Subharmonic Almost Periodic Functions of Slow Growth

0.21 MB·English

by  S. Favorov

Subharmonic Almost Periodic Functions of Slowly
Growth
7
0
0 Favorov S.Ju., Rakhnin A.V.
2

n
a
J Abstract

8
We obtain a complete description of the Riesz measures of almost periodic subharmonic
2
functionswithatmostoflineargrowthonC;asaconsequencewegetacompletedescription

] ofzerosetsfortheclassofentirefunctionsofexponentialtypewithalmostperiodicmodulus.
V
C 2000 Mathematics Subject Classification: Primary 31A05, Secondary 42A75, 30D15
. Keywords: subharmonicfunction,almostperiodicfunction,Rieszmeasure,zeroset,entirefunction
h
t of exponential type
a
m
Bohr’s Theorem (see [2], or [10], Ch.6, 1) implies that each almost periodic function1
[
on real axis R with the bounded spectrum§is just the restriction to R of an entire almost
1
periodic function f of exponential type. Moreover, f has no zeros outside of some strip
v
4 Imz H if and only if supremum and infimum of the spectrum f also belongs to the
1 | | ≤
spectrum. In [9] (see also [10], Appendix VI) M.G.Krein and B.Ya.Levin obtained a
8
1 complete description of zeros of functions from the last class. Namely, a set ak k Z in a
0 horizontal strip of a finite width is just the zero set of an entire almost perio{dic}fu∈nction
7
f of exponential type if and only if the set is almost periodic and has the representation
0
/
h a = dk +ψ(k), k Z, (1)
t k
a ∈
m
where d is a constant, the function ψ(k) is bounded, and the values
:
v
k
i
X S = lim [ψ(k +n) ψ(k)] (2)
n
r − k2 +1
r →∞ Xk<r
a | |
are bounded uniformly in n Z.
∈
It can be proved that almost periodicity of a yields representation (1) and a finite
k
limit in (2) for every fixed n Z. Also, one ca{n o}btained a complete description of zero
∈
sets for the class of entire functions of exponential type with almost periodic modulus and
zeros in horizontal strip of finite width: we should only replace the S by ReS .
n n
Observe that every entire function of exponential type bounded on R has the form
z
f(z) = Ceiνz lim 1 , ν R (3)
r→∞ aYk <r(cid:18) − ak(cid:19) ∈
| |
([10], Ch. 5; for simplicity we suppose 0 a ); so we have an explicit representation
k
6∈ { }
for functions from the classes above-mentioned.
1Explicitdefinitionsofalmostperiodicityforfunctions,measures,anddiscretesetssee[10],Ch.6andAppendix
VI,[11], or §3 of the present paper.
1
Notethatoneoftheauthorsofthepresent paper obtainedin[5]acomplete description
of zero sets for holomorphic almost periodic functions on a strip and on the plane without
any growth conditions. An implicit representation for a special case of almost periodic
holomorphic functions was obtained earlier in [3]. Besides, it was proved in [3] that zero
setsofholomorphicfunctionswiththealmostperiodicmodulusonastrip(orontheplane)
are just almost periodic discrete sets. This result is a consequence of more general one:
every almost periodic measure on a strip is just the Riesz measure of some subharmonic
almost periodic function on the strip.
In 3 of our paper we obtain a complete description of the Riesz measures for almost
§
periodic subharmonic functions of the normal type with respect to the order 1 (note that
it is the smallest growth for bounded on R subharmonic function). In particular, we
consider the case of periodic subharmonic functions. As a consequence, we get a complete
description of zero sets for the class of entire functions of exponential type with the almost
periodicmoduluswithoutanyadditionalrequirements ondistributionsofzeros. Notethat
representation (1) with a bounded function ψ(k) is incorrect here, therefore methods of
paper [9] do not work in our case. The integral representation from [3] creates an almost
periodic subharmonic function with a given almost periodic Riesz measure, but does not
allow to control the growth of the function, therefore it isn’t fit for our problem as well.
