Subharmonic almost periodic functions
A.V. Rakhnin and S.Yu. Favorov
Department ofMechanicsandMathematics,V.N.KarazinKharkov NationalUniversity
4SvobodySq.,Kharkov, 61077,Ukraine
E-mail:[email protected]
ReceivedJanuary21, 2005
Weprovethat almost periodicityin thesense ofdistributions coincides
with almost periodicity with respect to Stepanov's metricfor the class of
subharmonic functions in a strip fz2C : a<Imz<bg. We also prove
thatFouriercoecientsofthese functionsarecontinuousfunctions inImz.
Further,ifthelogarithmofasubharmonicalmostperiodicfunctionisa
sub-harmonicfunction,thenitisalmostperiodic.
Mathematics SubjectClassication 2000: 42A75,30B50.
Key words: subharmonic functions,almost periodic functions inthe senseof
distri-butions,Fouriercoecients.
Subharmonic almost periodic functions were introduced in [2] in connection
withinvestigationofzerodistributionofholomorphic almostperiodicfunctionsin
astrip. Inthis paperalmostperiodicitywasdened inthesense ofdistributions,
namely as almost periodicity of the convolution with a test function. However,
subharmonic functions logjf(z)j, where f(z) is a holomorphic almost periodic
function,wereconsidered much earlierinpapers [5]and[6 ],where theimportant
point was to prove almost periodicity of such functions in the sense of
distri-butions. In [4] this was extended to a subharmonic uniformly almost periodic
function whose logarithm isa subharmonic function.
Inthispaperweprovethatsubharmonicalmostperiodicinthesenseof
distri-butionsfunctionsarealmostperiodicintheclassicalsense,ifweconsiderStepanov
integral metric instead of the uniform metric. Therefore the classes of
subhar-monic almost periodic in the sense of distributions functions and subharmonic
Stepanov almostperiodicfunctions arethesame.
Now the FourierBohr coecients of such functions can be dened in the
usual way. For a horizontal strip these coecients are functions depending on
Imz. In this paperwe prove that these coecients depend continuously on Imz,
exponentialsumswithcontinuouscoecientsinStepanovmetric. Thusweprove
that subharmonic almost periodic functionsare Stepanov almost periodicin the
sense of thedenitionin[8].
In [2 ] it was proved that exp(u), where u is a subharmonic almost periodic
in the sense ofdistributions function, is also almost periodic inthe sense of
dis-tributions. Moreover, for an almost periodic function logjf(z)j, where f(z) is
aholomorphic function,jf(z)jisuniformlyalmostperiodic. Conversely,weprove
that the logarithm of a subharmonic almost periodicfunction is an almost
peri-odicfunction, provided it is a subharmonic function. Thus we obtain a stronger
than theone in[4 ],aswell astheconverse to theresultin[2].
We start withthefollowing denitions and notations(see [1, p. 51]).
Denition 1. A continuous function f(z) (z=x+iy),dened on R+iK,
where K isa compact subset of R (it isallows that K=f0g), is called uniformly
almost periodic (Bohr almost periodic), if from any sequence ft
n
gR one can
choose a subsequence ft
n 0
g such that the functions f(z+t
n 0
) converge uniformly
on R+iK.
Equivalent denitionisthe following:
For any " > 0 there exists L(") > 0 such that each interval of length L(")
contains a realnumber withthe property
sup
z2R+iK
jf(z+) f(z)j<":
Denition2. Adistributionf(z)2D 0
(S)oforder0(Sisanopenhorizontal
strip) iscalled almost periodic, iffor any testfunction '2D(S) the convolution
Z
u(z)'(z t)dxdy
is uniformly almost periodic on the real axis.
Notethat accordingto [6 ], for an almostperiodicdistribution f(z) from any
sequence fh
n
g R one can choose a subsequence fh
n 0g, such that R '(z)f(z+ h n 0
)dxdy convergeuniformlyon every set
K
=f'(z+t):t2R;'2Kg ,where
K isa compact subset ofD(S).
Anysubharmonic functionislocallyintegrable,sowecan consideritasa
dis-tribution.
A class of subharmonic almost periodic functions inan open strip S will be
denoted byWAP(S).
