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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
(k > 1).
the Poisson integral are studied in the half-space R+
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 ′ (x0 ), then
lim

∧
reix →eix0

∂u(f ; r, x)
= f ′ (x0 ),
∂x
∧

where u(f ; r, x) is the Poisson integral for a circle, and the symbol reix →
eix0 means that the point reix tends to eix0 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 D1 f (x0 ) which is a first
symmetric derivative of f at the point x0 (see [2], p. 99 − 100), i.e.,
f (x0 + h) − f (x0 − h)
,
h→0
2h

D1 f (x0 ) = lim
then
lim

r→1−

∂u(f ; r, x0 )
= D1 f (x0 ).
∂x

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.

585
c 1998 Plenum Publishing Corporation
1072-947X/97/1100-0585$12.50/0 

586

S. TOPURIA

In [4] a continuous 2π-periodic function f (x) is constructed such that
D1 f (x0 ) = 0, but
∂u(f ; r, x)
lim
∧ ix

∂x
ix
re →e 0
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 R2+ 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 Dk u(f ; r, ϑ1 , ϑ2 , . . . , ϑk−2 , ϕ) were
studied in [8] (see also [9], p. 118), where u(f ; r, ϑ1 , ϑ2 , . . . , ϑk−2 ϕ) is the
Poisson integral for the unit sphere in Rk (k > 2), and Dk 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 R3 are given a detailed consideration in [10, 11, 12], but for
the half-space R3+ in [13], [14], [15]. In [14] it is shown that there exists
a continuous function of two variables f (x, y) ∈ L(R2 ) which, at the point
(x0 , y0 ), has the partial derivatives fx′ (x0 , y0 ) and fy′ (x0 , y0 ), but the inte;x,y,z)
;x,y,z)
and ∂u(f∂y
(u(f ; x, y, z) is the Poisson integral for R3+ )
grals ∂u(f∂x
of this function have no values at the point (x0 , y0 ) 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
k+1
(k > 1)).
integral u(f ; x, xk+1 ) (u(f ; x, xk+1 ) is the Poisson integral for R+
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).

PARTIAL DERIVATIVES OF THE POISSON INTEGRAL

587

1. Notation, Definitions, and Auxiliary Propositions
The following notation is used in this paper:
Rk is a k-dimensional Euclidean space (R = R1 );
x = (x1 , x2 , . . . , xk ), t = (t1 , t2 , . . . , tk ), x0 = (x01 , x02 , . . . , x0k ) are the
points (vectors) of the space Rk ;
k
P
xi ti is the scalar product;
(x, t) =
pi=1
|x| = (x, x); x + t = (x1 + t1 , x2 + t2 , . . . , xk + tk );

ei (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 ∈ Rk and an arbitrary
set B ⊂ M , the symbol xB denotes a point from Rk 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 (xM = x,
B \ i = B \ {i}); if B = {i1 , i2 , . . . , is }, 1 ≤ s ≤ k (il < ir for l < r), then
xB = (xi1 , xi2 , . . . , xis ) ∈ Rs ; m(B) is the number of elements of the set B;
e k ) is the set of functions f (x) = f (x1 , x2 , . . . , xk ) such that
L(R
f (x)

(1 + |x|2 )

k+1
2

∈ L(Rk );

= {(x, xk+1 ) ∈ Rk+1 ; xk+1 > 0};

Rk+1
+
u(f ; x, xk+1 ) is the Poisson integral of the function f (x) for the half-space
Rk+1
+ , i.e.,
Z
xk+1 Γ( k+1
f (t) dt
2 )
u(f ; x, xk+1 ) =
k+1
k+1 .
2 + x2
2
π 2
(|t
)
−
x|
k+1

k
R

In investigating the boundary properties of the partial derivatives
∂
and ∂ϕ
uf (r, ϑ, ϕ) of the spherical Poisson integral uf (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
k+1
in the manner as follows: if the point N ∈ R+
converges to the
Rk+1
+
P
0
0 2 −1/2
point P(x , 0) under the condition xk+1 ( (xi − xi ) )
≥ C > 0,1 then
∂

