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r.
174
2. Interpolation and the Algebras A p
o
IJ,(t)
= { n(t) -
if 0 < t < r, if t ;: r.-
nCr)
Then the first tenn we want to estimate corresponds to the canonical product of the sequence (un)n, the second to the sequence (vn)n. Let p = 1, then from [BG, §2.6.lO] (or [Lev, p. 12]) we have
'"'
I!':;;-(,)
I (z 0) I ~ ior -JL,(t) 1 JL,(t) t - dt + r , ~dt 00
log E
zn'
=
'"' L log IE
(z zn' I )
l' {l o
I~ 4
net) -dt+n(r), t
OO
r
v,(t) dt + 2r2 f2
r
n>n~)
1T 00
v,(t) dt }
,
= 4 { 2r21OO n;t) dt - nCr) } . So, certainly the sum can be estimated by 8
{1'
n;t) dt
+ r2100 n~t) dt} .
In case p > I, we have similarly
~ 2P { (p +2
P
~C
P
+1
l)r P - 1
l' nt~) l
{(p + l)r + p
{r P- 1
dt
OO
1
r
+ nCr) }
net) dt - ncr)}
t P+2
dt + rP+1lOO net) dt}. ior net) tP , tP+2
o
Proof of Theorem 2.6.3. Let us first prove that the conditions (i) and (ii) in part (1) are enough to conclude that f is of finite type in the case when p p. We denote P + alz p - I + ... + p =
=
Q(z) aoz
a
and note that E(u, p) = eUP / p E(u, p - 1).
Consider the quantity D := log
If(z)1 - Re [zp
{ao + ~p E z~}] I:;:n:;:n(r)
2.6. The Algebra Ap of Entire Functions of Order at Most p
=Re{alz P -
1
L
+ ... + a p + m logz} +
175
log
IE (:n' p -
I) I
l:5n:5n(r)
The first tenn in D is bounded by 0 (r P - I ) when p > I, and by 0 (log r) when p = 1, we shall write this as O(r P - I ). Using now the preceding Lemma 2.6.6, we have D
= O(r P - 1) + C -
(r P-
1
dt + rP+lj"" net) dt) . ior net) t t + P
P
r
P 2
We have net) ~ At P for some A > O. It is clear that the two integrals together are dominated by 2C p Ar P, so that D = O(r P ). Since S(r) is bounded, it follows that: log Ifez)1
= D + Re (zp
{ao+ ~s(r)}) =
O(r P ).
In other words, f has finite type dominated by 2Cp A + S, where S is an upper bound for lao + (\/ p)S(r)l. A similar argument can now be made for part (2), also when p = p. For £ > 0 given, we can find ro such that n (r) = uP for all r ~ roo Then the integral tenns in the earlier estimate of D can be bound by
C r PP
1 ro
1
0
net)
-
tP
dt
+ 2CP up.
Since the series L:n~1 z;P is convergent to -aop, for r ~ rl we have ao
1
+ - '""' ~ p
z-P < £' n -
l:5n:5n(r)
whence, 10glf(z)1 = O(r P -
I)
+ (2C p + i)uP ,
so that f is of type zero. Consider now the case where p = p - 1. Then the series L:n>1 IZnrp < and the infinite product P(z) has type zero by Lemma 2.6.5. Wehave
D = log If(z)l- Re[Q(z)
+ m logzJ
00,
= log IP(z)1 = o(r P ).
Therefore, in part (I) we immediately obtain that log If(z)1 = O(rP), the type of f is exactly laol in this case. In part (2), ao = 0 means that deg Q < p, then log If(z)1 = o(r P ). This last argument concludes the proof of the sufficiency of the given conditions. We now want to show these conditions are also necessary. In part (I), We already know that f being of finite type implies nCr) = O(r P ). The boundedness of S(r) is only a problem when p = p. Returning to the proof of the
2. Interpolation and the Algebras Ap
176
sufficiency, we see that we have shown that - Re
{zp
lao + ~ L z~l} = -log If(z)1 + p I~n~n(r)
O(r P ).
For r ::: I we can estimate log If (z) I :'S log M (r) :'S B r P , but we need to estimate I log If (z) II. We can apply Lemma 2.6.4, with a = 3, and obtain that the measure of the set A(3, r) ~ aB(O, r), where log If(z)1 :'S -3Br P is 2rrrl(3, r), with l(3e, r) :'S Ur>r _0 A(3, r) we have Re
!. Therefore,
{zp [aD +.!.
in the complementary of the set
L z~l} = O(r P).
p I~n~n(r)
We want to conclude from this that S(r) remains bounded. If S(r) is not bounded, then the expression between square brackets is not bounded. This means that for some sequence of values rj ~ 00, we have for some I{Jj E [0,2rr[ and Rj > 0
4}
The measure of the set {O E [0, 2rr [: cos(O + I{Jj) ::: is 2rr /3, hence the set {z: Izl = rj, Rjrj cos(O + I{Jj) ::: iRjrfJ cannot be contained in A(3, rj), and this leads to a contradiction. We conclude that S(r) is bounded, as we wanted to show. (The reader will find an alternate proof of this fact in Exercise 2.6.1.) The necessity of the given conditions in part (2) is proved as follows: In the case p p - 1, we have already seen that the type of f is laol. Hence, 0 when f is of type zero. Let us now consider the case p = p. Since f is of type zero, then nCr) = o(r P ), and as proved earlier this leads to
ao = (*)
=
- Re {zp
lao +.!. L z~l} p I~n~n(r)
= -log If(z)1
+ o(r P ).
Let e > 0 be given, then we have ro > 0 such that the term o(r P ) is bounded by er P for r ::: ro, and we can also assume log M(r) :::: er P for r :::: rD. As above, the measure of the set A(3, r) £ aB(O, r) where
I log If(z)II ::: 3er P ,
Izl = r :::: ro,
2.6. The Algebra Ap of Entire Functions of Order at Most p
177
is at most (rr /2)r. We claim that lim sup
(lo + -1 L
p
P l:sn:Sn(r)
r->oo
::: 8e.
Zn
If not, there are rj ~ 00, Rj > 8e, ({Jj E [0, 2rr [ such that for
Re
{Zp [(l0 + ~
L z~l} =
Rjr; cos(8
+ ((Jj)
::::
Z
= rje iB ,
4Rjr;
P l:sn:Sn(r)
>
4er;
on a subset of aB(O, rj) of linear measure (2rr/3h. But, any point of this set belongs to the exceptional set A(3, rj). Using the expression (*) we obtain a contradiction, so that the inequality (**) is true. We need to show that (**) implies that the series l:n>l z;;P is convergent and has the sum -(loP. Consider a finite sum
L
1 p'
l::on::oN Zn
and let r =
IZNI,
then
L
l:Sn:sN
1
zP n
L
l:Sn:sn(r)
if r :::: ro. Hence, lim sup N--+oo
~
z~
::: L
1',I=r Zn
1(l0 + ~ P
It is now clear that
L
1
n~l
p
I
liP:::
L
nCr)
-P- ::: e
r
41::: ge.
l:sn,,;N Zn
= -(lop·
Zn
This concludes the proof of Lindelof's theorem.
o
From the above proof we can extract the following corollary of Theorem 2.6.3:
2.6.7. Proposition. Let the canonical product P(z) = TIn>l E(z/zn, p) represent a/unction in the space Ap, and PN(z) = TIl:sn:SN E(zjzn, p). Then PN
N--+do
P
in the space Ap.
Proof. It is well known that the partial products PN converge to P uniformly Over compact sets. Let VN(t) be the counting function of the sequence (Znh,,;n:sN.
2. Interpolation and the Algebras Ap
178
and let n(t) be the counting function of the whole sequence. First, consider the case p fJ N. Then we have p < p < p + I and some constant K = K(p) > 0, log IPN(z)1 ~ Kr P
-
0. (Here we use n(r) = O(r P ) as in the proof of the necessity of conditions (i), part (2) of Theorem 2.6.3.) As in Theorem 2.6.3 we rewrite
logIPN(')1
~R{: ".E" ,;,} + "I", log IE (~, I) I + L IE (:n ' I' p-
log
p)
n>VN(r)
and we conclude that
for some constant K = K(p), C 1 > 0, A > 0, as defined earlier. The case p E N*, p = p - I, is treated the same as the case p concludes the proof of the corollary.
fJ N. This 0
In the last chapter (Definition 1.3.24) we found that the space Expo(C) of entire functions of infraexponential type, i.e., order 1 and type 0, is isomorphic to the space of infinite order differential operators. For that reason it is interesting to consider the spaces of functions of minimal type. 2.6.8. Definition. Let Ap,o denote the Frechet space of entire functions of order p, p > 0, and type zero. The norms are given by
Ilfllm
W1m }, = max{lf(z)lezee
It is easy to see that Ap,o is an FS space.
mEN·.
2.6. The Algebra Ap of Entire Functions of Order at Most
179
p
2.6.9. Proposition. For p rt N, if the canonical product P(Z) = TIn>1 E(z/zn, p) belongs to Ap.o, and we let P N denote its partial products, then -
PN N~60P, in Ap.o. The same statement if true if p E N* and the genus
p
=p-
1.
Proof. It is the same as that of the first part of the preceding proposition. Note that in the case pEN', P = p - I, the products PN E Ap.o. 0
2.6.10. Remarks. (1) In the case p
E N*, we have a difficulty when the genus p p. In this case, the canonical factor E«z/zn), p) does not belong to Ap.o. Hence, in general, we have the same problem for the partial products. On the other hand,let us consider the case of an even function I E Expo(C), 1(0) = l. Then the zeros appear in pairs ±ako and if we order them so that
=
then
1 L-=O, n~1
so that if Ln~l (l/Izn J)
I(z) =
g
E
Zn
= 00, the canonical product has genus p =
(~, 1) =
D{(1 - :J (1 + :J ez/a,
1 and
e- Zja, }
This time, the partial products
IN(z)
= J~L
(1 - (:J 2)
do belong to Expo(C), they converge uniformly over compact sets, and for every (1IINllm)N~l is a bounded sequence. This implies that
mE N*,
IN N-+601 in Expo(C). Note that if we denote by iN (z) = TIk>N(1 - (z/ak)2), then iNez) E Expo(C) and iN -+- 1 in Expo(C). (2) Similarly to the last part of the previous remark we have that if P(z) = 1L.~1 E(z/zn' p) is a canonical product of a function in Ap, p > 0 (resp. Ap.o, P ¢ N*), then PN(Z)
=
II E ( : ,p) -+- I
n>N
in Ap (resp. Ap,o).
n
2. Interpolation and the Algebras A p
180
(3) A number of the preceding properties, as well as the problems about localizability of ideals in Ap and Expo(IC), and algebraic properties of closed ideals, that shall be considered below, can be generalized to the spaces Ap(IC), where p is radial and satisfies the doubling condition (see [RuT], [KT2], [Bra], [Nil], [KrD. We now return to the question of localizable ideals in Ap. 2.6.11. Proposition. Every ideal in Ap is localizable.
Proof. From the previous section we know that it is enough to prove that if I is a closed ideal without zeros, then 1 = A p. Let f be a nonzero function in I. If I(a) = 0, then there is gEl such that g(a) "# O. As we have done before, we can write
[ g(Z) - g(a)] fez) z-a
= g(z) [/(Z)] _g(a) I(z) . z-a
z-a
The two functions in square brackets belong to Ap, hence g(a)[/(z)/(z - a)] E I. Since g(a) "# 0, I(z)/(z - a) E I. If a "# 0, it is also true that for any pOlynomial Q of degree::::: p, [/(z)/(l - (z/a»]e-Q(z) E I. Therefore, if I has finitely many zeros, we conclude that 1 E I. If I has infinitely many zeros we have
I(z)
= zmeR(z) IT E k~l
(!.., p) , Zn
p ::::: p, deg R ::::: p. We can eliminate zm and eR(z) by the previous remark and assume I coincides with the infinite product. The Remark 2.6.10(2) shows that the sequence III
converges to the function 1 in Ap. Hence, 1 E I.
o
Let us prove that the same result holds for the space Expo(IC). 2.6.12. Proposition. Every ideal in Expo(1C) is localizable.
Proof. Let us assume first that 1 is a closed ideal in Expo(lC) such that V (1) == 0. Let IE I\{O}. Multiplying I by j (j(z) = I(-z», if necessary, we can assume I is an even function. The same reasoning as that of the previous proposition, shows that if Zo is a zero of multiplicity m, then I(z)/[(z - zo)m], as well as f(z)/[(z2 - z~)m], belong to I. Therefore we can assume that 1(0) = 1, that
fez)
= II k~l
(1 -(: r) k
2.6. The Algebra Ap of Entire Functions of Order at Most p
181
(why is there no exponential factor?), and that for any n ::: 1,
Since fn ~ 1 in Expo(C), we conclude that I = Expo(C). We now observe that the proofs of Propositions 2.5.3 through 2.5.6 and 2.5.8 can be carried on verbatim for the space Expo(C). It follows that every ideal in Expo(C) is localizable. 0 As a consequence of the localizability of the ideals in Ap one can easily prove that if I is closed ideal in Ap there are two functions fJ, 12 E I such that I = i(fl, h) (see Exercise 2.6.1). In fact, one can do much better, but at the cost of substantial amount of work. Squires [Sq I] proved that one can even find a pair g" g2 E I such that I = I (gl, g2) i.e., they are jointly invertible in Ap. As expected. this has interesting consequences about interpolation. We shall explain this work in the rest of this section. Recently Braun [Bra] has extended the result of Squires to the algebras Ap(C), p a radial weight satisfying the doubling condition. The reader can safely skip the proof of the following proposition in a first reading:
*2.6.13. Proposition (Squires). Given a multiplicity variety V there is a pair invertible.
S; V (g), g E Ap,
ft, hEAp such that V = V(fl. h) and ft, h are jointly
Recall that fl. h being jointly invertible means that if hEAp, V(h) ;2 V, then there are gl, g2 E Ap such that h = gIll + g2/z'
Proof of Proposition 2.6.13. Let V =
(Zk. mdk::I' then its counting function n v is given by nv(r) = Llz'I~" mk. The condition V S; V(g) implies
nv(r) .:'S ng(r) = O(r P ). Hence, if p ¢ N*, we have V = V(f), for a single function f E Ap, as a consequence of Lemma 2.6.2. Moreover, every function in Ap is invertible (see Proposition 2.2.14). Henceforth, we shall assume p E N*. The main part of the proof consists in finding fl E Ap, and 8, C > O. such that S(ft, 8, C) = {z E C: Ifl(z)1 < Ce- W }
has the following properties. Let Z counting multiplicities), then:
= (Zk Jk::I, Z (fl) = set of zeros of fl
(i) there are two disjoint open sets SI, S2. such that
(without
2. Interpolation and the Algebras A"
182
(ii) S2 is the union of disks B(zo, r) with the property that
d(B(zo, r), SI) ~ Kdzol-P for some convenient constant K I > 0; and (iii) for every Zk E Z, the multiplicity of Zk as a zero of fl is exactly mAo Once fl has been found we conclude the proof as follows: Let cp be a COO function which is equal to I on S2, 0 on SI, 0 :::: cp :::: I, and satisfies everywhere
I~~(Z)I:::: K2IzI
P,
for some K2 > O. This is possible due to condition (ii). The function Bcpjaz vanishes on SI U S2, hence the function (l!fI)(BcpjBz), which is well defined outside Z(h), can be extended to be Coo everywhere. Therefore, there is a Coo solution u of the equation BujBz = (l/fd(Brp/Bz), and the function h := cp - ufl is entire. We note that
hl(Z(fI)\Z) = 1
and
hlV
= O.
Moreover, the multiplicity of any ZI E Z(h)\Z as a root of the equation h(zl) = I is at least equal to the multiplicity of ZI as zero of fl. These properties of f2 follow from the fact that u is holomorphic on SI U S2. Hence
V(f ..
h) =
V.
What we need to know is that hEAp, for that we need to choose u satisfying the correct growth conditions, and, furthennore, we would like fl, h to be jointly invertible. Let us estimate 1(l/fd(ocp/Bz)l. Since supp(Bcp/Bz) n S(fl, e, C) = 0, then IfI (z)1 ~ ee- CW on supp lacp/Bzl, therefore
_1_1
acp Ifl(z)1 az
(Z)I O. From here we conclude there is a choice of u satisfying the estimate .
l
lu,2e-MW
C (1
+ Iz12)2
dm
0, as well as the equation au/az = (l/fl)(acp/az). It is possible to obtain a pointwise estimate for u from these two properties. Recall the following lemma from [BO, Exercise 3.2.8].
2.6.14. Lemma. Let n be an open set in C and let K be a compact subset ofn, there is a constant C > 0 such that for every function v of class C I in n
s~Plvl5 (s~pl:~I+ llv'dm).
2.6. The Algebra Ap of Entire Functions of Order at Most p
183
Proof. Let X E D(Q), X = 1, in a neighborhood of K, 0 ~ X ~ 1. Apply Pompeiu's formula [BO, §2.IAJ to the function XU for a E K, then v(a)
= x(a)v(a) =
1 -. 2m
1.
aX v(z) -::-(z)--dz 1\ dz az z- a
Q
1.
I av dz 1\ dz + -. X(z)-=-(z) . 2m!2 az z- a
Since Iz - al :::: .5 > 0 on supp(aX/az), we can estimate the first term by const. J!2 Ivl dm. Using that 1/(z - a) is locally integrable, we can estimate the second 0 integrand by sUP!2lav/azl. We apply this lemma to Q = B(O, 1), K =
to},
v(z) = u(zo
+ z), for a fixed
Zo E C, and obtain
lu(zo)1
~C(
sup Izo-zl:" 1
IfI a~1 + r 1
az
} B(zo.1)
IUldm).
On the other hand, for some constants A', A" > 0, we have sup Izo-xl:"1
I~ a~ I ~ It
az
sup
AeBlzlP
=
A' eBlzolP
Izo-zl:,,1
and
X
(1
I 12 -MlziP U e
B(zo.l) (1
+ Iz12)2
dm
) 1/2
< A" eM'lzolP
-
with M' = M /2. It follows that lu(z)1 ~ C1ec,W and 12 E Ap. Let hEAp be such that V(h) 2 V. The function h(l - h) E Ap and V(h(l- 12» 2 V(/t). Therefore, there is gl E Ap such that h(l- h) gtfl. Hence h = gtfl +hh,
=
which shows that fl, 12 are jointly invertible. We shall now construct fl. The idea is to add zeros to Z, with convenient multiplicities so that these new zeros are sufficiently distant from Z and the corresponding multiplicity variety V' satisfies V' 2 V and the conditions of Lindelofs Theorem 2.6.3. We call these new zeros {aj )j~1 (not necessarily distinct). There are three conditions to be satisfied: there is a constant K :::: 0 such that: (i)
(ii)
IRe { IZkl:"r L m: + lajlY E a~} I~ K; zk j
IIm{ L m: + L IZklY Zk
(iii) nv,(r) = O(rP).
~}I ~ K; and
lajlyaj
2. Interpolation and the Algebras A p
184
The reason to separate the real and imaginary part of LindelOf's condition is that for p odd we shall need to place the aj on the real and imaginary axes respectively to satisfy those two conditions, while for p even we need to place them on the rays (J = 0, (J = 7r / p, for the condition (i), and on the rays (J = 7r /2p and (J = 37r 12p, for the condition (ii). We shall show in detail for the case of p odd how to place zeros on the real axis to achieve the boundedness condition. The other cases, being entirely analogous, are left to the reader. The proof is a bit delicate; it takes six steps and follows the blueprint of the proof of the Cartan-Boutroux lemma [Lev], [BG, §4.5.13]. For n ~ I, let An = {z E C: 2n- 1 ::: Izl ::: 2n }, Bn = B(O, 2n ), and denote the points of Z n An by Vn = (b7h~j~An' repeated according to their multiplicities. Therefore, for AO = nv(l), we have Ao
+ AI + ... + An =
nv(2n).
For a disk C we denote by R(C) its radius. First Step. Let 0 < "f/ ::: 1/10 be a fixed constant that will be chosen later. Following Cartan- Boutroux one can find disjoint disks Cf, i = 1, ... , in, such that 2H:= R(Cn = "f/2n+ 1
L
l~i~in
and the number of points of Vn interior to q is exactly (HIAn)R(Cf). Set lln := HI An = "f/2n 1An. It is shown in [BG, §4.S .13] that the q have the property that if z 1- UI~i~ln q, then any disk B(z, kiln), kEN", contains at most k - I points of Vn • We replace the disks q by concentric disks Cf of radii R(Cf) = R(Cn + 311n. For z rt U1 IWp,I-P'
Z
E
An_I U An and
(;2 f'· n- 3
I ::;k::;p,
We have Pn
1 = n(2n)::::: log 2
1
2"+1 n(t)
2"
--dt ~ t
log M(2 n + l ) log 2
,
2.6. The Algebra Ap of Entire Functions of Order at Most
189
p
YJ) logM(2n+l) log Ig(z)l::: ( log 8e log 2 ' outside the omitted disks and the B(a}, 20k). If we choose 17 > 0 smaller if necessary, we can assume that the sum of the radii of the omitted disks and the disks B(aj, 20k ) which intersect An-I U An is less than 2n-2. Therefore, there is an R, 3 x 2n - 2 < R < 2n such that the last inequality holds on the circle Izl = R. Let us now choose () such that Ih(Re i9 )1 = M(h, R) = sUPlzl=R Ih(z)l. Since M(R, f) = M(R) ::: M(2 n +I ), we have log M(h, R) ::: log M(R) -log Ig(Re i9 )1
(!L)} .
::: log M(2 n+ l ) {I - _I-log log 2 Be
On the other hand, h has no zeros in Bn and h(O) = 1, hence the Borel-Caratheodory lemma [BG, §4.5.9l allows us to conclude that 2r
log Ih(z)1 ::: ---log M(h, R),
Izl ::: r < R.
R-r
Let r = 2n-l, then loglh(z)1 ::: -4IogM(h, R),
Izl ::: 2n- l .
We conclude that for Z E A n- Io outside the omitted circles and B(aj,28k ) we have log If(z)1 ::: -Cllog M(2 n +l ) ::: -C2IzI P - e3, for some el, e2, e3 > O. This shows that if e = e- c.., the set SUI, e, e) has the three required properties, namely, we let S2 = Up B(aj, 20k ) and SI its complement. This concludes the proof of Proposition 2.6.13. 0 2.6.16. Corollary. Given V = (Zk, mkh~1 ~ V(g), g E Ap. The necessary and sufficient condition for V to be an interpolation variety for Ap is the existence of fl' hEAp, and constants e, e > 0 such that V = V(fl, h) and Ifl(m,l(zk)1 mk!
+ I fi m,l (zt> I > mk!-
(CliP)
eexp -
Zk
•
In the context of this corollary, let us introduce another kind of interpolation problem, to distinguish it from those of Section 2.2. We shall call it the universal interpolation problem. The remainder of this section is based on [SqIl, [Sq2J. 2.6.17. Definition. A multiplicity variety V = (zt. mk)k>1 in Q is called a universal interpolation variety for Ap(Q) if for every sequence
2. Interpolation and the Algebras Ap
190
Yk.j(k ::: I, 0
~
j
~
mk - 1) for which there are constants A, B > 0 so that
We note that this concept only coincides with that of interpolation variety if mk = O(exp(Bp(Zk») for some B > 0, for instance, if p(z) = IzlP or p(z) = I Imzl + log(l + Izl). If V = (Zb mkh~l = V(j[, ... , fn), the functions ft, " ., fn are jointly invertible in Ap(C) and V is a universal interpolation variety for Ap(C), then there is F E Ap(C) such that
2.6.18. Proposition.
F(mkl(Zk) ---=1 md
(k ::: 1)
and V(F) 2 V.
We start with two preliminary lemmas:
2.6.19. Lemma. If V is a universal interpolation variety for Ap(C), then for every C > 0 there are constants A, B > 0 and functions fk,j E Ap(C) such that: (i) fi:j(zd = 0 except when k (ii) fk(,jj(Zk) = j!; AeBp(z) (iii) IA,(z)1 ~ (C ( exp P Zk
= I and i = j;
»
Proof. Let Ap,u(V):= {(Yk.j)k,j: 3A, B > 0, IYk.jl ~ Aexp(Bp(zd)} and p: Ap(C) ~ Ap,u(V), p(!) = f(j) (Zk)/j !, which is surjective by hypothesis. For a given C > 0, let D:= ((Yk,j) E Ap,u(V): IYk,jl ~ exp(Cp(Zk»}, topologized by the distance induced by the norm II(Yk,j)11 = sup{lYk.jle-Cp(z.J}. k,j It is easy to see that D is a complete metric space. Let Un := If E .*'(C): If(z)1 ~ n exp(np(z»)}. One shows, as in Lemma 2.2.6, that p(Un ) D is closed in D. By hypothesis, D Un>! (P(Un) D), hence one of the terms has a nonempty interior. Therefore, for some nand e > 0
n
=
n
we have It follows that p(Un/e)
n D = D. Hence there are entire functions .h,j such that
l.frc.j(z)1 ~ ~enP(Z), e
2.6. The Algebra Ap of Entire Functions of Order at Most p
if I = k and i otherwise.
h.j(Zt) = {exP(Cp(Zd)
j !
191
0
= j,
The functions fk,j := exp( -Cp(Zk».h,j satisfy the conditions (i), (ii), and (iii) of the lemma. 0
= V (fl, ... , fn). with fl, ... , fn jointly invertible in and V = (Zb mkh::: I is a universal interpolation variety for Ap(IC), then there is a constant Co > 0 such that 2.6.20. Lemma. If V Ap(IC),
L
e-COP(Zk)
0, such that
On the other hand, there are Ao, Bo > 0 such that for any Z E Bj, p(Z) ::::: AOP(Zj) + Bo. Therefore, for some, 8J, 82 > 0,
r
lB
e-C1P(z)
dm 2: "[(dJ81 exp(-C3P(Zj» 2: 82exP(-Cop(Zj».
j
Hence
o We can go back to the proof of Proposition 2.6.18.
Proof of Proposition 2.6.18. Let us consider the series F(z) = Lmk(Z - Zk)!k.mk-1(Z), j:::l
with the functions obtained in Lemma 2.6.12 for a value C > 0 to be chosen below. If this series were uniformly convergent over compact sets and FE Ap(IC), then we could differentiate it term by term and evaluate it at Zk. We would then
2. Interpolation and the Algebras A p
192
obtain
F(md (zd = Am, -I (Zk) = l. mk! (mk - I)! Let us prove the convergence of the series and estimate F. IF(z)1
:s Lmk(lzl + IZkDIAm,-I(Z)1 k:::1
:s IzlA exp(Bp(z»
mk exp( -Cp(Zk»
L k:::1
+ Aexp(Bp(z»
Lmklzklexp(-Cp(Zk». k:::1
Since we are in the conditions of Theorem 2.2.10, it follows that the property (*) given there holds. In the course of the proof that the property (*) just mentioned implies that V is an interpolation variety, we proved that there are constants E, F > such that
°
for every
k::: 1.