Furthermore,for 1<<<+1 we dene
S
[;]
ImS =fImz :z2Sg;
and for functions u, v, which are integrable on horizontal intervals in S
[;] , we denote d [;] (u;v):= sup z2S [;] 1 Z 0 ju(z+t) v(z+t)jdt:
Denition 3. A function f(z) integrable on horizontal intervals in an
open horizontal strip S is called Stepanov almost periodic, if from any sequence
fh
n
gR one can choose a subsequence fh
n 0
g anda function g(z) such that the
functionsf(z+h
n
0)converge tog(z) inthe topology denedby seminormsd
[;] ,
;2ImS.
A classof asubharmonicStepanovalmostperiodicfunctionsinanopenstrip
SwillbedenotedbyStAP(S). SincesuchfunctionsareStepanovalmostperiodic
on every liney=const,for u2StAP(S)there existsthemeanvalue
M(u;y):= lim T!1 1 2T T Z T u(x+iy)dx:
To eachsuchu we can associate FourierBohr series
u(z) X 2R a (u;y)e ix ; where a (u;y):=M(ue ix ;y):
areFourierBohr coecients.
Denition 4. A function u(z) 0 is called logarithmic subharmonic in
a domain GC, if the function logu(z) is subharmonic in this domain.
It iseasy to seethat alogarithmic subharmonic function issubharmonic.
Weprove thefollowing theorems:
Theorem 1. u(z)2WAP(S) if and onlyif u(z)2StAP(S).
Theorem 2. Let u(z) be a logarithmic subharmonic function in a strip S.
Thenlogu(z)2WAP(S) if andonlyif u(z)2WAP(S).
Theorem 3. Letu(z) be asubharmonicalmost periodicfunctioninastripS.
Thenits FourierBohr coecientsare continuous in ImS.
FromTheorem3and Besselinequalityfor FourierBohr coecientsitfollows
f2R :a
(u;y)60g)itismostcountable,whichalsofollowsfromTheorem1.12
in[6].
Theorem 4. Subharmonic function u(z) in an open horizontal strip S is
almost periodic if andonly ifthere existsa sequence of nite exponential sums
P m (z)= Nm X n=1 a (m) n (y)e i (m) n x ; (1) where n 2 R, a (m) n
(y) 2 C(ImS), which converges to the function u(z) in the
topology dened by seminorms d
[;]
,; 2ImS.
Moreover, P
m
(z), m=1;2;:::are subharmonic functions in S.
To provethe theoremsabove we usethefollowing propositions:
Proposition1. Convergence ofsubharmonicfunctionsinD 0 (G)isequivalent to the convergence in L 1 l oc (G)(see [7]).
Proposition2. Weaklimitofsubharmonicfunctionsissubharmonicfunction
(see [7]).
WedenotebyG
theGreenpotentialofameasureforthediskB(R ;0), i.e.
G (z):= Z B(R;z 0 ) log jR 2 zj R jz j d():
Lemma 1. Let measures
n
converge weakly to a measure in a
neighbor-hood of the disk B(R ;0) ,and (@B(R ;0))=0. Then for any t
1 >0,t 2 >0 such that t 2 1 +t 2 2 <R 2 , lim n!1 sup y2[ t 2 ;t 2 ] t1 Z t1 jG n (z) G (z)jdx=0; (2) where z=x+iy. Pr o of. Denote n = n . We have sup y2[ t2;t2] t 1 Z t 1 j G n (z) G (z)jdx sup y2[ t2;t2] t 1 Z t 1 Z B(R;0) log jR 2 zj R d n () dx + sup y2[ t 2 ;t 2 ] t 1 Z t 1 Z B(R;0) logjz jd n () dx: (3)
The condition (@B(R ;0)) =0 implies that the restrictions of the measures
n
to the disk B(R ;0) converge weakly to the restriction of the measure on
the disk, and the function log (jR 2 zjR 1 ) iscontinuous for jxjt 1 , jyjt 2 ,
2B(R ;0) . Thus therst term on the right-hand sideof (3) issmall. Without
lossofgenerality,wecan assumethatR<1=2,sothatforz; 2B(R ;0) wehave
logjz j<0.