∂ϑ uf (r, ϑ, ϕ)

i∈B

∧

we shall write N (x, xk+1 ) → P(x0 , 0). When B = M , we have an angular
xB

∧

convergence and thus we write N (x, xk+1 ) → P(x0 , 0). Finally, the notation
N (x, xk+1 ) → P(x0 , 0) means that the point N (x, xk+1 ) remaining in Rk+1
+
converges to P(x0 , 0) without any restrictions.
1 Here

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

588

S. TOPURIA

∂
∂
uf (r, ϑ, ϕ) and ∂ϕ
uf (r, ϑ, ϕ) have dihedral-angular
It is known that ∂ϑ
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

f (xB + x0B′ + uei ) − f (xB + x0B′ )
u
(u,xB )→(0,x0 )
B

lim

by:
(a) fx′ i (x0 ) for B 6= ∅;
(b) Dxi (xB ) f (x0 ) for i ∈ B ′ ;
(c) Dxi (xB ) f (x0 ) for i ∈ B.
2. Denote the limit
lim

(u,xB )→(0,x0B )

0
f (xB + x0B′ + uei ) − f (xB + xB
′ − uei )
2u

by:
6 ∅;
(a) Dx∗i (x0 ) for B =
(b) Dx∗i (xB ) f (x0 ) for i ∈ B ′ ;
(c) Dxi (xB ) f (x0 ) for i ∈ B.
The following propositions are valid:
(1) If B2 ⊂ B1 , the existence of Dxi (xB1 ) f (x0 ) implies the existence of
Dxi (xB2 ) f (x0 ) and Dxi (xB1 ) f (x0 ) = fx′ i (x0 ). The converse does not hold.
(2) The existence of Dxi (xB1 ) f (x0 ) implies the existence of Dxi (xB2 ) f (x0 )
and their equality.
(3) The existence of Dxi (xB ) f (x0 ) implies the existence of Dxi (xB1 ) f (x0 )
and Dxi (xB ) f (x0 ) = Dxi (xB\i ) f (x0 ) = fx′ i (x0 ).
(4) If fx′ i (x) is a continuous function at x0 , then for any B ⊂ M all
derivatives Dxi (xB ) f (x0 ) exist and
Dxi (xB ) f (x0 ) = fx′ i (x0 ).
Indeed, by virtue of the Lagrange theorem
f ′ [x + x0B′ + θ(x)uei ]u
f (xB + x0B′ + uei ) − f (xB + x0B′ )
= xi B
=
u
u
= fx′ i [xB + x0B′ + θ(x)uei ], 0 < θ < 1.
Hence we conclude that statement (4) is valid.
(5) There exists a function f (x) for which Dxi (x) f (x0 ) exist, but on an
everywhere dense set in the neighborhood of the point x0 there are no fx′ i (x).

PARTIAL DERIVATIVES OF THE POISSON INTEGRAL

589

(6) If the function f (x) has finite derivatives
Dx1 (x2 ,...,xk ) f (x0 ), Dx2 (x3 ,...,xk ) f (x0 ), . . . , Dxk−1 (xk ) f (x0 )

at the point x0 , then its continuity at x0 with respect to the argument xk
is a necessary and sufficient condition for f (x) to be continuous at x0 (see
[12], p.15).
(7) The existence of the derivatives Dx1 (x2 ,...,xk ) f (x0 ), Dx2 (x3 ,...,xk ) f (x0 ),
. . . , Dx1 (x2 ,...,xk ) f (x0 ) and fx′ k (x0 ) implies the existence of the differential
df (x0 ) (see [12], p. 16).
e k ).
In what follows it will be assumed that f ∈ L(R
Lemma. The equalities
Z
(ti − xi )f (t − ti ei + xi ei )dt
= 0,
I1 =
k+3
2 + x2
2
)
(|t
−
x|
k+1
Rk
Z
(k + 1)xk+1 Γ( k+1
(ti − xi )2 dt
2 )
I2 =
k+1
k+3 = 1
π 2
(|t − x|2 + x2k+1 ) 2
k
R

hold for any (x, xk+1 ).
Proof. We have
I1 =

Z

f (x + t − ti ei ) dS(tM \i )

Z∞

−∞

Rk−1

ti dti
2
(|t|2 + xk+1
)

k+3
2

= 0.