Using that estimate and the fact that from the properties of p we have for some L, K >0,
Izl
:s Lexp(Kp(z»,
we obtain IF(z)!
:s 2A exp«B + K)p(z»
Lexp(-(C - E - K)p(Zk». k:::1
°
Choose C > so that C - E - K ::: Co, Co the constant of Lemma 2.6.20, then the series converges uniformly and IF(z)1
:s A' exp«B + K)p(z».
This concludes the proof of the proposition.
o
2.6.21. Remark. Let us remark that we have also shown that if V:::; V (fl, ... , In), with II, ... , In jointly invertible, then V is a universal interpolation variety if and only if V is an interpolation variety. In order to find necessary and sufficient "geometric" conditions for a multiplicity variety to be an interpolation variety, we need to introduce a little extra notation. Given two positive constants K I , K2 and a point zo, we denote r(zo) the largest positive number such that if z E 8(zo, r(zo» then p(z) :s KIP(zo) + K2· The hypotheses on p guarantee that we can find Kl> K2 > 0 so that r(zo) ::: 2 for every zoo We fix these values henceforth. If V = (Zk. mk)k:::l, we denote rk := r(zk). We also denote n(zt, r, V) the number of points of V, counted with multiplicity, in B(z*, r)\{zd. If V :;: V(h). we write n(z*, r, h) instead of n(zl, r, V).
2.6. The Algebra Ap of Entire Functions of Order at Most p
193
2.6.22. Proposition. Let V = (Zk> mkk~1 = V(fl, ... , fn), where fl, ... , fn are jointly invertible in Ap(C). Assume V is an interpolation variety for Ap ( 0 such that for every k ::: 1 we have
1 r,
(1)
o
n(z
t V)
/." t
(2)
dt ~ Cp(Zk) mk < -
+ D,
Cp(zd + D . logrk
Proof. Let F be the function obtained in Proposition 2.6.18. For every k ::: 1 we have _1_
2;ri
r
J2-:.I=rk U; -
d~
F(O
=
F(mk)(zd
Zk)m,+1
I
Since F E Ap( 0, A > 0, one has
I: : Ig(md(Zk) mk!
(k :::: 1).
ee-A1z!lp
It is enough to prove it for IZkl :::: 2, and also to assume that g(O) = 1 (Why?). We can apply the minimum modulus theorem ([BG, §4.5.l4], [Lev, p. 21]) to obtain constants £1 > 0, A I > 0 (independent of k) such that for Z E B(O, 21zk I) and outside a finite collection of exceptional disks, Bk • l , ••• , Bk,jk the sum of whose radii does not exceed ~IZkl, one has
Ig(z)1 ::::
£le-A11z,IP.
It follows that there must exist a value r, !IZk I < r < IZk!. such that CJB(zt, r)
n Bk,j
= 0,
1 ::: j :::
ik.
Apply now Jensen's formula, [BG, §4.4.30] to g in the disk B(zt, r). One obtains log I
g<md(Zk) mk!
1+ mk logr +
1
r n(V,
Zk,
t
0
t)
1
dl = -2 7r
1
2".
log Ig(Zk
0
.
+ re '8 )1 de.
From this identity, (1'), (2'), and the choice of r, it is clear that log I
g(md(zd mk!
I :::: -AlzklP - B
o
for some A, B > O.
2.6.24. Remark. [Sq2J. For the weight p(z) = I Imzl + 10g(1 + Izl), and a multiplicity variety of the form V = V (h), given by h slowly decreasing for Ap(C), one has that V is a (universal) interpolation variety if and only if
1
P(Z') n(V Zk t) , , dt::: Cp(Zk)
o
and
+D
t
Cp(Zk) + D mk < - - ' - - - - log P(Zk)
for some C, D > 0, independent of k. One can find necessary and sufficient geometric conditions interpolation, generalizing Proposition 2.6.2 for arbitrary radial weights satisfying the doubling condition [BL]. The correct expressions to consider are just the integrated Nevanlinna functions N (V, Zk> r) = (n(V, Zb 1)/1) dt. In this form the criteria obtained extend also for meromorphic functions and to several complex variables. This can be generalized to other radial weights, but the necessary and
J;
2.6. The Algebra Ap of Entire Functions of Order at Most p
195
sufficient conditions do not coincide, though they are close to each other. This is a phenomenon characteristic of functions of infinite order of growth [BL]. In [GR] similar conditions for spaces of functions with prescribed indicator of growth have been found. Interpolation in the space of functions of infraexponential type has very interesting applications to the theory of Dirichlet series, as we shall see in Chapter 6. A classical theorem of Polya- Levinson [Levs, Theorem XXXI] states the following: Let V = (Zk}n~1 be a sequence of nonzero complex numbers such that Re Zn > 0,
. -n = 0, lIm
n~oo Zn
I1m Zn I =
ll.m
0
IZn I
n-'oo
and there exists c > 0 such that
Then there is an even function
f
of infraexponential type such that V S; V (f),
-log If'(zn)1 = o(lznl) and for
Z
as h --+
00
such that Re z > 0, dist(z, V) ::: c 18, we have -log If(z)1 = o(lzl)
(tt)
asizi --+
00.
Note that (t) implies that points Zn are distinct. One can prove that the main point (t) of the statement is exactly the condition that V is an interpolation variety for functions of infraexponential type. In [Vi2] it has been proved that the conditions on Rezn > 0 and I Imznl/lznl --+ 0 are not necessary. In fact, given any sequence V = (Zn In:::1 , n(r) = o(r), and {znln:::1 satisfies V, then there is a function f of infraexponential type such that (i) V 5; V (J); (ii) all zeros of f are simple; and (iii) V (J) is an interpolating variety in the space of functions of infraexponential type. (The condition (tt) also holds when dist(z, V(J» ::: c/8.) We provide a proof of this result in Chapter 6.
It is possible also to find necessary and sufficient geometric conditions for a variety to be interpolating in space of type zero for any radial weight satisfying the doubling condition [BLV]. EXERCISES
2.6.
I. Let j be an entire function, j(O) = 1. For r > 0, let ai, ... ,an denote the zeros of f in B(O, r) repeated according to multiplicity. (a) Show that
1121r e- i8 2" 7r
0
log If(re i9 )1 de
= ir/,(O) + i L"( ~ _ -) k=1
ak
ak r
•
Use this formula to show that if f is an entire function of exponential type then SI (r) Lld tl:5r ak' is bounded as r ~ 00.
=
2. Interpolation and the Algebras A p
196 (b) Let P E N*, find an explicit expression for -
I
27T
121< e-'p& . 10gl/(re,e)ldO . 0
=
involving Sp(r) L:1a,IY a;p. Use this expression to show that Sp remains bounded as r ~ 00, whenever lEAp. (Hint: Divide out the zeros of I to obtain a nonvanishing holomorphic function g in B(O, r). All this exercise requires is to compute the Fourier coefficients of the hannonic function Re(log g) (see also [BG, Exercise 4.6.9)).
2. Let E p , Ex be respectively the spaces of entire functions of order at most p (resp. finite order). For 0 < p < 00, nEW consider the Banach spaces Ep .• of all entire functions such that 11/11. = sup(l/(z)1 exp(-lzIP+'/.»
1 F•. Ep is an FS space and is a DFS space. (Hint: Use Proposition 1.4.8.) (b) Show that every closed ideal in Ep (resp. E",,) is principal. (Hint: Find g such that V(g) = V(I).)
3. For 0 < p < such that
00,
n
E
II/lIn
N*, let Ap.n be the Banach space of all entire functions
= sup (I/(Z)I exp ( -~IZIP))
0, E,>, €k < 00. Show that one can find a sequence of distinct Wk E C, 0 < IWk I /' 00, su~h that the set {Wk: k ~ I} is disjoint from {z.: n ~ I}, Iw;P z,;tl < €" and if the series En:::k z;;P converges to the value 01, and (Un)n"l is the sequence given by Un = Zn if n 1: nk(k ~ 1), Un, = Wko then En,,1 u;;P = S. (c) Let II E [\{O}. Use parts (a) and (b) to find 12 E Ap (resp. Ap.o) such that V(/I) n V (h) = V. Conclude that I = 1(11. h)· PEW, then there are two functions II,
6. Let the canonical product of a function I of finite order be I (z) ITn"l E(z/zn, p). Let PR(z) := zm Il1',I:::R E(z/zn, p), R > O.
zme Q
=
2.6. The Algebra Ap of Entire Functions of Order at Most p (a) Assume
f
E
A p, show that e- Q fI PR e- Q f
.
lIm - - g PR
R-co
E
A p, and that for every g
=g
.
In
197
E
Ap one has
Ap •
(b) Let [ be a closed proper ideal in A p , V = V(l), f E [\{Ol, g E l(V). Write P R as a product PH = Pk P; such that V(P;) = V n B(O, R). Show that fI Pk E I. Use this to prove that gEl. (c) Prove parts (a) and (b) in the case Ap.o, P f/. N, and also when p E N* but P = p-1. (d) Let f E Ap.o, p = p, Q(z) = aoz P + QI (z), deg QI ::: P - I. Show that
pf
exp (zp - ~
z;;P )
E
Ap.o
p l'nl"R
R
and
holds in Ap.o. Use (c) and (d) to prove that all ideals in Ap.o are localizable.
7. Let V = V(jl, ... .jn), fI, ... ,fn jointly invertible in Ap(lC). Show that V is a universal interpolating variety if and only if V is an interpolating variety.
=
8. Let p(z) Izl P , r > 0, kl > \, k2 > 0, compute the function r(z) > 0 that has the property: p(t;)::: kIP(z) +k2 if and only if t; E B(z, r(z».
9. Suppose V C, D>O
= (Zk> mdk~1
1 rl
o
satisfies condition (I) of Proposition 2.6.22, i.e., for some
n(zk> t, V)
---'-''---'-dt ::: C(P(Zk)
t
+ I)
(k ::: 1).
Show that there are e > 0, A > 0 such that for any j =1= k IZk - Zj I ~ u-AP(' 0 or Reaj = 0 and Imaj > O. Accordingly. we can find 8 > 0 sufficiently small so that the numbers Aj = e- i8 aj satisfy ReAj > 0.1 :::: j :::: m. It is clear that e- i8 V = L'~j~m QAj also has a Q-dimension equal to p. Let m, •...• m P be a Q-basis of this space. Assume Aj =
L
aj/m/.
'~I~p
with ajl E Q. We shall first find another basis Mi •...• Mp of e- i8 V so that mk E L'~19 Q+ M/. for 1 :::: k :::: p. For that purpose. let us choose rational numbers tl,k so that for every I they approximate Reml for 1 :::: k :::: p. Assume further that det(t/k)/,k =J:. O. Then we can detennine M, . ...• Mp from the system of equations
m, = L
tlkMko
1 :::: I :::: p.
'~k~p
This indicates that the Aj can be written as
Since we want
tlk
approximately equal to Reml. then the rational numbers
bjk := L'~I~p aj/tlk are approximately equal to Re(L'~I~p aj/ml) = Re Aj > O. In other words. if we choose tlk sufficiently close to Reml. 1:::: k :::: p ••we can guarantee that bjk E Q+ and. clearly. this can be done so that simultaneously
the detenninant det(t/k}J,k =J:. O.
3.1. The Ring of Exponential Polynomials We now let /-Lk LI!'Ok!'OP Q+/-Lk.
= ei8 Mb
203
I ~ k ~ p. They form a basis for V and Cij
E
0
Let us now return to the situation of Proposition 3.1.1.
3.1.7. Proposition. Let fez) = LO::;j!'Om Pj(z)e Uj ; and g(z) = LO!'Oj::;n Q/z)e PjZ be exponential polynomials of the form (*), with more than one term. Assume that g divides f in E and that Cil, ... , Ci m are linear combinations with nonnegative rational coefficients of p Q-linearly independent complex numbers /-LJ, ••• , /-Lp. In this case, the frequencies /3t, ... , 13m are also Q-linear combinations of /-LI, .•• , /-Lp with nonnegative coefficients. Proof. From Proposition 3.1.1 we conclude that
We want to show that rjk ::: 0 for every j, k, 1 ~ j ~ n, I ~ k ~ p. To fix ideas let us assume rj I < O. We choose among the f:3j, those for which rj I is minimal; among these, those for which rj2 is minimal, etc. One finds in this way, a certain f:3io, denoted 13*, whose coefficients with respect to /-LI, ••• , /-Lp are all successively minimal. Let us write
13* =
UI/-LI
+ ... + Up/-Lp,
Uj
E
Q,
1
~
k
~
p.
We have UI ~ rjo, 1 < O. Let 0 = Yo -< YI -< ' .. -< Yr, and Rk be polynomials such that
We augment the collection /-LI, ••• , /-Lp to a basis /-LI, ., . , /-Lq of the Q-vector space generated by Cil, ... , Ci m, 131, ... , f:3n, YI, ... , Yr' Among Yo, ... , Yr we find Y * whose successive coefficients with respect to /-L I, ••• , /-Lq are minimal. Let y* = VI/-LI + ... + vq/-Lq, with VI ~ 0 since Yo = O. As in Proposition 3.1.1, we can see that 13* for some i, 1 ~ i ~ m. Then
+ y* = Cii
+ VI )JLI + ... + (up + vp)JLp + Vp+1 JLp+1 + ... + Vq/-Lq. = ... = Vq = 0 and UI + VI < 0, which is a contradiction with the
Cii = (u I
Clearly Vp+1 hypothesis that every oti has nonnegative coefficients with respect to JLl, .•. , JLp. Therefore, it must be true that 'ik ::: 0, 1 ~ j ~ n, 1 ~ k ~ p. 0
3.1.8. Remark. It follows from Proposition 3.1.7 that the Yj are also Q-Iinear combinations of the JLk with nonnegative coefficients. We shall accept the following result of Ritt about factorization in the ring Eo. The proof is entirely algebraic and rather long (see [Ril], [Go], [Schi]).
204
3. Exponential Polynomials
3.1.9. Theorem. Every normalized exponential sum f E Eo which is ¥= 0, 1 can be written in one and only one way (up to reordering) as a finite product of normalized exponential sums
where the ai are simple, any two of them have no nonzero common frequency, and the lrj are irreducible in Eo.
3.1.10. Remarks. (1) Let f(z) = 1 + ale"": + ... + ame"'m z , 0 -< al -< •.. -< am. If all the aj, 1 ::s j ::s m, are Q-linear combinations of J-tl, •.. , J-tp with nonnegative coefficients, and f = It ... ft is the decomposition of the previous theorem, then the frequencies of all the jj are Q-linear combinations of J-t I, ... , J-tp with nonnegative coefficients. (2) If P = rank of the additive group generated by al, ... , am, then it coincides with the dimension of the Q-vector space V defined in Lemma 3.1.6. Changing, if necessary, the J-tk obtained in the lemma by a submultiple, we can assume l::sj::sm. aj E N* J-tk>
2:
I~k~p
(3) Every nonzero exponential sum form
f
can be written in a unique way in the
with ae"'z E U(Eo), ai, ... , as normalized, simple, sharing no common nonzero frequencies, and lrl, ... , lrr, normalized irreducible exponential sums. We recall some classical results of commutative algebra that we will use in the sequel, see, e.g., [AM]. Let A be a commutative integral domain with identity and let Jt be the field of quotients of A. Denote A* := A\{O}, U(A) := the group of invertible elements of A, Jt* := the multiplicative group of the nonzero elements of Jt, and A[X], resp. Jt[X], the rings of polynomials in the variable X, with coefficients in A and Jt, respectively. We assume A satisfies the condition (~)
Any two elements of A * admit a g.c.d.
3.1.11. Definitions. (1) The content of a nonzero polynomial f E A[X] is a g.c.d. of the coefficients of f. It is denoted c(f). (2) A polynomial f E A[X] will be said to be primitive if c(f) is invertible. For every f E A[X] we shall write f = c(f)j, with j E A[X] primitive. The product of two primitive polynomials is primitive. If f, g E A[X]\{O}, then c(fg) = ac(f)c(g), a E U(A). Let us recall finally that there is a family P S; A[X] such that every element of P is primitive in A [X] and irreducible in Jt[X] and, moreover, any f E Jt[X]
3.1. The Ring of Exponential Polynomials
205
can be written in a unique way in the fonn
I
= a II pn(p), peP
with a E .It, n(p) E N, and only a finite number of n(p) are different from zero. It follows that if I E A[X], then
I
cU) II pn(p),
=
peP
and the decomposition is unique. We shall now show that Eo verifies the property (Ll) and that E is isomorphic (as a ring) to Eo[X].
3.1.12. Lemma. Every exponential polynomial lEE can be written in a unique way in the lorm I(z) = lo(z)
+ II (z)z + ... + II(z)/,
where 10, ... , II E Eo. In other words, E is isomorphic to Eo[X].
Proof. Let I(z) = LI 0 (unless f is just an exponential), hence v > 0 also. The following consequence of Proposition 3.1.21 is an explicit form of the Lojasiewicz inequality for functions in E:
3.1.22. Theorem (Global Lojasiewicz Inequality). Let f E E*, and let V = {z E C: fez) = O} =1= 0. There exists a positive constant C such that
f
I (z)l:::
C d(z, V)" H(') (l + Izl)1' e -,
where v=9dof, /-L=v+8max'~J:"'Nm). Moreover, C can be explicitly estimated in terms of the frequencies and coefficients of f. Proof. Let z ¢ V and consider g = l/f in B(z, d(z, V». Let us choose r = inf(l, d (z, V». We can therefore apply the previous proposition and obtain eH(z)
----
0, independent of I. In fact, in this case I E E(a, m)\{O} cannot be a polynomial, hence
is also a nonn in £(a, m). Moreover, the previous proof yields for r
~
1
which shows the existence of the constant c. The main point of the argument of Grudzinski, to improve the previous Lojasiewicz inequality of Theorem 3.1.22, is the following observation. Let L z (f) be the exponential polynomial defined by Lz(f)(W) := I(w
+ z),
which also lies in E(a, m). Moreover, the leading coefficients of are related by identities
I
and L:(f)
As an immediate corollary of these identities, we obtain: 3.1.27. Corollary. For any r
E
]0, 1],
I
E
E(a, m), and z
E
C, we have
M(f, z, r) ~ erl" max {laj(f)leReajZ} ~ e(f)rl"eH(z), l~j~N
=
where c c(a, m) > 0 is independent of f, and the constant e(f) depends on f, but c(f) > 0 if f =1= O. Let I E £(a, m) be a nonzero exponential polynomial such that f(O) = O. Let v be the multiplicity of the origin as zero of f. Let 0 < r S 1 be given and fix p E ]0. r] such that f has no zeros in 0 < Izl s p. The monic polynomial Qj(z) = ZV has exactly the same zeros as f in 0 S Izi s p, hence f/Qj is a holomorphic nonvanishing function in B(O, p). Therefore, we have inf1z l9 If(z)/Qj(z)1 > O. Since maxlzl=p Ig(z)1 is also a nonn in E(a. m), if IIf - gil is sufficiently small. the function g E E(a. m) does not vanish on Izi = p and has the same number of zeros v as f in B(O, p). Moreover, if Qg is the monic polynomial of degree v whose roots coincide with those of g in B(O. p). we have that Qg is continuous as a function of g, for g close to f. Namely, recall that if ZI •••• , Zv
3.1. The Ring of Exponential Polynomials
are the roots of g, then for k
E
213
N* the Newton sums of the roots are given by
L zf = -1-1 2:n:i
O:'Olsv
zkg'(z) dz g(z) ,
Izl=p
and the coefficients of Qg are fixed polynomials in the Newton sums. Therefore, one can find a fixed number K > 0 such that inf Izl:'Op
IQg(z) g(z) I >
K
> 0
for any g in E(a, m), which is sufficiently close to f. Now let h be an arbitrary function holomorphic in B(O, r). We have M(gh, 0, r)
~ M(gh, 0, p) = M
( Qg . h·
~g ,0, p) ~ KM(Qg . h, 0, p).
In order to estimate this last quantity, let us prove the following simple lemma: 3.1.28. Lemma. There is a constant C v > 0 such that for any monic polynomial Q of degree v and any p > 0, there exists pi E [p /2, p] for which
IQ(z)1
if lzl
~ Cvpv
= p'.
Proof. (This lemma is also a consequence of the Cartan-Boutroux lemma [BO, §4.5.13].) Decompose the annulus p/2:s lzl :s p into v + I concentric annuli using circles such that the difference of the successive radii is exactly p/2(v + 1). One of these annuli contains no roots of Q. Let pi be the radius of the median circle of one such annuli. Let Q(z) = (z - ZI)" • (z - zv). For lzl = pi we have
Iz - zjl ~
p
4(v
+ 1)
for every j. Hence pV
IQ(z)1 ~
(4(v
+
1))"'
o
Izl = p'.
Returning to the previous inequalities we have M(gh, 0, r)
~
KM(Qg . h, 0, pi)
~
KCvpv M(h, 0, pi)
~
KCvp"lh(O)I.
Since K and p depend on f, all one can conclude is that there is a constant c(f) > 0 such that for every gin E(a, m) sufficiently close to f. one has M(gh, 0, r)
~
c(f)rVlh(O)I,
where c(1) has been chosen so that c(f)r V = KCvpv. Moreover, since r and v :s J.t, we also have M(gh, 0, r)
~
:s
1
c(l)rillh(O)I.
The unit sphere of E(a, m), IIfll = I, is compact, hence there is a constant Co = coCa, m) > 0 such that for any g. IIgll = I, we have M(gh, 0, r)
~
corlLlh(O)I.
3. Exponential Polynomials
214
f
Finally, we can conclude that for any M(fh, 0, r)
~ collfllr
l1
lh(O)1
~
E(a, m),
E
Co
C~~XN laj(f)I) r
l1
lh(O)1
for any h holomorphic in B(O, r). This inequality leads almost immediately to the following:
3.1.29. Proposition ([Grudl]). Let f minl-oj-oN laj (f)I. Then, for any z a neighborhood of B(z, r), we have Co
E(a, m), 0< r ::; I, and let K(f) = C and any holomorphic function g in
E E
M(fg, z, r) ~ K(f)r l1 e H(::)jg(z)l.
Proof. Let h = L:(g), then h(O) = g(z), and M(fg, z, r) = M(L:(f)h, 0, r)
Now, aj(L::(f»
~ Co (max la;(L:(f)I) r l1 lh(O)I. [-o'-oN
= aj(f)ecx}: #- 0 if and only if aj
is a frequency of f. But
max{Reajz: a, is a frequency of f} = H(z).
Hence,
and we obtain the stated inequality.
o
One should compare this statement with Proposition 3.1.21(ii). Here one can take JL = dOf and, moreover, there is no factor (1 + Izl)M,. The trade off is the very unpredictable behavior of the constant Co (see Remark 3.1.21). As a further corollary we obtain a global Lojasiewicz inequality without denominator.
3.1.30. Proposition ([Grudl]). Let f be a nonzero exponential polynomial with frequencies a[ -< 0i2 -< ... -< aN, V = {z E C: f (z) = OJ, and let d(z, V) = inf{l, d(z, V)}. There is a constant c = c(f) > 0 such that
If(z)1
~ c(d(z, V»dO j eH(z).
Proof. It is enough to consider fez) = L[-oi-oN Pi (z)eCX,Z ,mj = deg Pj , Pj #- O. Then JL = Iml + N - 1 = dOf, and the same proof of Theorem 3.1.22 applies, 0 replacing Proposition 3.1.21 by Proposition 3.1.28. We shall now show that the exponential polynomials are not only slowly decreasing in the sense of the previous chapter, but that one can give a very precise description of the size of the components of S(f, e, C) (see Definition 2.2.13).
3.1.31. Proposition. Let f be a nonzero exponential polynomial. For every s > 0, C > 0, one can find s[, C[ > Osuch thatfor every connected component 0 of
3.1. The Ring of Exponential Polynomials
215
S(f, cI, C1) = {Z E C: If(z)1 < cle- Cilz1 } one has IZI - z21 < ee- Cizil for every Zl, Z2 E
O.
°
Proof. We can assume c :s 1. Fix Z E 0 and let < r :s 1, to be chosen later (depending on z). Consider l; E aB(z, r) such that If(nl = M(f, z, r). As pointed out after the proof of Lemma 4.2.11, the Minimum Modulus Theorem yields P E ]~r, 2r[ such that for every w, Iw - l;1 = p, one has If(w)1
log If(nl > -810gM
(
f ) f(l;),l;,4er .
Hence, inf Iw-,I=p
If(w)1 > If(l;W - M(f, C;, 4er)8
We know that for some C > 0, J1 > 0,
If(l;)1 = M(f, z, r):::
Crf.LeH(z),
and, for some D > 0, M(f, l;, 4er) :S M(f, z, 5er):s D(l
+ IzI)MeH(zl.
Therefore,
'w~~f=p If(w)1
Er"e H (:) ::: (l
+ Izl)N
for some E > 0, N ::: 0, v > O. Let us now choose r = r(z) = ce-Clzlj6ec. We are going to choose CI > 0, C 1 > 0 so that for any z E C one has C
_ Er(z)"eH(z) e- C11 - 1 < ---....,-;I (1 + Izl)N .
To verify that this choice is possible, let us observe that there is a > 0 such that for every z E C H(z) :::
Since r(z)
-alzl.
= ee-Clzlj6eC, we have to choose £1,
CI so that
£" E e(C,-a-"C)lzl cI 0, C > 0, such that for every pair (k, j), k 1= j.
IZk -
Zj
I ~ ee- CI : kl •
What is the corresponding statement for p(z) = Iz IP, P > I? Could V be an interpolation variety when 0 < p < I if f has more than one frequency? 4. Let f be an exponential polynomial, V = V(f) = (z .. md.(':I' p ::c: I. Show that the following two statements are equivalent: (i) There is A > 0 such that IZk - Zj I ~ e-A such that for every k, If(m"(zk)1 ~ Ee-BlzAi P • Given p > I find an exponential polynomial f satisfying (ii) for this value p but for no p' < p.
°
5. Let f be an exponential polynomial with purely imaginary frequencies, and p(z) = Ilmzl +Iog(l + Izl), V = V(n = (Zk>mdk?l. Show that V is an interpolation variety
3.2. Distributions of Zeros of an Exponential Polynomial
217
for Ap(C) if and only if there are e > 0, A > 0, such that for every pair (k, j), k #- j. Equivalently, show this condition can be replaced by the existence of 0 > 0, B > 0, such that for every k I/(md(Zk)1 ~ oe-Bp(zd. 6. Let 1 be an exponential polynomial with purely imaginary frequencies, icxj' al < a2 < ... < a., and ml, ... , m., the degrees of the corresponding coefficients, and let J.L > max{mk!(ak - al): 2:5 k :5 n}. Show that if Imz > J.Lloglzl,
Izl» 1,
then I(z) #- 0. (Hint: Consider e- ia1z I(z).) Conclude that there is A > V
= V(j)
0, there is an r > 0 such that
Veep, /-L, H) n B(O, r)C
~
T(e).