Let">0beanarbitraryxednumber. Wedenotelog
"
jz j=maxflogjz
j;log"g. This function is continuous for jxj t
1 ,jyjt 2 , 2B(R ;0) and for any">0. We have sup y2[ t 2 ;t 2 ] t1 Z t1 Z B(R;0) logjz jd n () dx sup y2[ t 2 ;t 2 ] t1 Z t1 Z B(R;0) log " jz jd n () dx + sup y2[ t 2 ;t 2 ] t 1 Z t 1 Z B(R;0) jlogjz j log " jz jjdj n j()dx:
The rst term on the right-hand side of this inequality is small when n is
su-ciently large. Then
sup y2[ t 2 ;t 2 ] t 1 Z t 1 Z B(R;0) j logjz j log " jz jjdj n j()dx = sup y2[ t 2 ;t 2 ] Z B(R;0) Z [ t1;t1]\fx:jx+iy j"g (log" logjz j)dxdj n j() Z B(R;0) " Z " (log" logjxj)dxdj n j()2"j n j(B(R ;0)):
Notethat since ">0 isarbitrary,andmeasures
n
weaklyconverge to zero, one
can choosea constant C2R withj
n
j(B(R ;0))<C. Thelemma isproved.
Lemma 2. Let u
n
(z) be a sequence of subharmonic functions in a domain
GC converging to a function u 0 (z)6 1 in D 0 (G),and let sup z2G 0 u n (z)W(G 0 )<1
for any subdomain G 0
G. Then for any rectangle [a;b][;]G,
lim n!1 sup y2[;] b Z a ju n (z) u 0 (z)jdx=0: (4)
Pr oof. For every diskB(z
0
;R )Gwe have thefollowingrepresentation
u n (z) = G n R (z;z 0 )+H R (z;z 0 ;u n ) n=0;1;2:::; where n
arethe Riesz measures of the functionsu
n (z), G n R (z;z 0 ) isthe Green
potential of the measure
n inthe diskB(z 0 ;R ), and H R (z;z 0 ;u n
) arethe best
harmonicmajorantsof thefunctionsu
n
(z)inthis disk. Conditionsof thelemma
imply that
n
convergeweaklyto themeasure
0
. Without lossof generality,we
can assumethat(@B(z
0
;R ))=0,andusingLemma 1 weconclude thatfor any
t 1 ;t 2 , t 2 1 +t 2 2 <R 2 sup y 0 t 2 yy 0 +t 2 t 1 +x 0 Z t 1 +x 0 jG n R (z;z 0 ) G 0 R (z;z 0 )jdx !0;
when n! 1. From this itfollows that thefunctions H
R (z;z 0 ;u n ) converge to the function H R (z;z 0 ;u 0 ) in D 0 (B(z 0
;R )). Now using the meanvalue property,
Harnak inequality,andobviousinequality
H R (z;z 0 ;u n )W(B(z 0 ;R ))<1; n=0;1;2:::;
we obtain uniform convergence of H
R (z;z 0 ;u n ) to the function H R (z;z 0 ;u 0 ) in the rectangle [ t 1 +x 0 ;t 1 +x 0 ][ t 2 +y 0 ;t 2 +y 0
]. Covering the rectangle
[;][a;b]bya nitenumber ofsuch rectangles, we prove thelemma.
P r o o f o f T h e o r e m 1. Inclusion StAP(S) WAP(S) is obvious.
