On Existence of Boundary Values of Polyharmonic Functions
ON EXISTENCE OF BOUNDARY VALUES OF POLYHARMONIC
FUNCTIONS 1
M.L. GORBACHUK and S.M. TORBA
ABSTRACT. In trigonometric series terms all polyharmonic functions inside the unit
disk are described. For such functions it is proved the existence of their boundary
values on the unit circle in the space of hyperfunctions. The necessary and sufficient
conditions are presented for the boundary value to belong to certain subspaces of the
space of hyperfunctions.
The purpose of this paper is to find necessary and sufficient conditions for a solution of
the equation ∆mu = 0 inside a domain to have a limit on the boundary of the domain in
various functional spaces. We consider the simplest situation where a domain is the unit disk
K = {z = reit, 0 ≤ r < 1, 0 ≤ t ≤ 2π}. The case of m = 1 has been investigated during
20th century by a lot of mathematicians (we refer for details to [1 - 5]. The case of m = 2 was
considered in [6]. For an arbitrary m the problem of existence of boundary values in the space
L2(∂K) (∂K - is the unit circle) was discussed in [7].
1. Denote by D = D(∂K) the set of all infinitely differentiable functions on ∂K. We
say that a sequence ϕn ∈ D converges to ϕ ∈ D, n → ∞, and write ϕn
→ ϕ, if for every
k ∈ N0 = {0, 1, 2, . . .}, the sequence ϕ
n (t) converges to ϕ
(k)(t) uniformly in t ∈ ∂K. Let also
A = A(∂K) be the set of analytic functions on ∂K. The convergence in A is introduced in the
following way: a sequence ϕn ∈ A converges to ϕ in A (ϕn
→ ϕ) if there exists a neighbourhood
U of ∂K in which all the functions ϕn(t) converge to ϕ(t) uniformly on any compact set from
For a number α > 0 we put
Aα = {ϕ ∈ D
∣∣∃c > 0 ∀N0 max
|ϕ(k)(t)| ≤ cαkk!}.
The linear set Aα is a Banach space with respect to the norm
‖ϕ‖Aα = sup
|ϕ(k)(t)|
It is not hard to show that if α < α′, then Aα ⊆ Aα′ ,
A = ind lim
and the dense continuous embeddings
A ⊂ D ⊂ Lp(∂K) = Lp, 1 ≤ p < ∞,
1Mathematics Subject Classification. Primary 35J30
Key words and phrases. Polyharmonic equation, solution inside of a domain and its boundary value, hyperfunc-
tion, Fourier series.
Supported by CRDF and Ukr. Government (Project UM 1-2507-OD-03)
http://arxiv.org/abs/0704.0260v1
hold.
Let D′ and A′ are the spaces of continuous antilinear functionals on D (distributions) and
A (hyperfunctions), respectively (see [8]). In the following, < F, ϕ > denotes an action of the
functional F ∈ A′ (F ∈ D′) onto ϕ ∈ A (ϕ ∈ D). By convergence in A′ (in D′) we mean the
weak one, that is, Fn
→ F (Fn
→ F ) if for any ϕ ∈ A (ϕ ∈ D), the number sequence < Fn, ϕ >
converges to < F, ϕ >.
As ek(t) = e
ikt ∈ A (k ∈ Z), the Fourier coefficients ck(F ) =< F, ek > can be determined for
F ∈ A′. It is known (see e.g. [9]) that
ck(F )e
ikt A
and one can easily verify that the below assertion is valid.
Proposition 1 The following equivalence relations hold:
F ∈ D ⇐⇒ ∀α > 0 ∃c > 0 |ck(F )| ≤ c|k|
F ∈ A ⇐⇒ ∃α > 0 ∃c > 0 |ck(F )| ≤ ce
−α|k|;
F ∈ D′ ⇐⇒ ∃α > 0 ∃c > 0 |ck(F )| ≤ c|k|
F ∈ A′ ⇐⇒ ∀α > 0 ∃c > 0 |ck(F )| ≤ ce
α|k|.
Moreover, the series
ck(F )e
ikt converges to F in the corresponding space. The sequence
{Fn}n∈N, whose elements Fn belong to one of the spaces D,A, D
′ or A′, converges to F in this
space if and only if the constants c and α in the above estimates for |ck(Fn)| do not depend on
n and for any k ∈ Z, ck(Fn) → ck(F ), n → ∞.