3.2.1. Lemma. The intersection o/two half-strips V(cp) , f-L), H) is an unbounded set if and only if ep) = cpz and f-L) = f-L2·
n V (ep2, J.L2,
H2)
218
3. Exponential Polynomials
Figure 3.1
T(e)
Figure 3.2
3.2. Distributions of Zeros of an Exponential Polynomial
219
Proof. From the previous observation, if V = V(CPI, /11, Ht} n V(cpz, /12, H2 ) is unbounded, then
lim arg z ZEV
Izl .... oo
1r
1r
= CPI + -2 = CP2 + -. 2
Moreover, /11
=
.
Re(ze-i\O\)
lIm
log Izl
ZEV
Iz I"" 00
.
=
hm ZEV
Izl .... oo
Re(ze-\02)
log Izl
o
= /12.
We shall now prove a very famous result of P6lya concerning the distribution of zeros of exponential polynomials. Following the work of Dickson [Dill we shall give simultaneously the result for a larger class of functions, those that are asymptotically exponential polynomials (AEP) (see also [RoJ). 3.2.2. Definition. We say that f f E .Tf(C\K) and has the form fez)
=
E
L
AEP if there is a compact set K such that Ajzmj(l +Cj(z»e WjZ ,
l:sj:sn
for a finite collection of distinct frequencies WI, •.. , Wn E C, 2 and the functions Cj are holomorphic in K C and satisfy
~ n, Aj E
C*,
mj EN,
lim Cj(z) = O.
Izl .... oo
Note that any exponential polynomial with more than one frequency is in AEP. It is enough to write the polynomial coefficients Pj as Pj(z)
where
mj =
deg P, Cj(z) =
= ajZmj (l
Pj(z)/ajZmj -
+ Cj(z», 1, which is holomorphic for Izi
»
1.
3.2.3. Definition. For a function f E APE represented by (*), we denote by P := P (f), the P61ya polygon of f, the convex hull of the set WI, ... , Wn , and p := {z E C: Z E PCf)}' the set of conjugates. In what follows we shall assume that WI, ••• ,Wn are indexed in such a way that the following geometric condition holds: the points WI, ... ,wa are the vertices of P and, moreover, the oriented segments LI := [WI, ci}z], ... , L a- I := [Wa-I, w,,], L" := [w", WI], are precisely the successive segments of the posititively oriented boundary ap. The remaining points are ordered arbitrarily. Note that some may still be in aP. Let us denote by hk the ray that starts at the origin and has the direction of the outer normal to aP at any point in the open segment ]Wb Wk+1 [, 1 ~ k ~ a-I, h" is the one that corresponds to [wa , wd. We make a further assumption on the choice of WI. if '!/fl, ..• , '!/fa are the arguments of the points in hI •... , h",
220
3. Exponential Polynomials
respectively, then 0 ::: 1/11 < 1/12 < ... < 1/1" < 2rr. Note that for z E hk one has Re(wkz) = Re(wk+lz) for 1 ::: k ::: a - I (with the obvious extension to the case k = a). Let 51 := {z E c: 1/1" ::: arg z ::: 1/11 + 2rr} and, for k = 2, ... , a,
5k := {z
E
C: 1/Ik-l ::: argz ::: 1/Id.
Fora fixed 0 verifying 0 < 20 < min{1/I1 let us define SI
-1/1" + 2rr, 1/Ik -1/Ik-1 (k =
+ 0 ::: arg z ::: 1/11 + 2rr E C: 1/Ik-1 + 0::: argz ::: 1/Ik - O}
:= {z E C: 1/1"
Sk := {z
2, ... , a)),
O},
(2::: k ::: a).
Note that because n :::: 2 one has 0 < 0 < rr /2. Finally, let us denote
Tk := {z E C: 1/Ik -
(J :::
argz ::: 1/Ik
(See Figure 3.3.)
Figure 3.3
+ O}
(1 ::: k ::: a).
3.2. Distributions of Zeros of an Exponential Polynomial
221
3.2.4. Lemma. (a) If z E Sk. ~ E P, then Re«wk - ~)z) ~ o. (b) Ifz E Sk. ~ E P,then Re«wk -~)z) ~ Izllwk -~lsinO. (c) If z E Sk n Tk and Wj ¢ Lb then there is a value eo > 0 such that Re«wk - Wj )z) ~ Iz Ilwk - Wj I sin 00 . (d) If z E Sk+1 n Tk and Wj ¢ L k . then there is a value 01 > 0 such that Re«wk+1 - Wj)z) ~ Izllwk+1 - wjl sine l .
Proof. Recall that (·1·) denotes the Euclidean scalar product on ]R2 and that for any a, b E C, Reab = (alb) = (alb). We shall give the proof of the lemma for 2 :::: k :::: a - I, leaving it to the reader to adapt it to the cases k = I, k = a.
Part (a). We observe that if h-I denotes the ray perpendicular to hk-h chosen so that the angle from h-I to h k- I is n/2, and h, the ray perpendicular to hk. so that the angle from hk to h is n /2, then the sector Sk is complementary to the two quadrants determined by h-I and hk-I and by hk and [k. The point ~ E P and, therefore, ~ - Wk belongs to the closed sector determined by h-I and h. This is just the sector opposite to Sk. Hence, the angle between any z E Sk and ~ - Wk is at least n /2. In other words, Re(z(~ - Wk» = (zl~ - wd ::::
o.
This is equivalent to statement (a).
Part (b). Here the angle between z E Sk and ~ - Wk is ~ nl2 + 0, hence (zl~ - Wk) :::: -lzll~ -
wkl sinO.
Part (c). Since Wj ¢ Lko one can immediately see there is a Of > 0 such that
n, _ _ 3n 1/Jk + "2 + 0 :::: arg(wj - wd :::: 1/Jk-1 + T· On the other hand, z E Sk n Tb so that 1/Jk-1
+ e < 1/Jk -
0 :::: arg z :::: 1/Jk.
Hence, the angle between z and Wj - Wk lies between n/2 + 0' and 3nl2 - O. Letting 00 = min{O, O'}, we obtain (c).
Part (d). This time there is Oil > 0 such that ./,
'f'k+1
n + "2 :::: arg (Wj_
_
)
- Wk+1 ::::
./,
'f'k
n + "2 -
nil (7
,
and
1/Jk:::: argz:::: 1/Jk +0 < 1/Jk+1
-e.
Choosing 01 = mintO, Oil) we can proceed as before.
3.2.5. Lemma. (a) Ifz E Sk and j =1= k, then zmjeWjZ = e(z)zm'e w*,. (b) liz E Tk n Sk and Wj If Lb then zmjeWjZ = e(z)zm'eWkZ .
D
222
3. Exponential Polynomials
n Sk+l and Wj f/ Lb then zmjeWjZ = e(z)zmt+JeWk+Jz, where e(z) denotes a quantity, different in each case, such that
(c) If z E Tk
lim e(z) z~oo
= 0.
whenever z is restricted to the corresponding region. Proof. (a) By Lemma 3.2.4(b), we have
which vanishes at 00. (b) It is the same as part (a) using Lemma 3.2.4(c), and (c) Just use Lemma 3.2.4(d).
3.2.6. Lemma. For f
E
APE represented by (*) and any z fez) = Akzmk(l + e(z»e WkZ ,
where lim e(z)
z_oo
eo
instead of e. 0
E Sb
= 0.
ZESt
Proof. This is an immediate corollary of Lemma 3.2.5(a), and the definition (*) ~f.
0
3.2.7. Corollary. Under the same conditions of the previous lemma, there is r > such that
°
for all z
E
Sk
n B(O, r)c.
if
Proof. From Corollary 3.2.7 we have If(z)1 > (IA k I/2)lz mk e""zl whenever z Sk n B(O, ro)c. From Lemma 3.2.4(a), we have
E
T
> 0, such that
1e~~Zz~mk
~
le~ zz~mk
f(z)1 :::
if z E
n B(O, ro)'
and
3.2.8. Lemma. There are ro > 0, ~ E ft, then
Sk
T.
IAkl f(z)1 > -2- exp[Re«wk -
~)z)l :::
IAkl 2'
0
We remark that this lemma shows that all the zeros of f, of large absolute value, must lie within some sector Tk. The condition on T is just 0 < T < infk IAd/2. On the other hand, the definition of Sk depends on e, hence the value ro of this lemma depends on e. We proceed to study what happens in the sector Tk •
3.2.9. Lemma. Let A be the set of indices j such that Wj ELk. For have fez) = LAjzmj(1 + Bj(z»eWjZ , jElk
Z E
Tko we
3.2. Distributions of Zeros of an Exponential Polynomial
223
with lim £j(z) =0.
zET,
z~oo
Proof. This is a corollary of Lemma 3.2.5(b) and (c), the functions £j are the same as in (*) for j =I k, k + 1, those with these two indices will change. D In order to continue our study of zeros, it is convenient to introduce, for j E i k , the following quantities: 1:j
= Wj + mjei1/l"
which both depend on the frequencies and the degrees of the corresponding coefficients. To simplify the notation we rotate the coordinates and assume 1/Ik = 11/2. Let Qk denote the convex hull of the points Wb Wk+!, and 1:j, j E ik' We choose a new indexing of the vertices of Qb while traversing 8Qk in the positive sense. Start at Wk. We denote Wk,1 = Wb 1:k,l = Wk + imk (it could coincide with Wk,1 if mk 0), and denote the successive vertices 1:k,2, ... , 1:k,(71-1, 1:k.al = Wk+l + imk+l' Denote Wk.al := Wk+l' If ak > 2, all the mj that appear in 1:k,/ for 2 .::; I .::; ak - 1, are positive. We denote Wk,/ the corresponding projection onto the horizontal segment Lk = [Wb Wk+l]. The other Wj, j E iko are indexed arbitrarily as 1:k,/ = Wk.l + imk,l, with ak < I .::; nk. For 1 .::; j .::; ak-I. the segments [1:k,/, 1:k,/+)] are denoted Lk,I and the slopes f.Lk,l. Then
=
mk,l - mk,/+1
f.lk.l
= _
Wk,l - Wk,l+1
=
mk,l -
mk,l+1
Cik.1 - Cik,/+1
= Rewk.l = Rewk,l.
with
Cik.1
... >
Cik,a,.
Note that with our conventions Due to the convexity of Qko these slopes satisfy
ak,l
> CiU >
-cx:; < f.Lk,1 < f.Lk,2 < ... < J-Lk,a-l < 00.
Let z = x +iy, for H > 0 we define Uk•1 := {z E C: y 2: 0 and x + J-Lk.llog Izl > H}, Uk.l
:= {z E C: Y 2: 0, x + J-Lk,l-llog Izl < -H and x + f.Lk,llog Izl > H},
for I = 2, ... , ak
-
1, and
Uk.a, :=
{z E C: y 2: 0, x + J-Lk.a, log Izl < -H}.
Let us also introduce V k./
:=
V(O, f.Lk,/,
H) = {x E C: y 2: 0, Ix +J-Lk,I!oglzll.::; H}.
Figure 3.4 depicts these regions for 1 .::; I .::; ak - 1. The Uk,1 together with the V k .1 cover Tk (see Exercise 3.2.1). For R » 1, the different sets Vk,1 n B(O, Rr, Uk,! n B(O, r)', are connected and disjoint. Remark further that the "median" of Vk,1 is the curve x + J-Lk,/!og Izl = 0, which is asymptotic to the curve x + J-Lk./!Og Y = 0. We keep from now until Remark 3.2.22 the assumption that 1/Ik = 11/2 and the preceding notation.
3. Exponential Polynomials
224
Vk.3
Vk2
"tk.o,_1
"tk•O,
Figure 3.4
3.2.10. Lemma. There is v > 0 such that for every 1 and use the hypothesis that Lk,j-l when j = 2, . , , ,(Tk. We obtain mk.p -
Lk,p
Lk,p
¢
< -v.
Finally, if ak,p - ak,j = 0, we must have that mk,p < mk,j' Choosing v smaller if necessary, we have in any case, 1£(z)1 < Izl-v. (b) If j = 1, ... ,Uk - 1 and rk,p E Lk,j, then mk,p - mk,j = iJ-k,j (ak,p - ak,j)' On the other hand, we have x + iJ-10g Izi = 0 (with the same choice of iJas in (a» and x + iJ-k,j log Izl :::: H. Hence (iJ-k.j - fl-) log Izi :::: H and, since iJ-k,j - iJ- > 0, Iz I :::: exp(H /(iJ-k,j - iJ-», Because rk,p E Lk,j, we have ak,p .::: ak,j and therefore,
= Iz 111-' j (a, p-ct, j )-I1-(ct k p-ctk j) .::: eH(ct,p-akj)
=
e-HiaLP-akjl
so that
o
The other case is verified in exactly the same way.
3.2.12. Lemma. There exists r > 0 such that if z (a) Izm ,
(b)
jeW! jZ
I ::::
Izmk" ew"lz I;
E
Uk,j
n B(O, rY
then:
and
Izmkjew'.jZI :::: Izmk'"kewk,u,zl.
Proof. (a) If j = 1, the result is obvious. If j = 2, Lk, I E Lk,l, and we can apply (b) of the previous lemma with p = 1, j = 2. Hence
3. Exponential Polynomials
226
If 3 S j S ab then Lk,I ¢ Lk.j-I, and by Lemma 3.2.1l(a), we have zm, I e"'l.I z = £(z)zm"le""jZ, hence the inequality (a) holds if Izl» 1. The proof of (b) is similar, with Lk,rI, replacing Lk,l. 0
3.2.13. Lemma. If z
E
Uk,j n Tko then
f(z) = LAk,pzm,p(l +£k,p(z»e""p Z , where the sum is taken over the indices p such that: (i) Lk,p E Lk,I when j = 1; (ii) Lk,p E Lk,j n Lk,)-I when 2 S j S ak - 1; (iii) Lk,p E Lk,rI,-1 when j = ak; and, as always, lim £k,p(Z) = 0, Izl-H)O
ZEU, jnT,
Proof. It is an immediate consequence of Lemmas 3.2.9 and 3.2,Il(a).
0
3.2.14. Lemma. There are constants ro > 0, Ho > 0, such that if r ::: ro, and H ::: Ho, then for any z E Uk,j n Tk n B(O, r)f one has If(z)l> !IAk,jllzm,jeW,jZI.
Proof. We use the asymptotic development of f given in the previous lemma, and then use Lemma 3.2.11(b). We obtain If(z) I ::: IAk,jZm'j eW'j (1 = IAk,jZm, jeW, j x
+ £k,j (z»1 -
L IAk,pZm"p (l p,;,j
+ £k,p (z»e W, I pZ
I
(11 +£k,;(z)l- L
II~k'Pllll +£k,p(z)lzm"p-m'le(W,p-W,j)zJ) k,J
::: IAk,jZm'le"",jl {II +£k,j(z)l-
L II~k'Pllll +£k,p(Z)le-HIWl.p-Wkll} , ph
k,J
It is clear that by choosing r and H sufficiently big, the required estimate holds.
o
From now on we will assume r ::: ro ::: 1 and H ::: Ho, so that the conclusion of Lemma 3.2.14 holds. Note that the regions Tko and correspondingly the value ro, depend on the opening 9. On the other hand, we can take Ho independent of e. Namely, it is enough to choose it so that
"lax ,J
(E
IAk,pl e-HOlw"P-W"jl) S p,;,j IAk.j I
~,
3:2. Distributions of Zeros of an Exponential Polynomial
227
since we can take ro so that for Izl :::: ro, z E Tko we have
ISk,)(z)1 :::: ~ forallk,j.
3.2.15. Lemma. Let 0 < P, then
i"
< infk ,) IAk,) 1/2. If z E Uk,) n Tk, Iz I :::: ro, and ~ E le-~z f(z)1 > .,
Proof. If z E Uk,) Lemma 3,2.12(a) le-~z f(z)1
n Tk n Sk we have from the previous lemma and from :::: !IAk,jzmk,je"'k jZe-~ZI :::: !IAk,)zmk.le"'k,'Ze-~zl IAk,) II Imk 'I > . > -z -
2
= IA;,)IIZlmk.leRe«wk-~)Z)
'
where the previous to last inequality uses Lemma 3.2.4(a). If z E Uk,) n Tk n SHI, the reasoning is the same, with mk,! replaced by mk'''k'
o
3.2.16. Proposition. For r » 1,0 < i" < infk ,) IAk,)1/2, by switching the conditions ~ E P and z E (Uk,) Vk,jY n B(O, r)C, then le-~z
f(z)1 > .,
Hence, If(z)1 > uH(z), where H (z) is the supporting function of P. Proof. We fix some () > 0, as we have done above and let Tk = Tk «(}), Then, by Lemma 3.2.8, the inequality le-~z f(z)1 > i" holds if z E Sk, r :::: ro » 1. This value ro depends on the choice of (), but if we make it sufficiently large the conclusion of the previous Lemma 3.2.16 also holds for z E Uk,) n Tk for some k, j. On the other hand, if Z E Vk,j and Izl > rio then z E Tk; if z E Uk,) Vk,) either z does not belong to any Tko i.e., it belongs to some Si, or it belongs for Izl :::: r[, to some Uk,) n Tk. In both cases the inequality le-{Z f(z)1 whenever z rf. Uk,) Vk ,). Recalling that
>.
ma~Re(~z) ~EP
= max Re(wz) = max(wlz) = H(z), WEP WEP
concludes the proof of Proposition 3.2.16.
o
A consequence of the proposition is that all the zeros of f with large absolute value lie in Uk,j Vk,j' Note that because the Vk,j do not depend on e, the rl can be considered independent of e. The reader should verify that for those
228
3. Exponential Polynomials
segments Lb which contain no Wj in their interior, everything we did above works, it is even simpler. We also suggest comparing Proposition 3.2.16 with Corollary 3.1.27, we get more precision on the constants, and not only about local maxima of f, albeit only for r ~ rl. We shall presently study the behavior of f in one of the half-strips Vk.j.
3.2.17. Lemma. Let z (a)
If 'k,p f/
L~.j,
E
Vk,j' Then
then
for some function 8(Z) --. 0 as z --. rt Lk,j, then we also have
00
(z
E
Vk,j)'
(b) If'k,p
same condition on £ as that in the previous item, (c) If,,,p E Lk,j' then
Proof. Since z E Vk,j we have x + /lk,j log Izl = K, for some K such that IK I :s H. By Lemma 3,2.10, if 'k,p f/ Lk,j, we have Izm, p-m'.j e(w, p-OJ, J): I = Iz 1m, p-m, j eX (a, p-a, J)
= Izlm, p-m, j-J.l., j(a, p-ex, j)e(a, p-OI., j)K
This proves part (a). Part (b) is entirely analogous. If 'k,p E Lk,j then mk,p - mk,j = /lk,j(CXk,p - CXk,j), so that the previous computation shows that
o 3.2.18. Lemma. For r
»
1, one has:
(a) Izm"ew"zl:s Izm'iew'Jzl,ijl:sj :sak-l;and (b) Izm'u'ew,u"1 :s Izm, ieWkjZI, ij2:s j :s (J'k,
for every z E Vk,j n B(O, r)C, Proof. (a) If j = 1, the estimate is clear. If2:s j :s ak - 1, then we can use Lemma 3.2.17(a). (b) Similar proof, using Lemma 3.2,17(b).
3.2.19. Corollary. If z
E
fez) =
Vk.j
n Tb then
L fk,pELIt. j
A k • p(1
+ sk,p(z»zmk·peWk pZ,
'k,1
rt L k ,; and 0
3.2. Distributions of Zeros of an Exponential Polynomial
3.2.20. Lemma. For r
229
» 1, if t; E P and z E Vk,j n B(O, rY, then Izmk,je(wkJ-nzl:::: 1.
Proof If z
E Vk,j
Similarly, if
Z E
n Sk use Lemmas 3.2.18 and 3.2.4 to conclude that Izm!,j errol j-I;)z I :::: Izl m !" :::: 1. Vk,j
n Sk+I, the same argument holds with
mk,l
by mk,lJk'
0
3.2.21. Lemma. Let z f(z)
replaced
E Vk,j
= zmk"eWk.j
n Tb
L
then
Ak,p(l +£k,p(z»e(WkP-Wkj)(Z+/LkjLogz),
fJ..,pELk,j
where hp -+ 0 as z -+ 00 within Yj,k value of the logarithm.
n Tb
and
Log z represents the principal
Proof When
Lk,p E Lk,j, we have mk,p - mk,j = JLk,j(Wk,p - Wk,j)' We then 0 apply Corollary 3.2.19, and we are done.
3.2.22. Remark. If we do not make the assumption that 1{!k = 71: /2, and hence that Im(wk,p - Wk,j) = 0, then we can introduce CPk = 1{!k - 71:/2 in the expansion of f as follows:
2::
f(z)=zm'jeWk,jZ
A k,p(I+£k,p(z»
f1c,p EL k,j
x exp[ei'l" (Wk,p - Wk,j)(ze-i'l"
+ JLk,j logz)].
In this case, ei'l'k(wk.p - Wk,j) E JR, and f becomes, up to a multiplicative factor, a sum of exponentials in the variable ze-i'l'k + /-Lk.j log z, with coefficients almost constant. It also follows that the Vk,j coincide with V(cpt, JLk,j, H). In fact, the main relation to remember is that 1{!k = Arg(wk+l - wd - 71: /2, so that one has CPk = Arg(wk+1 - Wk) - 71:, hence CPk = - Arg(wk,p - Wk.j) (modulo 71:).
=
= +
3.2.23. Proposition. Let g(z) 2:1::;j::;n AjeWjZ , z x iy, Aj "# 0, Wj E JR, WI < W2 < ... < Wn • For H > 0, h > 0, and YI E R consider the rectangles R := {z E C: Ixl :::: H, Iy - YII :::: hI. There is a constant Ho :::: 0, independent of YI and h, such that every zero of g lies in the strip Ixl :::: Ho. Moreover. if H :::: Ho. the number N (R) of zeros of g in the rectangle R satisfies
/N(R) -
~(wn - WI)/ :::: n -
1.
G(z) = (l/AI)e-W1Zg(z) = 1 + 2:2::;j::;n Bje yjz • Bj = AdAI. and Yj = Wj - WI. 0 < Y2 < ... < Yn' Both functions G and g have the same zeros. For Re z = x > 0 we have
Proof Let
3. Exponential Polynomials
230
with lim.l->oo !31 (z) = O. Hence, if x » 1 we have II + !31 (z) I ::: ~, and G cannot have any zeros. Similarly, G(z) = (1 + !32(Z», where !32(Z) -')0 0 as x -')0 -00. Hence, there is Ho > 0 such that if IRe zl > Ho, G(z) #- O. Clearly, if H ::: Ho the number NCR) is independent of H. Let us fix £ > O. Choose H ::: Ho such that ne and I Arg(l + !3j(z))l < 4 on the lines x = ±H, j = 1,2. For a given h > 0, YI E R let us choose 8> 0, 0 < 8 < min{h, ne/2Yn}, with property that G has no zeros on the boundaries of the rectangles RI := [- H, H] X [YI - h - 8, YI + h + 8] and R2 := [-H, H] x [YI - h + 8, YI + h - 8]. This is possible because the zeros of G are isolated. Let us investigate now the variation of the argument of G along aRI. In the vertical portion x = - H, it is at most n e /2. On x = H, it is at most ne/2 + (2h + 28)Yn. On the horizontal sides, the function Re G(z) has the form Re G(z) = 1+
L
EjeYjX,
2sjSn
On the other hand, by Rolle's theorem, a function of the form K (x) = LI:::J:om Fje aJ '\, Fj E R*, al < a2 < ... < am, can have at most m - 1 real zeros (counting multiplicities). We show this by induction on the number of frequencies. It is clear for m = 1. If it is true for m - 1 frequencies, then we consider e- atX K(x) = L(x), L has the same number of zeros as K does. Its derivative L' is an exponential sum of the above type with m - 1 terms, hence L' has at most m - 2 zeros. Since between any two zeros of L there must be one of L', the claim is correct. Return to the function Re G(z) on a horizontal segment of aRlo We have that either Re G(z) == 0 on that segment or it has at most n - 1 zeros. In the first case, the variation of arg G along the segment is zero since G does not vanish on this segment. In the second case, the variation of arg G does not exceed ~ (n - 1)2n. Therefore, the total variation of arg G along aR I does not exceed ne
+ 2hYn + 28Yn + (n
-
1)2n.
Hence, by the argument principle N(R) ~ N(R I ) ~ n - 1 +
-hYn + e. n
By a similar reasoning hYn N(R)::: N(R2) ::: - n
n
+1-
e.
o
Since e > 0 was arbitrary, we obtain the proposition. The reader can easily verify that the example g(z) mate in Proposition 3.2.23 is optimal.
= 1+e
Z
shows the esti-
3.2. Distributions of Zeros of an Exponential Polynomial
231
We continue toward our goal of understanding the localization of the zeros of f E AEP. We are now trying to prove a sort of Lojasiewicz's inequality (see Proposition 3.1.30). We are still keeping the previous notation, i.e., 1/Ik = If /2. Let us prove a few technical observations about the regions Vk • j , 1 S j S ako which were stated at the beginning of this section. (I) If z = x + iy E Vk,j, then liml:l-+oo Iy/xl = 00. In fact, if J1.k.j = 0 the result is immediate. If J1.k,j :f= 0, let J1. = J1.k,j and let K (z) = x + 4J1.log(x 2 + yZ) E [-H. HI. Therefore, when Izl -+ 00 we also have Ix I -+ 00 and the sign of x must be the opposite to that of J1.. We can solve for y / x and obtain
(~r = x- 2 exp [2KJ1.(Z)
-
~]
The term 2K(z)/J1. is bounded, while -2x/J1. -+ Iy/xl -+ 00 as Izl -+ 00, Z E Vk,j' (2) If Z E Vk,j, then liml:l-+oo Arg Z = If /2. Namely, for large Izl we have 0 < Argz < be that Argz -+ If/2 when Izl-+ 00. (3) The curves x
+ J1.k.j log Izl = ±H
If
-1.
+00 as Izl -+ 00, hence
and, since Iy/xl -+
00,
it must
are a symptotic to the curves x
+
J1.k.j logy = ±H.
Let us show this for x + J1.log Izl = H, J1. = J1.k.j :f= 0, the case J1. = 0 being trivial. Let + (J1./2) log(x; + y2) = H and X2 + (J1./2) log y = H, then we have
Xl
1J1.1 ( 1+ IXI-x21=Tlog since XI/Y -+ 0 as Izi -+
00,
(Xly )2)
-+0
by the previous item (2).
