+ (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

Transcription

1 GEORGIAN MATHEMATICAL JOURNAL: Vol. 5, No. 4, 1998, 85-4 BOUNDARY PROPERTIES OF SECOND-ORDER PARTIAL DERIVATIVES OF THE POISSON INTEGRAL FOR A HALF-SPACE +1 + (k > 1) S. TOPURIA Abstract. The boundary properties of second-order partial derivatives of the Poisson integral are studied for a half-space 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,,..., k} (k ) and B M, B = M \ B. For every x and for an arbitrary B M the symbol x B denotes a point from 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,..., i s }, 1 s k (i l < i r for l < r), then x B = (x i1, x i,..., x is ) R s ; m(b) is the number of elements of the set B; L( ) is the set of functions f(x) = f(x 1, x,..., x k ) such that L( ); R+ k+1 = {(x 1, x,..., x k+1 ) +1 ; x k+1 > f(x) (1+ x ) k+1 } (half-space); x = x1 x x k = k function f(x) for the half-space +1 + is i=1 u(f; x, x k+1 ) = x k+1γ( k+1 ) k+1 ; the Poisson integral of the x 1 f(t) dt ( t x + x k+1 ) k+1 As in [6], we use the following generalization of dihedral-angular it introduced by O. P. Dzagnidze: if the point N +1 + tends to the point Mathematics Subect Classification. 1B5. Key words and phrases. Poisson integral for a half-space, angular it, derivatives of the integral, differential properties of the density X/98/7-85$15./ c 1998 Plenum Publishing Corporation

2 86 S. TOPURIA P(x, ) by the condition x k+1 ( (x i x i ) ) 1/ C >, 1 then we shall write N(x, x k+1 ) xb P(x, ). i B If B = M, then we have an angular it and write N(x, x k+1 ) P(x, ). Finally, the notation N(x, x k+1 ) P(x, ) means that the point N tends to P arbitrarily, remaining in the half-space Let u R, v R. We shall consider the following derivatives of the function f(x): 1. The it f(x B + x B + ue i ) + f(x B + x ue B i) f(x B + x ) B (u,x B ) (,x ) u B is denoted by: (a) D x i f(x ) if B = ; (b) D x i (x B ) f(x ) if i B ; (c) D x i (x B ) f(x ) if i B.. The it [ f(xb + x + ue B i + ve ) f(x B + x + ue B i) (u,v) (,) uv x B x B f(x + B x + ve B ) f(x B + x ) ] B uv is denoted by: (a) D xi x f(x ) if B = ; (b) D xix (x B )f(x ) if {i, } B ; (c) D [xix ] (x B )f(x ) if {i, } B; (d) D [xi]x (x B )f(x ) if i B, B.. The it (u,v) (,) x B x B [ f(xb + x + ue B i + ve ) f(x B + x + ue B i ve ) 4uv f(x + B x ue B i + ve ) f(x B + x ue B i ve ) ] 4uv is denoted by: (a) D x i x f(x ) if B = ; (b) D x ix (x B ) f(x ) if {i, } B ; (c) D [x i x ] (x B ) f(x ) if {i, } B; (d) D [x i ]x (x B ) f(x ) if i B, B. 1 Here and in what follows C denotes absolute positive constants which, generally speaking, can be different in different relations.

3 4. The it POISSON INTEGRAL FOR A HALF-SPACE +1 + (k > 1) 87 ρ x B +x B x 1 S ρ S ρ (x B +x B ) f(t) ds(t) f(x B + x B ) ρ /k where S ρ (x) is the sphere in with center at x, and radius ρ, and the (k 1)-dimensional surface area S ρ (x) is denoted by: (a) f(x ) if B = ; (b) x f(x ) if B = M; (c) xb f(x ) if B and B M. Remark. The λ-derivatives of a function of two variables have been studied by the author in [1, ] and hence are not considered here. The following propositions are valid: (1) For any B M, the existence of D x i (x B) f(x ) implies that there exists D xi (x B\i )f(x ) and D xi (x B )f(x ) = D x i (x B\i ) f(x ) = D x i f(x ). If there exists f x i (x ), then D x i f(x ) exists too, and they have the same value. () If in the neighborhood of the point x there exists a partial derivative f x i (x) which is continuous at x, then D x i (x) f(x ) exists too and D x i (x) f(x ) = f x i (x ). Indeed, if we apply the Cauchy formula for the function f(x + ue i ) + f(x ue i ) f(x ) and for u with respect to u, then we have f(x + ue i ) + f(x ue i ) f(x ) u = = f x i (x + θ(x)ue i ) f x i (x θ(x)ue i ), < θ < 1. θ(x)u Now applying the Lagrange formula, we obtain = θuf x i (x θue i + θ 1 θue i ) θu f(x + ue i ) + f(x ue i ) f(x ) u = = f x i (x θue i + θ 1 θue i ), < θ < 1, which by the continuity of f x i (x) implies that Proposition () is valid. Note that the continuity of the partial derivative f x i (x) at x is only a sufficient condition for the existence of the derivative D xi (x B )f(x ) for any B M.,