We make use of a subharmonic analogue of representation
∞n(0,t) n(z,t)
log f(z) = − dt νy +log C , (4)
| | Z t − | |
0
for functions of the form (3) (see [6] or review [7], p.45); here n(c,t) is a number of zeros
in the disc z : z c t . We obtain this analogue in 2 of our paper. Also, we get
a complete{descr|ipt−ion|o≤f th}e Riesz measures for bounded§on R subharmonic functions
with at most of linear growth on C.
Here we base on a subharmonic analogue of (3) as well. Of course, this analogue
can be obtained by repeating all the steps of the proof (3) for entire functions in [10],
nevertheless we prefer to give a short proof in 1, using Azarin’s theory of limit sets for
§
subharmonic functions [1]. The idea of the proof belongs to prof. A.F.Grishin, and the
authors is very grateful to him.
§1
In this section we prove the following theorem:
Theorem 1 . Let v(z) be a subharmonic function on C such that
v+(z) = O( z ) as z (5)
| | → ∞
and
supv(x) < . (6)
x R ∞
∈
Then
v(z) = lim (log z w log+ w )dµ(w)+A y +A . (7)
1 2
R Z | − |− | |
→∞
w<R
| |
2
Here z = x+ iy, µ = 1 v is the Riesz measure of v, A ,A R, and the limit exists
uniformly on compact s2uπb△sets in C. 1 2 ∈
Note that the condition (5) means just v is at most of normal type with respect to the
order 1.
Our proof of Theorem 1 based on Azarin’s theory of limit sets [1]. Thus, if a subhar-
monic function v satisfies (5), then
a) the family v (z) = t 1v(tz), t > 1 is a relatively compact set in the space of
t −
distributions (C); in other words, for every sequence of functions from this family there
′
D
is a subsequence converging to a subharmonic function.
Note that the convergence in (C) is the weak convergence on all functions from the
′
class of infinitely smooth compaDctly supported functions (C); moreover, the class of
D
subharmonic functions is closed with respect to this convergence.
b) If
v = lim v (z), (8)
t
∞ t
→∞
then the Riesz measure µ of the function v satisfies the equality
∞ ∞
µ = lim µ , (9)
t
∞ t
→∞
where µ (E) = t 1µ(tE) for Borel subsets of C; limits (9) exists in the sense of weak
t −
convergence on continuous compactly supported functions on C. Moreover, in this case
there exists
dµ(w)
lim = . (10)
R Z w 6 ∞
→∞
1 w<R
≤| |
If a subharmonic function satisfies (5) and (8), then it is called a completely regular
growth (with respect to the order 1).
In what follows we need the simple criterion of compactness for family of subharmonic
functions (see[1]).
Lemma A. A family u of subharmonic functions on C is a relatively compact set
α
in the space of distributio{ns } (C) if and only if
′
a) sup sup u (z) < Dfor all compacta K C,
b) inf αsupz∈K uα(z) >∞ for some compact⊂set K C.
α z∈K0 α −∞ 0 ⊂
Also, we need the following variant of Fragmen-Lindelof Theorem.
Theorem FL. If a function v is subharmonic in a neighborhood of the closure of upper
half-plain C+ = z = x+iy : y > 0 and satisfies conditions (5), (6), then for all z C+
{ } ∈
v(z) supv(x)+σ+y,
≤ x R
∈
with σ+ = limsupy 1v(iy).
−
y +
The proo→f of∞this statement is the same as for holomorphic on C+ and continuous on
C+ functions (see, for example, [8], p.28).
First let us prove a subharmonic analogue of Cartwright Theorem (the holomorphic
case see, for example, [10], Ch. V).
3
Theorem 2 . Let a subharmonic function v on C be satisfied (5) and (6). By definition,
put
v(iy)
σ = limsup . (11)
± y y
→±∞ | |
Then v(z) is a completely regular growth; moreover, the function v from (8) has the
∞
form
σ+y, y 0,
v (z) = ≥ (12)
∞ (cid:26) σ− y , y < 0,
| |
Remark. From Theorem FL it follows that if a subharmonic function v on C satisfies
the conditions of Theorem 2 with σ+ 0 and σ 0, then v is a constant.