(4) For 2 E Vk.j denote W = z + J1. Logz, J1. = J1.k.j, Logz the principal branch of the logarithm. For 8 > 0, there is rl ::: I such that if IZI - 221 ::: 8, and also zj, z2 E Vk. j n B(O, rr)c, then IWI - w21 ::: 8/2. Let Yj = Imzj. Vj = 1m Wj. If IYI - Y21 ::: 8, then we can use that by (2) VI - V2
= YI
- Y2
+ J1.(ArgzI
- Arg z2) -+ YI - Y2,
as Izl -+ 00, to conclude that IWI - w21 ::: IVI - v21 ::: 8/2 if rl If IYI - Y21 < 8, let q = YI - Y2. We have
:!. = Z2
YI Y2
(Xl + YI
i)
(X2 Y2
+
i) -I = (I + !L) (Xl + i) Y2
YI
In this case, we have XI/YI -+ 0, X2/Y2 -+ 0 and Y2 -+ ZI/22 -+ 1. Therefore, WI -
W2
= 2\
- 22
+ J1.(LogzI
- Logz2) = ZI - Z2
»
(X2 Y2
1.
+ i)-I
00 as rr -+ 00, hence
+ J1. Log (;~)
232
3. Exponential Polynomials
and Log(zl/z2) --+ 0 as r1 --+ IWI - w21 ;::: 8/2 if rl » 1.
00.
Hence, in this case, we can also conclude that
E AEP, Z = Z(n = (z E IC: Izl > R, fez) = O}, and H the supporting function of the P6lya polygon P of f. For any 8 > 0 there exist T > 0, r > 0, such that for Izl > r if d(z, Z) ;::: 8, then
3.2.24. Proposition. Let f
If(z)1 ;:::
reH(z).
Proof. From Proposition 3.2.16 we know the existence of Ho > 0 and TO > 0 such that if we fix H ;::: Ho there is ro > 0, so that Iz I > ro, z if Uk,j VA,j, implies that If(z)1 ;::: ToeH(z). Therefore, it is enough to consider z E Vk,j' Izl --+ 00. Following Remark 3,2.22, we can expand f as follows: fez)
L
= zm, jeW' jZ
Ak,p(1
+ Bk.p(Z»
T/.. pEL/.. j
x exp[ei'l" (Wk,p - Wk,j)(ze-i'l"
+ f.1k.j log z)],
and recall that the "frequencies" ei'l" (Wk,p - Wk.j) are real. Since, by Lemma 3,2.20 we know that for I'; E P, z E Vk,j, and Izl has Iz m , j e(Wk j-~)Z I ;::: I, we can just consider
»
1, one
Then we can change variables and study the sum gl(W)
=
L
Ak,p(l
+ 11k,p(w»exp({J/..,pw),
fl.,pELk,)
where W = ze-i'l'k + f.1k.j log Z, {Jk,p = ei'l" (Wk,p - Wk,j), and 11k,p (w) = £k,p (z), Clearly, 11k,p(W) --+ 0 as Iwl--+ 00, and if d(z, Z(f» ;::: 8, we have dew, Z(gl» ;::: 8/2 as long as Izl » 1. Here Z(n = (z: Izl > R, fez) = OJ, Z(gd = (w: gl (W) = OJ, the variables z, W being restricted to z E Vk,j' Izl > R, and W to the corresponding region. Note that as a consequence, I Re wi ~ Hand 1m W --+ 00, as z --+ 00, We are thus led to introduce the following exponential sum, with real frequencies:
where the summation takes place over the same set of indices as above. From what we have just said, if z --+ 00 in Vk,j, then gl (w) - go(w) --+ 0, Using Remark 3,1.23, we conclude that if Z(go) = (w E C: go(w) 0< 80 ~ 1, then dew, Z(go» ;::: 80 and I Rewl ~ H implies Ig(w) I
;::: c80
= OJ,
and
3.2. Distributions of Zeros of an Exponential Polynomial
233
for some c > 0, v :::: I. Therefore. there is r :::: 1 such that Iwl :::: r, I Re wi ::: H, and dew, Z(go» :::: 80 • then C
Ig) (w)1 ::::
280'
In fact, it is enough to choose r so that IgI(w) - go(w)1 ::: (c/2)8 0 for Iwl :::: r. As a consequence, we see that if WI E Z(gl), IRe wII ::: H, IWII :::: r, we must have d(w), Z(go» ::: 80 . Hence, if Iwl:::: r, I Rewl :::: r, and dew, Z(gl» :::: 8/2 we will have dew, Z(go» :::: 8/2 - 80 , Take 80 = 8/4, then 8/2 - 80 = 80, so that dew, Z(go» :::: 80, and thus 1!t(z)1 = IgI(w)1 ::::
~ (~r.
for z E Vk • j • Izl» 1, and d(z, Zen) :::: 8 (since this would imply that dew, Z(gl» :::: 8/2). It follows that:
If(z)e-~;;I
=
Izm'je(wkj-~)zfl(z)l:::: ~ (~r
for z E Vk,j. Izi » 1, d(z, Z(f» :::: 8, and ~ E Proposition 3.2.24.
P.
This concludes the proof of 0
Let us define. with the preceding notations for a function
f
E
AEP, ex > 0.
s > 0, H > 0, R.,j (ex, S, H) :=(z E C : Im(ze-i 0 there is ex (e) > 0 such that ifex :::: ex(e). s > 0,
IN(Rk,j(ex. s. H» - 2:
IWk,j+1 -
Wk,jll ::: nk,j - 1 + e,
where nk,j is the number of frequencies lying in the segment Lk,j [Wk,j, Wk,j+!l.
=
Proof. The statement (a) is a consequence of Proposition 3.2.16. The proof of (b) is a refinement of the proof of Proposition 3.2,24. Using the notation introduced
3. Exponential Polynomials
234
there, let f3k.p = e irp , (Wk,p - Wk,j) gk,j (w)
E
JR, and
2:
=
Ak,pe P, pW.
lk,pELk,j
°
Assume H > so that the conclusion of Proposition 3.2.23 holds for every gk,j' Note that the quantity If3k,HI - f3k,j I = IWk,HI - Wk,j I is the distance between the furthermost frequencies of gk,j' Choosing rI » I we can guarantee that for different choices of (k, j) the sets V (cpt, J-Lk,j, H + 1) n B(O, r])C are disjoint. (Recall Vk,j = V (cpt, J-Lk,j, H).) Note that rI can also be chosen so that for z E V(cpt, J-Lk,j, H + I) n B(O, rI)C we can take arg Z E [CPk, CPk + rr] and that for a convenient choice of ao Im(ze- iCP ,)
+ J-Lk,j argz :::: ao >
0.
Therefore, for a :::: ao, the regions Rk,j(a, s, H) and Rk,j(a, s, H + 1) cover Vk,j n B(O, rIY and V (cpt, J-Lk,j, H + 1) n B(O, r])C, respectively. Once we fix 8 E ]0, rr/2[ we can also choose aI » 1 so that for a :::: aI, S > 0, Rk,j(a, s, H
+ 1) £
Q := V(cpt, J-Lk,j, H
+
1)
n Tk (8) n B(O, rd c .
From Remark 3.2.22 we have fez)
2:
= zm, jeW,,) 'Ct
Ak,p(1
+ ek,p(Z» exp[f3k,p(ze- irp, + J-Lk.j logz)]
pEL" j
=
in the region Q (arg z E [cpt. CPk + rr]). The transformation Fk,j: z 1--+ W + iv := ze- irp, + J-Lk,j logz maps Q conformally into the half-strip lui:::::: H + 1, v :::: ao. The regions Rk,j(a, s, H + 1) are mapped conform ally onto the rectangles U
R(a,s, H
+ 8) = [-H -8, H +8]
+i[a,a +s],
for a :::: ab 8 > 0, 0:::::: 8 :::::: 1. Let fk,j(Z):= f(z)z-m'je-w'jZ and hk,j be defined by hk,joFk.j = fk,j' Hence hk,j(w) = gk,/W) + l1k,j(W), with l1k,j(W) -+ as lui:::::: H + 1 and v -+ 00. We clearly have for a :::: ab s > 0, 8 :::::: 1, that
°: :
Nj(Rk,j(a, s, H
+ 8»
°
= Nj,/Rk,j(a, s, H + 8» = N h, j(R(a, s, H + 8»,
where the index of N indicates the function whose zeros we are counting. Moreover, these values are independent of 8, since all the zeros of f lie in Vk,j' It follows from Proposition 3,2.23, applied to the gk,j, that there is a constant M > 0 such that for any a E JR, < 8 :::::: 1,
°
Ngkj(R(a,
8, H
+ 8)) :::::: M.
Hence, for any a, there is a horizontal line segment A, in R(a, s, H that dew, Z(gk,j» :::: 2(M
+ 1)
+ 8)
such
3.2. Distributions of Zeros of an Exponential Polynomial
235
for any point W EA. (The line segment A depends clearly on 8, ar, and (k, j).) It is enough to divide the rectangle in parallel horizontal strips of equal width 1/(M + 1), and take as A the bisectrix of a strip void of zeros of gk.j' Fix (k, j), s > 0, and ar ~ arl + 1, let Rex,s be the rectangle whose vertical sides lie in u = ±(H + 8) and whose horizontal sides are given
by the line segments Al and A2 corresponding to 'R(ar - 8,8, H + 8) and + s, 8, H + 8), respectively. We have that ROI,s 2 'R(ar, s, H) and that, for every point W E a ROI,s one has
'R(a
8
dew, Z(gk.j» ~ 2(M From 3.1.3 we conclude that there is r
+ 1)
= r(8)
Igk.j(w)1 ~ r
> 0 such that
on aROI,s.
If v ~ vo» 1, we have IYJk,j(w)1 :5 r/2, hence, for ar ~ ar2 ~ arl
+ 1, we have
Ng,/ROI,s) = Nh, ,(ROI,s)'
We remark that the choice of ar2 depends on 8, and eventually on E. The height of ROI .s lies between sand s + 28, and now from Proposition 3.2.23 we infer that
lN g,
J
SI
H, and as soon as ar » I, the number Nj('Rk,j(ar, s, H» = NrC'Rk.j(ar,s, HI»' (2) Taking E = 1 in the statement of Theorem 3.2.25, one gets the bound INjC'Rk,j(ar, s, H» -
for every s > 0, once a
»
1.
2:
IWk,j+1 - Wk,j1i :5 nk,j
236
3. Exponential Polynomials
=!.
Taking £ for those s such that (s/2rr)lwk.j+l -wk,jl sharper bound nk.) - 1 for the above quantity. In any case. we get
E
N* one gets the
We shall give below some consequences of the P6lya-Dickson theorem. 3.2.27. Lemma. Let fez) = Ll:: 0, let
'R~.p(a, s, H)
= (z E
V~,p: Im(ze-i 0 we can continue to the point we find qo, ... , qs-' E F, rs E F, and either rs = 0 or 81R (rs ) < W n , and one can write W~f
= (W~-lqO + w~-2ql + ... + qs-,)g + r s,
which proves the lemma in the case UI = W, = O.
3.2. Distributions of Zeros of an Exponential Polynomial
239
Let us now consider the case of two arbitrary values UI, WI. The functions F(z) := f(z)e-U\Z, G(z) := g(z)e-W\Z, are now in the previous situation. Hence, there are 1/1 E 6, q, rEF, such that either r = 0 or 8~(r) < 8IR(G), and
1/IF
= qG + r.
It follows that
1/I(z)f(z)
= [q(z)e(U\-w,)Z]g(z) + eU\Zr(z).
Clearly, either eUtZr = 0 or 81R(eU\Zr) = 8IR(r) < 81R(G) = 8IR(g).
0
3.2.31. Theorem. Let f, gEE, g =1= 0, and assume the coefficients of g are relatively prime. If there is R :::: 0, and an angular sector S of opening strictly bigger than T( such that fig is holomorphic in S n (B(O, R)Y, then there is h E E such that f = gh. Proof. Rotating coordinates if necessary, we can assume that the sector is S = {z E C: - 17 < Arg z < T(}, with 0 < 17 < T(. The proof proceeds in three stages. First, we shall show that if f / g is holomorphic in {1m z > 0, Izi > R}, then fig = q /1/1, with q, 1/1 E E, and 1/1 has only purely imaginary frequencies (i.e., 1/1 E 6* in the notation of the previous lemma.) We show later that q /1/1 being holomorphic in {z E R, Imz > 0, then there are q E E, X E 6* such that xf = gq. Proof of Lemma 3.2.32. Replacing z by z + i R we can assume f / g is holomorphic in 1m z > O. This transformation is compatible with the condition 1/1 E 6*. We show first the following:
£g is holomorphic in Imz > 0, then 81R(g) .:::: 81R(/)' In fact, let 8R(g) = Wn - W], 8R(f) = up - UI as in Lemma 3.2.30. The convexity of peg) and P(f) tells us that they have vertices PI, ... , Pn (resp. aI, ... , ap ), such that Re Pn = Wn > ... > Re PI = WI (resp. Re ap = up > If
... > Real = UI), hence these vertices span the upper part of peg) and P(f), respectively. For 1 .:::: k .:::: n, let Ok be the line through 0, with the direction of the outer normal to peg) along [Pb Pk+d. Let 1/Ik be the angle it makes with the positive real axis, then 0 < 1/Ik < T(. Choose (J > 0 such that 0 < 1/Ik - (J < 1/Ik + < T(. Let Tk «(J) be the sector of bisectrix Ok and opening 2(). Since fig is holomorphic in Tk«(}), we can apply Proposition 3.2.28 to conclude there is a side [aj" ajl+t] of pC!) which is parallel to LBko .Bk+JJ and whose length laA+t - ah I :::: IPk+l Pkl. Clearly, the length of the respective prOjections on the real axis preserves
e
3. Exponential Polynomials
240
Figure 3.5 the same size relation, let us say PIII.[t3k.t3HIJ ~ pIII.[aj"ajk+,J. Therefore, 8I11.(g) =
L
PIII.[t3b
PHd ~ ~
L
pIII.[aj" aj,+,]
L
pIII.[aj, aj+d
= 8R(f).
I~j~p
Appealing now to the preceding Lemma 3.2.30, we infer there are X E 6*, and exponential polynomials q and r such that
xl
= qg+r,
and, either r = 0, or 8I11.(r) < 8I11.(g). We can rewrite the previous equation as r I -=X--q, g g
which shows that r / g is holomorphic in {1m z > OJ. From what we have just shown we must have We conclude that r = 0. Hence,
xl =
qg,
o
as we claimed.
3.2.33. Lemma. Let h E E, 1/f E 6*, and assume there are 1], 0 < 1] < ']'(, and R > 0, such that h/1/f is holomorphic in {z E C: Izl > R, I Argzl < 1]}. Then there are b E C[z]*, q E E, such that bh
= q1/f.
3.2. Distributions of Zeros of an Exponential Polynomial
241
Proof of Lemma 3.2.33. Let l/f(z) = b l (z)eii'\z + ... + bn(z)eiA,Z, Al < A2 < ... < An, bj E (:[z]. The P61ya polygon P(l/f) is a segment of the imaginary axis, [-iAn, -iAIl. Proposition 3.2.28 states that there is a side [ab ak+Il of the P6lya polygon P(h) which is parallel to the imaginary axis and lak+1 - akl ::: IAn Ad. Let us change to new variables, ~ = iz, h'(~) = h(z), l/f'(~) = l/f(z). In this case, IAn - Ad = 8JR(l/f*) and lak+1 ~ akl :s 8JR(h*), hence 8JR(l/f*) :s 8R(h*).
Copying tbe proof of Lemma 3.2.30, we can show there is an s E N* and two exponential polynomials q*(n, r*(~), such that b~(-inh'(~)
= q*(~)l/f*(n
+ r*(n
and, either r* = 0, or 8JR(r*) < 8R(l/f*). Let r(z) = r*(iz). Then r(z)/l/f(z) is holomorphic in {z: Izl > R, I Argzl < 17}. From the first part of the proof we can conclude that 8JR(l/f*) :s 8JR(r*). This shows that r* O. Hence,
=
b~(z)h(z)
3.2.34. Lemma. Let f, g, h
E and c(f), c(g), and c(h), respectively, the
E
g.c.d. aftheir coefficients. If fg
o
= q*(iz)l/f(z).
= h, then c(f)c(g) = c(h).
=
=
Proofof Lemma 3.2.34. Let fez) LI~i~m Ai(z)e"'; g(z) L1:5j~n Bj(z)e PJ ; and h(z) = LI~k~p Ck(z)e YkZ • We can assume that al -< ... -< am, fiI -< ... -< f3n, and YI -< ... -< Yp· Consider first the case where c(f) = c(g) = I. We want to show that then the Ck must be coprime, i.e., c(h) = I. Let WI, .•. , Wq be the different values taken by ai + f3j, 1 :s i :s m, 1 :s j :s
n. Assume
WI
-
oo
2]I"
Proof. This follows from properties of functions of completely regular growth. See [Lev, p. 288]. 0 Another way to express this corollary is that
n(V, zo, r) =
e(p)
~r
+ 0(1),
so that, in particular, for every Zo there is a constant C :::: 0 such that
n(V, zo, r)
e(p)
~ ~r
+ C,
which is due originally to P61ya. In [Tu] it is shown that if f is an exponential polynomial one can prove that the inequality holds with a constant independent of zoo Tunln also showed that such an inequality has many applications in number theory and elsewhere. The original estimate of Tunin has been improved by Tijdeman [Ti], Waldschmidt [Wa 1], and others. By a very ingenious argument, generalizing Rolle's theorem to holomorphic functions, Voorhoeve [Voor] obtained the following sharpening of Turan's estimate for the number of zeros of an exponential polynomial f: n(V, Zo,
r) ~ inr + 2(JOf, ]I"
where n = max{IO!jl: obtained
O!j
frequency of
n(V, Zo, r)
fl.
Using a geometric argument, he also
e(p)
~ ~r
+ vdof,
where v denotes the number of vertices of the P61ya polygon. Let us start by recalling some notation. If f is a meromorphic function in an open set U of the complex plane, f ¢. 0, then for Zo E U, v(zo) = v(zo, f) is the index of the first nonvanishing coefficients of the Laurent development of f about Zo, i.e., if V(zo) < 0 then Zo is a pole of f and -v(zo) is the order of this pole, if v(zo) > 0, then it is the order of Zo as a zero of f, finally, if V(Zo) = 0, then Zo is neither a zero nor a pole of f. Following [Voor] we introduce a generalization of the total variation of the argument of f along a closed interval [a, b) of the real axis.
3.2. Distributions of Zeros of an Exponential Polynomial
245
3.2.38. Definition. Let f 1= 0 be a merom orphic function in neighborhood of [a, b] and Xl < X2 < ... < Xm the collection of zeros and poles of f in la, b[. We set A(a, b, f):=
lb Im(f'(~)lf(~»1 d~ I
L
+JT{
IV(Xbf)I+4Iv(a,f)1+4IV(b,f)I}.
l:,:k:,:m
Let V be a complex neighborhood of [a, b], f E M(V), and [a, b) £ ]C, d[ c c V, then if we denote by ~ I, ... the collection of zeros and poles of f in ]C, they are the only poles of 1'(z)lf(z) on ]e, and there is a domain V, ]e, d[ £ V £ V and a function g E Je(V) such that
,;n
dr.
1'~)
L
=
f()
V;~~'
l:,:j:,:n
f) ~J
dr.
+ g(z)
(z E V).
On the real axis the sum is real valued, hence, Im(f'(~)lf(~»
=
Img(~),
and we conclude that Im(f' If) can be considered as a continuous function in ]C, dr. In particular, A(a, b, f) is well defined. It is also clear that A is additive with respect to intervals, i.e., if e < r < s < t < d, then A Cr. t, f)
=
ACr, s, f)
+ A(s, t,
f).
Another useful property is the following. For YJ real, IYJI« 1, the function fry(z) := fez + iYJ) is meromorphic in a fixed complex neighborhood, still denoted V, of ]e, dr. Moreover, we can assume that for 0"# YJ, fry has no zeros or poles on the real axis, so that ACa. b, fry)
=
lb Im(f~(~)1fry(~»1 d~. I
3.2.39. Lemma. lim A(a, b, fry) = A(a, b, f).
ry--+O
Proof. By the additivity of the functional A we can assume that or poles in ]a, b[, so that for some YJo > 0 we have
1'(z) ex f( z) = z-a
ex
= v(a, f), f3 = v(b, f),
and g holomorphic in
a-Y/o::::Rez::::b+y/o,
Then, for
~ E
f3
+ --b +g(z), z-
JR, exYJ
IImzl::::YJo.
f has no zeros
246
3. Exponential Polynomials
For 0 < c « 1, a < ~ < a + c, the last two tenns are bounded for all values of rJ, 0 < IrJl < rJo· So that
+ c, fry) =
A(a, a
l
aH
I Im(f~(~)lfry(~»1 d~
1
!J~
a +o
=O(c)+larJl
rJ2+(~-a)2
a
= O(c) + lal arctan(c/lrJl). Similarly,
+ 1.81 arctan(c/ll7l).
A(b - c, b, fry) = O(c)
Since
= Img(o, we also have
Im(f'(~)lf(~»
1 IIm(f'(~)lf(~»ld~ 1 b
b e -
=
a
a+e
IImg(~)ld~ + O(c).
Therefore, we are led to consider
1 1 b-o'lm(f~(nlf~(~))! -lb-& IImg(~)ld~1 a+&
a+e
:: 1 b
8
-
I
a+8
Im(j~(~)1f~(~»
1
+
1 b
-
a+8
8
rJ2 +
(~ -
g(~)1 d~
1
d~
b- e
::::: larJl a+e
- 1m
d~
h- 8
a)2 + l.8rJl aH
I Im(g(~ + irJ)
-
rJ2 +
(~ -
b)2
g(~»1 d~
We infer that IA(a, b, f~) - A(a, b, j)1 :::::O(c) + O(lrJD
+ (lal + If we let IrJl
+ O(lrJl/c 2 )
I.8DI arctan(e/l rJ I) - n,/21.
= O(e 3 ), then all the tenns above are O(c), which proves that lim A(a, b,
~->o
f~)
o
= A(a, b, j).
3.2.40. Corollary. Let f, g be two nontrivial meromorphic functions on [a, b], and let k E Z, then:
(i) IA(a, b, j) - A(a, b, g)1 ::::: A(a, b, fg) ::::: A(a, b, j) + A(a, b, g); (ii) A(a, b, fk) = IkIA(a, b, j).
Proof. For 0 < rJ « 1 we have that neither in [a, b]. Moreover,
f~
nor
g~
have any zeros or poles
3.2. Distributions of Zeros of an Exponential Polynomial
247
and
(f;}'fl; = k(f~/I~), hence the identity (ii) is clear for I~ instead of I, and the estimates (i) are a consequence of the triangle inequality for I~, g~. Using the continuity at 1'/ = 0 proved in the previous lemma, we conclude the corollary holds. 0
3.2.41. Proposition. Let I be a meromorphiclunction on [a, b) such that neither I nor f' have zeros or poles at the endpoints 01 the interval. Then A(a, b, f) ::: A(a, b, I') where e(z):= I Arg(f'(z)/I(z»1 argument of ~ ).
(Arg~
+ B(a) -
B(b),
is, as usual, the principal value
01 the
Proof. Note first that I Arg ~ I is a continuous function in C* with values in [0, rr]. Hence, 19 is a continuous function in a neighborhood of a and of b. Moreover, for small T} real, e~(z) := I Arg(f'(z)/I~(z»1 = e(z
+ i1'/),
so that e~(a)
e~(b)
--+ e(a),
--+ e(b)
as
1'/ --+ O.
Therefore, by Lemma 3.2.39, if we can prove the inequality for I~, 0 < 1'/ « 1, it follows for I. Therefore, we assume that I and f' have neither zeros nor poles on [a. b). In other words, we can assume that f' /1 is a holomorphic function on [a, b]. and thus Im(f'/f) has only finitely many zeros on [a, b]. Let a < XI < ... < xn < b be the zeros ofIm(f' If) in la, b[, denote a = Xo, b = Xn+l, which could be zeros of Im(f'/f) or could be not. Whenever Im(f'(~)/I(~» :::: 0 we have Arg(f'(O/I(~»::::O and conversely, so that in any interval]xt,Xk+l[ we have with
ak
= ±1, and
for the sarne value
ak.
Consequently, for
~ E ]Xk, Xk+I[,
~ ee~) = ak :~ Im(Log(f'(~)/I(~)))
(:~ LOg(f'(~)/I(;»)
=ak1m =
ak
Im(f"en/I'(;) - I'(~)/f(~».
Using this identity we obtain,
l
X, + I
x"
I Im(f'(~)/f(~»1 d~ = ak
l
x' + 1 Im(f'(~)/f(~»d~
Xk
3. Exponential Polynomials
248
1 Xk
=ak
+J
~
Im(j"(~)/f'(~»d~-
lx'+J q
d
-8(~)d~ d~
so that A(a, b, f) =
=
~ 1:k+1 I Im(j'(~)/f(~»1 d~
ta 1:'+1 adm(j"(~)/f'(~» d~
~ A(a, b,
1') + 8(a) -
- (8(b) - 8(a»
o
8(b).
3.2.42. Corollary. Let f be meromorphic and not constant on [a, b], then
°
A(a, b, f) ~ A(a. b, f')
Proof. For < T} « 1, the functions From the previous proposition A(a, b, f~) ~ A(a, b, f~)
since
°
~ 8~(~) ~
n. Letting
T} ~
f~, f~
+ 8~(a) -
°
+ n.
have no zeros or poles at a and h. 8~(b) ~ A(a, b, f~)
we obtain the corollary.
+ n,
o
Corollary 3.2.42 is the key to good estimates. It allows us to estimate the variation of the argument of solutions of differential equations, as shown by the following result:
3.2.43. Proposition. Let f he a nontrivial meromorphic function on [a, b] satisfying a nontrivial differential equation of the form
!!... [1/In!!... dz dz
[...
!!...
dz
[1/I1!!... f] ... ]] dz
= 0,
where 1/110"" 1/In are meromorphicfunctions on [a, b]. (For simplicity we shall drop the brackets in the future applications of this proposition.) Then A(a, b, f) ~
L
A(a, b, 1/Id
+ nn.
l:::k::5n
Proof. If n = 0, i.e., df/dz = 0, then f == c =1= 0, and it is easy to see that A(a, b, c) = 0 (see Exercise 3.2.3). The right-hand side is also zero in this case. For n ::: 1, let g = 1/11 f'. If g == 0 then we are in the previous case, so we can assume g is nontrivial. We can apply induction to obtain A(a, h, g) ~
L 2skSn
A(a, b, 1/Ik)
+ (n -
l)n.