4 88 S. TOPURIA () If in the neighborhood of the point x there exists a derivative f x ix (x) which is continuous at x, then there exists D [xix ] (x)f(x ) and D [xi x ] (x)f(x ) = f x i x f(x ). The continuity of f x i x (x) at the point x is only a sufficient condition for the existence of D [xix ] (x)f(x ). (4) If for the function f(x) at x there exists D xi x f(x ), then at the same point there exists Dx i x f(x ), and their values coincide. (5) If the function f(x) at the point x has continuous partial derivatives up to second order inclusive, then at the same point there exists f(x ), and f(x ) = f(x ) (see [7], p. 18). In what follows it will always be assumed that f L( ). 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: (k + )(t i x i ) t x xk+1 ( t x + xk+1 ) k+5 (k + )(t i x i ) t x x k+1 ( t x + xk+1 ) k+5 (k + 1)x k+1 Γ( k+1 ) k+1 dt i =, f(t t i e i + x i e i ) dt =, (k + )(t i x i ) t x x k+1 ( t x + x k+1 ) k+5 t i dt = 1. Theorem 1. (a) If at the point x there exists a finite derivative D x i (x) f(x ), then u(f; x, x k+1 ) (x,x k+1 ) (x,) xi = Dx i f(x ). (1) (b) There exists a continuous function f L( ) such that for any B M, m(b) < k all derivatives D x i (x B ) f() =, i = 1, k, but the its do not exist. Proof of part (a). Let x =, C k = u(f; x, x k+1 ) x i = C k x k+1 u(f;, x k+1 ) x k+1 + x i k+1 (k+1)γ( ) k+1. It is easy to verify that (k + )(t i x i ) t x x k+1 ( t x + xk+1 ) k+5 f(t) dt =

5 POISSON INTEGRAL FOR A HALF-SPACE +1 + (k > 1) 89 = C k x k+1 = C k x k+1 (k + )ti t x k+1 ( t + x k+1 ) k+5 (k + )ti t x k+1 ( t + x k+1 ) k+5 f(x + t) dt = f(x + t t i e i ) dt. By Lemma 1 this gives u(f; x, x k+1 ) x i D x i (x) f() = 1 c kx k+1 [(k + )ti t xk+1 ]t i ( t + x k+1 ) k+5 [ ] f(x + t) + f(x + t ti e i ) f(x + t t i e i ) Dx i (x) f() dt, t i which implies that equality (1) is valid. Proof of part (b). Assume D = [ t 1 < ; t < ; t < ]. Let f(t) = { t1 t t if (t 1, t, t ) D, if (t 1, t, t ) CD. We can easily find that D x i (x ) f() =, i, = 1,,, i. However, for the given function we have u(f;, x 4 ) x 1 = 4x 4 = 4x 4 6t 1 t x 4 ( t + x 4 )4 t 1 t t dt = D 6ρ sin ϑ cos ϕ ρ x 4 (ρ + x 4 )4 ρ sin ϑ cos ϑ sin ϕ cos ϕ ρ sin ϑ dρ dϑ dϕ = = C x 4 (4 sin ϑ cos ϑ sin ϑ dϑ sin ϑ cos ϑ sin ϑ dϑ sin ϕ sin ϕ dϕ ) sin ϕ dϕ = = 4C 1 ) ( Γ( Γ 9x 4 ) ( sin ϕ ) sin ϕ dϕ = 8