Further, if we use the spherical coordinates, ρ, θ1 , . . . , θk−2 , ϕ, we shall have
Z
(k + 1)xk+1 Γ( k+1
ti dti
2 )
I2 =
k+1
k+3 =
2
2
2
2
+
π
(|t|
)
x
k+1
Rk
Z 2 2
2
2
2
(k+1)xk+1 Γ( k+1
2 ) ρ sin ϑ1 sin ϑ2 · · ·sin ϑi−1 cos ϑi
×
=
k+1
k+3
2 + x2
2
π 2
(ρ
)
k+1
k
k−1

×ρ

=

k−2

sin

R
k−i

ϑ1 · · · sin

k−2

Z∞ k+1
2(k+1)Γ( k+1
ρ
dρ
2 )
π

k−1
2

(1 + ρ2 )

0
2

=

2(k +

k+3
2

1)Γ( k+1
2 )

π

k−1
2

z }| {
Zπ Zπ
· · · sink ϑ1 sink−1 ϑ2 · · ·sink−i+2 ϑi−1 ×
0

k−i−1

× cos ϑi sin

ϑi−1 · · · sin ϑk−2 dρdϑ1 · · · dϑk−2 dϕ =

0

ϑi · · · sin ϑk−2 dϑ1 dϑ2 · · · dϑk−2 =
√
k−2
kΓ( k2 ) π
π 2
·
·
= 1.
k
2(k + 1)Γ( k+1
2 ) kΓ( 2 )

590

S. TOPURIA

2. Boundary Properties of the Integral

∂u(f ;x,xk+1 )
∂xi

The following theorem is valid.
Theorem 1.
(a) If a finite derivative Dxi (x) f (x0 ) exists at the point x0 , then
∂f (x0 )
∂u(f ; x, xk+1 )
=
.
∂xi
∂xi
(x,xk+1 )→(x0 ,0)
lim

(1)

(b) There is a continuous function f ∈ L(Rk ) such that for any B ⊂ M ,
m(B) < k, at the point x0 = (0, 0, . . . , 0) = 0 all derivatives Dxi (xB ) f (0) =
0, i = 1, k, but the limits
∂u(f ; 0, xk+1 )
,
xk+1 →0+
∂xi
lim

i = 1, k,

do not exist.
Proof.
(a) Let x0 = 0, Ck =

(k+1)Γ( k+1
2 )
π

k+1
2

. It is easy to check that

∂u(f ; x, xk+1 )
= Ck xk+1
∂xi

Z

Rk

(ti − xi )f (t) dt

(|t − x|2 + x2k+1 )

k+3
2

.

By virtue of the lemma we have
∂u(f ; x, xk+1 )
− Dxi (x) f (0) =
∂xi
Z
i
h f (x + t) − f (x + t − t e )
ti2
i i
= Ck xk+1
f
(0)
dt =
−
D
xi (x)
k+3
ti
(|t|2 + x2 ) 2
Rk

k+1

= I1 + I2 ,
where
I1 = Ck xk+1

Z

Vδ

,

I2 = Ck xk+1

Z

,

CVδ

Vδ is the ball with center at 0 and radius δ. Let ε > 0. Choose δ > 0 such
that
Œ f (x + t) − f (x + t − t e )
Œ
Œ
Œ
i i
− Dxi (x) f (0)Œ < ε
Œ
ti
for |x| < δ, |t| < 2dl.
Hence
Z
t2i dt
|I1 | < Ck xk+1 ε
k+3 <
(|t|2 + x2k+1 ) 2
Vδ

PARTIAL DERIVATIVES OF THE POISSON INTEGRAL

< Ck xk+1 ε

Z

t2i dt
(|t|2 + x2k+1 )

Rk

k+3
2

= ε.

591

(2)

It is likewise easy to show that
lim

(x,xk+1 )→(x0 ,0)

(3)

I2 = 0.

