A proof of Khavinson's conjecture in $\mathbf{R}^4$

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by  David Kalaj

1
A PROOF OF KHAVINSON’S CONJECTURE IN R4
DAVIDKALAJ
Abstract. The paper deals with an extremal problem for bounded harmonic functions in the
unit ball of R4. We solve the generalized Khavinson problem in R4. This precise problem was
6 formulated by G. Kresin and V. Maz’ya for harmonic functions in the unit ball and in the half–
1 space of Rn. We find the optimal pointwise estimates for the norm of the gradient of bounded
0 real–valued harmonic functions.
2

n
a
J 1. Introduction and statement of the main results

3
1 Inthispaperweconsiderthesharppointwiseestimatesforthegradientsofreal–valuedbounded
harmonic functions. We will first recall the known estimates of this type in the plane and in the
]
V space.
C For every fixed z = (x,y) ∈ R2+ there holds the optimal gradient estimate
. 2 1
h (1.1) V(z) 6 V ,
t |∇ | πy| |∞
a
m where V is an arbitrary bounded harmonic functions in the upper half–plane H2 = R2, and
+
[ V = sup V(z). Using the conformal transformation of the unit disk B2 onto R2 one
| |∞ z∈R2+| | +
1 easily derives
v
4 1
7 (1.2) U(z) 6 U ,
4 |∇ | π1 z 2| |∞
3 −| |
3 forx B2, whereU isharmonicandboundedintheunitdisc[2]. Thisclassical resultisimproved
∈
0 in the recent paper of D. Kalaj and M. Vuorinen [3]. Their form of the above inequality says that
.
1 41 U(z)2
0 (1.3) U(z) 6 − .
6 |∇ | π 1 z 2
−| |
1
This relation requires that U is bounded by 1 in the disc B2. The proof of the inequality (1.3)
:
v
lies on the classical Schwarz lemma for bounded analytic functions.
i
X Recently G. Kresin and V. Maz’ya [8] proved the following generalization od (1.1):
r
a 4 (n 1)(n 1)/2 Γ(n/2) 1
(1.4) V(x) 6 − − V .
|∇ | √π nn/2 Γ((n 1)/2)xn| |∞
−
Here, V is a bounded harmonic function in the half–space Rn+, |V|∞ = supy∈Rn+|V(y)| , and
x = (x,x ) Rn is fixed. These optimal poitwise estimates arise arise while proving Khavinson
′ n ∈ +
conjectureinthehalfspacesetting. Inordertoformulate theconjectureweintroducethenotation
we need.
For every fixed x let (x) denote the optimal number for the gradient estimate
C
(1.5) U(x) 6 (x)U ,
|∇ | C | |∞
12010 Mathematics Subject Classification: Primary 47B35
Key words and phrases. Harmonic mappings, Schwarz lemma.
1
2 DAVIDKALAJ
where U is harmonic and bounded in Bn or Rn. Similarly, for every v Rn, v = 1 denote by
+ ∈ | |
(x,v) the optimal number for the gradient estimate in the direction v, i.e., the smallest number
C
such that the following relation holds
U(x),v 6 (x,v)U
|h∇ i| C | |∞
for every bounded and harmonic U. Since
U(x) = sup U(x),v ,
|∇ | v ∂Bn|h∇ i|
∈
it follows that
(1.6) (x) = sup (x,v).
C v C
It turned out that the variational problem (1.6) is a hard problem, especially in the unit ball
setting. The generalized Khavinson conjecture states that
Conjecture 1.1. For x Bn we have
∈
(1.7) (x) = (x,n ),
x
C C
where n = x/x is the vector normal to the boundary at x.
x
| |
In 1992, Khavinson [6] obtained the optimal estimate in the normal direction of the gradient
of bounded harmonic functions in the unit ball in R3. In a private conversation with K. Gresin
and V. Maz’ya he believed that the same estimate hold for the norm of the gradient, i.e., that
the above conjecture is true for the unit ball B3.
In their recent paper [8] and in their book [9], G. Kresin and V. Maz’ya considered the Khavin-
son problem from a more general aspect including harmonic functions with Lp-boundary values
(1 6 p 6 ). They formulated the generalized Khavinson conjecture and proved it for bounded
∞
harmonic functions in Rn. In this context we have n = e for all x Rn. After replacing (x)
+ x n ∈ + C