We prove the opposite inclusion. We consider arbitrary substrip S
[;]
, ; 2
ImS and a sequence fh
j
g R. Since u(z) is a subharmonic almost periodic
distribution, there exists a subsequence fh
j
k
g such that for some subharmonic
(clearly also almostperiodic)function v(z) and for any'2D(S
[;] ), uniformly in t2R, lim k!1 Z S (u(z+h j k +t) v(z+t))'(z)dxdy=0: (5)
Now we will show that the functions u(z+h
k
) converge to v(z) inthe topology
dened by seminorms d
[;]
, ; 2 ImS. Assuming the contrary, there exist
"
0
>0,;2ImS such thatfor aninnitesequence k 0 d [;] (u(z+h j k 0 );v(z))>" 0 ;
and thereforethere existsasubsequence ft
k 0g2R suchthat sup y2[;] 1 Z 0 ju(z+h j k 0 +t k 0) v(z+t k 0)jdx>" 0 : (6)
Passing to asubsequence (ifnecessary), we can assumethat u(z+h j k 0 +t k 0 )!w(z); v(z+t k 0) !w 1 (z)in D 0 (S [;] ): Lemma 2 implies sup y2[;] 1 Z 0 ju(z+h j k 0 +t k 0) w(z)jdx!0; k 0 !1; and sup y2[;] 1 Z 0 jv(z+t k 0) w 1 (z)jdx!0; k 0 !1;
and thus inequality(6)implies
sup y2[;] 1 Z 0 jw(z) w 1 (z)jdx" 0 : (7)
Onthe other hand,using(5), for anytestfunction '(z),
Z S [;] (w(z) w 1 (z))'(z)dxdy= lim k 0 !1 Z S [;] (u(z+h j k 0 +t 0 k ) v(z+t k 0))'(z)dxdy = lim k 0 !1 Z S [;] (u(z+h j k 0 ) v(z))'(z t k 0)dxdy=0; andthusw(z)=w 1
(z)almosteverywhere. Sincew(z)andw
1
(z)aresubharmonic
functions, thenw(z) w
1
(z), which contradicts (7). The theoremisproved.
Toprove Theorem 2 weneed thefollowing lemmas.
Lemma 3. Let '(t) be a function continuous in [ c;c]. Then for any ">0
there existsÆ, depending on 'and",such that forany two integrable on compact
set K functions f;g: K ![ c;c] the inequality
Z
K
jf(x) g(x)jdm<Æ
implies the inequality
Z
K
Proof. Choose >0suchthatjt 1 t 2 j< impliesj'(t 1 ) '(t 2 )j< " 2m(K) , and denote A 1 =fx2K:jf(x) g(x)j <g; A 2 =fx2K:jf(x) g(x)j g:
Noticethat m(A
2 ) 1 R A 2 jf(x) g(x)jdm, andtherefore Z K j'(f(x)) '(g(x))jdm Z A1 j'(f(x)) '(g(x))jdm+ Z A2 j'(f(x)) '(g(x))jdm m(A 1 )" 2m(K) + 2supj'(t)j Z K jf(x) g(x)jdm:
Choosingsuitable Æ,(8)follows. Thelemma isproved.
Lemma 4. Let u
n
(z) be a sequence of uniformly bounded from above
lo-garithmic subharmonic functions in a domain G C, converging to a function
u
0
(z) 60 in the sense of distributions. Then the functions logu
n
(z) converge to
the function logu
0
(z) in the sense of distributions.
Proof. Thefunctionsu
n
(z) arelogarithmic subharmonic,andinparticular
subharmonic. UsingProposition 1,u
n (z) converge tou 0 (z) inL 1 l oc (G).
Next, these functions are uniformly bounded from above by some constant
V >0, bounded from below by 0, and the function l
"
(t)= logmaxf";tg is
con-tinuous in the interval [0;V]. Lemma 3 implies that for xed " the functions
l
" (u
n
)(z) converge to the function l
" (u 0 )(z) in L 1 l oc
(G), and thus in the sense of
distributions. From Proposition 2it follows that thefunctionsl
" (u
0
)(z) are
sub-harmonicfor all",andtheir monotonelimit when"!0,i.e. thefunctionlogu
0 ,
is alsosubharmonic.
Nowwe consideradiskB(z
0
;r)G. Fromtheconvergence inL 1 (B(z 0 ;r)) ofthesequenceu n
(z)itfollowsthatthesubsequencefu
n 0
(z)gconvergesuniformly
on every xedcompact set K
1 B(z
0
;r) withpositive Lebesguemeasure. Since
thefunction logu
0
(z) issubharmonic and not identically 1 onK
1 , sup z2K 1 (logu 0 (z))C 0 ; or sup z2K 1 (u 0 (z))e C 0 :
Thus forall n 0 >n 0 sup z2K1 (u n 0(z))e C0 1 ;
and sup z2K 1 (logu n 0(z))C 0 1; 8n=0;1::: :
Sincethe functionsu
n
(z)areuniformlybounded fromabove oncompact subsets
ofG, itfollows thatthefamily flogu
n
0(z)g iscompactinD 0
(G). Thereforethere
existsasubsequencelogu
n 00 (z)whichconvergesinD 0 (G)(andalsoinL 1 l oc (G))to
some subharmonic functionv(z) inG.