Equalities (2) and (3) imply the validity of (1).
(b) Let D = (0 ≤ t1 < ∞; 0 ≤ t2 < ∞; . . . , 0 ≤ tk < ∞). Define the
function f as follows:
( √
k+1
t1 t2 · · · tk if (t1 , t2 , . . . , tk ) ∈ D,
f (t) =
0
if (t1 , t2 , . . . , tk ) ∈ CD.
Clearly, f (t) is continuous in Rk and Dxi (xB ) f (0) = 0, i = 1, k, for any
B when m(B) < k.
If in the integral
Z
(ti − xi )f (t) dt
∂u(f ; x, xk+1 )
= Ck xk+1
k+3
∂xi
(|t
− x|2 + x2k+1 ) 2
k
R

we use spherical coordinates, then for the considered function we shall have
Z
ti f (t) dt
∂u(f ; 0, xk+1 )
= Ck xk+1
k+3 =
2
∂xi
(|t| + x2k+1 ) 2
k
R

Z∞

= Cxk+1

0

= Cxk+1

Z∞
0

ρ

k
k+ k+1

ρ

k+3
2

> Cxk+1

ρk

(ρ2 + x2k+1 )

dρ

(ρ2 + x2k+1 )

p

k−1

> Cxk+1

k+3
2

xZk+1
0

xZk+1
0

k
k+ k+1

ρ

xk+3
k+1

dρ

=

ρk−1 dρ =
k

ρk+ k+1 dρ
(ρ2 + x2k+1 )

k+3
2

ρk−1 dρ >

C
.
√
xk+1

k+1

Hence
lim

xk+1 →0+

∂u(f ; 0, xk+1 )
= +∞.
∂x

Corollary 1. If finite derivatives Dxi (x) f (x0 ), i = 1, k, exist at the point
x , then
lim 0 dx u(f ; x, xk+1 ) = df (x0 ).
0

(x,xk+1 )→(x ,0)

592

S. TOPURIA

Corollary 2. If f has a continuous partial derivative at the point x0 ,
then
∂u(f ; x, xk+1 )
= fx′ i (x0 ).
lim 0
∂xi
(x,xk+1 )→(x ,0)
Corollary 3.
(a) If f is a continuously differentiable function at the point x0 , then
lim

(x,xk+1 )→(x0 ,0)

dx u(f ; x, xk+1 ) = df (x0 ).

(b) There exists a differentiable function f (t1 , t2 ) at the point (0, 0) such
that df (0, 0) = 0 but the limits
∂u(f ; x1 , x2 , x3 )
∂x2
(x1 ,x2 ,x3 )→(0,0,0)

∂u(f ; x1 , x2 , x3 )
,
∂x1
(x1 ,x2 ,x3 )→(0,0,0)
lim

lim

do not exist.
Proof of assertion (b) of Corollary 3. We set D = [0, 1; 0, 1]. Let
(p
5
t31 t32 for (t1 , t2 ) ∈ D,
f (t1 , t2 ) =
0
for (t1 , t2 ) ∈ [−∞, 0; 0, ∞[∪] − ∞, ∞; −∞, 0[,
and continue f onto the set ]0, ∞; 0, ∞[\D so that f ∈ L(R2 ). It is easy to
check that f (t1 , t2 ) is differentiable at the point (0, 0) and
ft′1 (0, 0) = ft′2 (0, 0) = 0.
Let (x1 , x2 , x3 ) → (0, 0, 0) for x1 = 0, x3 = x22 , x2 > 0. Then for the
considered function
3x3
∂u(f ; 0, x2 , x3 )
=
∂x1
2π

Z∞ Z∞

3x22
2π

Z∞ Z∞

3x2
= 2
2π

Z∞ Z∞

=

0

0

[t21

0 −x2

0 −x2

3x2
> 2
2π

t1 f (t1 , t2 ) dt1 dt2
=
+ (t2 − x2 )2 + x24 ]5/2

t1 f (t1 , t2 + x2 ) dt1 dt2
=
(t21 + t22 + x24 )5/2
t1

p
5
t31 (t2 + x2 )3 dt1 dt2
>
(t21 + t22 + x42 )5/2

2
2
p
Z2x2 Z2x2 p
t1 5 t31 5 x32 dt1 dt2
>
(t21 + t22 + x42 )5/2

x22 x22

PARTIAL DERIVATIVES OF THE POISSON INTEGRAL
2

2

Z2x2 Z2x2

3/5

3x2 x
> 2 2
2π

x22 x22

=

593

6/5

x22 · x2 dt1 dt2
=
(4x)24 + 4x42 + x42 )5/2

1
√ →∞
162π 5 x2

for x2 → 0 + .