with (x,e ) in (1.5), they obtained (1.4).
n
C
M. Markovi´c in a recent paper [11] proved the conjecture when x is near the boundary, i.e., if
1 ǫ 6 x < 1. Therefore, in (1.5) one can replace (x) with (x,n ), if x is near 1. In this
x
− | | C C | |
paper we prove the conjecture for n = 4, i.e. we prove the following theorem
Theorem 1.2. For x B4 we have
∈
(1.8) (x) = (x,n ),
x
C C
where n =x/x is the vector normal to the boundary at x.
x
| |
A reformulated version of Theorem 1.2 is the following theorem, whose proof follows directly
from Theorem 2.9 and relation (2.1) below.
Theorem 1.3. Let n= 4. Then we have the sharp inequality for every x B4, r = x :
∈ | |
r√4 r2 2+r2 +4 1 r2 tan 1 r√4 r2
|∇u(x)| 6 (cid:16) − (cid:0) (cid:1)π(1 (cid:0)r2−)r3(cid:1) − h r2−−2 i(cid:17)|u|∞, u∈ h∞(B4).
−
Here and in the sequel, h (B4) is the Hardy space of bounded harmonic functions on the unit ball
∞
4
B (cf. [1]).
A PROOF OF KHAVINSON’S CONJECTURE IN R4 3
Corollary 1.4. For the decreasing diffeomorphism C :[0,1] 3√3, 16 , defined by
→ 2π 3π
h i
r√4 r2 2+r2 +4 1 r2 tan 1 r√4 r2
C(r)= − − − −2+−r2 ,
(cid:16) (cid:0) (cid:1) π(cid:0)r3 (cid:1) h i(cid:17)
we have the sharp inequality for every x B4, r = x :
∈ | |
C(r)
u(x) 6 u u h (B4).
∞
|∇ | 1 r2| |∞ ∈
−
Remark 1.5. Observe that for R4, Kresin - Maz’ya inequality (1.4) reads as
+
3√3 1
(1.9) V(x) 6 V .
|∇ | 2π x4| |∞
Proof of corollary. We need to prove that C(r) is a decreasing function. We have that
( 2+r)r(2+r) 6+r2 4√4 r2 3+r2 tan 1 r√4 r2
C′(r) = −(cid:16) − (cid:0)− (cid:1)π−r4√4 −r2 (cid:0)− (cid:1) − h −2+−r2 i(cid:17).
−
So C(r)6 0 if and only if
′
r(4 r2) 6 r2 r√4 r2
v(r) = − − +tan−1 − > 0.
4(3−r2)(cid:0)√4−r(cid:1)2 " −2+r2 #
Since
r4√4 r2
v′(r) = 2(3 −r2)2 > 0,
−
and v(0) = 0 and the claim follows. (cid:3)
2. The technical lemmas
Let r = x . For n > 3, let ω be the area of Sn 1. Markovi´c in [11] proved that
n −
| |
1
(2.1) (x) = supC(z,r),
C 1 r
z>0
−
where
4ω 2n 1 1 1 Ψ (zt)+Ψ ( zt)
n 2 − r r
(2.2) C(z,r) = − − dt.
ωn (1+r)n−1√1+z2 Z0 (1−t2)4−n
Here
p
z+√z2+1−α2r
(2.3) Ψr(z) = 1−αr n−βr +nzw−βrw2 wn−2dw,
(1+w2)n/2+1(1+k2w2)n/2 1
Z0 r −
and
1 r r(n 2) (n (n 2)r)
k = − , α = − , β = − − .
r r r
1+r n 2
Further, in the same paper he has showed that the conjectured equality (1.7) is equivalent to the
equality
(2.4) supC(z,r) = C(0,r).
z>0
Our goal is to prove (2.4) for n =4.
4 DAVIDKALAJ
2.1. Explicit representation of Ψ for n = 4. Let us recalculate the integrand in (2.3):
w2 4+ 1( 4+2r) 1(4 2r)w2+4wz
Q(w) = 2 − − 2 −
(cid:0) (1+w2)3 1+ (1(−1+r)r2)w22 (cid:1)
(1+r)2w2 2+r (cid:16)2w2 +rw2+(cid:17)4wz
= −
(1+w2)3((1+r)2+( 1+r)2w2)
(cid:0) (cid:1)
−
(1+r)2 (1+r)2(1+4r)
=
r(1+w2)3 − 4r2(1+w2)2
(1+r)4 ( 1+r)2(1+r)4
+ −
16r3(1+w2) − 16r3((1+r)2+( 1+r)2w2)
−
(1+r)2zw (1+r)4zw
+
r(1+w2)3 − 4r2(1+w2)2
( 1+r)2(1+r)4zw 1+r2 4zw
+ − − .
16r3(1+w2) − 16r3((1+r)2+( 1+r)2w2)
(cid:0) (cid:1)−
By elementary integration and since
1 1 w 5+3w2
dw = +3tan 1[w]
Z (1+w2)3 8 ((cid:0)1+w2)2(cid:1) − !
while
1 1 w
dw = +tan 1[w]
−
(1+w2)2 2 1+w2
Z (cid:18) (cid:19)
we obtain
32r3
R(w) = Q(w)dw
(1+r)2
Z
4rw 1+w2+r 1+w2 4r 1+r2+(1+r)2w2 z
= − −
(1+w2)2
(2.5) (cid:0) (cid:0) (cid:1)(cid:1) (cid:0) (cid:1)
( 1+r)w
+2 1+r2 tan−1[w]+2 1+r2 tan−1 −
− − 1+r
(cid:20) (cid:21)
(cid:0) (cid:1) (1+r)2(cid:0)+( 1+(cid:1)r)2w2
+ 1+r2 2zlog − .
− 1+w2
(cid:20) (cid:21)
(cid:0) (cid:1)
Thus we have
A PROOF OF KHAVINSON’S CONJECTURE IN R4 5
Lemma 2.1. For r (0,1) and z > 0 we have
∈
(1 r)(1+r)3
Ψ (z) = −
r 64r3 ×
r 4+r2(4+r) z+4 1+r2 z3+ 2+r2+2 1+r2 z2 √4 r2+4z2
−
(cid:16)(cid:0) (cid:1) (cid:0) ((cid:1)1+z2)(cid:0)(1 r2) (cid:0) (cid:1) (cid:1) (cid:17)
(cid:18) −
r 2rz+(r2 2)√4 r2+4z2
+4tan 1 − − −
−  (cid:16) ( 2+r2)2 4( 1+r2)z2 (cid:17)
− − −
 