3.2. Distributions of Zeros of an Exponential Polynomial
249
On the other hand, the previous corollary and Corollary 3.2.40(i) and (ii) yield
+ n = A(a, b, g1/f j l) + n A(a, b, g) + A(a, b, 1/fjl) + n A(a, b, g) + A(a, b, 1/f1) + n
A(a, b, f) ~ A(a, b, 1') ~
=
~
L
o
A(a, b, 1/fk) + nn.
One can prove, using this proposition, that if rpl, ... ,rpm are linearly independent meromorphic functions on [a, b], and CI, ... ,Cm are arbitrary constants such that f = 2:: c)rp) ¢:. 0, then an upper bound for A(a, b, f) is independent of the values of the constants. The same holds for an upper bound of the total number of zeros and poles of f on [a, b] (see Exercise 3.2.8). We are now ready to go back to exponential polynomials. We recall from Section 3.1 that we denote
L
fez) =
p)(z)eC(jZ,
1:S):Sn
3.2.44. Proposition. Let a < b, N (a, b, f) be the number of zeros (counted with multiplicity) of the exponential polynomial f in the segment ]a, b[. Set J:= maxIma).
/ := minIma), J
J
Then (b - a)/ = n(dOf - N(a, b, f)
~
lb
~
(b - a)J
Im(f'(x)/f(x» dx
+ n(dOf -
N(a, b, f).
Proof. We can assume the indices have been chosen so that
It is easy to prove that
where D = d/dz (see Exercise 3.2.6 and the proof of Lemma 3.1.2). We are using the notation from the statement of Proposition 3.2.43. Further, one can write D2 = DID, D3 = DIDID, ... , where 1 is the constant function 1 acting as a multiplication operator. Hence (*), has the form described in Proposition 3.2.43, applied to ! (x )e-a,X, where we have a total dO! functions 1/fko some
250
3. Exponential Polynomials
1/Ik == 1, and the others 1/Ik(X) = exp«ak - ak+')x). From Proposition 3.2.43 we conclude that
A(a, b, !(x)e- Ci1X
):5 L
A(a, b, e(Ci.-Cik+il x )
+ ndO!,
l:sk:Sn-l
L
=
(b - a)1 Im(ak - ak+t>1
+ 77:dO!,
l:sk:Sn-'
since A(a, b, 1) = 0 and A(a, b, eWX ) = (b - a)1 Imwl (Exercise 3.2.3). Since 1m ak is increasing, we have
L
1Im(ak - ak+,)1 =
L
(Imak+l - Imak)
= 1m an and hence,
A(a, b, !(x)e- Ci1X ) :5 ndO!
Imal
+ (b -
=J -
/,
a)(J -I).
On the other hand,
lb
Im(f'(x)/!(x» dx
:51
b
1Im«f'(x)/!(x» - a,)1 dx
+ (b -
:5 A(a, b, !(x)e-a1X ) - n N(a, b, f) :5 ndO!
+ (b -
a)J -
77: N(a,
a) Imal
+ (b -
a)/
b, f),
where in the third inequality we used the definition of the functional A and the fact that N(a, b, f) = N(a, b, !(x)e- a1X ). This is half of the inequalities we wanted to prove. To prove the other half, we observe that it is also true that so that As earlier, we have
lb
Im(f'(x)/J(x» dx ::: (b - a) Iman ::: (b - a)J
-lb
1Im«f'(x)/!(x» - an)1 dx
+ 77: N(a, b, f) -
A(a, b, l(x)e-a,X),
which proves (b - a)/
+ nN(a, b, f) -
77:dO!
:51 Im(f'(x)/!(x»dx, b
o
as desired.
3.2.45. Corollary. The number N(a, b, f) o! zeros o! J in a closed interval [a, b] oJthe real line satisfies the inequality -
N(a, b, f) :5 dOl
+ 2n1 (b -
a)(J - I).
3.2. Distributions of Zeros of an Exponential Polynomial
°
Proof. For s > we have N(a, b, f) :'S N(a - s, b proposition we have (b - a + 2£)1
+ rrN(a -
s, b
+ s, f)
251
+ s,
f). From the previous
-rrdof
:'S (b - a
+ 2£)J + rrdof -
rr N(a - s, b
+ £,
f),
so that 2rr N(a, b, f) :'S 2rr N(a -
Letting
£ -+
£,
b
:'S 2rrdof + (b - a
+ £, f)
+ 2s)(J -I).
0+ we obtain the required inequality.
D
We can obtain from Proposition 3.2.44 inequalities holding in any line segment of the complex plane.
3.2.46. Corollary. Let a, bEe and denote by [a, b] the line segment joining these two distinct points. If f has no zeros on la, b[, then
1m
(1
f'(z) dZ) :'S rrdOf
[a.b)
f
(z)
+ max Im«b -
a)aj)
)
::::: rrdOf
+ max{lm(i(b -
a»: Z E P},
where P is the P6lya polygon of f. Proof. Let g(n = f (a
+ (b -
ag) =
L
p) (1;) exp«b -
a)aj1;),
'~j~n
for some new polynomials
h(n =
eaaj pj(a
+ (b -
a)l;),
of degrees mj as before. The exponential polynomial g satisfies dOf N(O, 1, g) = 0, and
r' g'(~)
Jo
g(n
d~ =
=
r'
Jo
= dOg,
!'(a + (b - a)~) d (b - a) f(a + (b - a)~) ~
r J[a,b]
!'(z) dz.
fez)
Hence, from the second inequality of Proposition 3.2.44 we infer that 1m
(1
[a,b]
Since P
ff'(Z) dZ) ::::: rrdOf
+ max{lm«b -
(z)
= cv{a, ... , an}, we have maxj{Im«b -
a)aj)}.
)
a)aj)} ::::: maxZEP 1m (i(b - a». D
As a consequence, we obtain an estimate of the number of zeros of f in any square with sides parallel to the axes.
252
3. Exponential Polynomials
3.2.47. Theorem. Let Q be the square
Q := (z
E
C: a
~
~
Re z
b, c
~
1m z
~
d)
and N Q the number of zeros of the exponential polynomial f in Q. Further, let t:.y:= max (Re(a, -aj)). IS.j:::n
Then, NQ ~ 2d °f
+ 2rr1 {(b -
a)t:.x
+ (c -
d)t:.y}.
Proof. For 0 < e« 1 we have that if Qe:= {z E C: a -e ~ Rez ~ b+e, c - e ~ 1m z ~ d + e}, then BQe contains no zeros of f. (If BQ contains no zeros of f we can let e = 0.) Rouche's theorem states that N
r
< N = _1_ f'(z) dz = _11m QQ, 2rri laQ, fez) 2rr
(r
/,(z) dZ) . laQ, fez)
We divide BQe into four segments and apply Corollary 3.2.46 to each of them. We obtain 1m
(1
/,(z) dZ) aQ, fez)
~ 4rrdOf + max (lm«b - a + 2e)aj» I:::j:::n
+ I:::j:::n max (Im«a -
b - 2e)aj»
+ l:::J:::n max (lm(i(d -
c - 2e)aj»
+ max (lm(i(c -
d - 2e)aj»
l:::J:::n
= 4rrdOf
+ (b
- a
+ 2e)Llx + (d -
c + 2e)t:.y.
o
It is clear that this inequality proves the theorem.
We now obtain the promised Tunin type inequality. Let, as earlier, denote by V = V(n = the multiplicity variety of f, let n(V, Zo, r) be the number of points of V in 8(zo, r), and let Q = maxj lajl. First, we state a result that the reader should compare with Proposition 3.2.23.
3.2.48. Corollary. If f has real frequencies, then the number Ns of zeros of f in the strip S := (z E C: c ~ 1m z ~ d) can be bounded by Ns
~
2dof
(c - d)
+ -2rr - - l:::i.j:::n max (a,
- aj).
Proof. It is easy to see that there is R > 0 such that f has no zeros in the region I Rezl > R. Let Q = (z E C, -R :5 Rez :5 R, c :5 Imz :5 d), and apply the previous theorem. 0
3.2. Distributions of Zeros of an Exponential Polynomial
3.2.49. Corollary. For any
Zo E
253
C, r > 0, we have
n(V, zo, r) ::::: 2dof
4Q
+ -r. 7r
Proof. Let Q = {z E C: Xo - r ::::: Re Z B(zo, r) ~ Q, so that
:::::
Xo
+ r, Yo -
r ::::: 1m Z
Yo
:::::
+ r}. Then
Clearly, Llx ::::: 2Q,
Lly ::::: 2Q,
hence, n(V, zo, r) ::::: 2dof
4Qr
o
+ -. 7r
Finally, we need a little geometric argument to obtain Voorhoeve's strengthening of the P6lya estimate involving the perimeter R(P) of the P6lya polygon of f.
3.2.50. Lemma. Let P and Q be convex polygons with exactly v vertices aI, ... , a v and f31, ... , f3v, respectively, where their indexing corresponds to the positive orientation of the respective boundaries. Let av+1 := aJ, f3v+1 := f31. Assume that 1
~
k
~
v.
Then,
L I~k~v
max{lm(z(ak+1 - ad)} = zeQ
L I~k~v
=
L
max{lm(z(f3k+1 zeP
.Bk»}
Im(.Bk(ak+1 - ad)
I~j~u
=L
Im(akCf3k+1 - f3k».
I~k~v
Proof. Let HQ be the supporting function of Q and evaluate it at the exterior unit normal nk to the side [f3k. f3k+d of Q, i.e., consider HQ(nd
= max Re(Znk) zeQ
as, in fact, the maximum is attained at any point of [f3k, f3k+I]. The condition Arg(ak+1 - ad = Arg(f3k+1 - f3k) indicates that
nk
=-
i(ak+1 - ak) . lak+1 - akl
254
3. Exponential Polynomials
Since H Q is a homogeneous function. we have HQ(-i(ak+l - ak»
= Re(-i(ak+l
= Re(-i(ak+l
- adPd
- ak)PHl)
= Im(Pk(ak+l - ak» = Im(Pk+l(ak+l - ak», and also Similarly, max Im(z(.Bk+l - f3d) = Im(ak(f3k+l - f3d) = Im(ak+l (f3kH - f3d). ZEP
Moreover, because of the periodicity of the indices,
L
L
Pk+lak+l =
Pkak,
and for a, bEe, Imiib = - 1m ba.
Consider now the expression we want to identify 'L....J " maxlm(z(aHl - ak» ZEQ
l~k~v
= '" L....J
Im(Pk(ak+l - ak»
l~k~v
L
=
Im(Pk+l (ak+1 - ak»
l~k~v
= 1m
(L
Pk+lak+1 -
l~k~v
= 1m
(L
Pkak -
I~k~v
L
= 1m (
L
Pk+lak)
l~k~v
L
PkHak)
I~k~v
(Pk - Pk+I)ak)
I~k~v
L
=
Im(ak(f3k+l - f3d)
l~k9
= '" L....J
max Im(z(f3k+1 - f3d).
l~k~v
ZEQ
This long chain of identities proves the lemma.
o
We remark that each sum in the previous lemma is invariant under translation of P or Q. e.g., for any Zo, Zl, Z2 E C we have
L l~k~v
max Im«z - zO)(ak+l - ak» ZEQ
=
L l~k~v
Im«ak - ZI)(f3k+l - 13k»
255
3.2. Distributions of Zeros of an Exponential Polynomial
3.2.51. Theorem. Let v be the number of vertices of the P61ya polygon P of the exponential polynomial f and V = V(f). Then, for any Zo E C l(P) n(V, Zo, r):::: 2rr r + v dOf·
Proof. If v = 1, fez) = e a ,! PI (z), dOf = ml = deg PI and l(P) = 0, so that n(V, zo, f) :::: dOf is best possible. If v = 2, then all the frequencies lie in a straight line in the complex plane. Moreover, considering f(z)e- Cl1Z we can assume that al = O. Rotate the variables, z = (an/lanl)~. Then, in the variable ~ we have an exponential polynomial with real frequencies, say 0 = WI < W2 < ... < W n • Then l(P) = 2wn and we can apply Corollary 3.2.48, which states that in any strip S of the form S = {~ E C: c :::: 1m ~ :::: d) we have ° (c - d) N~::::2df+~wn.
Clearly, B(zo, r) ~ S when c
= Imzo -
r, d
n(V, zo, r) :::: 2dof
= Imzo + r. Thus, l(P)
+ 2rr
r.
Assume now v ~ 3. Let us index the frequencies of f in such a way that ai, .... cl u are the vertices of P in the positive direction, we set av+1 = al. We can find a polygon Q, which circumscribes B(zo. r) and such that if PI, ... , Pv are the vertices of Q, then Arg(uk+1 - ad = Arg(pk+1 - Pk), with Pv+1 = PI. (See Figure 3.6.) If CJQ contains any zeros of f, we replace B(zo. r) by B(zo, p) for p > r and let Q still be the corresponding polygon. We can do this with p so close to r so that n(V, zo, r) = n(V, zo, p), and if we prove the estimate for p we will be done by letting p -+ r. Hence, we may as well assume that CJ Q contains no zeros of f. Therefore, if N Q denotes the number of zeros of fin Q, we have n(V, Zo, r) < N Q
-
1 = -2Jri
1 aQ
1'(z) -dz fez)
av-I
Figure 3.6
= -2rr1 1m
1 aQ
1'(z) -dz fez)
3. Exponential Polynomials
256
::::
L
~dol + _I 2
2n
I: 0 such that
Q, i.e., A is an algebraic number, then for any fJ
I A-~I n
257 > 0,
>_8 - n2+~'
see [Ba]. Note that in this case the frequencies of fare ±i ± iA, which are all algebraic numbers if A E Q. Moreover, the coefficients of f written as an exponential sum are also algebraic numbers. So that for f(z) = sinzsinAz, A E Q, guarantees that the zeros are well separated for any weight. If we consider f(z) = sin(z - a) - sin(J2z - a), a > 0, then the frequencies are again algebraic, but the coefficients are combinations of sin a and cos a with algebraic coefficients, which in general are transcendental numbers. On the other hand the zeros form the sequences 2krr 2a + (2k + l)rr - - - and kEZ. I-J2 1+J2 Again, depending on the choice of a we can make the zeros well separated or not. Moreover, note that from the P6lya-Dickson Theorem 3.2.25 it follows that when the frequencies of an exponential polynomial f are purely imaginary, then there is a positive constant A such that the zeros of f lie in the logarithmic strip
IImzl:::: A(l +log(l + Izl)) so that, if the zeros of f are well separated for the weight p(z) = 11m zl log(l + Izl), it follows that there are 8 > 0, N > 0, such that IZj - zkl
~
+
8
(Izkl
+ IZjl)N
for any two distinct zeros Zj, Zk. These considerations led to the following conjecture of Ehrenpreis [Eh5], [BY2]: 3.2.52. Conjecture. Let f be an exponential polynomial with algebraic frequencies and with polynomial coefficients in Q[z], then the zeros are well separated for the weight p(z) = Izl. That is, there are constants 8 > 0, A > 0, such that for any pair of distinct zeros Zj and Zk we have IZj - zkl ~ 8e-Alzkl.
Moreover, if the frequencies are purely imaginary, then the zeros are well separated with respect to the weight p(z) = 11m zi + log(l + Izl). That is, there are constants 8 > 0, N > 0, such that
for any pair of distinct zeros Zj and Zk. It is very little we know about this conjecture. In trying to prove it, it is natural to consider the Q-vector space r generated by the frequencies ai, ... , an
3. Exponential Polynomials
258
of I. The dimension of r as a Q-vector space is called the rank of r. The Conjecture 3.2.52 has been shown to the true when rank r = I, and also when rank r = 2 under the further assumption that I is an exponential sum (see [BY2]). This conjecture is equivalent to the fact that if the frequencies and the coefficients of f are algebraic, then V (f) is an interpolation variety for the space Exp(C), or for the space F(£'(IR». Another consequence of the conjecture is that if II, h are exponential polynomials with algebraic coefficients and frequencies, without any common zeros, then I E /(fI, 12) in Ap(C) (for p(z) = Izl or p(z) = lIm zl + log(l + Izl». We shall mention some related questions about mean periodicity in Chapter 6. It would seem strange that one could relate the separation properties of the zeros of f to the arithmetic nature of its frequencies on coefficients, but, in fact, following the work of Siegel and Shidlovsky one has quite a bit of knowledge about the zeros of f. For instance, they are not only transcendental numbers but one can even measure how closely can they be approximated by algebraic numbers. The reader is referred to [Va] and the references therein for some results of this type. Nevertheless, some simple looking arithmetic conjectures are still open and may be related to Conjecture 3.2.52. For instance, the conjecture of Schanuel stating that if (XI, ••• ,(Xn E C are linearly independent over Q, then the degree of transcendency of the field k generated by (X I, ... , (Xn, e'" , ... , e'" , is at least n, which has not yet been solved for n ::: 2 (see [Wa2]). EXERCISES
3.2.
1. Using the notation from the proof of the P6lya-Dickson theorem show that: (a) If z = x + iy E V,,)' then lim:~oo Iy/xl = 00. (b) For z E V. j, Iim:~oc Arg z = 7T /2. (c) Show that the curves of equation y + /-Lkj Arg z c in VA,j are asymptotic to y C - /-Lk,j7T/2 as Izl --+ 00. Similarly, the curve x + /-LA,j log Izl H is asymptotic to x + /-LA,j logy = H. (d) Let w Z + /-Lj,A logz, z E VA,j' Show that if Iz, - z21 ~ 8> 0, then for z), Z2 large one has IWI - w21 ~ 8/2. (e) For some R» 1, (U Uu) U (U Vk,I) 2 (B(O, R)' n (y ~ OJ).
=
=
=
=
2. Assuming the Conjecture 3.2.52 is correct, show that if fl' 12 are exponential polynomials with algebraic frequencies and algebraic coefficients, which have no common zeros, then I E l(f" 12), the ideal generated by flo 12 in Ap(lC)(p(z) = Izl or I Imzl + 10g(1 + Izl). 3. Let A(a, b, f) be the functional defined in Definition 3.2.38. Evaluate it in the following cases: (i) fez) == C E IC; (ii) fez) = e'''; (iii) fez) = e'P r2. 0
4.1.2. Proposition. Let K be a convex compact subset of n, U a convex open set, K oo lim -s
1- e
L / T, Inl:::N
\
S
1.
+ 2rrln -
~
).
On the other hand, from the diagram following Definition 3.3.12 we see that / T,
\
s
1.
+ 2rrm -
~
) = rrT(s
+ 2rrin)
= C(T)(s + 2rrin) = B(f)(s Thus, F(s)
as we wanted to prove.
=
f(O) -2-
. + N->oo hm
+ 2rrin)
L
B(s
= B(s
+ 2rrin).
. + 2rrtn),
Inl::::N
o
266
4. Integral Valued Entire Functions
4.1.8. Example. Let T = [(-I)k/k!}8 k , so that K = {O}. Then 3'(T)(n = ~k / k!, C(z) = 1fT(z) = l/zk+l and 1 ) 1 dk G(T)(z) = ( - - - k! dl;k 1 - ze~ ~=o For k = 0 we have G(T)(z) = 1/(1 - z) and I
G(T)(e- S ) = - - = I - e- S
4+ N-+oo lim" ~
Inl:::N
S -
1 2.
(s ¢ 21fiZ).
1f1n
For k 2: 1 we have G(T)(z) = Pk(Z)/(l - z)k+l, where Pk is a polynomial of degree k. Moreover,
1
-S
G(T)(e
)
(d
k
= k! d~k 1 -
=L (s nEZ
1)~=o = ~ d k
(-ll
dsk
el;-s
1 21fin)k+l
(
1 1- r
) S
'
(s ¢ 21fiZ),
which can be developed as for
Res> 0, Res < O.
More generally, the G-transform of T = 8~k) is of a rational function of the form Pk(z)/(l - zea )k+l, with Pk a polynomial of degree k.
4.1.9. Remark (On the Inversion Formula for the G-Transform). Let y be a loop in e-fl\e- K of index 1 with respect to the points of e- K , we denote by OO1(g), the Mellin transform of g E Jr o(lC\e- K ), the entire function
1
g(w) 1 OO1(g)(z) := - - . dz, 21f1 y w z+ 1
where W Z = eZ Log w. The inversion formula of G can be obtained from the commutativity of the following diagram (see Proposition 4.1.2): ;j
4.1. The G-Transfonn
267
where Exp(K) :=
If
E
-*'(C) : "Is > 0, 3C. 2:: 0
such that If(z)1 ~ C.eHdz)+'lzl, z E IC}. We also consider Exp(Q) := UKccn Exp(K). We would like to find now some explicit formulas for the inverse 9)1-1 of the bijective map 9)1: -*'o(Q(K» ~ Exp(K). Let f = ;reT). For 0 < ({J+ < 7r/2 and -7r/2 < ({J- < 0 we denote by y+ = y+(a, ({J+) (resp. y_ = y_(a, rp_» the half-lines of origin a E] - 1, O[ making an angle ({J+ (resp. rp_) with the positive real axis. Let y be the broken line y_ y+. As we shall now see, for Izi sufficiently small we have the integral representation G(T)(z) =
~ 21
1 y
!(w) e- i7CW z W dw
Slll7rW
(0
£ -
°
1, since cos qi+ > 0. Therefore, the integrand satisfies If(w)e- i 1!"W z WI ,- ' - - - - - - < C e po
I sinrrwl
-
,
for some c' > 0 and I) > 0 that can be chosen independent of r, (), as long as 1. This confirms that the integral over y+ converges uniformly and absolutely for all z E 8(0, ro)\{O}. The same is true for y_. On the other hand, consider the vertical segment r n of Yn, w = n + + i u, (n + 4- a)tancp_ .:::: u .:::: (n + 4- a) tanqi+. We have
o< r «
4
If(w)1 .:::: C,e A (n+(1/2)+lul) .:::: C;e Bn
for some A, 8 > O. Furthermore,
and
I sin rrwl Isin(rr(n + 4) + iu)1 = I cosh ul 2:
e1!"lul
2'
so that
for some C, 8' > O. Clearly, the upper bound tends to zero when n 0< r < ro « 1. As a consequence, we obtain that for 0 < Izl < ro « 1 1 -:
1
21
y
, z Wdw = "L....., f(n)zn _.f(w) _ _ e-I1!"W
sm rrw
n~O
~ 00
if
= G(T)(z),
since the series also converges for small Izl. Note that the value of the integral we obtained does not depend on the choice of arg z. On the other hand. for z = 0, the value of the integral obtained setting Ow = 0 does not necessarily coincide with G (T) (0) unless f (0) = O. We are going to use these remarks to show that the integral converges and represents G(T) for all z ¢ (e- L U CO}). L := K +] - 00,0].
269
4.1. The G-Transfonn
For that purpose we consider separately the convergence of the integral over f/ (e- L U (O}), z = re ilJ , then the integral over y+ is absolutely convergent if there is an £ > 0 such that y+ and over y_. Let z
On the other hand, the definition of HK(z) = sup{Rezt: t E K} implies that a + i/3 E K if and only if for every rp with -rr ~ rp ~ rr one has
Thus, a + i/3 E L if and only if the inequalities (**) hold for -rr /2 ~ rp ~ rr /2_ Since z f/ (e- L U (O}), then the points -log Izl - i argz f/ L for any choice of argz. Clearly, if a + i/3 f/ L, then a + i/3 + t f/ L for any t :::: O. Therefore, for the given rp+(O < rp+ < rr/2), since the inequality (**), with rp = rp+, determines the supporting line of L slope tan(rr/2 - rp+), we can choose a determination of arg z such that the point - log Iz I - i arg z lies on the open side of this line and the set L on the other. (Namely, if argz E ] - rr, rr] does not work, we can choose argz E] - (2n + l)rr, -(2n - l)rr] for n» 1.) (See Figure 4.2.) It is clear that we can find a half-strip R+ = {a + i/3: a > aD > -00, /31 < /3 < /32, /32 - /31 < 2rr} such that -log Iz I - i arg z E R+ and there is £ > 0 such that for every a + i/3 E R+
a cosrp+ - /3 sinrp+ :::: Hde iifJ +) + 2£. The image of R+ by the map a + i/3 1-+ e-(a+i{3) will be the open angular sector S+ = (t E C, 0 < Izl < e-ao , -/32 < argt < -/3d and z E S+. Since (**) holds now for every point t E S we see that the integral over y+ defines a holomorphic function of t in the sector S+. In exactly the same way we construct a sector S_ such that the integral over y_ defines a holomorphic function of t. Now Z E S = S+ n S_ and we have a holomorphic function defined in S, which for 0 < Izl < ro coincides with G(T)(t) since, as we said before,
7t
-7t
Figure 4.2
4. Integral Valued Entire Functions
270
for those values the integral representation does not depend on the choice of arg l;. We conclude that the integral representation is now valid everywhere in C\(e- L U {O}) because G(T) is holomorphic and single valued there. In the same fashion, if we take ifJ+ E lrr, rr/2[, ifJ- E l- rr, -rr/2[ and consider the path y (ifJ+, ifJ-, a) = y defined as before, then the integral
_~ { !(w) 21
i
(_z)W
dw
smrrw
y
+ [0, oo[ (see Figure 4.3).
represents G(T)(z) outside e- A , A = K
1t
1t/2 - (1\
L
Figure 4.3
4.1.10. Remark. If carrier of T. Then,
f =
J(T), T E Jt"(n), let K denote the minimal convex
hj(O) = lim
log If(r)1 r
r----)-oo
= supRez ZEK
and the radius of convergence of the series Ln>o f(n)zn is exactly exp( -hj(O». Hence, lim sup yilf(n)1 = exp(hj(O» n.... oo
and · sup log If(n)1 I1m n----+oo
n
= h j (0) .
This is a theorem originally due to P61ya which can be easily derived from the previous Remark 4.1.9. We are going to use the G-transform to study some classical results about holomorphic functions. The first question to study is that of the convergence of
4.1. The G-Transform
271
the Newton series of an entire function. (An excellent reference for this subject is [Ge].) We recall from [BG, p. 237] that if f is an entire function, its nth divided difference An = An (f) is given by An =
(~) (_l)n- j f(})·
L: O~J~n
The Newton series
f is the series " z(z - l) ... (z - n ~An , n~O
+ 1)
n.
.
Clearly, if f is a polynomial, the series terminates and its sum coincides with f. The empty product has the value one, as always. Let U = {z E Q: le z - 11 < I}. This set is an unbounded open convex subset of Q whose boundary is the curve
x
= log cos y + log 2,
Iyl
0 there exists Co ~ 0 such that (z E C).
If for some pEN, fen) = O(lnIP),for all nEZ, then there is a E [0, Jr[ and C > 0 such that If(z)1 ~ C(1
+ Izj)Peollrnz l
(z E C).
4.1.16. Corollary. If f is an entire function of exponential type zero such that for some pEN, fen) O(jnI P ) (n E Z), then f is a polynomial of degree at most p.