Theorem 2. If a finite derivative Dxi (xM \i ) f (x0 ) exists at the point x0 ,
then
∂u(f ; x, xk+1 )
∂f (x0 )
=
.
lim
∧
∂xi
∂xi
(x,xk+1 )→(x0 ,0)
xi

Proof. Let x0 = 0. By virtue of the lemma we have the equality
∂u(f ; x, xk+1 )
− Dxi (xM \i ) f (0) =
∂xi
Z
h f (t) − f (t − t e )
i
ti (ti − xi )
i i
− Dxi (xM \i ) f (0) dt =
= Ck xk+1
k+3
ti
(|t − x|2 + x2 ) 2
k+1

Rk

where I1 = Ck xk+1

R

Vδ

= I1 + I2 ,
R
, I2 = Ck xk+1
.
CVδ

Let ε > 0 and choose δ > 0 such that
Œ
Œ f (t) − f (t − t e )
Œ
Œ
i i
− Dxi (xM \i ) f (0)Œ < ε
Œ
ti

for

|r| < δ.

Now,

|I1 | < Ck xk+1 ε

Z

|ti (ti − xi )| dt

(|t − x|2 + x2k+1 )

Rk

< Ck xk+1 ε

Z

t2i dt

+ Ck xk+1 ε|xi |

(|t|2 + x2k+1 ) 2
Z
|ti | dt

Rk

2
(|t|2 + xk+1
)

= ε + Cxk+1 ε|xi |
=ε+

+

k+3

Rk

Cxk+2
k+1 |xi |ε
xk+3
k+1

Z∞
0

k+3
2

=

ρk dρ
(ρ2 + x2k+1 )

Z∞
0

<

k+3
2

ρk dρ
(1 + ρ2 )

k+3
2

k+3
2

=


C|xi | ‘
ε.
= 1+
xk+1

594

S. TOPURIA

Hence we obtain
lim

I1 = 0.

∧

(x,xk+1 )→(x0 ,0)
xi

In a similar manner we prove that
I2 = 0.

lim

∧

(x,xk+1 )→(x0 ,0)
xi

Theorem 3. If at the point x0 there exist finite
Dxi (xM \i ) f (x0 ) and Dxj (xB ) f (x0 ), i 6= j, B = M \ {i, j}, then

derivatives

∂u(f ; x, xk+1 )
∂f (x0 )
=
,
∧
∂xi
∂xi
(x,xk+1 )→(x0 ,0)
lim

xi

∂f (x0 )
∂u(f ; x, xk+1 )
=
.
∧
∂xj
∂xj
(x,xk+1 )→(x0 ,0)
lim

xi

Proof. Let x0 = 0. By virtue of the lemma we have
Z
∂u(f ; x, xk+1 )
(tj − xj )f (t) dt
= Ck xk+1
k+3 =
∂xj
(|t − x|2 + x2k+1 ) 2
k
R
Z
(tj −xj ){[f (t) − f (t − ti ei )]+[f (t−ti ei )−f (t−ti ei −tj ej )]}dt
= Ck xk+1
=
k+3
2
) 2
(|t − x|2 + xk+1
Rk
Z
(tj − xi )[f (t) − f (t − ti ei )]
= Ck xk+1
dt +
k+3
2 + x2
2
)
−
x|
(|t
k+1
Rk
Z
(tj − xi )[f (t − ti ei ) − f (t − ti ei − tj ej )]
dt = I1 + I2 ,
+Ck xk+1
k+3
2
(|t − x|2 + xk+1
) 2
k
R

where
I1 = Ck xk+1

Z

Rk

= Ck xk+1

Z

(|t −

ti (tj − xj )
x|2

+

k+3
x2k+1 ) 2

ti (tj −xj )

k+3
2

·

f (t) − f (t − ti ei )
dt =
ti

i
h f (t)−f (t−t e )
i i
−Dxi (xM \i ) f (0) dt+
ti

(|t−x|2 +x2k+1 )
Z
ti (tj − xj ) dt
′
′′
+ Ck xk+1 Dxi (xM \i ) f (0)
k+3 = I1 + I1 .
2
2
2
(|t − x| + xk+1 )
k
Rk

R

PARTIAL DERIVATIVES OF THE POISSON INTEGRAL

595

It is easy to see that I1′′ and
Z
Œ
Œ
|ti (tj − xj )|
Œ
Œ f (t) − f (t − ti ei )
′
−Dxi (xM \i ) f (0)Œdt.
|I1 | < Ck xk+1
k+3 Œ
2
2
ti
(|t − x| + xk+1 ) 2
k
R

Hence we obtain

lim

I1′ =

∧

lim

∧

I1 = 0.

