{k+5}{2}} ×

×[\frac{f(t) + f(0) − ( ∑_{ν=1}^{k} t_{ν} \frac{∂}{∂t_{ν}})f(0) − \frac{1}{2}( ∑_{ν=1}^{k} t_{ν} \frac{∂}{∂t_{ν}})^{2}f(0)}{|t|^{2}} dt.

Proof of part (b). Consider the function equal to \sqrt[3]{(2t_{1} − t_{2})^{2}(t_{2} − \frac{1}{2}t_{1})^{2}} when (t_{1}, t_{2}) ∈ D = \{(t_{1}, t_{2}) ∈ ℝ^{2}: 0 ≤ t_{1} < ∞; \frac{1}{2}t_{1} ≤ t_{2} ≤ 2t_{1}\} and to 0 otherwise, which is continuous in ℝ^{2}, differentiable at the point (0, 0), and has all partial derivatives of any order equal to zero, but

\frac{∂^{2}u(f; 0, 0, x_{3})}{∂x_{1}∂x_{2}} = \frac{15x_{3}}{2π} ∫_{0}^{∞}dt_{1} ∫_{\frac{1}{2} t_{1}}^{2t_{1}} \frac{t_{1}t_{2} \sqrt[3]{(2t_{1} − t_{2})^{2}(t_{2} − \frac{1}{2}t_{1})^{2}}}{(t_{1}^{2} + t_{2}^{2} + x_{3}^{2})^{7/2}} dt_{2} >

> cx_{3} ∫_{x_{3}}^{2x_{3}}t_{1}^{2}dt_{1} ∫_{t_{1}}^{\frac{3}{2} t_{1}} \frac{\sqrt[3]{(2t_{1} − \frac{3}{2}t_{1})^{2}(t_{1} − \frac{1}{2}t_{1})^{2}}}{(t_{1}^{2} + 4t_{1}^{2} + x_{3}^{2})^{7/2}} =

= \frac{c}{x_{3}^{6}} ∫_{x_{3}}^{2x_{3}}t1^{\frac{13}{3}} dt_{1} = \frac{c}{\sqrt[3]{x_{3}^{2}}} → ∞ as x_{3} → 0 + .

Theorem 8.

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

lim_{x_{k+1}→0+} \frac{∂^{2}u(f; x^{0}, x_{k+1})}{∂x_{i}∂x_{j}} = D_{x_{i}x_{j}}^{∗}f(x^{0}).

(b) There exists a continuous function f ∈ L(ℝ^{k}) such that we have D_{x_{i}x_{j} (\overline{x}_{M\\{i,j\}})}^{∗}f(x^{0}) = 0, but the limits

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

do not exist.

Proof of part (b). This is given for the case k = 2. Assume that D_{1} = [0, 1; 0, 1], D_{2} = [−1, 0; 0, 1]. Let

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

and extend f(t_{1}, t_{2}) continuously to the set ℝ^{2}\(D_{1}∪D_{2}) so that f ∈ L(ℝ^{2}). It is easy to verify that D^{∗}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{∂^{2}u(f; x_{1}, x_{2}, x_{3})}{∂x_{1}∂x_{2}} = \frac{15x_{3}}{2π} \{ ∫_{0}^{1} ∫_{0}^{1} \frac{(t_{1} − x_{1})t_{2}\sqrt{t_{1}t_{2}} dt_{1} dt_{2}}{[(t_{1} − x_{1})^{2} + t_{2}^{2} + x_{3}^{2}]^{7/2}} +

+ ∫_{−1}^{0} ∫_{0}^{1} \frac{(t_{1} − x_{1})t_{2}\sqrt{−t_{1}t_{2}} dt_{1} dt_{2}}{[(t_{1} − x_{1})^{2} + t_{2}^{2} + x_{3}^{2}]^{7/2}} \} + o(1) =

= cx_{1}\{ ∫_{−x_{1}}^{1−x_{1}}∫_{0}^{1} \frac{t_{1}t_{2}\sqrt{t_{2}(t_{1} + x_{1})} dt_{1} dt_{2}}{(t_{1}^{2} + t_{2}^{2} + x_{1}^{2})^{7/2}} −