Notethatfor anycompactsetKGandforany">0wehavethefollowing
inequality:
Z
K
jmaxflogu
n
0(z);log"g maxfv(z);log"gjdxdy Z
K jlogu
n
0(z) v(z)jdxdy:
Hence, the functions maxflogu
n 0
(z);log"g converge to the function
maxfv(z);log"g inL 1
l oc
(G)for any ">0.
On the other hand, as was shown above, l
" (u n )(z) converge to l " (u 0 )(z) in L 1 l oc
(G). Thus almost everywhere (and, since the functions are subharmonic,
everywhere)
maxfv(z);log"g=maxflogu
0
(z);log"g: (9)
Since a set on which a subharmonic function equals to 1 has
Lebesgue measurezero,then"!0impliesthatmes(fz 2G:v(z)<log"g)!0 ,
mes(fz 2G:logu
0
(z)<log"g)!0,andv(z)=logu
0
(z)almosteverywhere,and
henceeverywhere. Thus thesequenceof thefunctionslogu
n 0 (z)converges tothe function logu 0 (z) inD 0 (G)andinL 1 l oc (G).
If for some subsequence of the functions logu
n
j (z), "
0
> 0 and compact set
K 0 2G Z K0 jlogu n j (z) logu 0 (z)jdxdy" 0 ; (10)
then, using the above construction of the sequence u
n
j
(z), we have that some
subsequence of the sequence flogu
n j g converges to logu 0 (z) in L 1 l oc (G), which
contradicts (10). The lemmais proved.
P r o o f o f Th e o r e m2. From Proposition 3 in [2 ] it follows thatthe
inclusion logu2WAP(S) impliesthatinclusion u2WAP(S). We aregoing to
show the oppositeinclusion. Letu(z) 2WAP(S) and fh
n
gR be anarbitrary
sequence. Passing to a subsequence if necessary, we can assume that for some
subharmonic function u 0 ,uniformlyint2R, lim n!1 Z S (u(z+h n ) u 0 (z))'(z t)dxdy=0: (11)
Toprove the theoremit issucient to verifythat uniformlyint2R lim n!1 Z S logu(z+h n +t)'(z)dxdy= Z S logu 0 (z+t)'(z)dxdy: (12)
Assuming thatthis fails, forsome ">0and some sequencet
n !1, Z S logu(z+h n +t n )'(z)dxdy Z S logu 0 (z+t n )'(z)dxdy ": (13) Hereu 0
(z)isalogarithmicsubharmonicfunctionwithu
0
(z)2WAP(S). Passing
toasubsequenceandusingalmostperiodicityofthefunctionu
0 (z),wecanassume that lim n!1 Z S u 0 (z+t n )'(z)dxdy= Z S v(z)'(z)dxdy (14)
for some subharmonic in the strip S function v(z). Since the limit in (11) is
uniform int2R, (14)implies lim n!1 Z S u(z+h n +t n )'(z)dxdy= Z S v(z)'(z)dxdy:
Now Lemma 4 implies that both integrals in (13) have the same limit
R
logv(z)'(z)dxdy, when n ! 1, which is impossible. Thus (12) holds and
Theorem 2 isproved.
Proof of Theo rem3. Without lossofgenerality,wecan assumethat
S is a stripwith nite width. Let S
0
be an arbitrary substrip, S
0
S. Since
the function u(z) isalmost periodic, its Riesz measure := 1
2
u isalso almost
periodicinthesenseof distributions.
Denote K(w)= 1 2 logje w 2 1j; where 0< < max y 1 ;y 2 2ImS (y 1 y 2 ) 2 :
Notethatthe kernelK(w) isa subharmonic function whichis bounded from
aboveinS andits restriction to S
0
satisestheequation
where Æ(w) is astandard Diracmeasure. Denote
V(z)= Z
S
K(w z)'(Im w)d(w); (16)
where '0isa test function onImS such that'(y)=1 fory 2ImS
0 .
Denote P
n
= f(n 1=2;n+3=4)ImSg S. We are going to show that
V(z) isa subharmonicfunction ineveryP
n . Fixing n 0 2Z,we have Z S K(w z)'(Im w)d(w)= Z [n 0 1;n 0 +1)ImS K(w z)'(Im w)d(w) + X n2Znfn0 1;n0g Z [n;n+1)ImS K(w z)'(Im w)d(w): (17)
Every term in the right hand side of (17) is obviously a subharmonic function.