(x,xk+1 )→0

(x,xk+1 )→0

xi

xi

Now we shall show that
lim

∧

(x,xk+1 )→0

I2 = Dxj (xM \{i,j} ) f (0).

xj

Indeed,
I2 = Ck xk+1

Z

Rk

|tj (tj − xj )|

(|t − x|2 + x2k+1 )

k+3
2

×

h f (t − t e )−f (t − t e − t e )
i
j j
i i
i i
−Dxj (xM \{i,j} ) f (0) dt+Dxj (xM \{i,j} ) f (0).
×
tj

This readily implies

lim

∧

(x,xk+1 )→0

I2 = Dxj (xM \{i,j} ) f (0) =

∂f (0)
.
∂xj

xj

Finally, we obtain
lim

∧

(x,xk+1 ) →

∂f (0)
∂u(f ; x, xk+1 )
=
.
∂xj
∂xj
0

xi xj

By a similar reasoning we prove
Theorem 4. If at the point x0 there exist finite derivatives
Dx1 (x2 ,x3 ,...,xk ) f (x0 ), Dx2 (x3 ,...,xk ) f (x0 ), . . . , Dxk−1 (xk ) f (x0 ), fx′ k (x0 ), then
∂u(f ; x, xk+1 )
∂f (x0 )
=
,
∧
∂x1
∂x1
(x,xk+1 )→(x0 ,0)
lim

x1

lim

∧

(x,xk+1 ) →

∂f (x0 )
∂u(f ; x, xk+1 )
=
,
∂x2
∂x2
(x0 ,0)

x1 x2

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

596

S. TOPURIA

∂f (x0 )
∂u(f ; x, xk+1 )
=
.
∧
∂xk
∂xk
(x,xk+1 )→(x0 ,0)
lim

Corollary. If at the point x0 there exist finite derivatives Dx1 (xM \1 ) f (x0 ),
Dx2 (xM \{1,2} ) f (x0 ), . . . , Dxk−1 (xk ) f (x0 ), fx′ k (x0 ), then
lim

∧

dx u(f ; x, xk+1 ) = df (x0 ).

(x,xk+1 )→(x0 ,0)

Theorem 5.
(a) If at the
Dx∗i (xM \i ) f (x0 ), then

point

x0

there

exists

a

finite

derivative

∂u(f ; x − xi ei + x0i ei , xk+1 )
= Dx∗i f (x0 ).
∂xi
(x−xi ei +xi ei ,xk+1 )→(x0 ,0)
lim
0

(b) There exists a continuous function f (x) such that Dx∗i (xM \i ) f (x0 ) = 0,
but the limit
∂u(f ; x, xk+1 )
lim
∧
∂xi
0
(x,xk+1 )→(x ,0)
does not exist.
Proof. (a) Let x0 = 0. The validity of (a) follows from the equality
∂u(f ; x − xi ei + x0i ei , xk+1 )
− Dx∗i (xM \i ) f (x0 ) =
∂xi
Z
h f (t + x − x e ) − f (t + x − x e − 2t e )
ti2
i i
i i
i i
−
= Ck xk+1
k+3
2 + x2
2ti
2
)
(|t|
k+1
Rk
i
−Dxi (xM \i ) f (x0 ) dt.
(b) We set D1 = [0, 1; 0, 1], D2 = p − 1, 0; 0, 1]. Let
p √

t2
for (t1 , t2 ) ∈ D1 ,
pt1 √
f (t1 , t2 ) =
−t1 t2 for (t1 , t2 ) ∈ D2 ,


0
for t2 ≤ 0

and continue f (t1 , t2 ) onto the set R2+ \ (D1 ∪ D2 ) so that f ∈ L(R2 ). It
is easy to check that Dt∗1 (t2 ) f (0) = 0. Let x01 = x02 = 0 and (x1 , x2 , x3 ) →
(0, 0, 0) so that x2 = 0 and x3 = x1 . Then for the constructed function we
have
Z
3x3
(t1 − x1 )f (t1 , t2 ) dt1 dt2
∂u(f ; x1 , x2 , x3 )
=
=
∂x1
2π
[(t1 − x1 )2 + (t2 − x2 )2 + x23 ]5/2
R2