=
Proof. The distribution R from the preceding Proposition 4.1.14 will have
support at the origin.
0
275
4.1. The G-Transfonn
There are elements of Exp(Q) such that their zero sets contain the complement of a set of the fonn Z\pZ, for some p E N*. For instance, if w == e21ri / 3 is the cubic root of unity, consider the function fez) := (l
with g
E
Exp(t Q ). Then
f
+e +e WZ
E
Z. Let g =
f(3n
~(T),
f
)g(z)
Exp(Q) and it verifies
E
f(3n) = 3g(3n),
for every n
W2Z
+ 1) =
f(3n
+ 2) =
0
then
= ~«8
+ 8w + Owl) * T)
and the function G(z)
= G«o + 8w + 8wl) * T)(z) = Z(f)(z)
satisfies the functional equation G(wz) = G(z).
In fact, for Iz I « 1 we have G(z) =
L
f(n)zn =
n~O
and G(wz) =
L
f(3k)z3k
k~O
L f(3k)(wz)3k = L f(3k)z3k = G(z). k~O
k~O
Let .ifo := UK .ifo(Q(K» and let K be a convex compact subset of Q. We would like to detennine all the G E .if0 such that for some w E C" one has G(wz) = G(z),
Z
E Q(K).
4.1.17. Lemma. If G
E .ifo\{O} and WE C* satisfy G(wz) = G(z) for all Q (K), then w is a root of unity.
Proof. We have G = G(T), T G(wz)
=L n~O
E .if'(Q)\{O},
~(T)(n)(wzt
=L
z
E
and
~(T)(n)zn
for
Izl« 1.
n~O
If J(T)(n) = 0 for n ~ 1, then G is a constant, hence G == O. Therefore, there is n ~ 1 such that J(T)(n) t= O. We conclude that wn = 1. 0
For G, w as in Lemma 4.1.17, let p = inf{n E N*: w n = I}. Then we can assume w = e 21ri / p in the equation G(wz) = G(z). Let us denote by Sk the angular sector 21T 21T} , Sk:= { ZEO unz- n be the Laurent series expansion at 00 of a function G hoiomorphic D = C\E, where E is a compact set with 7:(E) < 1. If the coefficients Un are integers, then G is a rational function.
in
Proof. Let £ > 0 be such that 7:' = 7:(E) + 2£ < 1 and let y be the previously found rectifiable curve such that E ~ Int(y), 7:(Int(y» :::; 7:(E) + £. Let (Tkk~.o (To == 1) be a sequence of Chebyschev polynomials for Int(y).
4.2. Integral Valued Entire Functions
289
Ho
We are going to transform the sequence of Hankel determinants of (un)n?,:O with the help of the sequence (Tkh?':o. As we have seen above, if Vj,k
= -I . 27rl
1 y
dz G(z)1j(Z)Tk(Z)-, z
then H~ = det(vp)O:::j,k:::n'
Let M = max zEy IG (z) / z I and L be the length of y. Since the Tk are Chebyschev polynomials for Int(y), there is a constant A > 0 (independent of k) such that for z E Int(y) ITk(z)1 .::::
+ e/
A(r(Int(y»
.::::
A(rY.
Therefore,
and
From the Hadamard inequality for determinants we have
IHnl o -
0 there is C. > 0 such that If(z)1 :::; C.eHdz)+elzl (z E IC),
4.2. Integral Valued Entire Functions
293
has the form
L
fez) =
Pj(z)cJ.
I:5j:5N
where Pj E Q[z] and Cj are algebraic integers belonging together with their e Z Iogcj). conjugates to e K , 1 ~ j ~ N. (cJ
=
Proof. Let T be the analytic functional carried by K such that f = ~(T). Recall that t is such that (t. h) = (T. h). where h(z) = h(-z). Then is carried by -K and ~(T) = j. The function G(T) is holomorphic in (C\e K and has the Laurent expansion
t
about z = 00. From the previous results of this section we conclude that G(T) is a rational function, all whose poles ZI •... , ZN are algebraic integers. which belong to e K together with all their conjugates, and G(T)(z) = ao
+ alz + ... + as_lz s- I bo + bIZ + ... + bsz s
ao •...• as-I. bo.···. bs E Z. bs = 1. Applying the inversion formula 6.1.3. we have v f(l;) =
v
~(T)(l;)
= - -1.
1
v {L dz G(T)(z)eogz_
27r1
=
L
Z
y
P/l;)e-{LOgzj
•
I:5j:5N
When we compute the integral by residues we see that the polynomials Pj have algebraic coefficients. Namely. this follows the usual rule of computation of the residues of a quotient. In fact the coefficients are in the field generated over IQ by Zl •..•• ZN and their conjugates. 0
4.2.15. Corollary. Let K and f be as in the preceding proposition. If there is an integer m
~
2 such that K £ {z E Q: le z
-
ml
< I},
then fez) = P(z)m Z for some polynomial P
E
Q[z].
Proof. The compact e K cc B(m, 1), hence r:(e K ) < 1. Moreover, by Lemma 4.2.12, m is the only algebraic integer that together with its conjugates belongs to B(m. 1) = {z: Iz - ml < I}. The remark at the end of the proof of Proposition 0 4.2.14 implies that the polynomial P has rational coefficients.
294
4. Integral Valued Entire Functions
4.2.16. Corollary. If K and f satisfy the hypothesis of the Proposition 4.2.14 and if, moreover, K S; {z E Q: le'l < I},
then fez)
=
L
Pk(z)cZ,
l:::,k:::,N
there the Ck are roots of unity, Pk E Q[z].
Proof. There is an £ > 0 such that e K S; B(O, 1) n {z: Rez > -1 + £}, so that reeK) < I. We can apply Lemma 4.2.12 to P(z) = z and obtain that the Ck are roots of unity and moreover the Pk have coefficients in the field Q(w), where w m = 1 for some mEN'. 0
4.2.17. Corollary. Let f, K be as in Proposition 4.2.14, and K S; {z 1I :::: I} then
fez) = Po(z)
+
L
E Q:
Ie: -
Pk(Z)(I +e2rrir,),
l:::,k:::,N I
I
-
rk E] - 2' 2[nQ, Pk E Q[z].
Proof. There is £ > 0 such that e K S; B(l, I) n {z: Rez > £}, which ensures reeK) < I. We can now apply Lemma 4.2.12 to P(z) = z - I. The nonzero roots of (p(Z»m = 1 have the form 1 + e2rrik/n, -n/2 < k < n/2. 0 One can prove that if ao > 0 is such that r(exp(B(O, ao») = 1, then ao E ]0.834,0.845[. (See [Pi].) With this notation, we can prove the following result:
4.2.18. Corollary. Let f be an entire function of exponential type such that
If(z)1 < Me alzl for some M > 0 and 0:::: a < ao. Assume feN') S; Z. Then fez) =
L
Pk(z)c",
l:::,k:::,N
where Ck are algebraic integers in B(O, a) (together with their conjugates) and Pk E Q[z]. Moreover,
. .
(l)ifa 1; (iii) the function fzo defined by fzo(z) := Zo
z
roo e-tzo/zT3,(f)(tzo)dt
Jo
is holomorphic in K Zo and it coincides with f on the segment Z = AZo, 0 < A < min{1, R/lzoll.ln particular, f and fzo define a holomorphicfunction, still denoted fzo' in the open set B(O, R) U K zoo Proof. There is nothing to prove in (i), it is just a change of variables. On the other hand, if Res:::: 1 we have le-stT31(f)(tzo)1 :'S e-tlT3,(f)(tzo)l. This immediately guarantees the condition (ii) (see [BG, § 1.2.9]). To prove (iii), we use that the earlier description of Kzo tells us that Re(zolz) > 1 is equivalent to z E Kzo. Hence, fzo is holomorphic in Kzo by (ii) and coincides with f on the segment AZo, 0 < Amin{1, Rltzt} by (i). Therefore, fzo = f on B(O, R) n Kzo,
and the conclusion of the lemma is correct.
D
This lemma justifies the definition of the Borel polygon B(f) as the set of those of l; E C such that there is a compact neighborhood V~ of l; and a nonnegative function g~ E L' ([0, oo[) with the property that for every z E VI; the inequality a.e. t > 0, holds.
5.1.3. Proposition. The sets M(f) and B(f) coincide. Proof. Let us show that M(f) S; B(f). Let Zo E M(f), then Kzo Cc D(f), therefore we can find a positively oriented circle y in D(f) enclosing K zo ' For ~ E y, we have l; :f. 0 and l; t---+ Re(zog) is a continuous function strictly
smaller than 1 - 8, for some 0 < 8. Choosing y conveniently we can make 8 as small as necessary. Hence, we can write for;; E y __ 1_ =
1 - zoll;
1
00
0
e-Ue(zo/I;ludu.
5. Summation Methods
302
We denote still by Then
I the analytic continuation of the series
I(zo)
Let A := max y mate
-1-.1y
=
bn
I/(n/~1
I(t;) (
t;
2::n~o anz n
roo e-Ue(zog)u dU)
io
< 00, hence for
(S-, u)
E Y
to D(f).
d~.
x [0, oo[ one has the esti-
which allows us to interchange the order of integration so that I(zo)
=
ioroo e-
U
(-1-.1y
1(t;)e(zom U dt;) duo
2:rr1
t;
Let 0 < e < R be such that B(O, e) S; Int(y). In this case,
-1-.1 2m
I (S)e(zo/nu dt;
t;
y
= _1_.
(
2m il~l=e
I(t;)e(zom u dS-
S-
= 13 1(f)(uzo).
where the last identity is an elementary computation of residues. Moreover, this identity yields the estimate. 113, (f)(uzo) I :s A C~:) e(H)u.
Finally, we conclude that I(zo)
=
1
00
e- U 13, (f)(uzo) du,
and the integrand is bounded by the integrable function A[C(y)/2:rr]e- 8u • It is clear we can choose a compact neighborhood Vzo of Zo such that for all z E V:o' K: CC Int(y) and Re(zg) :s 1-8 for Z E Vzo ' S- E y. Under these conditions, all the previous reasonings are valid and, hence, I(z) =
1
00
e- t 13, (f)(tz) dt
for all z E V: o' and the estimate e- u 113, (f)(uz)1
:s A e~:) e-~u
holds for z E Vzo. u E [0,00[. This shows that Zo E B(f). We conclude that M(f)
~
B(f).
Let us now prove the other inclusion, B(f) S; M(f). Let Zo E B(f), and a > small enough so that B(zo, a) ~ B(f) and there is a nonnegative integrable function gzo such that e- t l13 1(f)(tz)1 ::: gzo(t), a.e. t > 0, z E B(zQ,a). Consider the open set
°
U zo := (
U KZ) zeB(zo,a)
\{O}.
5.1. Borel and Mittag-Leffler Summation Methods
303
From Lemma 5.1.2 we obtain a collection of analytic continuations fz of f, defined on B(O, R) U Kzo ' z E B(zo, a)\{O}. We claim that Vzo U B(O, R) S; D(f). In fact, the open set Vzo U B(O, R) is clearly star-shaped with respect to the origin, and the different fz define, together with f, a single valued holomorphic function in this set. Indeed, if KZI n KZ2 -I 0, then this intersection is fZ2 in the nonempty open set B(O, R) n k ZI n k Z2' so that connected anc! fZI fZI = fZ2 in K ZI n K z,. Due to the maximality of the set D (f), we conclude that Vzo U B(O, R) S; D(f). In particular, Kzo S; D(f), so that Zo E M(f). That is, B(f) S; M(f). 0
="
We can rephrase these results by means of the following proposition: 5.1.4. Proposition. Let f (z) = L:n>O anz n have radius of convergence < R < 00, 8 1(f)(z) = L:n>O anz nliz!. The function f has an analytic continuation still denoted f, -to the Borel polygon B(f). B(f) is an open set which can be characterized as B(f) = M(f) = (z E C: Kz S; D(f)}. This analytic continuation is given by
°
Z E
B(f).
This procedure to obtain the analytic continuation f in B(f) is called the Borel summation method. In order to obtain a similar explicit analytic continuation to the whole star of holomorphy D(f) we must replace the exponential function by the Mittag-Leffler function Ep. We proceed to give a succint description of Ep and its properties. (The reader will find a detailed study of this function and its applications in [Dr].) 5.1.5. Definition. For p > 0, the Mittag-Lerner function Ep of index p is the entire function Ep(z) =
L
Zk
f(l
+ k/ p) .
k":O
It is easy to see that E 1 (z) = e Z and that E p is a function of exponential type, for instance, from the decay rate of its coefficients (see [BO, Chapter 4]). We need somewhat more precise asymptotic information. For future use we shall restrict ourselves to p ~ 1. Let us recall that the entire function 1/ r can be expressed in terms of the Hankel integral I r(z)
1
= 2Jri
1
Ya.
e'
~ dt.
where Ya,a is the Hankel contour in Figure 5.1, the indices satisfy 0 < a, Jr/2 a < Jr, and t Z = eZ Logt (see [BO, Exercise 5.3.3]).
o all z", with radius of convergence and p 0:: 1 we denote by Mp(f) the open set, Mp(f) := Iz E C*:
5.1.12. Lemma. D(n
Kf
Method~
°
< R
0,
00
as above,
C; D(f)}.
= Up~1 Mp(f).
Proof. It is left to the reader. The main point is to observe that K r: lies in an 0 angular sector of opening 7r / p. As before, we denote by Bp(f) the family of those l; E C such that there is a compact neighborhood V~ of l; and a function & ELI ([0, oo[), gl; 0:: 0, such that for every ;; E VI: we have e- I " tP-1jBp(f)(tz)j ~ gl;(t)
a.e.
t > 0,
with
The function Bp(f) is sometimes called the Mittag-Leffler transform of f. One verifies without difficulty the following lemmas and propositions, whose proofs we leave to the reader:
order p of
5.1.13. Lemma. Bp (f) is an entire function of order p and finite type. 5.1.14. Lemma. Let Zo on [0,
00[.
E
C· be such that t
Then:
(i) For every z = AZo, 0oanz n is the Taylor series about z = 0 of the function Ll 0, such that
if
11/II:so 1/10,
if
1/IO:SO 11/II:so
and qJ(I;)(-r)/; 1
(c) Choose
Let k
e21r1~ _
1
I
< k r Pe- R(1rsinl/to-r-E)
-
2
ro so that r < Jr sin 1/10 < Jr,
1/10,
0 < ro < 1, (-log ro) cos 1/10 < ro, I; ErR, one has
= max(k l , k 2 ) and show that for any 0 < r 0 such that It I > A implies Jr - r + (t/ltIWI + e(t) > 11 and
I
1-
rp(l;)zl; P -~ltI e2r 0 be such that :rreTo :rreTO } K -C { -2- < argz < 2:rr - 2 .
Show that if 0 < f3 < 1 and 0
~
eT ~ eTo, then for z E K
z" 1 l L---= nl7" 1- z
fJ-ioo
";::0
Deduce that
fJ +ioo
(1 _ ~-17~)
e 21ril
-
1
z\d~.
zn I lim " 'n17" = 17-+0+~ 1-z n~O
uniformly on every compact subset of C\[I, 00[. (d) Let fez) Ln;::o anz" be a power series with finite nonzero radius of convergence. Show that the series
=
converges uniformly on every compact subset of D(f).
5.1. Borel and Mittag-Leffler Summation Methods
315
9. Let !p be a holomorphic function in 1C\(alo ... , aN j, aj E IC\Z, such that there is a value Ro > max/ laj I: I :::: j :::: N j with the property that for any e > 0 there is CE > 0 so that if Izl ::: Ro then l!p(z) I :::: CEeElzl. Use the results of this section and Exercise 5.1.8 to prove that: (a) The series fez) = Ln>orp(n)zn has an analytic continuation to IC\[I, 00[. (b) Let No E N be such that all aj satisfy Re(aj) > -No, and define . ~ ct>(z) := 2:n:1 ~ Res
(rp(~)z{
)
e27ri~ _ I' aJ
I~J~N
which depends on the choice of argz. Show that if z E 1C\[I, 00[,0 < argz < 2:n:, and -No - 1 < fJ < -No then the function f of part (a) satisfies fez)
()
~ !P -n _ ct>(z) _
=-
~
zn
l~j~N
(c) Let z E
I-
00, -1[,
1
1Hioo
= -(No + 1/2), and ~ = fJ+
fJ
(7)Z'
!P .'-'
, e 2m'-1 /3-100
d~.
it, then, for some e(f3) > 0,
-z){ I : : Izllle-7rltl+E(Il)JIl2+t2, l !p(~)( e 2m , - 1
with the additional property that if we let No -+ -00 then e(f3) -+ 0. Conclude that
11
1l+ ioo Il-ioo
!p(~)(-z){ d~1 e27rt{ -
1
::
21zlll {e E (lllJ211l1
Deduce that for Izl > 1 fez)
r
lill
10
e-7rt dt
rp(-n) = - 2: -zn
+
1
00
e- 7rt +E(lll../'ir
dt}.
IPI
ct>(z).
n~1
(d) Let n = /z E IC: Izl> I, z fj [I, choice of arg z in n, let fez) := -
ooD
rp(-n) 2: - zn
and 0 < argz < 2:n: for ZEn. With this
CI>(z)
(z En).
n:::l
Prove that for x > lone has F(x
+ iO) -
F(x - iO)
= -ct>(x) + ct>(e 27ri x) = 2:n:i 2: Res(!p(~)x~,aj). l,!!:j~N
Let FI (z) := F(z) + 2:n:i Ll~h,N Res(!p(~)z{, aj). Compare the boundary values FI (x ± iO) and F(x ± iO) for x > I. (e) Show that if rp is an entire function of exponential type zero, then FI F f, and this is a function holomorphic in IC\/lj, which vanishes at 00. (f) Let k E N*, rp(z) = Z-k. In this case,
= =
zn
fez)
. (1
2m Res
;;k
z~
e27ril; _
=:E kn ' n~J
0) _~ 0, and this estimation holds everywhere if C2 is sufficiently large. Let then
1
Iv(z)12 + Iz12)2
----'.,-'--:-::--::-e
e (l
It follows that if ql(z) := q(z)
-2q
+ log(l +
d
m < 00.
Iz1 2), then
11/12e-2q, dm
00
since
tP
t
/--'>00
nr],
I~I:::: ~})*:::: 1/I(z),
1/1 is upper semicontinuous. Hence, hi,p(z) :::: 1/I(z),
On the other hand, for z = I we can take t = 2 j , and we get f(2 j
= g(2 j
)
) -
v(2 j )h(2 j
)
=g(2 j ) = exp(1/I(2 j )). So, for those t log If(t)1 = 1/I(t)
= tP1/IO).
tP
It follows that hi,pO) 2: 1/1(1),
whence, hi,p(l) =
1/10). This concludes the proof of the proposition.
0
5.2.7. Lemma. Let u be a subharmonic function in C that is homogeneous of degree p > O. There is a decreasing sequence (U n )n2:1 of subharmonic functions, homogeneous of degree p, and Coo in C\{O}, such that Un converge toward u everywhere. Proof. We have shown in [BG, Chapter 4J that we can always approximate u by Coo subharmonic functions. The problem is to keep the homogeneity. This requires a slight modification of the usual construction. For that purpose, let q; be a standard radial function in C, i.e., q; E V(8(0, I», 0:::: q;:::: I, Icq;dm = I, q;e(z) = e- 2q;(Z/e), let
k k
ve(z):=
and define us(z):=
u(z - w)q;e(w)dm(w),
u(z -Izlw)q;,(w)dm(w).
It is easy to see that U e is Coo in C\ {OJ and it is homogeneous of degree p. We know [BG, Chapter 4J that the Ve are Coo, subharrnonic, and converge decreasingly to u as e -+ 0+. On the other hand, for Izl = 1, we have us(t) = vt(z),
so that ue(z) -+ u(z) decreasingly on Izl = 1 when e -+ 0+. The homogeneity now implies that U e -+ U decreasingly everywhere. (Note that for z = 0 we have u(O) = us(O) = 0.)
5.2. The Lindelof Indicator Function
321
The only thing to verify is that the for each r > 0 the function Ur(Z):=
121< u(z -Izlreiil)d(l
is a subharmonic function of z. If z Ur(z)
are subhannonic. Let us show first that
Ue
= re ia , we have
= 121< u(z =
1
2IC
= lim
e--+O
ze- ia re i9 )d(l
u(z - zre if3 )d{3
1
ve(z - zre if3 ) d{3.
2IC
0
Each of the last integrals is easily seen to be subhannonic by direct computatioll of the Laplacian. Since Ur is a decreasing limit, if follows that it is also q;e(r)r dr = 1, so that Us is an average subharmonic. Finally, q;e(r)r ~ 0 and of subhannonic functions. It foHows that U e is subhannonic. 0
f;
We are trying to prove that every subharmonic function that is homogeneous of degree p > 0 is the LindelOf indicator function h f.p of some entire function. Since we know that indicator functions are continuous, we need to show first that any such subharmonic function is also continuous. S.2.S. Lemma. Every subharmonic function u that is homogeneous of degree p > 0 is a continuous function in the whole complex plane. Proof. It is clear u is continuous at z
= O. Therefore,
its enough to prove that
in every angular sector of the form S9.a = {z E C*: I arg z -
(I
I
0 so that hl,p(I;) < 1/1(1;) - e, By continuity, there is a 8 > 0 so that hl,p(z) < 1/I(z) - el2 for all Z E B(I;, 8). There is a point zno of the sequence such that zno E B(I;, 8). Since the same point appears infinitely often, there will be an index n! ~ no such that Let
hl,p(I;)
! En,
1/1(1;),
11;1= l.
and It follows that
EXERCISES
I
E
En,. This is a contradiction. The theorem is hence proved.
o
5.2.
1. Let rp be a subhannonic function, homogeneous of degree p > O. Let E", be the space defined in the text, Show it is a Frechet space. Prove that fEE", if and only if hl,p 5 rp. 2. Let f be a holomorphic function of order p and finite type in the angular region 91 5 argz 59,.,92 - 91 < 7rlp, and h = hl.p, the indicator function of f of order p. (a) Show that for 91 < 9 < 92 the following inequality is correct: h(9) - h(91 ) < h(92 )
sinp(9-91)
-
-
h(9 1)
sinp(92 -91 ) . (92-9) -9-1 ) seep +h(O,)smp - 2 - secp (9 -2 2
(0-9 -2-
1)
.
324
5. Summation Methods
(b) Let 92 - 91 ::: A < 7r/p and let k := max {lih(8)/(sinPA)21 : 81 ::: 9::: 92}. Show that for 8 1 < 8 < 92 h(8) - h(el ) h(8) - h(el) -'--"---'-'-'- < sinp(9-8d - sinp(92 -81)
Conclude that
8
t-+
h(9) - h(9d sinp(8 -8d
+ k (82 -
+ k(8 -
il) f7
•
el)
is an increasing function. Deduce the existence of the right and left derivatives and h~(ed. (c) Show that h~(91)::: h~(e}). (d) Show that if 90 is a local maximum of h and 18 - 80 1 ::: 7r / P then
h~(91)
h(8) ::: h(90 ) cos peo•
3. (a) Let
I
be an entire function of order 0 < p < 1 and finite type, so that I(z)
(ak
i= 0, c i= 0, m
k~1
::: 0). Show that if Co
colzl"'
(I -..:..)
= cz"' IT
ak
= Icl,
IT (1 - l:L) : : I/(z)1 ::: colZI"' ITk~1 (1 + -IIZI ) . k~1 lakl akl
(b) Let
II (z)
:= coz m
IT (1 __z_) , k~1
lakl
and let hI be the indicator function of order p of It- Show that hI (7r) (c) Let mer) = min{l/(z)l: Izl = r}, M(r) = maxll/(z)l: Izl = r}. Show that
= sup~ hI (0).
· logm(r) I· log III (r)1 I1m sup > 1m sup HOO logM(r) - ' .... 00 logl/l(-r)1 . log III (r)1 > I IIDSUP -
' .... 00
hI (ll')r P
hI (0) =- > COSlrp. h l (ll) -
(Use part (d) of the previous exercise.) (d) For E > 0 prove the existence of a strictly increasing sequence converging to 00 and such that
(rj)j~I' rj >
0,
(j ::: 1).
4. Let
I
be a holomorphic function of order p and finite type in the angular sector
a::: Argz ::: {J. Assume -7r < a < 0 < {J
0 there is a sequence r. -+ 00 such that log I/(r.)1 ::: (hl(O) - y)r:. (b) Prove that if 0
0, the inequality log IIPn(z)1 2: -2yH (~) (rn
+ 2e8rn )p(r'+Mrnl
holds in Izl ::: 8rn , outside an exceptional set of finitely many disks, the sum of whose radii does not exceed w8rn • (d) Conclude that in the interval (1 - 8)rn ::: r ::: (l + 8)rn one has the inequality
(~)
log I/(r)1 2: {hj(O) - y - 3yH
(l
+ 2e8)2P }
r:,
with the possible exception of values of r lying in a finite collection of subintervals, the sum of their length is at most 2w8rn. (e) Since for rn ::: r ::: (1 + 8)rn, r: 2: (1 + 8)-P r P, and for 0 < y « 1, 0 < 8 « 1, hj(O) - y - 3yH
(~)
(1
+ 2e8)2p
2: (hj(O) - £)(1
+ 8)2 p ,
prove that log I/(r)1 2: (hj(O) - £)r P for Tn ::: r ::: (1
+ 8)rn' with the exception of a set of points r lying in a set of measure
::: 2wiJrn.
Prove the same result for any ray Arg z
= 8 E la, ,8[.
5. Let I be an entire function of order p and finite type, 1(0) :f. O. Let nCr) be the number of zeros of I in Izl < rand N(r) J~(n(t)/t)dt. (a) Show that
=
N(r) = _1_
rP
r
2r
log I/(rei6)1 d8 _ log 1/(0)1
211: TP Jo
rP
and conclude that N(r)
1
lim sup - - ::: -2 r--+oo rP 7r (b) Prove that lim sup -nCr) ::: -ep r~oo rP 211: *(c) For p > 0, let
ak
= 22k / p , k 2:
1 1
.
nCr)
h j (8)d8.
2 "
h j (8)d{}.
0
I, and
Show that I is an entire function of order p, h j hmsupr~oo rP
2 "
0
= 1 = -ep
211:
({})
= 1/ pe, and
1z"
h j (8)dB.