6 9 S. TOPURIA = C ( arcsin 8 + x 4 arcsin 8 arcsin 8 + arcsin 8 ) = = C x 4 (I 1 + I + I ). Clearly, I 1 <, I >, I >. Further, I = arcsin 8 ( cos ϕ 8 ) sin ϕ dϕ. Since I 1 + I >, we have I 1 + I + I >. Finally, we obtain u(f;, x 4 ) x 1 + as x 4 +. Corollary. If the function f has a continuous partial derivative f(x) x i at the point x, then equality (1) is fulfilled. Theorem. (a) If at the point x there exists a finite derivative D xi (x M\i )f(x ), then u(f; x x i e i + xi e i, x k+1 ) (x x ie i+xi ei,x k+1 ) (x,) x = Dx i f(x ). i (b) There exists a function f L( ) such that D xi (x M\i )f(x ) =, but the it u(f; x, x k+1 ) (x,x k+1 (x,) x i does not exist. Proof of part (a). Let x =. By Lemma 1 we easily obtain u(f; x x i e i, x k+1 ) xi Dx i (x M\i ) f() = = 1 c [(k + )ti kx t x k+1 ]t i k+1 ( t + x k+1 ) k+5 [ f(t + x xi e i ) + f(t t i e i + x x i e i ) f(t t i e i + x x i e i ) t i D x i (x M\i ) f() ] dt.

7 Proof of part (b). have = x POISSON INTEGRAL FOR A HALF-SPACE +1 + (k > 1) 91 u(f; x 1, x, x ) x 1 This will be given for the cases i = 1 and k =. We = x R 4(t 1 x 1 ) (t x ) x ( t x + x )7/ f(t) dt = 4(ρ cos ϕ x 1 ) (ρ sin ϕ x ) x [(ρ cos ϕ x 1 ) +(ρ sin ϕ x ) +x f(ρ cos ϕ, ρ sin ϕ)ρ dρ dϕ. ]7/ Let N(x 1, x, x ) (,, ) so that x = and x = x 1. Then = x u(f; x,, x ) x 1 4ρ cos ϕ 8ρx cos ϕ ρ sin ϕ+x (ρ ρx cos ϕ+x ) 7/ f(ρ cos ϕ, ρ sin ϕ)ρ dρ dϕ. Using Lemma 1, for any x 1, x, and x we have 4(t 1 x 1 ) (t x ) x dt ( t x + x + ) 7/ 1 dt =, R which, in particular, yields 4ρ cos ϕ 8ρx cos ϕ ρ sin ϕ + x ρ dρ dϕ = (ρ ρx cos ϕ + x ) 7/ = Therefore 4ρ cos ϕ 8ρx cos ϕ ρ sin ϕ+x (ρ ρx cos ϕ + x ) 7/ f(ρ cos ϕ, ρ sin ϕ)ρdρdϕ=. = 4ρ cos ϕ 8ρx cos ϕ ρ sin ϕ + x (ρ ρx cos ϕ + x ) 7/ ρ dρ dϕ =. () In the interval ϕ we have > x x 4ρ cos ϕ 8ρx cos ϕ ρ sin ϕ + x (ρ ρx cos ϕ + x ) 7/ ρ dρ dϕ > 5ρ cos ϕ ρ (ρ + ρx + 4x ) 7/ ρ dρ dϕ=x 5ρ cos ϕ ρ (ρ + x ) 7 ρ dρ dϕ =

8 9 S. TOPURIA Hence x + x = 4 x ρ dρ (ρ + x ) 7 > cx x ρ rho (ρ + x ) 7 = c x. 4ρ cos ϕ 8ρx cos ϕ ρ sin ϕ + x (ρ ρx cos ϕ + x ) 7/ ρ dρ dϕ = +. () Equations () and () imply x + x then 4ρ cos ϕ 8ρx cos ϕ ρ sin ϕ + x (ρ ρx cos ϕ + x ) 7/ ρ dρ dϕ =. (4) Next, we define f(t 1, t ) as follows: 1 for t 1 >, t >, 1 for t 1 <, t >, f(t 1, t ) = for < t 1 <, t, for t 1 =, < t <. Clearly, for this function Dx 1 (x ) f(, ) = f(t 1, x 1 ) + f( t 1, x ) f(, x ) (t 1,x ) (,) t =. 1 However, by () and (4), u(f;x 1,,x ) +, as (x x 1, x, x ) (,, ). 1 Lemma. For any (x 1, x,..., x k+1 ) the following equalities are valid: (t i x i )(t x )f(t t i e i + x i e i ) ( t x + x k+1 ) dt =, k+5 (t i x i )(t x )f(t t i e i t e + x i e i + x e ) ( t x + x k+1 ) dt =, k+5 (k + 1)(k + )x k+1 Γ( k+1 ) k+1 (t i x i ) (t x ) ( t x + xk+1 ) k+5 t i dt = 1. Theorem. (a) If at the point x there exists a finite derivative D [xi x ] (x)f(x ), i, u(f; x, x k+1 ) = D xix (x,x k+1 ) (x,) x i x f(x ).