PARTIAL DERIVATIVES OF THE POISSON INTEGRAL

= Cx3

š Z0 Z1
−1 0

+

Z1 Z1
0

0

= Cx1

” 1+x
Z 1Z1
x1

−x1

= Cx1

0

0

√
•
t1 (t1 + x1 ) t2
dt
dt
1 2 + o(1) =
(t21 + t22 + x12 )5/2

−x1 0

x1

0

p
√
t1 (t1 − x1 ) t2
dt1 dt2 +
(t21 + t22 + x21 )5/2

p

š Zx1 Z1

1−x
Z 1Z1

+

p
√
(t1 − x1 ) −t1 t2 dt1 dt2
+
[(t1 − x1 )2 + t22 + x23 ]5/2

p √
›
(t1 − x1 ) t1 t2 dt1 dt2
+ o(1) =
[(t1 − x1 )2 + t22 + x23 ]5/2

1−x
Z 1Z1

+

p
√
t1 (t1 + x1 ) t2
dt1 dt2 +
(t12 + t22 + x21 )5/2

p
p
√
√
›
t1 [ (t1 + x1 ) t2 − (t1 − x1 ) t2 ]
dt
dt
=
1 2
(t12 + t22 + x12 )5/2

= Cx1 (I1 + I2 ) + o(1),
where

I1 =

I2 =

Zx1 Z1

t1 [

0 0
1−x
Z 1Z1
x1

0

p
p
√
√
(t1 + x1 ) t2 − (x1 − t1 ) t2 ]
dt1 dt2 > 0,
(t21 + t22 + x12 )5/2

√ √
√
t1 4 t2 ( t1 + x1 − t1 − x1 )
dt1 dt2 >
(t12 + t22 + x21 )5/2

√
√
Z2x1 Z2x1 √
t1 4 t2 ( t1 + x2 − t1 − x1 )
>
dt1 dt2 >
(t21 + t22 + x21 )5/2
x1 x1

>

Z2x1 Z2x1

x1 x1

597

√
√ √
√
x1 4 x1 ( 2x1 − x1 )
2−1
1
dt1 dt2 =
· p
.
2
5/2
4
128
(9x1 )
x51

Thus, by the chosen path, we obtain
C
∂u(f ; x1 , 0, x1 )
> √
,
4
∂x1
x1

598

S. TOPURIA

which yields
path.

∂u(f ;x1 ,0,x1 )
∂x1

→ +∞ when (x1 , x2 , x3 ) → (0, 0, 0) by the chosen

By a similar reasoning we prove
Theorem 6.
(a) If at the point
Dx∗ i (x) f (x0 ), i = 1, k, then
lim

(x,xk+1 )→(x0 ,0)

x0

there

exists

a

finite

derivative

∂u(f ; x, xk+1 )
= Dx∗i f (x0 ).
∂xi

(b) There exists a continuous function f (x) such that for any B ⊂ M ,
m(B) < k, all derivatives D∗xi (xB ) f (0) = 0, i = 1, k, but the limits
∂u(f ; 0, xk+1 )
xk+1 →0+
∂xi
lim

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 (x0 ) at the point x0 , then
dx u(f ; x, xk+1 ) = df (x0 ).

lim

∧

(4)

(x,xk+1 )→(x0 ,0)

(b) there exists a continuous function f which has partial derivatives of
any order, but the limits
∂u(f ; x0 , xk+1 )
xk+1 →0+
∂xi
lim

do not exist.
Proof.
(a) By virtue of the lemma we have (x0 = 0)
∂u(f ; x, xk+1 ) ∂f (0)
−
=
∂xi
∂xi
k
k
P
P
∂f (0)
Z
f (t) − f (0) −
(ti − xi )
|tν |
∂xi ti
ν=1
ν=1
dt.
= Ck xk+1
k+3 ·
k
P
(|t − x|2 + x2k+1 ) 2
|tν |
Rk
ν=1

PARTIAL DERIVATIVES OF THE POISSON INTEGRAL

599

This equality implies
∂f (0)
∂u(f ; x, xk+1 )
=
,
∂x
∂xi
i
(x,xk+1 )→0
lim

∧

i = 1, k.