−
≤ ≤
The proof is based on the following lemma
Lemma 1 . Let u < 0 be a subharmonic function on C+. Then for every R < and
∞
r (0,R/2)
∈
u(ir) C
u(z)dm (z), (13)
r ≤ R3 Z 2
z iR<R
| − |
where C is an absolute constant and m is the plain Lebesque measure.
2
Proof
Since Poisson formula for the disc B(iR/2,R/2) = z : z iR/2 < R/2 , we have
{ | − | }
2π
1 (R/2)2 (R/2 r)2
u(ir) u(iR/2+eiθR/2) − − dθ.
≤ 2π Z (R/2)2+(R/2 r)2 R(R/2 r)cos(π/2+θ)
− − −
0
Using the inequality u < 0, replace the interval of integration by [π/4,3π/4]. We obtain
r 1 iR R
u(ir) sup u + eiθ (14)
≤ (R r)4 (cid:18) 2 2 (cid:19)
θ [π/4,3π/4]
− ∈
Since for all θ [π/4,3π/4]
∈
√2
B(iR,R/2) B(iR/2+eiθR/2,(1+ )R/2) C+,
⊂ 2 ⊂
we have
iR R 4
u( + eiθ) u(z)dm (z)
2
2 2 ≤ πR2(1+ √2)2 Z ≤
2
z iR/2 eiθR/2 (1+√2)R/2
| − − |≤ 2
8
u(z)dm (z)
2
≤ πR2(3+2√2) Z
z iR<R/2
| − |
Then replace the average over the disc B(iR,R/2) by the average over the disc B(iR,R).
So the assertion of the Lemma follows from (14).
4
Proof of theorem 2.
Without loss of generality it can be assumed that supv(x) = 0. Put u(z) = v(z) σ y.
+
x R −
From Theorem FL it follows that u(z) < 0 on C+, the∈n
u(iy)
limsup = 0 (15)
y
y +
→ ∞
Fix z C+. Let ϕ, 0 ϕ 1, be an infinite differentiable and compactly supported
0
function∈on C+, depend≤ing o≤nly on z z . Apply Lemma 1 for the function u (z) =
0 t
| − |
u(tz)t 1. If suppϕ B(iR,R), then we get
−
⊂
u(itr) C C
u (z)dm (z) u (z)ϕ(z)dm (z)
tr ≤ R3 Z t 2 ≤ R3 Z t 2
z iR<R C+
| − |
Since (15), we see that for each ε > 0 and t > t(ε) there is r (0,R/2) such that
∈
u(itr) εtr. We obtain
≥ −
εR3
u (z)ϕ(z)dm (z) .
t 2
Z ≥ − C
C+
Hence,
u (z)ϕ(z)dm (z) 0 as t .
t 2
Z → → ∞
C+
Therefore, v (z) = u (z)+σ y σ y in the space (C+). Similarly, v (z) σ y in
t t + + ′ t
the space (C ), where C =→z = x + iy : y <D0 . Each limit function→for−v (−z) is
′ − − t
D { }
always subharmonic, therefore we get (12). So limit (8) exists. Theorem is proved.
Consequence. The Riesz measure of the limit function v (z) equals
∞
σ +σ
+
−m1(x),
2π
where m being the Lebesgue measure on R.
1
Proof of theorem 1. From Jensen-Privalov formula for subharmonic function, we
get the estimate for r 1
≥
µ(B(0,r)) C r, (16)
1
≤
where constant C depends only on v. Using Brelot-Hadamard Theorem for subharmonic
1
function (see, for example, [12]), we obtain that there is a harmonic polynomial H(z) of
degree 1 such that
z z
v(z) = log z w dµ(w)+ log 1 +Re dµ(w)+H(z). (17)
Z | − | Z (cid:16)(cid:12) − w(cid:12) w(cid:17)
w<1 w 1 (cid:12) (cid:12)
| | | |≥ (cid:12) (cid:12)
Denote by v0(z) the first integral in (17). Since (10), we get
z
v(z) = v0(z)+ lim log 1 dµ(w)+A x+A y +A . (18)
0 1 2
R Z (cid:12) − w(cid:12)
→∞1 w<R (cid:12) (cid:12)
≤| | (cid:12) (cid:12)
5
The application of Theorem 2 yields that the function v is a completely regular growth,
hence the measures µ converge weakly to the measure
t
σ +σ
+
µ = −m1. (19)
∞ 2π
Let µ be the restriction of the measure µ to C B(0,1). Obviously, measures µ weakly
′ \ ′t
converge to the measure µ as well. Therefore,
∞
z A
v (z) = v0(z)+ lim log 1 dµ (w)+A x+A y + 2. (20)
t t R Z (cid:12) − w(cid:12) ′t 0 1 t
→∞w<R (cid:12) (cid:12)
| | (cid:12) (cid:12)
Pass toa limit in(20) ast inthe space (C). First, by Theorem 2, the functions
′
→ ∞ D
v (z) converge to the function v (z) from (12). Since v0(z) = O(log z ) as z , we
t
see that v0(z) 0. By (16), w∞e obtain µ (B(0,R)) C R for all|t| 1|a|nd→R∞> 0.