− ∫_{x_{1}}^{1+x_{1}}∫_{0}^{1} \frac{t_{1}t_{2}\sqrt{(t_{1} − x_{1})t_{2}} dt_{1} dt_{2}}{(t_{1}^{2} + t_{2}^{2} + x_{1}^{2})^{7/2}} \} + o(1) =

= cx_{1}\{ ∫_{−x_{1}}^{x_{1}} ∫_{0}^{1} \frac{t_{1}t_{2}\sqrt{t_{2}(t_{1} + x_{1})} dt_{1} dt_{2}}{(t_{1}^{2} + t_{2}^{2} + x_{1}^{2})^{7/2}} +

+ ∫_{x_{1}}^{1−x_{1}}∫_{0}^{1} \frac{t_{1}t_{2}[\sqrt{t_{2}(x_{1} + t_{1})} − \sqrt{t_{2}(t_{1} − x_{1})}]}{(t_{1}^{2} + t_{2}^{2} + x_{1}^{2})^{7/2}} dt_{1} dt_{2} −

− ∫_{1−x_{1}}^{1+x_{1}}∫_{0}^{1} \frac{t_{1}t_{2}\sqrt{(t_{1} − x_{1})t_{2}} dt_{1} dt_{2}}{(t_{1}^{2} + t_{2}^{2} + x_{1}^{2})^{7/2}} \} + o(1) =

= Cx_{1}(I_{1} + I_{2} − I_{3}) + o(1).

We can readily show that

I_{1} = ∫_{0}^{x_{1}} ∫_{0}^{1} \frac{t_{1}t_{2}[\sqrt{t_{2}(t_{1} + x_{1})} − \sqrt{t_{2}(x_{1} − t_{1})}]}{(t_{1}^{2} + t_{2}^{2} + x_{1}^{2})^{7/2}} dt_{1} dt_{2} > 0, I_{3} = O(1). (5)

Next,

I_{2} > ∫_{x_{1}}^{2x_{1}}∫_{0}^{x_{1}} \frac{t_{1}t_{2}[\sqrt{t_{2}(x_{1} + t_{1})} − \sqrt{t_{2}x_{1}}]}{(t_{1}^{2} + t_{2}^{2} + x_{1}^{2})^{7/2}} dt_{1} dt_{2} =

= ∫_{0}^{x_{1}}t_{2}^{3/2} ∫_{x_{1}}^{2x_{2}} \frac{t_{1}(\sqrt{x_{1} + t_{1}} − \sqrt{x_{1}})}{(t_{1}^{2} + t_{2}^{2} + x_{1}^{2})^{7/2}} dt_{1} >

> ∫_{0}^{x_{1}}t_{2}^{3/2} ∫_{x_{1}}^{2x_{2}} \frac{x_{1}(\sqrt{2x_{1}} − \sqrt{x_{1}})}{(4x_{1}^{2} + x_{1}^{2} + x_{1}^{2})^{7/2}} dt_{1} > \frac{c}{x_{1}^{2}} .  (6)

Therefore for x_{2} = 0 and x_{3} = x_{1} (5) and (6) imply \frac{∂^{2}u(f;x_{1},0,x_{1})}{∂x_{1}∂x_{2}} > \frac{c}{x_{1}} ,

from which we obtain \frac{∂^{2}u(f;x_{1},x_{2},x_{3})}{∂x_{1}∂x_{2}} → +∞, as (x_{1}, x_{2}, x_{3}) → (0, 0, 0).

Theorem 9. If at the point x^{0} there exists a finite derivative D_{[x_{i}x_{j}] (x)}^{∗}f(x^{0}), then

lim_{(x,x_{k+1})→(x^{0},0)} \frac{∂^{2}u(f; x, x_{k+1})}{∂x_{i}∂x_{j}} = D_{x_{i}x_{j}}^{∗}f(x^{0}).

Part (a) of Theorem 3 is the corollary of Theorem 9.