9 POISSON INTEGRAL FOR A HALF-SPACE +1 + (k > 1) 9 (b) There exists a continuous function f L( ) such that for any B M, m(b) < k all derivatives D [xix ] (x)f() =, but the its do not exist. u(f;, x k+1 ) x k+1 + x i x Proof of part (a). Let x = and B k = verify that u(f; x, x k+1 ) = B k x k+1 x i x By Lemma we obtain u(f; x, x k+1 ) x i x D [xi x ] (x)f() = B k x k+1 k+1 (k+1)(k+)γ( ) k+1 (t i x i )(t x )f(t) dt ( t x + x k+1 ). k+5 t i t ( t + xk+1 ) k+5. It is easy to [ f(x + t) f(x + t ti e i ) + f(x + t t e ) + f(x + t t i e i t e ) t i t ] D [xi x ] (x)f() dt, which implies that part (a) is valid. Proof of part (b). Let k = 4 and D = ( 4 ti i=1 1, t i, i = 1, 4). We define f as f(t) = 5 t 1 t t 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 [xi x ] (x)f() =. But = cx 5 u(f;, x 5 ) = cx 5 x 1 x 1 1 D t 1 t 5 t 1 t t t 4 ( t + x dt + o(1) = 5 )9/ ρ 9/5 dρ (ρ + x 5 )9/ + o(1) + as x 5 +. Corollary. If the function f at the point x has a continuous derivative f x i x (x), then u(f; x, x k+1 ) = f(x ). (x,x k+1 ) (x,) x i x x i x The following theorem is valid by the corollaries of Theorems 1 and.

10 94 S. TOPURIA Theorem 4. If the function f is twice continuously differentiable at the point x, then (x,xk+1 ) (x,) d xu(f; x, x k+1 ) = d f(x ). Theorem 5. (a) If at the point x there exists a finite derivative D [xi ]x (x)f(x ), then u(f; x, x k+1 ) = D xi x (x,x k+1 ) (x,) x i x f(x ). x (b) There exists a continuous function f L( ) such that for every B M, m(b) < k 1 all derivatives D [xi ]x (x B )f() =, but the its do not exist. u(f;, x k+1 ) x k+1 + x i x Theorem 6. (a) If at the point x there exists a finite derivative D xi x (x M\{i,} )f(x ), then u(f; x, x k+1 ) = D xix (x,x k+1 ) (x,) x i x f(x ). x i x. (b) There exists a continuous function f L( ) such that for every B M, m(b) < k all derivatives D xi x (x B )f() =, but the its do not exist. u(f;, x k+1 ) x k+1 + x i x Theorem 7. (a) If the function f is twice differentiable at the point x, then dxu(f; x, x k+1 ) = d f(x ). (x,x k+1 ) (x,) (b) There exists a continuous function f L( ) such that it is differentiable at the point x = (, ) and has at this point all partial derivatives of any order; however the its do not exist. u(f;, x k+1 ), i, = 1, k x k+1 + x i x

11 POISSON INTEGRAL FOR A HALF-SPACE +1 + (k > 1) 95 Proof of part (a). Let x =. The validity of part (a) follows from the equalities = c k x k+1 [f(t) + f() ( k ν=1 = B k x k+1 u(f; x, x k+1 ) xi f() x = i [(k + )(t i x i ) t x x k+1 ] t ( t x + x k+1 ) k+5 t ν t ν )f() 1 ( k t ν ν=1 t u(f; x, x k+1 ) x i x [f(t) + f() ( k ν=1 (t i x i )(t x ) t f() x i x = t ν ) f() dt, ( t x + x k+1) k+5 t ν t ν )f() 1 ( k t ν ν=1 t t ν ) f() dt. Proof of part (b). Consider the function equal to (t 1 t ) (t 1 t 1) when (t 1, t ) D = {(t 1, t ) R : t 1 < ; 1 t 1 t t 1 } and to otherwise, which is continuous in R, differentiable at the point (, ), and has all partial derivatives of any order equal to zero, but u(f;,, x ) = 15x x 1 x > cx = c x 6 x dt 1 t 1 1 t 1 t 1 t (t 1 t ) (t 1 t 1) t 1 (t 1 + t + x )7/ dt > (t 1 t 1dt t 1) (t 1 1 t 1) 1 (t 1 + 4t 1 + = x )7/ x t 1 x 1 t1 dt 1 = c as x x +. x then Theorem 8. (a) If at the point x there exists a finite derivative D x i x (x M\{i,} ) f(x ), u(f; x, x k+1 ) = Dx x k+1 + x i x ix f(x ).