t → ′t ≤ 1 ≥
Therefore, we get uniformly in t 1,
≥
dµ (w) ∞µ (B(0,s)) µ (R)
′t = 2 t ds t′ 0. (21)
Z w 2 Z s3 − R2 →
| |
w R R
| |≥
as R . Also, by (10), uniformly in t 1, R R
′
→ ∞ ≥ ≥
dµ (w) dµ(w)
′t = 0, (22)
Z w Z w →
R w R Rt w Rt
≤| |≤ ′ ≤| |≤ ′
as R . For z < C and a sufficiently large w
→ ∞ | | | |
z z z 2
log 1 +Re | | .
− w w ≤ w 2
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) | |
(cid:12) (cid:12) (cid:12) (cid:12)
Therefore, taking into account (21) and (22), we obtain for all ϕ (C) uniformly in
∈ D
t 1
≥
z
 lim log 1 dµ (w)ϕ(z)dm (z) =
t′ 2
Z R Z (cid:12) − w(cid:12)
C  →∞w R (cid:12) (cid:12) 
 | |≤ (cid:12) (cid:12) 
z
= lim  log 1 ϕ(z)dm2(z)dµt′(w). (23)
R Z Z (cid:12) − w(cid:12)
→∞w R C (cid:12) (cid:12) 
| |≤ (cid:12) (cid:12)
Note that measure µ does not charge any circle w = R, therefore the restrictions of
∞ | |
measures µ (w) to any disc B(0,R) weakly converge to the restriction of the measure
t′
µ (w). The function log z w ϕ(z)dm (z) is continuous in the variable w, so we have
2
∞ | − |
R
lim log z w ϕ(z)dm (z)dµ (w) = log z w ϕ(z)dm (z)dµ (w). (24)
2 t 2
t Z Z | − | Z Z | − | ∞
→∞
w R w R
| |≤ | |≤
6
By the same reason for each δ > 0
lim log w µ (w) = log w µ (w). (25)
t
t Z | | Z | | ∞
→∞
δ w R δ w R
≤| |≤ ≤| |≤
Furthermore,
δ
µ(B(0,δt)) µ(B(0,st))
′ ′
log w µ (w) = logδ ds (26)
Z | | ′t t −Z st
w δ 0
| |≤
Since µ(B(0,r)) C r for all r > 0, we see that (26) tends to zero as δ 0 uniformly
′ 1
≤ →
in t 1. Combining (23), (24), (25), and (26), we get the equality
≥
z
v (z) = lim log 1 dµ (w)+A x+A y.
0 1
∞ R Z (cid:12) − w(cid:12) ∞
→∞w R (cid:12) (cid:12)
| |≤ (cid:12) (cid:12)
Take y = 0. Since v (x) = 0 and
∞
x
lim log 1 du = 0
R Z (cid:12) − u(cid:12)
→∞u R (cid:12) (cid:12)
| |≤ (cid:12) (cid:12)
for all x R, we obtain A = 0. Now the assertion of Theorem 1 follows from (18).
0
∈
§2
Here we get a complete description of Riesz measures for subharmonic functions with
at most of linear growth on C (i.e., do not exceed C( z +1) with C < ) and with some
additional conditions (bounded on R or with the com|pa|ct family of tran∞slations along R).