2. A function u(r, t) = u(reit) ∈ C2m(K) is called m-harmonic in K if it satisfies the
equation
∆mu(r, t) = 0, 0 ≤ r < 1, t ∈ [0, 2π]. (1)
Note, that no conditions on the behaviour of u(r, t) near ∂K are imposed.
Theorem 1. In order that a function u(r, t) ∈ C2m(K) be m-harmonic in K, it is necessary
and sufficient that the representation
u(r, t) =
(r2 − 1)j−1
ck(Fj)r
|k|eikt, Fj ∈ A
′, (2)
be admissible, where Fj are uniquely determined by u(r, t).
Proof. By Proposition 1,
∀α > 0 ∃cj = cj(α) > 0 ∀k ∈ Z |ck(Fj)| ≤ cje
α|k|.
So the series
ck(Fj)r
|k|eikt converges uniformly in the disk KR = {z ∈ C : |z| ≤ R} of radius
R < e−α and determines an infinitely differentiable function there. The direct check shows that
the functions (r2 − 1)j−1
ck(Fj)r
|k|eikt, j = 1, 2, . . . , m, satisfy (1) in KR. Since α > 0 is
arbitrary, these functions are solutions of the equation (1) inside K.
To prove the necessity, suppose at first m = 1. Let u(r, t) be a harmonic function in K.
Then for a fixed r < 1, u(r, t) is infinitely differentiable in t, and it may be written in the form
u(r, t) =
ck(r)e
ikt, ck(r) =
u(r, t)e−ikt dt, (3)
where the series and all its derivatives converge uniformly in t ∈ [0, 2π]. The coefficients ck(r)
are infinitely differentiable on [0, 1) and satisfy the equation
′′(r) + rck
′(r) = k2ck(r).
Hence,
ck(r) = r
|k|ck, ck ∈ C.
It follows from the convergence of the series in (3) that
∀r < 1 r|k||ck| = e
−α|k||ck| ≤ c,
where α = − ln r > 0 is arbitrary. By Proposition 1, ck are the Fourier coefficients of a certain
hyperfunction F1, and
u(r, t) =
r|k|ck(F1)e
ikt, F1 ∈ A
′. (4)
Thus, the representation (2) is valid when m = 1.
Assume the representation (2) to be true for an (m−1)-harmonic insideK function u(r, t) (m ≥
2), and we shall prove that such a representation holds for an m-harmonic function.
If u(r, t) is anm-harmonic function, then ∆u(r, t) is an (m−1)-harmonic one. By assumption,
there exist Ej ∈ A
′, j = 1, 2, . . . , m− 1, such that
∆u(r, t) =
(r2 − 1)j−1
ck(Ej)r
. (5)
If we choose ũ ∈ C2(K) so that
∆(u(r, t)− ũ(r, t)) = 0, (6)
then, because of (4), we shall have
u(r, t) = ũ(r, t) +
ck(F1)e
, F1 ∈ A
Let us find at first ũ(r, t) in the case where the equation (5) is of the form
∆u(r, t) =
r|k|ck(E1)e
ikt, E1 ∈ A
By using the identity
(r2 − 1)j
r|k|ck(F )e
(r2− 1)j−1(|k|+ j)+ (j− 1)(r2− 1)j−2
r|k|ck(F )e
for F ∈ A′, one can verify that the function ũ2(r, t) = u2(r, t), where
u2(r, t) =
(r2 − 1)
|k|+ 1
ck(E1)e
(ũ1(r, t) ≡ 0), satisfies (6). Set ck(F2) =
ck(E1)
4(|k|+1)
. By Proposition 1, F2 ∈ A
′. So, in the case
under consideration
u(r, t) =
(r2 − 1)j−1
r|k|ck(Fj)e
Suppose now that we know solutions ul(r, t) of the equations
∆u(r, t) =
(r2 − 1)j−1
r|k|ck(Ej)e
ikt, (8)
for all l ≤ s, s ≤ m− 2 is fixed. We show how to find a solution of the equation
∆u(r, t) =
(r2 − 1)j−1
r|k|ck(Ej)e
ikt. (9)
We put
us+2(r, t) = (r
2 − 1)s+1
4(s+ 1)(|k|+ s+ 1)
ck(Es+1)e
It follows from (7) and (8) that if u(r, t) is a solution of (9), then
∆(u− us+2)(r, t) =
(r2 − 1)j−1
ck(Ej)e
(r2 − 1)sr|k|ck(Es+1)e
ikt −
s(r2 − 1)s−1
|k|+ s+ 1
+ (r2 − 1)s
r|k|ck(Es+1)e
ikt =
(r2 − 1)j−1
r|k|ck(Ej)e
ikt + (r2 − 1)s−1
ck(Es)−
s · ck(Es+1)
|k|+ s + 1
eikt.