Theorem 10. If at the point x^{0} there exists a finite derivative D_{[x_{i}]x_{j} (x)}^{∗}f(x^{0}), then

lim_{(x−x_{j}e_{j}+x_{j}^{0}e_{j},x_{k+1})→(x^{0},0)} \frac{∂^{2}u(f; x − x_{j}e_{j} + x_{j}^{0}e_{j}, x_{k+1})}{∂x_{i}∂x_{j}} = D_{x_{i}x_{j}}^{∗}f(x^{0}).

Lemma 3. The following equalities are valid:

∫_{0}^{∞} \frac{2ρ^{2} − kx_{k+1}^{2}}{(ρ^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}} ρ^{k−1}dρ = 0,

\frac{(k + 1)x_{k+1}Γ(\frac{k+1}{2})}{k\sqrt{π}Γ(\frac{k}{2} )} ∫_{0}^{∞} \frac{3ρ^{2} − kx_{k+1}^{2}}{(ρ^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}} ρ^{k+1}dρ = 1.

Theorem 11.

(a) Let B ⊂M. If at the point x^{0} there exists a finite derivative \overline{Δ}_{x_{B}}f(x^{0}), then

lim_{x_{B} +x_{B′}^{0} ,x_{k+1})→(x^{0},0)} Δ_{x}u(f; x_{B} + x0_{B′}, x_{k+1}) = \overline{Δ}_{x_{B}}f(x^{0}).

(b) There exists a function f ∈ L(ℝ^{k}) such that \overline{Δ}f(x^{0}) exists but the

limit lim_{x,x_{k+1})→^{∧}(x^{0},0)} Δ_{x}u(f; x, x_{k+1}) does not.

Proof of part (a). One can easily verify that

Δ_{x}u(f; x, x_{k+1}) = −\frac{∂^{2}u(f; x, x_{k+1})}{∂x_{k+1}^{2}} =

= \frac{(k + 1)x_{k+1}Γ(\frac{k+1}{2})}{π^{\frac{k+1}{2}}} ∫_{ℝ^{k}} \frac{3|t − x|^{2} − kx_{k+1}^{2}}{(|t − x|^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}} f(t) dt.

Assume that D_{k} = \frac{2(k+1)Γ( \frac{k+1}{2} )}{\sqrt{π}Γ( \frac{k}{2} )} and θ = [0, π]^{k−2} × [0, 2π]. Passing to the spherical coordinates, we have

Δ_{x}u(f; x_{B} + x0_{B′}, x_{k+1}) = c_{k}x_{k+1} ∫_{0}^{∞} \frac{3ρ^{2} − kx_{k+1}^{2}}{(ρ^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}} ×

× ∫_{Θ} f(x_{B} + x0_{B′} + t)ρ^{k−1} sin^{k−2} ϑ_{1} ⋯ sin ϑ_{k−2} dρ dϑ_{1} ⋯ dϑ_{k−2} dφ =

= D_{k}x_{k+1} ∫_{0}^{∞} \frac{3ρ^{2} − kx_{k+1}^{2}}{(ρ^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}} ρ^{k−1}[\frac{1}{|S_{ρ}|} ∫_{S_{ρ}(x_{B} +x_{B′}^{0} )} f(t) dS(t)]dρ.

Hence by Lemma 3 we obtain

Δ_{x}u(f; x_{B} + x0_{B′}, x_{k+1}) − \overline{Δ}_{x_{B}}f(x^{0}) = D_{k}x_{k+1} ∫_{0}^{∞} \frac{3ρ^{2} − kx_{k+1}^{2}}{(ρ^{2} + x_{k+1}^{2})^{\frac{k+5}{2}}} ×

×[ \frac{\frac{1}{|S_{ρ}|} ∫_{S_{ρ}(x_{B} +x_{B′}^{0} )} f(t) dS(t) − f(x_{B} + x0_{B′})}{ρ^{2}/2k} − \overline{Δ}_{x_{B}}f(x^{0})]\frac{ρ^{2}}{2k} dρ.

For the proof of part (b) see [2], p. 16.