1+z z+r2z r√4 r2+4z2
+2 1 r2 zlog − − .
−  (cid:16) (1+r)2(1+z2) (cid:17)
(cid:19)
(cid:0) (cid:1)
 
Proof. In view of (2.5) and (2.3) we obtain
z+√z2+1−α2r
32r3 Ψ (z) = 32r3 1−αr Q(w)dw = R z+ z2+1−α2r R(0),
(1+r)2 r (1+r)2 1 α −
Z0 p − r !
which after some elementary transformations implies the lemma. (cid:3)
2.2. Explicit representation of C for n = 4. From Lemma 2.1 we have
Lemma 2.2. For r (0,1) and z > 0 we have
∈
(1 r)(1+r)3 r√4 r2+4t2z2 2+r2+2 1+r2 t2z2
Ψ (zt)+Ψ ( zt) = − −
r r − 16r3 2(1 r2)(1+t2z2)
(cid:18) −(cid:0) (cid:0) (cid:1) (cid:1)
r 2rtz+(2 r2)√4 r2+4t2z2
tan 1 − −
− −  (cid:16)( 2+r2)2 4( 1+r2)t2z2 (cid:17)
− − −
 
r 2rtz (2 r2)√4 r2+4t2z2
+tan 1 − − −
−  (cid:16)( 2+r2)2 4( 1+r2)t2z2 (cid:17)
− − −
 
rtz√4 r2+4t2z2
1 r2 tztanh−1 − .
− − 1+(1+r2)t2z2
" #
(cid:19)
(cid:0) (cid:1)
Using integration by parts for V = t and
r 2rtz+(2 r2)√4 r2+4t2z2
U = tan 1 − −
− −  (cid:16)( 2+r2)2 4( 1+r2)t2z2 (cid:17)
− − −
 
r 2rtz (2 r2)√4 r2+4t2z2
+tan 1 − − − ,
−  (cid:16)( 2+r2)2 4( 1+r2)t2z2 (cid:17)
− − −
 