Taking into account that Es, Es+1 ∈ A
′, we conclude, by Proposition 1, that there exists E ′s ∈ A
such that
s) = ck(Es)−
s · ck(Es+1)
|k|+ s+ 1
whence
∆(u− us+2)(r, t) =
(r2 − 1)j−1
j ∈ A
where E ′j = Ej as j = 1, . . . , s− 1. By assumption, we can find ũs+1(r, t) so that
∆(u− us+2 − ũs+1)(r, t) = 0.
Setting
ũs+2(r, t) = us+2(r, t) + ũs+1(r, t),
we arrive at the equality
∆(u− ũs+2)(r, t) = 0.
It is not hard to observe that for the desired function ũ(r, t) we have the formula
ũ(r, t) = ũm(r, t) = um(r, t) + um−1(r, t) + · · ·+ u2(r, t) =
(r2 − 1)m−1
4(m− 1)(|k|+m− 1)
ck(Em−1)e
ikt + . . .
+(r2 − 1)2
4 · 2(|k|+ 2)
ck(E2)e
ikt + (r2 − 1)
4(|k|+ 1)
ck(E1)e
u(r, t) =
(r2 − 1)j−1
r|k|ck(Fj)e
ikt, Fj ∈ A
where
ck(Fj) =
ck(Ej−1)
4(j − 1)(|k|+ j − 1)
Since for F ∈ A′
r|k|ck(F ) → ck(F ), r → 1, and |r
|k|ck(F )| < |ck(F )|,
we have, by Proposition 1, that
(r2 − 1)j−1
r|k|ck(Fj)e
ikt A
F1 if j = 1
0 if j > 1,
as r → 1. The elements Fj ∈ A
′ are determined uniquely by the function u(r, t) in the following
F1 = lim
u(r, ·), Fj+1 = lim
u(r, ·)−
(r2 − 1)p−1
r|k|ck(Fp)e
(r2 − 1)j
where the limit is taken in the space A′. This completes the proof.
Because of harmonicity in K of the functions
uj(r, t) =
r|k|ck(Fj)e
the representation (2) implies, in particular, the next assertion (cf. [4]).
Corollary 1. Let u(r, t) be an m-harmonic in K function. Then it admits a representation of
the form
u(r, t) =
(r2 − 1)j−1uj(r, t), (10)
where the functions uj(r, t) are harmonic in K.
When proving the theorem, it was also established the following fact.
Corollary 2. If u(r, t) is an m-harmonic in K function, then there exists its radial boundary
value u(1, ·) on ∂K in the space A′, that is,
u(r, ·)
→ u(1, ·) as r → 1.
3. Let Φ be a complete linear Hausdorff space such that the continuous embeddings
A ⊂ Φ ⊂ A′
hold. We say that F ∈ Φ is a boundary value on ∂K of an m-harmonic in K function u(r, t)
and write F = u(1, ·) if u(r, ·)
→ F as r → 1.
It is seen from Theorem 1 and Corollary 2 that every m-harmonic in K function has a
boundary value in A′. Moreover, each element F ∈ A′ is the boundary value of a certain m-
harmonic in K function. The natural question arises: under what conditions on an m-harmonic
in K function u(r, t) its boundary value u(1, ·) belongs to Φ?
Theorem 2. The boundary value u(1, ·) of an m-harmonic in K function u(r, t) belongs to the
space Φ if and only if the set {u(r, ·)}r<1 is compact in Φ.
Proof. Necessity. It is known that if r0 < 1, then u(r0, ·) ∈ A, and u(r, ·)
→ u(r0, ·) as
1 > r → r0. Since the embedding A ⊂ Φ is continuous, u(r, ·)
→ u(r0, ·) (r → r0). By
assumption, u(r, ·)
→ u(1, ·) if r → 1. So, the set {u(r, ·)}r<1 is compact in Φ.