Holomorphic analogues of the corresponding theorems were obtained earlier one of the
author in [6].
First prove some lemmas.
Lemma 2 . If a measure µ on C satisfies the condition
µ(B(0,R+1)) µ(B(0,R)) = o(R) as R , (27)
− → ∞
and the limit
lim (log+ z w log+ w )dµ(w)
R Z | − |− | |
→∞
w<R
| |
exists at some point z C, then the limit equals
∈
∞µ(B(0,t)) µ(B(z,t))
− dt.
Z t
1
7
Proof
For all z C and R ( z +1, ) we have
∈ ∈ | | ∞
log+ z w dµ(w) log+ w dµ(w) = (logR)µ(B(z,R))
Z | − | − Z | | −
w<R w<R
| | | |
R R
µ(B(z,t)) µ(B(0,t))
dt (logR)µ(B(0,R))+ dt+
−Z t − Z t
1 1
+ log+ z w dµ(w) log+ z w dµ(w) =
Z | − | − Z | − |
w<R, w z R w R, w z<R
| | | − |≥ | |≥ | − |
R
µ(B(0,t)) µ(B(z,t))
= − dt+
Z t
1
z w z w
+ log − dµ(w) log − dµ(w). (28)
Z (cid:12) R (cid:12) − Z (cid:12) R (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
w<R, w z R (cid:12) (cid:12) w R, w z<R (cid:12) (cid:12)
| | | − |≥ (cid:12) (cid:12) | |≥ | − | (cid:12) (cid:12)
If w < R and z w R or w R and z w < R, then we have
| | | − | ≥ | | ≥ | − |
z z w z
1 | | − 1+ | |.
− R ≤ (cid:12) R (cid:12) ≤ R
(cid:12) (cid:12)
(cid:12) (cid:12)
Therefore the integrand functions of las(cid:12)t two i(cid:12)ntegrals in (28) are O(1/R) as R .
→ ∞
The domains of integrations are subsets of the ring R z w R + z , hence, by
− | | ≤ | | ≤ | |
(27), these integrals tends to 0 as R . Lemma is proved.
→ ∞
Lemma 3 . Let a measure µ be satisfied (10), (16), and (27), and let
V(z) = lim (log z w log+ w )dµ(w). (29)
R Z | − |− | |
→∞
w<R
| |
Then V is a subharmonic function with Riesz measure µ and
∞µ(B(0,t)) µ(B(z,t))
V(z) = − dt+ log z w dµ(w) (30)
Z t Z | − |
1 w z<1
| − |
for all z C. Furthermore, the function
∈
2π
1
V˜(z) = V(z +eiθ)dθ (31)
2π Z
0
satisfies the equality
∞µ(B(0,t)) µ(B(z,t))
V˜(z) = − dt. (32)
Z t
1
8
Proof
It follows from (16) that the integral in (17) is a subharmonic function on C with the
Riesz measure µ; besides, it satisfies (5) (see, for example, [12], Ch.1). If, in addition, µ
satisfies (10), then the limit in (29) exists uniformly on bounded sets, and the function
V coincides with the integral in (17) up to a linear term; so it has the same properties as
well.
Using the equality 2πlog a+eiθ dθ = 2πlog+ a , we get
0 | | | |
R
V˜(z) = lim (log+ z w log+ w )dµ(w).
R Z | − |− | |
→∞
w<R
| |
and V(z) = V˜(z) + log z w dµ(w). Then Lemma 2 implies (32) and (30).
w z<1 | − |
Lemma 3 is proved. R| − |
Lemma 4 . Let a subharmonic on C function v be satisfied (5). Then the family of
translations v(z+h) h R is a relatively compactsubset in ′(C) if and only if the function
{ } ∈ D
1 2π
v˜(z) = v(z +eiθ)dθ (33)
2π Z
0
is bounded on R; this function is bounded simultaneously with the function
1
vˆ(z) = v(z +w)dm (w) (34)
2
π Z
w<1
| |
Proof . If the family v(z + h) h R is a relatively compact subset, then it is uniformly
bounded from above o{n compac}ta∈in C and the function v is uniformly bounded from
above on every strip y < H. Then by Lemma A there is a compact subset K of C such
0
| |
that
supv(z +h) C , h R.