0
6. Let I be an entire function of order p, 0 < p < 1, 1(0) = 1. Assume the only zeros of I are simple, located at z = -rn, 0 < rJ < r2 < ... , and such that for some A > 0, net) - AlP as t ~ 00,
326
5. Summation Methods
(a) Show that for Log(f(z»
Z E
Ie\] -
00,
0[,
(1 + f) = 10f"\og (1 + ~) dn(t) = [n(t) Log (1 + :)]00 + z roo ~ dt = z roo ~ dt. t 0 10 t(t+z) 10 t(t+z) =L
n~1
Log
n
(b) Show that given e > 0 there is A > 0 such that t > A implies
1
z
1
00
o
n(t) - At P I dt < t(t+z) -
1A
Izl
n(t)
+ ().. + e)tP dt + elzl
1""
tlt+zl
0
0
tP - - dt. tlt+zl
(c) Prove that using the principal branch of zP
z
1'" o
tP ---dt=rrzPcosecrrp. l(t+Z)
(d) Conclude that for -rr < (} < rr one has Log(f(re i8 as r --+
» ~ eip rrA(cosec rrp)r 8
P
00.
5.3. The Fourier-Borel Transfonn of Order p of Analytic Functionals If T is an analytic functional, the entire function ~p(T) defined by (Z E C),
is called the Fourier-Borel (or Mittag-Leffler) transform of order p of T. One can see that 'l:'
s' ~n) z:= ~ r (1+n)
(T)( )
Up
""
n
(T
n~O
Z ,
p
and conclude without difficulty that ~p(T) is of order at most p and finite type. The case p = I coincides with the Fourier-Borel transform considered in Chapter I, hence we shall restrict ourselves to p > I in this section.
5.3.1. Definitions. (1) For a nonempty subset K of C let Hp.K(z) := sup Lp(~z) ~EK
be the p-support function of K. We say that K is p-convex if K = (~ E
(2) For'" # K
~
c:
Lp(~z) ::: Hp.dz)
for every
Z E
q.
C, its p-polar set is the set K*P :=
R
E
C:
Lp(~z)
< 1 for every Z E K}.
327
5.3. The Fourier-Borel Transform of Order p of Analytic Functionals
5.3.2. Proposition. Let A, B, Ai (i
E l) be nonempty subsets ofe.
(i) A S; B implies B*P S; A*p. (ii) A S; (A *p)*p and A *p «A *P)*P)*P. (iii) For a> 0, (aA)*P = (l/aP)A*P. (iv) If A is stable by multiplication by every a > 0, then
=
A*P
= (~ E c: Lp(~z) = 0,
Vz E A).
(v)
Proof Item (i) and the first part of (ii) are evident. From them it follows that «A*P)*P)*P S; A*P. Moreover, from the first part of (ii), A*P S; «A*P)*P). So (ii) holds. Part (iii) is a consequence of the homogeneity of Lp. It is clear that
R
E
C:
Lp(~z)
= 0, Vz
On the other hand, if ~
E
A) S; (~ E
c:
Lp(~z)
0 and
Z E
E
Lp(~az)
I,
= A*P. A we have
< 1,
so that
Since in part (iv), a > 0 is arbitrary, we obtain that Lp(~z) ::: 0 for every ~ E A*P, Z E A. As pointed out in Lemma 5.1.10, for p > 1, the function Lp ~ 0, hence we conclude that A*P S; R E C: Lp(~z) = 0, Vz E A}. From Ai E UjEI Aj and (i) we conclude that nEI A? ;2 (UiEI Ai )*P. Conversely, if ~ E nEI P and Z E UiEI Ai, then Z E Aj for some j E I and, hence, Lp(~z) < 1 since ~ E AjP. Therefore, ~ E (UiEI Ai)*p. 0
A7
Let us remark that the definition of p-convexity implies that every p-convex set contains the origin and that, for every K S; C one has
5.3.3. Proposition. (i) If K is p-convex, then K = (K*P)*P and, moreover, then K*P = {~ E C: Hp.d~) < 1).
if K
is also compact,
(ii) If K is compact and K = (K*P)*P, then K is p-convex. (iii) The intersection of an arbitrary number of p-convex sets if p-convex. (iv) If K =f:. (0 there is a smaller p-convex set containing K. It is called the
p-convex hull of K and denoted cVp(K). Moreover, cVp(K) =
R
E
C: Lp(zi;) ::: Hp.K(z),
Vz E
K}.
5. Summation Methods
328
(v) For any two closed p-convex sets KI and K 2, the inclusion KI 5; K2 is
equivalent to Hp,K ::::: H p,K2 ' (vi) For any nonempty bounded set K the function
Z r-+
Hp • K (z) is continuous.
Proof. (i) From the previous proposition we know that K 5; (K*P)*P. Assume that K is p-convex. If there was I; E (K*P)*P such that I; rf. K, then one could find Zo E C such that Lp(~zo) > Hp.K(zo), Hence Lp(l;zo) > 0 and
Or, in other words,
Therefore, E K*P
Zo
(Lp (I;zo» l/p
whence, as I;
E (K*P)*P,
,
we have Lp
(~ (Lp(;;O»I/P)
< I,
which leads to the contradiction
It foUows that K = (K*P)*P. Let K be p-convex and compact and let B := (I; E C: Hp,K(1;) < I}. If I; E K*P and Z E K, then Lp(zl;) < 1. By the continuity of Lp and the compacity of K, it foUows that sup{Lp(l;z): z E K} < 1, so that I; E B. Conversely, if I; E B, then for any Z E K, Lp(zt;) ::::: Hp.K(i;) < 1, hence ~ E K*P. (ii) Let K be a nonempty compact set such that K = (K*P)*P. To show that
K is p-convex it is enough to show that A := (I; E C: Lp(zt;) ::::: Hp,dz), Vz E K} ~ (K*P)*P.
Let t;o rf. (K*P)*P. There is then Zo E K*P such that Lp(zol;o) :::: 1. On the other hand, since Zo E K*P we have Lp(zot;) < 1 for aU i; E K. By the compacity of K it follows that H p.K (zo) < 1. Hence i;o rf. A. (iii) Let K = K;, aU K; being p-convex. Let A be the set defined in (ii). We need to show that A 5; K. Let i; E A, then for any z E C, i E /,
nEI
Lp(zi;) ::::: Hp.dz) ::::: Hp.K,(z).
Hence, i; E K; for all iE/, and so i; E K. (iv) The existence of a smallest p-convex set containing K follows from (iii). The set A defined earlier clearly contains K. Moreover, H p •A = H p •K , which will show that A is p-convex. In fact, for z E C, we have Hp.dz)::::: Hp.A(Z)
=
sup Lp(zi;) ::::: Hp.K(z), S"EA
5.3. The Fourier-Borel Transfonn of Order p of Analytic Functionals
329
by the very definition of A. Finally, if C is a p-convex set, K S; C, then for ~ E
A, Z E C, Lp(~z) ~
Hp,K(Z) ~ Hp,C 0, from the convexity of the region I Arg z I ~ rr /2p we conclude there is a unique to E [0, 1] such that rp(t) = {
0 Re{(wl
+ t(W2 -
WI»P)
for 0 < t < to, - for to < t ~ 1.
The function cp is hence continuous and differentiable except at to, where it has right and left derivatives. Clearly for 0 ~ t < to, for to < t ~ 1. It follows that Icp(1) - cp(O) I = IL p(W2) - Lp(w,)1 =
~ p(lwIi
=
11'
+ IW2i)P-'lw, -
cpl(t)dtl
w21·
=
If Lp(w,) L p(W2) 0, then cp(t) == 0, and if Lp(WI) > 0 and L p(W2) > 0, then cp is differentiable everywhere and the same inequality holds. Let now z E K, ~" ~2 E C, and set WI = z~" W2 = Z~2' We infer that
and so
hence with Cp(K) := pmax{lzIP:
Z E
K). It follows that:
IHp,d~l) - Hp,K(~2)1 ~ Cp(K)(Rll
+ 1~2I)P-ll~1 -
~21,
which shows that Hp,K is Lipschitz on any bounded subset of C. The continuity 0 is now obvious.
5. Summation Methods
330
5.3.4. Definition. Let E be a non empty subset of S2. The conjugate set defined by
E- := S2 \ where 1/00
{I;-:
Z
E
E
E is
},
= 0, 1/0 = 00. For simplicity 0 = S2, .52 = 0.
5.3.5. Proposition. For any E £ S2, subsets of S2, i E J, we have
E= E. Furthermore,for any family E; of
o
Proof. Left to the reader. Recall from Section 5.1 that for any
Kt 5.3.6. Lemma. For any
Z E
IC* we have
= ({ E C*: Lp(zg) ~ I} U {OJ. Z E
IC*,
if =
(W E IC: Lp(wz) < I}.
Proof. Let us define Az := (w E IC: Lp(wz) < 1). First, let us show that if £ At. If W ¢ Az, then Lp(wz) ~ l. hence wz #- O. Let 1" = I/(wz), ~ = n, then Lp(z/~) = Lp(I/1") ~ 1, which means ~ E Kf. Since w~ = 1"WZ = 1, we have w ¢ if. Conversely, to show that Az £ if, let w ¢ if. Then there is ~ E Kf such that ~w = 1. By definition, Lp(z/n ~ 1, so that Lp(wz) ~ 1, which means that W ¢ Az• 0 5.3.7. Lemma. Let 0
#- B £ C, then
U Kf = (B*P)-. ZEB
Proof. In fact, let G = UzeB Kf, then
{; =
(U Kf) -
=
ZEB
= (V E IC:
n if ZEB
Lp(~z)
< 1, Vz E B)
= B*P. Hence,
o
5.3. The Fourier-Borel Transform of Order p of Analytic Functionals
331
5.3.8. Lemma. If K is p-convex, then
K=
U
Kf·
zeK*P
Proof. From the previous lemma we have
U Kf = «K*P)*P)~, zEK;
while (K*P)*P
=
o
K by the p-convexity.
5.3.9. Lemma. Let 0 f=. K be a compact set in C, e > O. Then K,.p :=
{~ E IC: Lp(z~)
::: Hp.K(z)
+ elzl P ,
Vz
E
IC}
is a compact neighborhood of K.
o
Proof. Left to the reader.
5.3.10. Theorem. Let p > 1, K a compact nonempty set in C, and T E JIf'(K). Then, for every 8> 0 there is a constant C, > 0 such thatJor all ~ E C
Conversely, if f is an entire function for which there is a p-convex compact set K such that for every e > 0 (V~ E C)
for some C, > 0, then there is an analytic functional T carried by K so that 'Jp(T) = f· Moreover, if f(~) = 2:n>O an~n is the Taylor series of f, then the function \II(~) := 2:n>O i(l + n;p)ang n+ l , which is holomorphicfor I~ I sufficiently large. has an analytic continuation to the set K C , and T can be defined by (T,h) = _1_.j
2m
\II(~)h(~)d~,
hE JIf(K),
y
where y is any piecewise C l loop in K C , which has index I with respect to K and lies in the domain of holomorphy of h. Proof. Since T is carried by K, and, for there is a constant AE > 0 such that
8>
0, K E /2.p is a neighborhood of K.
l'Jp(T)(z) I ::: A, sup{lEp(z~)I: We also know that for every
8'
~ E
K E /2.p}.
> 0 there is BE' > 0 such that
and hence. sup IEp(z-nI::: BE' exp (Hp.X 0 so that 8(8') < 8/2 accomplishes the desired estimate. Conversely, consider the integral rpCt;) := p
1
00
e- t - P t p- 1f(tt;") dt.
It converges uniformly on any compact subset of K*P. In fact, if Q c c K*P, there is 0 :::: J.L < 1 and A > 0 such that Hp,KCt;") :::: J.L and It;" IP :::: A for any t;" E Q. Choose 8 > 0 so small that v := J.L + sA < 1, then for t ~ 0 and t;" E Q we have so that e-tP tP-1If(tt;")1 :::: Cetp-1e-(I-V)tP.
We conclude that K*P S; Bp(rp) = Mp(rp) (see Section 5.1). Therefore we have Kf S; Bp(rp) for any Z E K*P. It follows from Proposition 5.1.15 that rp has an analytic continuation to UZEK'P Kf = K. Now let y be a piecewise C 1 loop of index 1 with respect to K and contained in the domain of holomorphy of h E JIf(K). We know that rp admits the Taylor development
!.. L r
rp(t;") =
p
n",O
(1 + ~)
ant;"n
p
in a neighborhood of t;" = 0 (and hence 0 is an interior point of K). Therefore the function 1{fCt;) = (p/r;)cp(l/t;) is holomorphic in K, i.e., in KC, and its Laurent development at 00 is the one required above. This means that we can in fact define the action of an analytic functional T on h by (T, h) := _1_. 2m
ir1{f(t;")h(r;) dt;". y
This defines an analytic functional T carried by K and its Fourier-Borel transform of order p is given by Jp(T)(z)
=
_1_. 2T(l
Taking as y the circle 8B(0, R), R
J
(T)(z) =
I: n,m",O
p
»
(~1 27f1
= I:anz n
=
irEp (zt;)1{f (t;") dt;". y
1, we have r(l
y r(1
+ n/ p) ant;"n-m-l dl;) zn + m/ p)
fez),
n",O
as we wanted to prove.
o
5.3.11. Corollary. Let p ::: 1 and T an analytic functional in C, then there is a smallest p-convex carrier of T.
5.4. Analytic Functionals with Noncompact Carrier
333
Proof. For p = 1 it was done in Theorem 1.3.5. For p > 1, let K be the intersection of all p-convex carriers of T. If ~p(T) = Ln>oan~n, consider V(O = Ln>or(l +nlp)an/~n+l. Then V has an analytic co;tinuation to K C • Therefore, we can represent T by (T,h)
1 .1h(nV(~)d~, = -2 7rl
hE .ff(K),
y
where y is any piecewise C 1 loop in K C , of index 1 with respect to K, and can 0 be taken as close to aK as one wants. EXERCISE
5.3.
1. Use Definition 1.3.12 to show that for p = 1 and T BI (ZC(T)
G))
=
E
.ff(C),
~I(T).
Verify the analogous identity for p > 1.
5.4. Analytic Functionals with Noncompact Carrier In this section we indicate how some of the preceding results about analytic functionals (with compact carriers) can be extended to a more general setting. We follow the work of Morimoto [Mol], [M02]. Let L = [a, oo[ + i [k l • k z] be a closed half-strip of the complex plane, with a E JR, -00 < kl < k2 < 00. For e > 0, e' > 0, < k' < 00, we define the space
°
Q(L, e, k', £') := {f E .ff(ie)
n e(L£): sup If(z)e(k'+£')z I 0 there is C = C(e, e') ~ 0 such that I(S, f)1 ~ C sup I/(z)e(k'H')zl ZELs
for every I E Q(L, e, k', ef). We shall frequently use the following version of the Phragmen-Lindelof principle: 5.4.1. Lemma. Let 0 < e < r, I alunction continuous on Lr and holomorphic such that there are n E N*, C > 0, with the property that in
L
sup I/(w)le- Ciwln
00
= {F
E
.;tf(C): 'ir > 0 sup IF(w)e-(k'+e')wl < oo}. weLr
These two spaces are FS. Lemma 5.4.1 shows that the second one is a closed subspace of the first. If 0 < e~ < e', we have the commutative diagram of continuous linear injections R(C\L; k',
ell - -...,
R(C\L; k', e')
I
r
R(C; k', e;J - - - - . - . R(!C;k',e;J
This diagram induces a continuous linear mapping among the quotient spaces R(!C\L; k', e;J R(C\L; k', e') --+ ~~-----R(C; k', ell R(C; k', e')
~~~-~~
5.4.2. Lemma. The preceding map is injective. Proof. Let F E R(C\L; k',
ell n R(C; k', e'). For 0
0 such that f E Q(L, co, k', c~), then for 0 < c < co, 0 < c' < eb, the integral
1
f(w)Fe,(w)dw,
Fe' E [F],
aL,
converges absolutely. Namely, there exists A 2: 0 such that If(w)1 :::: Ae-(k'+e~)Rew,
WE
Leo,
and, if we let B:=
sup
IF(w)e-(k'+e')WI
0 it satisfies the inequality
sup
15.,(w)e-(k'+e')wl < 00.
w 0, there is a constant C = C(8, 8') ~ 0 such that for every f E Q(L, 8, k', 8') j(S, f)l ~ C sup If(z)e(k'+e')zl. zeL€
338
5. Summation Methods
For any w
E
L2e one clearly has sup zeL,
e-(k'+e')z 1
W -
e(k'+e')z
and thus, A
,
,
sup ISe,(w)e-(k+e)WI
= sup
W¢L2s
1
Z
W¢L2'
I
-2 rr
I(
I
~-, £
Sz,
e-(k'+e')z)1
z
w -
~ -C- , 2rr £
o
which proves the proposition.
5.4.8. Proposition. Let SEQ' (L. k'), 0
0. In the angular sector bounded by the two half-rays ~o + S' 00 (that is, (~o + ~'t : t E [0, oo]}) and ~o + ('00, the function r ~ F(r)e- Wr is of exponential type and decreases exponentially on the boundary. By the Phragmen-LindelOf theorem it also decreases exponentially in the sector. Thus, one can use Cauchy's theorem to obtain
r
F(-r)e- rW dr
=
il;o+I;'oo
r
F(r)e- rW dr
il;o+I;"oo
o
as we wanted to prove. 5.4.18. Lemma. Let Few, ~o) in C\L by
~o
= -(k' + s'). One can define a holomorphic function
Few, ~o)
=
F(w, ~o,;;')
for
wE
wen·
For every s > 0, this function satisfies
sup IFew, ~o)e-(k'+e')wl
0 a function B(F)e' by B(F)e'(w)
= F(w, -(k' + e'»,
WE
C\L.
The family (B(F)e')e'>o detennines an element B(F) of H(C, k'), called Borel (or Laplace) transform of F. Denote B: Exp(] -
00,
-k'[
+ ilR, L)
~ H(C, k'),
F t-+ B(F).
5.4.21. Example. The function F(/;) =ezl; belongs to Exp(] - 00, -k'[ +ilR, L) if and only if Z E L. The Borel transfonn of F is the "Cauchy kernel" used in Definition 5.4.6, I e-(k'+e')(z-w) B(F)e'(w) = - . - - - 27r1
z-w
(w E C\L).
5.4.22. Proposition. The following identity holds:
= idExp(]_oo,-k'[+iIR,L)' 00, -k'[ + ilR, L), S = Int([B(F)]).
~o Int oB
Proof. Let F E Exp(] tion of Int we have, for Re /; < -k' - e',
~(S)(/;)
= -
r
JdL.
From the defini-
B(F)e,(w)e-WI; dw.
Decompose the path aLe into three parts following Figure 5.8. Denote 1;0 = -k' - e', and since in the portion I, we have 1m w k2, we can let Fe,(w, 1;0) = F(W,l;o, -i). Then
= k2 + e >
344
5. Summation Methods
W2.----+----~------------------
II
o WI = (a - e) wI ~__~____~~Ill________________
+ i(k l
-
e)
Figure 5.8
-1-.1 1
=
2ll"1
F(r)
(o-ioo
1 = --.
(le(~-r)w dW)
dr
I
e(~-r)Wl
F ( r ) - - dr. ~o-ioo t; - r
2ll" I
Similarly, if Imt; =1= 0 we have
re~w
In
F(w,t;o) dw
= re~w F(w,t;o, -1) dw In 1 =-
1
2ll"i
e«(-T)Wl -
e«(-T)W3
dr.
t; - r
(0- 1'00
Finally, we also have
r
Jm
e(W F(w,t;o)
dw =
r
Jm
e(W F(w,
= _1_. 2ll" 1
1
= -.
t;o. i) dw
r eW~ (1
JIll
1
2ll" 1
F(r)e- WT
e(,-T)Wl
F ( r ) - - dr. ,o+ioo t; - r
Thus, for Re l; < -k' - e', 1m l; =1= 0, we have ~(S)(t;)
1
= -. 2ll" I
1 c,
F(r)
e(~-r)Wl
l; -
dr) dw
(o+ioo
1
d"C "C
+ -2'
1
ll" I
where CJ, C2 are the paths suggested in Figure 5.9.
C2
e(~-")W3
F ( r ) - - dr. t; - r
5.4. Analytic Functionals with Noncompact Carrier
345
Figure 5.9
An application of Cauchy's theorem yields
1 11 1
e(~-r)WI {-F(~) F(r:)--dr:= ~ - r: 0
27ri
C1
27ri
e(~-r)W3 F ( r : ) - - dr: = c, ~ - r:
{O -F(~)
if if
Im~
if if
Im~
0,
0, > O.
Therefore, ~(S)(O
=
F(~)
whenever Re ~ < -k' - s'. This concludes the proof.
o
5.4.23. Proposition. The following diagram is commutative. Each arrow is a bijective linear map: Q'(L, k')
~ ,nt
C
~
•
ET:
-=.-k'[ +nR.L)
H(C,k').
(/njact, all the maps are topological isomorphisms [Mol] [Mo2].)
5. Summation Methods
346
Proof. We have already shown that C and Int are inverses of each other and J' oInt 0 B = id. It is enough to show B 0 J' = C to conclude the proof. Let J'(S)(I;), 1;0 = -(k' + e'). For WE Wen one has S E Q'(L, k'), F(I;)
=
F(w, 1;0) = -I . 2:;rl
1
~oH'oo
= _1_ / S 2:;ri \
F(r)e- Wr dr = -I. 2:;rl
f
1
(Sz, eZT)e- Wr dr
_oH'oo
er(z-w) dr) = __1_ / S
z, J~oH'OO
2:;ri \
z,
e-(k'+e')(Z-W»)
z-
W
0
= -Se'(W)'
We are now going to extend the results of Sections 4.1 and 4.2. Let 0:::: k' < 1, I; E e- L , then the function z ~ 1/(l -seZ) belongs to Q(L, k') (as the reader can easily verify). Thus, for any T E Q'(L, k') we can define a function G(T) by
_1_),
G(T)(I;) := / Tz , \ I -I;e z
which is hence defined for I; E Q (L) := C\r L and has the following properties:
5.4.24. Proposition. Let T E Q'(L, k'), 0 :::: kif < 1. Then: (i) Thefunction G(T) is holomorphic in Q(L). (ii) The Laurent development of G(T) at I; = 00 is given by
G(T)(l;) = -
L n:;:J
J'(T)(-n) ,
I;n
Proof. The first item is an immediate consequence of Morera's theorem. The second item follows from the development
which, for 11;1>
e-(a-£),
converges in Q(L. k').
o
If the half-strip L has width k2 - kJ < 2:;r, then Q (L) contains the angular sector A(-kJ, -k2 +:;r) := {I; E C*: - kJ < argl; < -k2 + 2:;r}. We are now going to describe the behavior of G(T) in this angular region.
5.4.25. Proposition. Let L = [a, oo[ + i[kJ, k 2] with k2 - kJ < 2:;r and let 0:::: k' < 1. If T E Q' (L, k'), then for every e such that 0 < 2e < 2:;r + kJ - k2 and every e' such that 0 < e f < 1 - k' there is a constant C ~ 0 with the property that C
IG(T)(~)I !S 1l;lk'+e'
5.4. Analytic Functionals with Noncompact Carrier
347
Figure 5.10
for every
Proof. From the continuity of T there is a constant C' e- L '/2. then
IG(T)(~)I =
1
I(Tz • -1--)1 ~ez
-
~
C'sup ZEL
I
~ c' sup _e_ _ ZEL'/2
e(k'+s'-l)z
I
e- Z -
(k' +s')z
~
0 such that if
I
1 - ~ez
I
~
~ C' (sup le- z _ ~ I-k'-S') zEL'/2
~
C' dist(~, e- L'/2)-k'-s' sup 11 _ zeLE/l
eZ~lk'+s'-I.
~
¢
348
5. Summation Methods
Since e- L '/2 S; A(-k2 - s12, -k\ + s12), for t; E Q(L f
)
we have
dist(t;. e- L '/2) ~ 1t;1 sin(sI2). On the other hand, for
Lf/2 and t; E A(-k\ + s, -k2 + 27r - s). we have
E
Z
_
s
larg(~e-)I ~
2(mod27r).
so that inf 11 - t;e=1
~
sin(sI2).
;eL E /2
Therefore. the inequality of the statement holds with
e = e'l sin(s 12).
0
Let us keep the hypotheses of the preceding proposition from now on. Denote 1io(Q(L). k') the space of all holomorphic functions q; in Q(L) such that: (i)
lim q;(t;) = 0; and
I~I-""
(ii) sup{lq;(t;g-k'+e'I: t; E A(-kl + s, -k2 + 27r - s)} < 00, whenever 0 < 2s < 27r + k. - k2, 0 < s' < 1 - k'. This space provided with the seminorms sup{Iq;(t;)I: 1t;1 > e-a+e} < sup{lq;(t;)t;k'+e' I: t;
E
A(-k\ + S, -k2 +
00,
27r - s)}
0, be sufficiently small so that 0 < 2s < 27r - k\ + k2' 0 < s' < 1 - k', and h E Q(L, s, k', s'). Then: (i) For p > 0 the function
h (z) = - 1 p 27ri
1
aL"
hew) dw 1 - e=-W
belonl:s to Q(L, 1) (and hence to Q(L, k'», with the contour aL e.p = aL f n {w: Re w ::: pl. (ii) For Z E I hew) h(z) = - . z-w dw. 2m iJL, I - e
If
1
5.4. Analytic Functionals with Noncompact Carrier
349
(iii) The limit lim hp = h
p-+oo
holds in the space Q(L, k').
Proof. (i) The fact that hp is holomorphic is clear. On the other hand, sup Ihp(z)eZI = sup ZeL,j2
ZeL,j2
1
~
iJL,
1
aL,
h(w)(e- Z - e-w)-I dwl p
Ih(w)1 dist(e- W , e L'/2)-1 dlwl
0 is fixed), i.e., /J.
I 1x+r/2
* f(x) = -
T
f(t)dt = 0,
x-r/2
where /J. = (l/T)X[-r/2.r/2j. This function f is periodic of period T (as it can be seen by differentiation of the above convolution equation) and has zero "mean" (average) over any interval of length T, whence the name mean-periodic.
353
6. Hannonic Analysis
354
6.1. Convolution Equations in R Recall that according to the Paley-Wiener theorem, F: £'(lR) ~ Ap(C) is an isomorphism where p(z) = I Imzl + Log(2 + Izl). As before, we denote by £(lR) the space of the COO-function on JR, with its usual Frechet topology. 6.1.1. Definition. Let 0 i= f.J- E £'(JR). We say that a function f E £(JR) is f.J-mean-periodic (or simply mean-periodic) if it satisfies the convolution equation
un
= {f E £(JR): f.J- * f 6.1.2. Proposition. The set of £(JR) which is invariant (under translations).
= O} is a closed subspace
Proof. We first recall that given h E JR one defines the translation operator Th by Th(f)(X) = f(x - h). To say that un is invariant under translations means that Th (un) ~ un for every h E lR. Since Th (f) 8h f, the fact that un is invariant is clear. The other property is also immediately apparent from the fact that the convolution operator f.J-*: £ (JR) ~ £ (IR) in continuous. 0
= *
6.1.3. Proposition. If f.J- i= 0 and f.J- i= c8a for any a E JR and c E C*, the space Ker f.J-* is a proper invariant subspace of £(IR).
un =
Proof. First we remark that un ~ Ker j1, where j1 is considered as a continuous linear functional on £(IR) (and not as a convolutor). This follows from the definition of convolution:
0= f.J-
* f(O)
= (j1, f).