Sufficiency. Let the set {u(r, ·)}r<1 be compact in Φ. Suppose r → 1. Then there exists
a subsequence rk → 1 such that u(rk, ·) converges in Φ (rk → 1) to a certain element F ∈ Φ.
Since Φ ⊂ A continuously, u(rk, ·) converges in A
′. Taking into account that u(r, ·)
→ u(1, ·) as
r → 1, we have u(1, ·) = F ∈ Φ which completes the proof.
In the partial case where Φ = L2(∂K), Theorem 2 was obtained in [7]. By using compactness
criteria for sets, one can find the sufficient conditions for the boundary value of a polyharmonic
function to belong to Lp(∂K), 1 ≤ p < ∞. For instance, the following assertion is valid.
Corollary 3. Let u(r, t) be an m-harmonic inside the disk K function. In order that u(r, t)
have a boundary value in Lp = Lp(∂K), it is necessary and sufficient that:
1) sup
0≤r<1
‖u(r, ·)‖Lp < ∞;
rei(t−τ)
− u(reit)
∣∣p dt → 0 (τ → 0) uniformly in r ∈ [0, 1).
Now we consider in more detail the case of L2. Let
F ∈ A′
∣∣∣∣∣
‖F‖Bj = sup
0≤r<1
(1− r2)j
r2|k||ck(F )|
The set Bj with norm ‖ · ‖Bj forms a Banach space.
Theorem 3. If u(r, t) is an m-harmonic in K function, then
0≤r<1
|u(r, t)|2 dt < ∞ ⇐⇒ F1 ∈ L2, Fj ∈ Bj−1 if 2 ≤ j ≤ m,
where Fj are taken from representation (2). Moreover, u(r, ·) → F1 (r → 1) weakly in the space
Proof. Assume that in the representation (2) F1 ∈ L2, Fj ∈ Bj−1 (j = 2, . . . , m). Then
|u(r, t)|2 dt =
r2|k|
∣∣ck(F1) + (r2 − 1)ck(F2) + · · ·+ (r2 − 1)m−1ck(Fm)
∣∣2 ≤
r2|k|(r2 − 1)2(j−1)|ck(Fj)|
2 ≤ c.
Conversely, let sup
0≤r<1
‖u(r, ·)‖L2 < ∞. Then, as was shown in [4, Lemma 7], each summand
in (10) is bounded, too:
0≤r<1
‖(1− r2)j−1uj(r, ·)‖L2 < ∞, j = 1, 2, . . . , m.
This is equivalent to the inequality
0≤r<1
(1− r2)2(j−1)
r2|k||ck(Fj)|
2 < ∞,
that is, F1 ∈ L2, Fj ∈ Bj−1 (j = 2, . . . , m).
It still remains to prove the weak convergence of u(r, ·) to F1 (r → 1) in L2. Since u(r, ·)
as r → 1 and eikt ∈ A (k ∈ Z), we have
u(r, t)eikt dt = lim
r|k|[c−k(F1) + (r
2 − 1)c−k(F2) + · · ·+ c−k(Fm)] =
ck(F1) =< F1, ek >=
F1(t)e
Thus, u(r, ·) → F1 (r → 1) weakly in L2 on a total set, and sup
0≤r<1
‖u(r, ·)‖L2 < ∞. It follows
from here that u(r, ·) → F1 (r → 1) weakly in L2. The proof is complete.
Let u(r, t) be a harmonic in K function. It follows from (2) that
‖u(r, ·)‖2L2 =
r2|k||ck(F1)|
In view of ‖u(r, ·)‖L2 ≤ c, the well-known Fatou lemma and the Lebesgue theorem on passage
to the limit yield
|ck(F1)|
2 < ∞, F1 = u(1, ·) ∈ L2, ‖u(r, ·)‖L2 → ‖u(1, ·)‖L2, r → 1.
Therefore the weak convergence of u(r, t) to u(1, t) implies the strong one. As was shown in
[7], in the case of m = 2 the boundedness of ‖u(r, ·)‖L2 does not guarantee the convergence of
u(r, ·) (r → 1) in L2.
We pass now to the Sobolev spaces
2 = W
2 (∂K) =
F ∈ A′
|k|2α|ck(F )|
, α ∈ R.