6 DAVIDKALAJ
and in view of the formula
rtz2 2+r2 2+2 2 3r2+r4 t2z2
− −
U = ,
′ (1+t2z2(cid:16))(cid:0)√4 r2(cid:1)+4t2z(cid:0)2 1+( 1+(cid:1)r2)2t(cid:17)2z2
− −
(cid:16) (cid:17)
which can be proved by a direct computation, we obtain
rtz2 2+r2 2+2 2 3r2+r4 t2z2
− −
U(t)dV = tU(t) t dt.
− (1+t2z2(cid:16))(cid:0)√4 r2(cid:1)+4t2z(cid:0)2 1+( 1+(cid:1)r2)2t(cid:17)2z2
Z Z
− −
(cid:16) (cid:17)
Similarly, for V = t2/2 and
1
rtz√4 r2+4t2z2
U1 = −z(1−r2)tanh−1 1+(1−+r2)t2z2
" #
we obtain
rz2 4+5r2 r4+ 4+7r2 4r4+r6 t2z2
U = − − − − ,
1′
(1+(cid:0)t2z2)√4 r2+4(cid:0)t2z2 1+( 1+r2)2(cid:1)t2z2 (cid:1)
− −
and then (cid:16) (cid:17)
t2rz2 4+5r2 r4+ 4+7r2 4r4+r6 t2z2
U (t)dV = U V − − − − dt.
1 1 1 1
Z −Z 2 (1+(cid:0)t2z2)√4 r2+4(cid:0)t2z2 1+( 1+r2)2(cid:1)t2z2 (cid:1)
− −
(cid:16) (cid:17)
Furthermore we have
rt2z2 4+3r2 r4 4 5r2+r6 t2z2
VU +V U = − − − − .
′ 1 1′ −2(1+t2z(cid:0)2)√4 r2+4t2z2(cid:0) 1+( 1+(cid:1)r2)2t2(cid:1)z2
− −
(cid:16) (cid:17)
Let
r√4 r2+4t2z2 2+r2+2 1+r2 t2z2
Y = −
2(1 r2)(1+t2z2)
−(cid:0) (cid:0) (cid:1) (cid:1)
and
X = VU V U +Y.
− ′− 1 1′
By using the formula
a at
dt = tanh 1
−
Z √b+ct2 1+ (c−ba2)t2 (cid:20)√b+ct2(cid:21)
(cid:16) (cid:17)
and the representation
4r 1+r2 t2z3
2 1 r2 zX = +r 1+r2 z 4 r2+4t2z2
− √4 r2+4t2z2 −
(cid:0) (cid:1)
(cid:0) (cid:1) − rz (cid:0) (cid:1) p r 3+r2 z
+ − ,
√4 r2+4t2z2(1+t2z2) − √4 r2+4t2z(cid:0)2 1+((cid:1)1+r2)2t2z2
− − −
(cid:16) (cid:17)
A PROOF OF KHAVINSON’S CONJECTURE IN R4 7
we obtain
r 1+r2 tz√4−r2+4t2z2+tanh−1 √4 rr2t+z4t2z2 −tanh−1 √r4(−r32++r42t)2tzz2
Xdt = h − i (cid:20) − (cid:21).
(cid:0) (cid:1) 2(1 r2)z
Z −
Now the formula
16r3
Ψ (zt)+Ψ ( zt)dt = Ydt+ Udv+ U dv
(1 r)(1+r)3 r r − 1 1
− (cid:18)Z (cid:19) (cid:18)Z Z Z (cid:19)
= Ydt+UV +U V VdU V dU
1 1 1 1
− −
Z Z Z
= UV +U V + Xdt,
1 1
Z
yields the following
Lemma 2.3. For t [0,1] we have
∈
t 1 1
(Ψ (zs)+Ψ ( zs))ds = (1+r)2 r 1+r2 tz 4 r2+4t2z2
r r − 16r3z 2 −
Z0 (cid:18)
(cid:0) (cid:1) p
1 rtz 1 r 3+r2 tz
+ tanh 1 tanh 1 −
− −
2 (cid:20)√4−r2+4t2z2(cid:21)− 2 "√4(cid:0)−r2+4(cid:1)t2z2#
r 2rtz+(2 r2)√4 r2+4t2z2
+ 1+r2 tztan 1 − −
− −  (cid:16)( 2+r2)2 4( 1+r2)t2z2 (cid:17)
− − −
(cid:0) (cid:1)
 
r 2rtz (2 r2)√4 r2+4t2z2
1+r2 tztan 1 − − −
− − −  (cid:16)( 2+r2)2 4( 1+r2)t2z2 (cid:17)
− − −
(cid:0) (cid:1)
 