For Rew2[n;n+1), Imw2supp',z2P
n0 ,n6=n 0 ,n6=n 0 1we have e (w z) 2 =e (Rew Rez) 2 +(Imw Imz) 2 e (jn n 0 j 3=4) 2 : Thus X n2Znfn 0 1;n 0 g Z [n;n+1)ImS K(w z)'(Im w)d(w) X n2Znfn0 1;n0g sup z2Pn 0 sup w2[n;n+1)supp' 1 2 logj1 e (z w) 2 j ([n;n+1)supp'):
Since the measure is almost periodic, ([n;n+1)supp') is bounded from
above uniformlyin n(see [2]), and therefore theseries (17) converges uniformly
inz2P
n
0
and thefunctionV(z) issubharmonic inP
n
0
,and also inS.
Now we are going to show that the function V(z) is subharmonic almost
periodicinS. We consideratestfunction (z) onS andverifythatthefunction
f(t)= Z
S
V(z) (z t)dxdy
isuniformly almostperiodicon the real axis. We have
f(t)= Z S 0 @ Z S K(w z) (z)dxdy 1 A '(Im w)d(w+t):
Note thatthefunction
(w):= Z
S
K(w z) (z)dxdy
is continuous in S, because the dierence K(w) logjwj is continuous in some
neighborhoodof zero. Moreover, (w)=O(e jwj
2
)when jRewj!1.
Sincethevalues ([n;n+1]ImS
0
) areuniformlybounded inn, then
Z S '(Im w+Imz)d(w+z) 1+jwj 2 C 1 <1 (18) uniformlyinz2S 0 .
We x " > 0 and choose a test function (t), 0 (t) 1 on R, and such
that(Re w)=1on the set
w: j (w)j> " C 1 (1+jwj 2 ) :
For allt2R we have
f(t)= Z S (w)(Re w)'(Im w)d(w+t) + Z S (w)(1 (Rew))'(Imw)d(w+t):
Property(18)impliesthatthesecondintegralintheequalityisnotgreaterthan"
forallt2R. Sinceisanalmostperiodicmeasure, therstintegral isanalmost
periodic function, and if is an "-almost period, then it is a 2"-almost period
for f. Thus, thefunction V(z) isasubharmonic almostperiodic, andinaddition
(15)impliesthatV(z)=2'(y)(z) inthesenseofdistributions. Considerthe
function
H(z):=V(z) u(z):
Thisfunctionis harmonicandalmostperiodicinthesenseofdistributions inS
0 .
Let '0 be a test function in thedisk B(";0), which dependsonly on jzjand
suchthat R
'(z)dxdy=1. Sincethe convolution R H(z)'(z+)dxdy isequal to H()insome stripS 1 S 0
,thentheremarkafterDenition2 impliesuniform
almostperiodicityofthefunctionH(z) inS
1
. SoitsFourierBohr coecientsare
continuous in ImS
1
and, since " is arbitrary, in ImS
0
. Thus it is enough show
thatthe FourierBohr coecientsof V(z),
a (V;y)=M(Ve ix ;y)= lim T!1 1 2T T Z T V(x+iy)e ix dx;
arecontinuous.
Wex">0. We have
K(w)=maxfK(w); 2logNg+minfK(w)+2logN;0g=K
1
(w)+K
2 (w);
where N <1will bechosenlater. Denote
V 1 (z):= Z S K 1 (z w)'(Im w)d(w); V 2 (z):= Z S K 2 (z w)'(Im w)d(w): Since for jw 2 j<1=2 we have K(w)=1=2logj1 1+w 2 2 w 4 2! +:::j=logjwj+(w);
where (w) is a continuous function, then K
2
(w) = 0 for jwj Æ
0
> 0 and N
suciently large. Moreover, ifj(w)jlogN,thenfor all w2C and y2ImS,
T Z T K 2 (z w)dx= T Z T
minflogjx u+i(y v)j+(z w)+2logN;0gdx
1 Z 1 minflogjNx Nuj;0gdx= C N ; (19)
with some constant C, 0 < C < 1. Now using the property that
([n;n+1]supp')areboundedandthefactthatK
2
(z w)=0forjz wjÆ
0 ,
we havethat for allT
1 2T T Z T V 2 (z)e ix dx Z jRewjT+Æ0 1 2T T Z T jK 2 (z w)jdxd(w) C 2TN Z jRewjT+Æ 0 '(Imw)d(w) C 2 N "; (20) ifN is suciently large.