The following statement is valid.
Theorem 4. The embeddings
2 ⊂ Bj ⊂ W
hold.
Proof. Since the function f(r) = (1 − r2)2jr2k, j, k ∈ N0, reaches its maximum at the point
r2 = k
, and
0≤r<1
f(r) =
k + 2j
)2j (
k + 2j
we have
0≤r<1
(1− r2)2j
r2|k||ck(F )| < cj
|ck(F )|
|k|2j
that is, W
2 ⊂ Bj .
Suppose now F ∈ Bj. Then, substituting z := r
∃c > 0 (1− z)2j
z|k||ck(F )|
2 < c.
Multiplying this inequality by (1− z)−α and then integrating along [0, 1), we obtain
|ck(F )|
(1− z)2j−αz|k| dz < ∞. (11)
If we put δ = 2j − α+ 1, we get for n ∈ N
(1− z)2j−αzn dz =
δ(δ + 1) . . . (δ + n)
Since
= 1 +
the relation an = O
is fulfilled. Taking in (11) α = 1− ε, ε ∈ (0, 1), we conclude that
|ck(F )|
|k|2j+ε
that is, F ∈ W
2 , which completes the proof.
The next theorem is devoted to the question on the existence of boundary values in the space
D′ of distributions.
Theorem 5. In order that an m-harmonic in K function u(r, t) admit a representation of the
form (2) with Fj ∈ D
′ (j = 1, . . . , m), it is necessary and sufficient that
∃α ≥ 0 ∃c > 0 sup
t∈[0,2π]
|u(reit)| ≤ c(1− r)−α. (12)
Proof. Let the inequality (12) hold. Then for p ∈ N, p > α, the function v(r, t) = (1−r2)pu(r, t)
is (m+ p)-harmonic in K, and it is not difficult to verify that
0≤r<1
‖v(r, ·)‖L2 ≤ 2πc.
By Theorems 3,4, the function v(r, t) may be represented in the form (2) where Fj ∈ W
−(j+p)
Since D′ =
W−α2 , we have Fj ∈ D
The necessity of condition (12) for m = 1 was proved in [5]. Namely, it was shown there that
for a harmonic function of the form
u(r, t) =
r|k|ck(F )e
ikt, f ∈ D′
there exists α ≥ 0 such that
|u(r, t)| ≤ c(1− r)−α.
If we take α = max
αj , where αj corresponds to Fj from (2), we obtain the estimate (12) for an
m-harmonic function (m is arbitrary).
Corollary 4. An m-harmonic in K function u(r, t) has a boundary value in D′ if and only if
it satisfies (12).
For a number β > 1 we put
G{β} = G{β}(∂K) = {ϕ ∈ D
∣∣∃α > 0 ∃c > 0 ∀k ∈ N0 max
|ϕ(k)(t)| ≤ cαkkkβ}. (13)
The linear space G{β} is endowed with the inductive limit topology of the Banach spaces G{β,α}
of functions ϕ ∈ D satisfying (13) with a fixed constant α. The norm in G{β,α} is defined as
‖ϕ‖G{β,α} = sup
|ϕ(k)(t)|
αkkkβ
It is evident, that
A ⊂ G{β} ⊂ D ⊂ L2 ⊂ D
′ ⊂ G′{β} ⊂ A
where G′{β} denotes the dual of G{β}.
Theorem 6. An m-harmonic in K function u(r, t) admits a representation of the form (2)
with Fj ∈ G
{β} (j = 1, . . . , m) if and only if
∀α > 0 ∃c = c(α) > 0 sup
0≤r<1
|u(r, t)| ≤ ceα(1−r)
, q =
β − 1
. (14)
The proof follows the scheme like that in Theorem 5 if to take into account that
F ∈ G′{β} ⇐⇒ ∀α > 0 |ck(F )| < ce
−α|k|1/β ,
and the series
ck(F )e
ikt converges to F in G′{β}-topology.
Corollary 5. In order that an m-harmonic in K function u(r, t) have a boundary value in the
space G′{β}, it is necessary and sufficiant that the condition (14) be satisfied.
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Institute of Mathematics
National Academy of Sciences of Ukraine
3 Tereshchenkivs’ka
Kyiv 01601, Ukraine
E-mail: imath@horbach.kiev.ua, sergiy.torba@gmail.com