1 rtz√4 r2+4t2z2
1+r2 2t2z2tanh−1 − .
− 2 − 1+(1+r2)t2z2
" #
(cid:19)
(cid:0) (cid:1)
By putting t = 1 in Lemma 2.3 and using (2.2), in view of
4ω 24 1 4 2π 8 32
4 2 −
− = · · = ,
ω (1+r)4 1 2π2(1+r)3 π(1+r)3
4 −
we obtain
Lemma 2.4. For r (0,1) and z > 0 we have
∈
2(1 r)
C(r,z) = − (h (z)+h (z)+h (z)),
1 2 3
πr3√1+z2
where
r 1+r2 √4 r2+4z2 tanh−1 √4 rr2z+4z2 −tanh−1 √r(4−3r+2+r24)zz2
h1(z) = 2(1 −r2) + h − 2(1i r2)z (cid:20) − (cid:21),
(cid:0) (cid:1) − −
8 DAVIDKALAJ
r 2rz (2 r2)√4 r2+4z2
h (z) = tan 1 − − −
2 −  (cid:16)( 2+r2)2 4( 1+r2)z2 (cid:17)
− − −
 
r 2rz+(2 r2)√4 r2+4z2
tan 1 − −
− −  (cid:16)( 2+r2)2 4( 1+r2)z2 (cid:17)
− − −
 
and
1 rz√4 r2+4z2
h3(z) = 2 −1+r2 ztanh−1 1+(−1+r2)z2 .
" #
(cid:0) (cid:1)
Finally we need the following lemmata
Lemma 2.5. Let
rz r 3+r2 z
L(z) = tanh 1 tanh 1 − .
− −
(cid:20)√4−r2+4z2(cid:21)− "√4(cid:0)−r2+4(cid:1)z2#
Then, L(z) 6 rz√4 r2.
−
Proof. By differentiating L we obtain
r 4 r2+ 4 3r2+r4 z2
L(z) = − − .
′
(1+z2)(cid:0)√4 r2+(cid:0)4z2 1+( 1(cid:1)+r(cid:1)2)2z2
− −
(cid:16) (cid:17)
Since
4 r2+ 4 3r2+r4 z2 2z 2 3+r2 1+r2 2z
∂ − − = + − − 6 0,
z((cid:0)1+z2) (cid:0)1+( 1+r2)(cid:1)2z2(cid:1) −(1+z2)2 (cid:0)1+( 1(cid:1)+(cid:0) r2)2z2(cid:1)2
− −
(cid:16) (cid:17) (cid:16) (cid:17)
it follows that (L(z) zr√4 r2) 6 L(0) r√4 r2 = 0. So L(z) 6 zr√4 r2. (cid:3)
′ ′
− − − − −
Lemma 2.6. Let
r√4 r2+r 1+r2 √4 r2+4z2
g (z) = − − .
1
√1+z2
(cid:0) (cid:1)
Then sup g (z) = g (0) = r√4 r2+r 1+r2 √4 r2.
z>0 1 1 − −
Proof. Since (cid:0) (cid:1)
rz r2+r4 ( 4+r2)(r2 4(1+z2))
− − −
g (z) =
1′ (cid:16) (1+pz2)3/2√4 r2+4z2 (cid:17)
−
and
(r2+r4)2 ( 4+r2)(r2 4(1+z2)) = 2r6+r8 16(1+z2)+4r2(2+z2)6 5 12z2 6 0,
− − − − − −
it follows that g (z) 6 0 for z > 0. Thus g (z) 6 g (0) for z > 0 what we needed to prove. (cid:3)
1′ 1 1
A PROOF OF KHAVINSON’S CONJECTURE IN R4 9
Lemma 2.7. Let
r 2rz+(2 r2)√4 r2+4z2
h (z) = tan 1 − −
2 −  (cid:16)( 2+r2)2 4( 1+r2)z2 (cid:17)
− − −
 
r 2rz (2 r2)√4 r2+4z2
tan 1 − − − .
− −  (cid:16)( 2+r2)2 4( 1+r2)z2 (cid:17)
− − −
 