Further,sinceK
1
(w)=O(e jwj
2
)forjRe wj!1,thenonecanchooseatest
function 0(t)1 onR suchthat (Re w)=1 onthe set
w: jK 1 (w)j> " C 1 (1+jwj 2 ) ; where C 1
is the constantfrom (18). We have
V 1 (z)= Z S K 1 (w)(Rew)'(Im w+Imz)d(w+z) + Z S K 1
(w)(1 (Re w))'(Im w+Imz)d(w+z)=V
3
(z)+V
4 (z):
Fromthechoiceof thefunction itfollows that
jV
4
(z)j"; for z2S
0
: (21)
Sincethekernel K
1
(w)is continuousandthefamily ofshiftsofatestfunction in
ImS
0
is a compact set, then (see the remark to Denition 2) thefunction V
3 (z)
is uniformly almost periodic in S
0
and it has continuous in ImS
0
FourierBohr
coecients(see[1,p. 145]). Thus,ify
1 ;y 2 2ImS 0 andjy 1 y 2 jÆ("),then(20) and (21)imply ja (V;y 1 ) a (V;y 2 )jja (V 3 ;y 1 ) a (V 3 ;y 2 )j+ja (V 4 ;y 1 )j+ja (V 4 ;y 2 )j +ja (V 2 ;y 1 )j+ja (V 2 ;y 2 )j5": Thus a
(V;y) arecontinuous. Thetheoremis proved.
Pr o o f o f Th e or e m 4. For P
m
(z) we choose BohnerFejer sums of
the function u(z)
P m (z):= lim T!1 1 2T T Z T u(z+t) (m) (t)dt= X k (m) a (u;Imz)e iRez : Here (m)
(t)is a sequence of BohnerFejer kernels(see. [3, p. 69]), and theset
fk (m) :k (m)
6=0g is nite for every m. Note that, according to Theorem 3, the
functionsa(u;y)are continuousiny2ImS.
We aregoingto show thatP
m
(z) aresubharmonic.
Notethatthe kernels (m)
(t)are nonnegative,bounded,and
lim T!1 1 2T T Z T (m) (t)dt=1:
Alsonotethatthesubharmonicalmostperiodicfunctionu(z)isboundedfrom
above in anysubset S 0
S. Thus, using Fatou's lemma, for any m=1;2;:::,
z2S,and suciently small ,
1 2 2 Z 0 P m (z+e i' )d' lim T!1 1 2T T Z T 1 2 2 Z 0 u(z+e i' +t) (m) (t)d'dt lim T!1 1 2T T Z T u(z+t) (m) (t)dt=P m (z):
Asitisshown in[6 ],for anytest function'(z)inS and forsome (depending
only on the spectrum of the function u(z)) sequence of BohnerFejer sums, for
m!1,uniformlyint2R Z S P m (z)'(z+t)dxdy! Z S u(z)'(z+t)dxdy: (22)
Now we are going to verify that it implies the convergence in the topology
dened by seminorms d
[;]
, ; 2ImS. Indeed, if itis not true, thenfor some
sequence x
m
!1 and some; 2ImS,"
0 >0, sup y2[;] 1 Z 0 ju(x m +iy+t) P m (x m +iy+t)jdt" 0 :
Since u2StAP(S),then, passingto a subsequence ifnecessary,one can assume
thatfunctionsu(z+x
m
)converge tosomefunctionv 2StAP(S) withrespectto
metric d [;] ,and therefore sup y2[;] 1 Z 0 jP m (x m +iy+t) v(t+iy)jdt" 0 =2: (23)
Moreover, accordingto Theorem 2,
Z S u(z+x m )'(z)dxdy ! Z S v(z)'(z)dxdy: Therefore,bysettingt= x m
in(22), wehave for anytest function '(z)
lim m!1 Z S P m (z+x m )'(z)dxdy= Z S v(z)'(z)dxdy:
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