12 96 S. TOPURIA (b) There exists a continuous function f L( ) such that we have D x i x (x M\{i,} ) f(x ) =, but the its do not exist. u(f; x, x k+1 ) (x,x k+1 ) (x,) x i x Proof of part (b). This is given for the case k =. Assume that D 1 = [, 1;, 1], D = [ 1, ;, 1]. Let t1 t for (t 1, t ) D 1, f(t 1, t ) = t1 t for (t 1, t ) D, for t and extend f(t 1, t ) continuously to the set R \(D 1 D ) so that f L(R ). It is easy to verify that D f(, ). Let x 1 = x = and (x 1, x, x ) (,, ) so that x = and x = x 1. Then for the constructed function we have u(f; x 1, x, x ) = 15x { 1 x 1 x { 1 x1 = cx 1 x 1 1+x 1 1 x 1 { x1 = cx (t 1 x 1 )t t1 t dt 1 dt [(t 1 x 1 ) + t + x ]7/ + (t 1 x 1 )t t1 t dt 1 dt [(t 1 x 1 ) + t + x ]7/ 1 } + o(1) = t 1 t t (t 1 + x 1 ) dt 1 dt (t 1 + t + x 1 )7/ t 1 t (t1 x 1 )t dt 1 dt (t 1 + t + x 1 )7/ 1 x 1 1 x 1 1 x 1 1+x x 1 } + o(1) = t 1 t t (t 1 + x 1 ) dt 1 dt (t 1 + t + x 1 )7/ + t 1 t [ t (x 1 + t 1 ) t (t 1 x 1 )] (t 1 + t + x 1 )7/ dt 1 dt } t 1 t (t1 x 1 )t dt 1 dt (t 1 + t + + o(1) = x 1 )7/

13 POISSON INTEGRAL FOR A HALF-SPACE +1 + (k > 1) 97 = Cx 1 (I 1 + I I ) + o(1). We can readily show that I 1 = x 1 1 Next, t 1 t [ t (t 1 + x 1 ) t (x 1 t 1 )] (t 1 + t + x 1 )7/ dt 1 dt >, I = O(1). (5) I > = > x 1 x 1 x 1 x 1 x 1 t / t / t 1 t [ t (x 1 + t 1 ) t x 1 ] (t 1 + t + x 1 )7/ dt 1 dt = x x 1 x x 1 t 1 ( x 1 + t 1 x 1 ) (t 1 + t + x 1 )7/ dt 1 > x 1 ( x 1 x 1 ) (4x 1 + x 1 + dt x 1 )7/ 1 > c x1. (6) Therefore for x = and x = x 1 (5) and (6) imply u(f;x 1,,x 1 ) x 1 x from which we obtain u(f;x 1,x,x ) x 1 x +, as (x 1, x, x ) (,, ). Theorem 9. If at the point x there exists a finite derivative D [x i x ] (x) f(x ), then u(f; x, x k+1 ) = D (x,x k+1 ) (x x,) x i x ix f(x ). Part (a) of Theorem is the corollary of Theorem 9. Theorem 1. If at the point x there exists a finite derivative D [x i ]x (x) f(x ), then u(f; x x e + x e, x k+1 ) = D (x x e +x e,x k+1 ) (x x,) x i x ix f(x ). Lemma. The following equalities are valid: (k + 1)x k+1 Γ( k+1 ) k Γ( k ) ρ kxk+1 (ρ + xk+1 ) k+5 ρ k 1 dρ =, ρ kx k+1 (ρ + x k+1 ) k+5 ρ k+1 dρ = 1. > c x 1,