Thus equality (4) is valid.
(b) Consider the function
q
1
4


 (2t1 − t2 )(t2 − 2 t1 ) for (t1 , t2 ) ∈ D = {(t1 , t2 ) :
f (t1 , t2 ) =
0 ≤ t1 < ∞; 12 t1 ≤ t2 ≤ 2t1 },


0
for (t1 , t2 ) ∈ CD.

This function is continuous in R2 , has partial derivatives of any order at the
point (0, 0) which are equal to zero, but
q
Z2t1 t 4 (2t − t )(t − 1 t )
Z∞
2
2
1
1
3x3
∂u(f ; 0, 0, x3 )
2 1
dt2 >
dt1
=
2
2
2
∂x1
2π
(t1 + t2 + x3 )5/2
0

1
2 t1

3

Z2x3
Z2 t1
t1 dt1
> Cx3
x3

t1
3

Z2x3
Z2 t1
> Cx3
t1 dt1
x3

t1

q

(2t1 − t2 )(t2 − 12 t1 )

q

(2t1 − 32 t1 )(t1 − 21 t1 )

4

4

(t21 + t22 + x23 )5/2

2 5/2
2
( 13
4 t1 + x3 )

dt2 >

dt2 >

3

Z2 t1 p
Z2x3
Z2x3
4
t21 dt2
C
5/2
= 4
t1 dt1 =
t1 dt1
> Cx3
x53
x3
x3

t1

x3

C
= √ → +∞ for x3 → 0 + .
x3
References
1. P. Fatou, S´eries trigonom´eetriques et s´eries de Taylor. Acta Math.
30(1906), 335–400.
2. A. Zygmund, Trigonometric series, I. Cambridge University Press,
1959.
3. N. K. Bari, Trigonometric series. (Russian) Fizmatgiz, 1961.
4. S. B. Topuria, Boundary properties of the differentiated Poisson integral in a circle. (Russian) Trudy Gruz. Politekhn. Inst. 3(176)(1975),
13–20.

600

S. TOPURIA

5. A. G. Jvarsheishvili, On analytic functions in a half-plane. (Russian)
Trudy Tbiliss. Mat. Inst. Razmadze 31(1966), 91–109.
6. S. B. Topuria, Boundary properties of the differentiated Poisson integral in a half-plane. (Russian) Soobshch. Akad. Nauk Gruz. SSR 78(1975),
No. 2, 281–284.
7. S. B. Topuria, Summation of the differentiated Fourier series by the
Abel method. (Russian) Dokl. Akad. Nauk SSSR 209(1973), No. 3, 569–
572.
8. S. B. Topuria, Summation of the differentiated Fourier–Laplace series
by the Abel method. (Russian) Trudy Gruz. Polytekhn. Inst. 7(147)
(1971), 17–24.
9. S. B. Topuria, Fourier–Laplace series on the sphere. (Russian) Tbilisi
University Press, Tbilisi, 1987.
10. O. P. Dzagnidze, Boundary properties of derivatives of the Poisson
integral for a ball and representation of a function of two variables. (Russian) Trudy Tbiliss. Mat. Inst. Razmadze 98(1990), 52–98.
11. O. P. Dzagnidze, Boundary properties of second-order derivatives of
the spherical Poisson integral. Proc. A. Razmadze Math. Inst. 102(1993),
9–27.
12. O. P. Dzagnidze, On the differentiability of functions of two variables and of indefinite double integrals. Proc. A. Razmadze Math. Inst.
106(1993), 7–48.
13. S. B. Topuria, Solution of the Dirichlet problem for a half-space.
(Russian) Dokl. AN SSSR 195(1970), No. 3, 567–569.
14. S. B. Topuria, Boundary properties of the differentiated Poisson
integral and solution of the Dirichlet problem in a half-space. (Russian)
Trudy Gruz. Polytekhn. Inst. 6(197)(1977), 11–30.
15. S. B. Topuria, Boundary properties of derivatives of the Poisson
integral in a half-space and representation of a function of two variables.
(Russian) Dokl. Akad. Nauk 353(1997), No. 3.
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Tbilisi, 1993.

(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