Corollary 1. If at the point x^{0} there exists a finite derivative \overline{Δ}f(x^{0}), then

lim_{x_{k+1}→0+} Δ_{x}u(f; x^{0}, x_{k+1}) = \overline{Δ}f(x^{0}).

Corollary 2. If at the point x^{0} there exists a finite derivative \overline{Δ}f(x^{0}), then

lim_{(x,x_{k+1})→(x^{0},0)} Δ_{x}u(f; x, x_{k+1}) = \overline{Δ}_{x}f(x^{0}).

Let δ > 0 and S_{δ} = ∏_{ν=1}^{k}[x_{ν}^{0} − δ; x_{ν}^{0} + δ].

Theorem 12. Let f′_{t_{i}}(t) ∈ L(S_{δ}). If f′_{t_{i}}(t) has a finite derivative D_{x_{i} (\overline{x}_{M\i})}f′_{x_{i}}(x^{0}) at the point x^{0}, then

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

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

\frac{∂^{2}u(f; x, x_{k+1})}{∂x_{i}^{2}} − D_{x_{i} (\overline{x}_{M\i})}f′_{x_{i}}(0) =

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

−D_{x_{i} (\overline{x}_{M\i})}f′_{x_{i}}(0)]dt + o(1).

The theorems below are proved analogously.

Theorem 13. Let f′_{x_{i}}(t) ∈ L(S_{δ}). If f′_{x_{i}}(t) has a finite derivative D_{x_{j} (\overline{x}_{M\j})}f′_{x_{i}}(x^{0}) at the point x^{0}, then

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

Corollary. Let f′_{x_{i}} and f′_{x_{j}} ∈ L(S_{δ}) for some δ > 0. If at the point x^{0} there exist finite derivatives D_{x_{i} (\overline{x}_{M\i})}f′_{x_{j}}(x^{0}) and D_{x_{j} (\overline{x}_{M\j})}f′_{x_{i}}(x^{0}), then

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

Note that by this corollary we have obtained the condition of the coincidence of mixed derivatives of the function f at the point x^{0}.

Theorem 14. Let f′_{x_{i}} ∈ L(S_{δ}) for some δ > 0. If at the point x^{0} there exists a finite derivative \overline{D}_{x_{i} (x)}f′_{x_{i}}(x^{0}), then

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

Theorem 15. Let f′_{x_{i}} ∈ L(S_{δ}). If f′_{x_{i}}(t) has a finite derivative \overline{D}_{x_{j} (x)}f′_{x_{i}}(x^{0}) at the point x^{0}, then

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

References

1. S. B. Topuria, Solution of the Dirichlet problem for a half-space. (Russian) Dokl. Akad. Nauk SSSR 195(1970), No. 3, 567–569.

2. S. B. Topuria, Boundary properties of the Poisson differentiated integral and solution of the Dirichlet problem in a half-space. (Russian) Trudy Tbiliss. Polytekh. Inst. 197(1977), No. 6, 11–30.

3. S. B. Topuria, Boundary properties of the Poisson differentiated integral in a half-space and representation of a function of two variables. (Russian) Bull. Georgian Acad. Sci. 152(1995), No. 2, 235–238.

4. S. B. Topuria, Boundary properties of derivatives of the Poisson integral in a half-space. (Russian) Bull. Georgian Acad. Sci. 152(1995), No. 3, 453–457.

5. S. B. Topuria, Boundary properties of the Poisson differentiated integral in a half-space and representation of a function of two variables. (Russian) Dokl. Akad. Nauk 353(1997), No. 3, 306–309.

6. S. B. Topuria, Boundary properties of first-order partial derivatives of the Poisson integral for a half-space. Georgian Math. J. 4(1997), No. 6, 585–600.

7. I. I. Privalov, Subharmonic functions. (Russian) Moscow–Leningrad, 1937.

(Received 3.11.1995; revised 15.02.1996)

Author’s address:

Department of Mathematics (No. 63)

Georgian Technical University

77, M. Kostava St., Tbilisi 380075

Georgia