Then sup h (z) = h (0) = 2tan 1 r√4 r2 .
z>0 2 2 − 2 −r2
−
h i
Proof. We have
2r 2 r2 z 2+r2+2 1+r2 z2
h (z) = − − − .
′2
(1+(cid:0)z2)√4(cid:1) r(cid:0)2+4z2 1+(cid:0)( 1+r2(cid:1))2z(cid:1)2
− −
So h (z) 6 0, and h (z) 6 h (0) for every z. (cid:16) (cid:17) (cid:3)
′2 2 2
By using the formula 1log 1+a = tanh 1(a), for a = rz√4 r2+4z2 < 1, we obtain
2 1 a − 1+(1−+r2)z2
−
Lemma 2.8. For z > 0 and 0 < r < 1 we have
1 rz√4 r2+4z2
h3(z) := 2 −1+r2 ztanh−1 1+(−1+r2)z2 6 h3(0) = 0.
" #
(cid:0) (cid:1)
2.3. Proof of the main result.
Theorem 2.9. For r (0,1) we have
∈
r√4 r2 2+r2 +4 1 r2 tan 1 r√4 r2
− − − r2−2
supC(z,r) = C(0,r) = − .
z>0 (cid:16) (cid:0) (cid:1)π(1+(cid:0) r)r3 (cid:1) h i(cid:17)
Proof. From Lemmas 2.4, 2.5, 2.6, 2.7 and 2.8 we obtain
2(1 r)
C(z,r) = − (h (z)+h (z)+h (z))
1 2 3
πr3√1+z2
2(1 r) r 1+r2 √4 r2+4z2 L(z)
6 − − + +h (0)+h (0)
πr3√1+z2 (cid:0) 2(cid:1)(1−r2) 2(1−r2)z 2 3 !
2(1 r) r 1+r2 √4 r2+4z2 rz√4 r2
6 − − + − +h (0)+h (0)
πr3√1+z2 (cid:0) 2(cid:1)(1−r2) 2(1−r2)z 2 3 !
(1 r) 2(1 r)
= − g (z)+ − (h (0)+h (0))
πr3(1 r2) 1 πr3√1+z2 2 3
−
(1 r) 2(1 r)
6 − g (0)+ − (h (0)+h (0))
πr3(1 r2) 1 πr3 2 3
−
2(1 r)
= − (h (0)+h (0)+h (0)) = C(0,r)
πr3 1 2 3
r√4 r2 2+r2 +4 1 r2 tan 1 r√4 r2
− − − r2−2
= − .
(cid:16) (cid:0) (cid:1)π(1+(cid:0) r)r3 (cid:1) h i(cid:17)
So sup C(z,r)= C(0,r) what we needed to prove. (cid:3)
z>0
10 DAVIDKALAJ
Remark 2.10. It seems that the same strategy for n = 4 does not work, because the integrand
6
that appear in definition of the function C is much more complicated.
References
[1] Sh. Axler,P. Bourdon and W. Ramey,Harmonic function theory, Springer, NewYork (1992).
[2] F. Colonna, The Bloch constant of bounded harmonic mappings, Indiana University Math. J. 38 (1989), 829–
840.
[3] D. Kalaj and M. Vuorinen, On harmonic functions and the Schwarz lemma, Proc. Amer. Math. Soc. 140
(2012), 161–165.
[4] D. Kalaj and M. Markovi´c, Optimal estimates for harmonic functions in the unit ball, Positivity 16 (2012),
771–782.
[5] D. Kalaj and M. Markovi´c, Optimal Estimates for the Gradient of Harmonic Functions in the Unit Disk,
Complex Analysis and Operator Theory 7 (2013), 1167–1183.
[6] D. Khavinson, An extremal problem for harmonic functions in the ball, Canadian Math. Bulletin 35 (1992),
218–220.
[7] G. Kresin and V. Maz’ya, Sharp pointwise estimates for directional derivatives of harmonic functions in a
multidimensional ball, J. Math. Sci. 169 (2010), 167–187.
[8] G. Kresin and V. Maz’ya, Optimal estimates for the gradient of harmonic functions in the multidimensional
half-space, Discrete Contin. Dyn.Syst.28 (2010), 425–440.
[9] G. Kresin and V. Maz’ya, Sharp real-part theorems. Springer, Berlin, Jan. 1, 2007. (Lecture Notes in Mathe-
matics, 1903). ISBN:978-3-540-69573-8.
[10] M.Markovi´c,Onharmonic functions andthe hyperbolic metric,IndagationesMathematicae26(2015),19–23.
[11] M. Markovi´c, Proof of the Khavinson conjecture near the boundary of the unit ball, arXiv:1508.00125v1.
University of Montenegro, Faculty of Natural Sciences and Mathematics, Dzordza Vaˇsingtona
b.b. 81000 Podgorica, Montenegro
E-mail address: [email protected]