2
≥ ∀ ∈
K0
Take d > sup z . Then for each h R there is a point z(h), z(h) < d, such that
K0| | ∈ | |
1
v(w+h)dm (w) v(z(h)+h) > C 1.
2 2
(d+1)2π Z ≥ −
w z(h) d+1
| − |≤
Further, we have
v(w +h)dm (w) v(w+h)dm (w)+(d2 +2d)π sup v(z).
2 2
Z ≤ Z
y<d+1
w z(h) d+1 w 1 | |
| − |≤ | |≤
Therefore, infh Rvˆ(h) > . Since v˜(h) vˆ(h), we see that the functions vˆ and v˜
are bounded un∈iformly fro−m∞below on R. I≥t is clear that these functions are bounded
uniformly from above on R as well.
On the other hand, if v˜(z) is bounded from below onR, then infh Rsupw=1v(h+w) >
; if v˜(h) is bounded from above on R, then v(h) is bounded from∈ abov|e|on R as well.
−∞
Since Theorem FL, we see that v(z) is bounded from above on every strip y < H. It
| |
follows from Lemma A that v(z + h) h R is a relatively compact set. Hence vˆ(x) is
bounded on R. { } ∈
Now we can prove the theorems mentioned above.
9
Theorem 3 . For a measure µ on C to be the Riesz measure for some subharmonic
function satisfying conditions (5) and (6) it is necessary and sufficient that the conditions
(10), (16), (27), and
∞µ(B(0,t)) µ(B(x,t))
sup − dt < . (35)
x RZ t ∞
∈ 1
be fulfilled.
Proof
If asubharmonic functionv satisfies (5)and(6), then, byTheorem 2, itsRieszmeasure
µ satisfies (10) and the measures µ converge to the measure µ = (σ +σ )(2π) 1m .
t + − 1
∞ −
The last measure does not charge any circle w = R, therefore (9) implies that
| |
µ(B(0,R)) = CR + o(R) as R . Hence we get (16) and (27). By theorem 1,
v(z) = V(z) + A y + A . So the→fun∞ction V is also bounded from above on R. Since
1 2
˜
Theorem FL, we see that the same is true for the function V from (31). Now Lemma 3
implies (35).
Conversely, if a measure µ satisfies (10), (16), and (27), then, by Lemma 3, the sub-
harmonic function V has the Riesz measure µ and satisfies (5) and (6). Theorem is
proved.
Theorem 4 . For a measure µ on C to be the Riesz measure for some subharmonic
function v with the property (5) such that the family of translations v(z + h) h R is a
relatively compact subset in (C), it is necessary and sufficient that th{e conditi}on∈s (10),
′
D
(16), (27), and
∞µ(B(0,t)) µ(B(x,t))
sup(cid:12) − dt(cid:12) < (36)
x R(cid:12)(cid:12)Z t (cid:12)(cid:12) ∞
∈ (cid:12)1 (cid:12)
(cid:12) (cid:12)
be fulfilled. (cid:12) (cid:12)
Proof . If a function v satisfies (5), and the function v˜ from (33) is bounded on R, then
v is bounded from above on R. Theorem 3 implies conditions (10), (16), and (27) for the
Riesz measure µ of the function v. By Theorem 1, v˜ equals V˜ from (31) on R up to a
constant term, therefore Lemmas 3 and 4 imply (36).
Conversely, if a measure µ satisfies (10), (16), (27), and (36), then V from (29) satisfies
(5), and the function V˜ is bounded on R.
Hence the assertion of Theorem 4 follows from Lemma 4.
§3
A continuous function F(z) on a closed strip z = x + iy : x R, y H with
{ ∈ | | ≤ }
H 0 is almost periodic if the family of translations F(z+h) h R is a relatively compact
≥ { } ∈
set with respect to the topology of uniform convergence on the strip; a function is almost
periodic on an open strip (in particular, on C), if it is almost periodic on every closed
substrip of a finite width.
A measure (maybe complex) µ on C is called almost periodic if for any test-function
ϕ (C) the convolution ϕ(w+t)dµ(w) is an almost periodic function in t R ([11]).
∈ D ∈
R
10