14 98 S. TOPURIA Theorem 11. (a) Let B M. If at the point x there exists a finite derivative xb f(x ), then x u(f; x x B +x B,x B + x, x k+1) (x B k+1) = xb f(x ).,) (b) There exists a function f L( ) such that f(x ) exists but the it x,xk+1 ) (x,) xu(f; x, x k+1 ) does not. Proof of part (a). One can easily verify that x u(f; x, x k+1 ) = u(f; x, x k+1 ) = = (k + 1)x k+1γ( k+1 ) k+1 x k+1 t x kxk+1 ( t x + x k+1 ) k+5 f(t) dt. k+1 (k+1)γ( Assume that D k = ) Γ( k and θ = [, ] k [, ]. Passing to ) the spherical coordinates, we have Θ x u(f; x B + x B, x k+1) = c k x k+1 ρ kx k+1 (ρ + xk+1 ) k+5 f(x B + x B + t)ρk 1 sin k ϑ 1 sin ϑ k dρ dϑ 1 dϑ k dϕ = = D k x k+1 ρ kx k+1 (ρ + xk+1 ) k+5 Hence by Lemma we obtain [ 1 ρ k 1 f(t) ds(t) S ρ S ρ (x B +x B ) ] dρ. x u(f; x B + x, x B k+1) xb f(x ρ kx k+1 ) = D k x k+1 (ρ + x k+1 ) k+5 [ 1 S ρ S ρ (x B +x B ) f(t) ds(t) f(x + B x ) ] B ρ ρ xb f(x ) /k k dρ. For the proof of part (b) see [], p. 16. Corollary 1. If at the point x there exists a finite derivative f(x ), then x k+1 + xu(f; x, x k+1 ) = f(x ).

15 POISSON INTEGRAL FOR A HALF-SPACE +1 + (k > 1) 99 Corollary. If at the point x there exists a finite derivative f(x ), then Let δ > and S δ = xu(f; x, x k+1 ) = x f(x ). (x,x k+1 ) (x,) k ν=1 [x ν δ; x ν + δ]. Theorem 1. Let f t i (t) L(S δ ). D xi (x M\i )f x i (x ) at the point x, then If f t i (t) has a finite derivative u(f; x, x k+1 ) (x,x k+1 ) (x,) x = f x (x ). i i x i Proof. Let x =. The validity of the theorem follows from the equality = c k x k+1 S δ u(f; x, x k+1 ) x i (t i x i )t i ( t x + x k+1 ) k+ ] D xi (x M\i )f x i () D xi (x M\i )f x i () = [ f ti (t) f t i (t t i e i ) dt + o(1). t i The theorems below are proved analogously. Theorem 1. Let f x i (t) L(S δ ). D x (x M\ )f x i (x ) at the point x, then If f x i (t) has a finite derivative u(f; x, x k+1 ) = f(x ). (x,x k+1 ) (x,) x i x x i x x Corollary. Let f x i and f x L(S δ ) for some δ >. If at the point x there exist finite derivatives D xi (x M\i )f x (x ) and D x (x M\ )f x i (x ), then u(f; x, x k+1 ) = f(x ) = f(x ). (x,x k+1 ) (x,) x i x x i x x x i x i x Note that by this corollary we have obtained the condition of the coincidence of mixed derivatives of the function f at the point x.

16 4 S. TOPURIA Theorem 14. Let f x i L(S δ ) for some δ >. If at the point x there exists a finite derivative D xi (x)f x i (x ), then u(f; x, x k+1 ) (x,x k+1 ) (x,) x i = f(x ) x. i Theorem 15. Let f x i L(S δ ). If f x i (t) has a finite derivative D x (x)f x i (x ) at the point x, then u(f; x, x k+1 ) = f(x ). (x,x k+1 ) (x,) x i x x i x References 1. S. B. Topuria, Solution of the Dirichlet problem for a half-space. (Russian) Dokl. Akad. Nauk SSSR 195(197), No., 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.. 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. 15(1995), No., S. B. Topuria, Boundary properties of derivatives of the Poisson integral in a half-space. (Russian) Bull. Georgian Acad. Sci. 15(1995), No., 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 5(1997), No., 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, I. I. Privalov, Subharmonic functions. (Russian) Moscow Leningrad, 197. (Received ; revised ) Author s address: Department of Mathematics (No. 6) Georgian Technical University 77, M. Kostava St., Tbilisi 875 Georgia