Hence, if f.J- i= 0, then VJt i= £(IR). On the other hand, if f.J- is not of the form c8a , then the entire function Ff.J- has zeros [BG, §4.S.11J. Let ).. be such a zero and f(x) = exp(i)..x), then f.J-
Therefore
un i=
* f(x) = Ff.J-()..)f(x) = O. o
(OJ.
6.1.4. Remark. This proof shows that the exponential eih of frequency).. lies in VJt if and only if ).. is a zero of :Ff.J-. There are other simple functions in namely, the exponential monomials xke ih , when).. is a zero of FJ-t of multiplicity m > k. In fact, one has the identity
un,
k
(f.J-t, (x - t/eiA(X-t») = e iAX
L j=O
= eih
(~) x k- j (-i)j (fLt, tje- iAt ) J
t (~) j=O
J
x k-
j (-i)j (FJ-t)(j) ()..).
6.1. Convolution Equations in R
355
From this identity we see that if the multiplicity of the zero A of :Ff.,L is m, then xke iAX E 9J1 for 0 ~ k ~ m - 1. Conversely, if for some k, A we have xke iAx E 9J1, the polynomial on the right-hand side of this identity is identically
zero, hence (:Ff.,L)(j)()..) multiplicity m > k.
= 0 for j = 0, ... ,k. Therefore, ).. is a zero of :Ff.,L of
6.1.5. Proposition. Let 9J1 be a closed invariant subspace of £(l~), and let (!JJt).L = {f.,L E £'(l~): (/L, /) = 0, V f E 9J1}. Then (!)]l).L is a closed ideal of the convolution algebra £'(l~). Moreover, (9Jt).L = {f.,L E £'(l~): f.,L f = O,for all
*
f
9J1}. If'J is a closed ideal of £'(R), then (J).L = If E £(R): (it, f) = 0, Vf.,L E 'J} is a closed invariant subspace of £(R) which coincides with the subspace of £(R) given by If E £(R): f.,L * f = 0, Vf.,L E 'J}. Furthermore, under these conditions, «(m).Lf).L = 9J1 and (((J).Lf).L = 3. E
Proof. It is evident that (m).L is closed even if 9J1 were not closed. Once we show that (9Jt).L = {f.,L E £'(R): f.,L * f = 0, Vf E 9J1} is true, it becomes clear that (m).L is an ideal of £'(R). Let f.,L E (fUt).L, f E 9J1 then (f.,L
* J)(x) = (f.,Lt, f(x -
= (f.,Lt, (LAJ)f(t») = 0, ~ {f.,L E £'(l~): f.,L * f = 0, Vf E 9J1}. The other t»)
since LAJ) E 9J1. Hence (fUt).L inclusion is evident. We also have, for f.,L E J and f E (J).L, (f.,L,
rAfn = (f.,L * 'l"x(J)(0) = (f.,L * 8x * J)(O) = (f.,L * Ox, /) = 0,
*
since f.,L 8x E'J for every x E R Hence rAJ) E (J).L. The identification of (J).L is similar to the previous case. The final statement is a consequence of the Hahn-Banach theorem [Hor]. 0 Now let 0 i: f.,L E £'(R) and consider 9J1 = Ker f.,L*. We can identify (9Jt).L immediately as follows. Let I = :F«fUt).L). Then 1 is a closed ideal in the algebra Ap(C). From Remark 6.1.4 we know that if (A, m) E V(:Ff.,L), the multiplicity variety associated to :Ff.,L, then xke iAX E 9J1 for 0 ~ k ~ m - 1. Hence, if v E (9Jt).L we must have v * xkeO. x O. This implies that :Fv vanishes at the point A with multiplicity at least m. Therefore :Fv = ¢:F/L for some entire function ¢. In the terminology of Chapter 4, if we let J be the algebraic ideal in Ap(C) generated by :Ff.,L and let (:Ff.,L) be the algebraic ideal generated by :Ff.,L in Jt"'(e), we see that 1 = Ap(C) n (:Ff.,L). In other words, I = Jloc' If we define 9J10 as the invariant closed subspace generated by the exponentials monomials xke iAx , with 0 ~ k ~ m - I and (A,m) E V(:Ff.,L), then the preceding argument shows that 10 := :F(Cmo).L) coincides with Jloc' The reason is that to show that 1 = Jloc we only used the exponential monomials in 9J1. It follows, from Proposition 6.1.5 that 9J1o = 9J1. That is, the exponential polynomials solutions of f.,L * f = 0 are dense in the family of all solutions.
=
356
6. Harmonic Analysis
Lemma 6.1.8 and Theorem 6.1.9 extend this result to arbitrary closed invariant subspaces 9R. Before we proceed, let us point out that there is another way to obtain invariant subspaces. 6.1.6. Definition. Let f E £(JR). We denote by 'r(f) the closure of the space spanned by Tx (f), X E R This space is the smallest closed invariant subspace of £(JR) containing f. 6.1.7. Proposition. A function f E £' (JR) is fL-mean-periodic for at least one fL 1= 0, fL E £'(R), if and only lj'r(f) 1= £(R). Proof. It is clear from the previous Proposition 6.1.5 that 'r(f) = £(JR) if and only if the ideal ('I(f)').l is the zero ideal. If not, let 0 1= fL E ('r(fr ).l. Using 0 the same proposition we conclude that fL * f = O. We have just seen that the exponential polynomial solutions of a single convolution equation fL * f = 0 are dense in the family of all solutions of the same equation. Of course, we could not hope to obtain Fourier-type expansions of the solutions unless this density property were true. We will show in detail how to accomplish this expansion in Theorem 6.1.10 under some restrictions on fL. What is not immediately apparent is whether a given function could not have two different Fourier expansions. For instance, the reader could ask himself what happens with a function f which is both I-periodic and 2-periodic. How about I-periodic and J2-periodic? More generally, we could ask what happens with the solutions of systems of homogeneous convolution equations. Is it possible for such a system to have some nonzero solutions and, at the same time, have no exponential solutions? This problem is a particular case of the spectral analysis problem, which asks to decide whether given a nonzero closed invariant subspace 9R there is or there is not an exponential monomial xke iAx E 9R. Let us denote by 9Ro the closure of the span of the exponential monomials in 9R. The spectral analysis problem becomes: Does 9R 1= 0 imply 9Ro 1= O? From the previous Remark 6.1.4 and Proposition 6.1.3 we know that when 9R = Ker fL* then 9Ro 1= to} if and only if fL is not of the form coa , in which case 9R = 9Ro = to}. We also have seen for 9R = Ker fL* that 9R = 9Ro always hold. This question can be posed for arbitrary nonzero closed invariant subspaces 9R. The spectral synthesis problem for 9R is: Is
9Ro = 9)1?
If this holds for every closed invariant subspace 9)1, we say that the spectral synthesis property holds for £(JR). It is a famous result of L. Schwartz [Schwl] that the spectral synthesis property holds, in fact for £(JR). One can generalize these concepts to cORn) and find the surprising fact that the spectral synthesis property does not hold for cORn), n ~ 2 [Gul], [Gu2].
6.1. Convolution Equations in IR
357
We are sure the reader recognizes the relation between this question and the Wiener-Tauberian theorem for the algebra LI(JRn). We need a few more definitions before proceeding to prove Schwartz's theorem. Let 9J1 and 9J10 be as above, denote by I, 10 the closed ideals in Ap(C) given by F«mn.L) = I, 10 = F«Mo).L). Let V be the multiplicity variety associated to I, it is called the spectrum of 9J1. We recall also that Iloc is a closed ideal. For a function j, its spectrum is simply the spectrum of'I(f). 6.1.8. Lemma. 10 = I loc ' Proof. We remind the reader that Iloc = Ap(C) n I (V), where I (V) is the ideal in )f(C) of functions vanishing on V. First we want to show that Iloc s:; 10. Let xk e iAX E !mo. Then, by Remark 6.1.4, F p,(A) = 0 with multiplicity > k for every p, E (9Jt).L, since 9J10 S:;!m. It follows that (A, m) E V for some m > k. Hence, if F1) E Il oc , Fv will vanish at the point A with order at least m. The same remark shows that v * xke iAx = O. Therefore, v E (9Jto).L and Fv E 10. On the other hand, if Fv E 10 and (A, m) E V, we need to show that F1) vanishes at A with multiplicity at least m. If that were the case we could conclude that FlJ E I(V), hence Fv E I loc ' By the definition of V, for every p, E (M).L we have that Fp,(A) = 0 with order at least m, hence for 0 :s k :s m - I, p, xke iAX = O. Thus, xke iAx E «(9Jt).L)").L =!m. In other words, xke iAX E 9J10 for O:s k :s m - 1. This forces any Fv E 10 to vanish at Ato the order at least m. That is, 10 ~ I loc ' 0
*
From this lemma we can conclude that 9J1 admits spectral analysis if and only if Iloc "# Ap(C), which is equivalent to the statement that V oF 0. On the other hand, from the same lemma we can conclude that !m admits spectral synthesis if and only if I = I loc . Namely, I = I loc if and only if I = 10. Equivalently, if and only if, (9Jt).L = (9Jto).L. 6.1.9. Theorem. E(JR) has the spectral synthesis property. Proof. We follow the ideas of the proof given by Schwartz [Schwl]. Let !m be a closed invariant nonzero subspace of E(JR.) and let I be the corresponding ideal of Ap(C). In order to prove I = I loc we can apply Lemma 2.5.5 to reduce this problem to the case where the spectrum V of !m is empty. In this case, all we need to prove is that I = Ap(C). Let us first prove an auxiliary lemma.
6.1.10. Lemma. Let I be a closed ideal in Ap(C) which is invariant under differentiation. Then, either I = (O) or I = Ap(C). Proof of Lemma 6.1.10. Assume that I is a proper ideal of Ap(C) and let I be the proper ideal of E'(JR) such that F(I) = I. Then, from the properties of the Fourier transform, we conclude that I is invariant under multiplication
6. Hannonic Analysis
358
by the function x, hence, by multiplication by polynomials. Since I ¥- {O}, there is IL ¥- 0 E I. As we can find cp E V(lR) such that cp * IL ¥- 0, thus cp * IL E In V(R), we might as well assume that IL E V(R). Further, using a translation if necessary, we can assume IL(O) ¥- O. Let R > 0 be such that supp IL C [- R, R]. The Weierstrass approximation theorem allows us to find a sequence of polynomials Pn such that: ~
0 on [-R, R]; (ii) J~R Pn (x) dx = 1; and (iii) for any e > 0, Pn(x) -l> 0 uniformly in [-R, -e] U [e, R] as n (i) Pn
-l>
+00.
Let f E Il., thus T:yf E Il. for any y E R Since PnlL E I we have
I:
0= (PnJL, T:yf) =
Pn(X)IL(X)T:y (f)(x) dx
Therefore, 0 = -cyf(O) = f(-y) and so, f I = e'(lR), a contradiction.
-l>
IL (O)(T:y (f) (0».
== o. This proves that Il. = {O}
and 0
It is clear now that to conclude the proof of Theorem 6.1.9 we need to prove that if cP E I then its derivative cp' E I also. Recall from [BO, §4.6.l5] the Hadamard canonical product of cp is
cp(z) = zneaz+p
g[(
1 - :k) ezfz'r' ,
where n ~ 0 denotes the multiplicity of Z = 0 as a zero of cp and mk are the multiplicities of the distinct zeros Zk ¥- 0 of q;. As cp has order one, we have mk L -IZkl2 - 0, O:::i:::m,-l
{;eBP(z
C::OI~-] lak,ll)
C2, the canonical injection Ap,c, (V) ~ Ap,Cl (V) is a compact map. It is also evident that if we restrict the value of C to a strictly increasing sequence converging to +00 the corresponding .c~-topology is stronger that the topology of the coordinatewise convergence for Ap(V), Then this topology is the same as the one obtained using all values of c. It follows that this topology is DFS. Since the map p is continuous and surjective from Ap(C) to Ap(V), and Ap(C) is also a DFS space, it follows from the Open Mapping Theorem that p is a topological isomorphism. To prove the second part, let us show that any sequence a = (ak,l) satisfying the estimates from (ii) defines a continuous linear functional on Ap(V) by b
~ (a, b) = L
L
aklbkl .
k!,:IO:;:/:;:mk- 1
In fact, we have I(a, b)1
:s L
L
laklbktl
k!': 1 0:;:1 :;:m,-I
:s IIbll e LeCP(Z;) k!': I
L
lak.li:S const.llbll e .
0:;:/ :;:m,-I
This shows the continuity on each space Ap,c(V), Conversely, given an element v E (Ap(V))', we define akJ = (v, E k,/),
where is the Let B fJk,1 = mk
:s
E k,/ is the sequence in Ap(V) whose only entry different from zero one whose index is (k, I), and the value for this entry is exactly one. > 0 be fixed and define bkl = eBp(Zk l (hi, where fJkl = Qkil ak/ if akl =1= 0, 0, otherwise. Since there are positive constants A 1 and A2 such that AI p(Zk) + A2 for every k (see Exercise 2.2.9) then Ibkl! = eBp(Zklmk
L
:s
(A1P(Zk)
+ A 2)e Bp (Zkl,
O:;:/:;:mk-I
Choose any c > B, then it follows that b = (bk,/) E Ap,c(V). (It is this reasoning that allows maxI lakJ I and L:/ lak,11 be interchangeable in the definition of (Ap(V»'.) Therefore, all the finite sequences
satisfy Since v is continuous on Ap(V) we have I(v, ben»)! :::: Mcllb(nlil c :::: Mcllbllc
0). When L = 0, /L is an ordinary differential operator with constant coefficients (up to a translation). In this case, one needs the values of f and a certain number of derivatives at one point to vanish. We will assume that L > 0 unless stated otherwise. 6.1.19. Proposition. Let f E £(IR) be the solution of two convolution equations /L * f = v * f = 0, and assume that f has two representations f(x) = (>..,m)eV(FJL)
(a,n)eV(Fv)
which converge in £(IR). Then each a in the second series for which Qa ¢. 0 is a zero of F/L and moreover Qa = Pal and a similar statement holds for ).. in the first series.
Proof. If Q,,(x) contains the nonzero tenn aa,IX'e iax let Va = Va,o be the distribution defined in Lemma 6.1.14. It is easy to see that Va
* f(x) =
Va
*(
L
Qp(x)e iPX ) = Qa(x)e iax .
(,B,n)eV(Fv)
Hence /L * (Qa (x)e iax )
= (/L * Va * f)(x) = 0,
which implies that F/L(a) = O. In the same way the zeros).. of F/L which appear in the first representation must be zeros of Fv. The fact that Per = Qa is due to the uniqueness of the coefficients in Corollary 6.1.15. 0
6.1.20. Corollary. If f has a convergent representation f(x) =
L
P>..(x)e iAX
A
in terms of v-mean-periodic exponential polynomials, then the only nonzero coefficients are those P>..(x)e iAX in 'r(f). In other words, P}" is different from zero if
6.1. Convolution Equations in lR.
367
and only if A. belongs to the spectrum of f with multiplicity bigger than or equal to (degree PA) + 1. Proof. It is enough to observe that in the previous proposition one does not need to know that f has a representation with respect to J1, to conclude that F J1, vanishes at A to the order (deg PA) + 1 when J1, * f = O. D
6.1.21. Examples. (l) If a function f E t' (lR) is periodic of period. > 0, then of the convolution equation J1,
* f(x) = «8, -
80)
* f)(x) = f(x -.) -
f(x)
f
is the solution
= O.
The variety V of F J1,(z) = e- in - 1 is exactly the collection of points 2krr I., k E Z, every point with multiplicity one. The function F J1, is slowly decreasing since it is an exponential polynomial and V is an interpolation variety. The distributions J1,k J1,k,O are given by
= (J1,b t) = i Pk,O (8, -8 0 , e- i(2rrk/,)x = -iPk,o
l'
1
00
t(s)e i(2rrk/')SdS)
t(s)e i(2rrk/,)sds.
Here, Pk,O = 1/(FJ1,)'(zk) = 1/(-iT) and the coefficient ak of ak
=
(J1,b /) =
~
r
• io
11'
=_ •
fC-s)e i(2rrk/,)sds =
~
•
1°
f is given by
fCs)e-2rrikS/Tds
-T
f(s)e-2rriks/rds.
0
This is the classical Fourier coefficient of f(x)
=
f, f is represented by the series
00
L
ake2rriks/T,
k=-oo
and the estimates from Theorem 6.1.11 are 00
L
eBp(Zk)lakl
O. Here P(Zk) = log(2 + 1(2rri/.)kI 2), hence we have ktoo
(2 + 12;i
k1
2 )
B lakl
o.
0 means that f is a solution constant coefficients, namely, The characteristic equation is
6. Hannonic Analysis
368
Ft-t(')..) = 2:~=o Ck(i')..)k = 0 and the exponential monomial solutions are x j eiAX if ').. is a root of F t-t of multiplicity bigger than j. This is the classical result of
Euler. (3) As a final example let us consider the difference-differential equation f'ex) = f(x
+ 1).
This equation corresponds to convolution with t-t = 80- 8_ 1 , its Fourier transform is Ft-t(z) = iz - eiz and (Ft-t)'(z) = i(t - eiz ). The zeros of Ft-t are all simple as we can thus see easily. Let V = V(Ft-t) = {ztJ, since (Ft-t)'(Zk) = i + Zb then V is an interpolation variety by Proposition 2.7.8. In fact we can describe reasonably well the roots Zk as follows. Let i z = pe i8 , then the equation defining V becomes {
-n < () k E Z.
p cos () = log p, p sin () = () + 2kn,
:s n,
The first equation implies that cos e > 0 for p > 1, this restricts () to the interval -n /2 < e < n /2. Furthermore, we can see that p2 _ (log pf {
tan()
= «() + 2krr)2, = e + 2kn. logp
From the first equation Pk ::::::: 2lkln. Hence, the second equation becomes tane :::::::
2krr , log21kln
which shows that for k ~ +00 one has a root ()k ~ n /2 and for k ~ -00 one has a root ()k ~ -n /2. Therefore, one can improve the estimate of Pk to Pk ::::::: 21kln
.
n
+ (slgnk)2"'
The corresponding roots Zk are such that
and arg Zk ::::::: Arctan (
2krr ) - ~. log21kln 2
It follows that the Zk lie in the lower half-plane. For k ~ +00, arg Zk ~ 0 and, for k ~ -00, argzk ~ -n. Furthermore, one can see IImzkl = Loglzkl hence p(zt> ::::::: Log IZk I : : : : log 2lkln. From these considerations we obtain that
any COO-solution of this difference-differential equation has the form 00
f(x)
=L
akeiz,x
k=-oo
with lakl = O(lkl- N ) for all N > O. We can also give an explicit formula to compute the coefficients ak' We have, after a simple computation, and setting
369
6.1. Convolution Equations in lR J-Lk
= J-Lk.O, 'Z
I +Zk
=. (I
{1°e "I(-s)ds -
Zk·
v
a/.. = (J-L/.., f) = -.-Zk,
[
-I
t
+ Zk)(IZk)N io
'}
I
- . 1(0) IZk
e-iz,s I(N)(s)ds _ !(N)
(0)] .
IZk
The last line is obtained using the equation and integration by parts. It shows the coefficients have the correct rate of decay. We have seen some properties of Fourier development and uniqueness for solutions of a homogeneous convolution equation and one could ask when is the inhomogeneous equation J-L * 1 = g solvable.
6.1.22. Proposition. Let 0 i= J-L is surjective
if and only if FJ-L
E £'OR). Then the operator J-L*: £(lR) -+ £(lR) is invertible.
Prooj. The surjectivity of J-L* is equivalent to the injectivity and closed range of the transpose operator IH: £'(lR) -+ £'(lR). By Fourier transformation the
last statement is equivalent to the fact that the multiplication operator by Fji in Ap(C) -+ Ap(C) must be injective and have closed range. The injectivity is obvious since J-L i= and the range is the principal ideal j generated by F ji in Ap(C). Since FJ-L is invertible then 1 = (FJ-L)Ap(C) is closed. Since 1 is closed if and only if j is closed we have proved the surjectivity. Conversely, if J-L* is surjective, then j is closed, and by the spectral synthesis property 1 = Iloe, which is precisely the definition of the invertibility for F J-L. 0
°
Before continuing the discussion of inhomogeneous convolution equations, let us consider the relation between the two concepts, slowly decreasing and invertibility for J-L E £' (lR). Let us recall that in Definition 2.2.13 we have defined 1 = F J-L to be slowly decreasing if and only if there exist constants e > 0, c > 0, and A I > 0, such that the connected components Oa of the set
SClII, e, c)
= {z E C:
are relatively compact and, for every z,
11(z)1
W E
< ee- cP (:)}
00' we have
p(Z) :::: AIP(w).
We have also proved in Proposition 2.2.14 that if 1 is slowly decreasing then the ideal 1 = 1 . Ap( 0, c > 0, Al > 0, A2 > 0, with the properties imposed earlier. Let a > 0 be such that '>Ir 2: O.
Assume x E IR and Ifex)1 < (a + Ixl)-I, then x E S(lfl, e, c). Let 0 be the connected component of S (I f I, e, c) containing the point x and let x + i Y be a point in 80 n (x + iIR). Then
Iyl :::
p(x
+ iy)
::: Atp(x) + A 2 ,
that is, and
If(x + iy)1 =
e-cp(x+iy)
2: 8(2 + Ixl)-Y
for convenient 8 > 0, y > O. From the Minimum Modulus Theorem [BG, §4.S.14J or Lemma 2.2.11 above, we have that for any R > 0 there is a value r, R/4 ::: r :5 R/2, such that mip.
1~-(x+ly)l=r
log If(nl 2: 9y log(2 + Ix\)
+ 9log8 -
Slog M,
6.1. Convolution Equations in
~
371
where M =
max
1~-(x+iY)I::::2eR
I/(~)I.
Let us choose R = 51YI, then for any ~ in B(x + iy, 2eR) we have that ::s 311yl, IRe~l::s Ixl + 301yl. Thus, for any such~,
Ilm~1
1/(s)1 ::s
::s (2 + Ixl)BI
C1eCzp(n
for a convenient constant Bl > 0, since p(~)
+ log(2 + 611yl + Ixl) ::s 31Allog(2 + Ixl) + 31A 2 + log(2 + 61A 1 log(2 + Ixl) + 61A2) ::s B o log(2 + Ixi). ::s 311yl
Therefore there exists
KO
> 0 such that
min
l~-(x+iy)l=r
log 1/(s)1
~ KO
log(2 + Ixl)
and since r ~ ~IYI there is x' E JR so that lx' - (x that lx' - xl ::s r ::s ~IYI. Thus Ix' - xl ::s ~(Al log(2
+ Ixl + A 2)
+ iy)1 =
r, then it follows
::s a log(2 + Ixl)
and I/(x')1 ~ (2
+ Ixi)-Ko
:::: (a
+ Ix'I)-a
for a > 0 conveniently chosen. Hence, we have just proved that (1) implies (2). To show (2) implies (3) given a point z = x + iy, using (2) choose x' E JR, lx' - xl ::s a log(2 + lxI), and I/(x') ~ (a + Ix'l)-a. We can assume that a:::: 1, thus Iz - x'i ::s a log (2 + Ixl) + Iyl ::s ap(z). Therefore z' = x' satisfies (3). We are going to prove that (3) implies (1). Given a point z we know there exists a point z' such that Iz - z'l ::s Cp(z) and I/(z')1 ~ e-Cp(z') for some C ~ 1. Let R = 51z - z'l, then for I~ - z'l = 2eR we have I Im~1 ::s I Imz'l
+ I~ -
+ 281z :5 11m zl + 291z -
z'l ::s I Imz'l
z'l z'l :5 30Cp(z).
Similarly, we have I~I ::s Iz'l
+ I~ -
z'l ::s Iz'l
+ 2eR ::s ::s
+ 281z - z'l Izl + 291z - z'l ::s Iz'l
Izl
+ 29Cp(z).
Hence,
+ log(2 + I~I) ::s 30Cp(z) + log(2 + Izl + 29Cp(z»
peS) = I Im~1
::s A1P(z)
for some Al > O. Applying the Minimum Modulus Theorem as done earlier we obtain a circle r of center z' and radius r ~ Rj4 so that Z E Int(r) and on r we have an inequality of the form
372
6. Harmonic Analysis
for a convenient choice of e > 0, c > O. Moreover, the same computation shows there is A2 > 0 such that for any two points w, Wi of Int(r) one has pew) s A2P(W'). 0 6.1.25. Lemma. If / is not slowly decreasing there is an unbounded/amily (gj)j in Ap(C) such that the/amity (jgj)j is bounded.
Proof of Lemma 6.1.25. Since / is not slowly decreasing then condition (2) cannot be satisfied for any choice of a ::: 1. Hence, for any j E N* there is Xj E JR. so that: (i) IXj I ::: e3j ; and (ii) for any X ElR, Ix-xjl sjloglxjl implies
For each j let k the function
1/(x)l:::: Ixl-J.
= [j log Ix) 11. Recall that sinc z = sin z/ z and let us define
hJ(z) = (sinc(rrz/j»j
=.r ((irrX[-j/rr,)/rr1)*j)
(z),
where the power represents repeated convolution. The family that will work is given by First, we observe that all the functions h j are of exponential type rr. Hence, the same is true for the gj. On the real axis, we have the following properties: (a) hj(O) = 1; (b) Ihj(x)1 :::: 1 (x E JR.); and (c) Ihj(x)1 :::: (j/rrlxl)j for Ixi
'# O.
We claim that the sequence (gj) is not bounded in A p (C). For that, it is enough to show that there is no value n > 0 such that Igj (Xj) I :::: IXj In for every j ::: 1. In fact, Igj(Xj)1 = eklhj(O)1
= ek ::: ~Ixjlj. e
We want to show now that the sequence /gj is bounded in Ap(C). All these functions are of exponential type:::: rr + (type of f). From the PhragmenLindelOf theorem we conclude that it is enough to prove that / gj are uniformly bounded on JR. by a function of the form (2 + Ixl)N for some N. For Ix - Xj I :::: j log IXj I we have from (b) that Igj(x)1 = eklhj(x - x})1 :::: ek ::::
On the other hand, for Ix -
x}
I :::: j log IXj I we have
IXj I
:::: Ix I + j log Ix} I
Ix)I}.
6.1. Convolution Equations in R
373
while
so that and Therefore, for the same range of x, Ix - x] I :::: j log IXj I, one has that I/(x)g] (x)1
For the complementary range,
::::
Ix -
IXjlJlxl- j :::: 2 j ::::
IxI-
Xj I ::: k. using (c) one has
I/(x)gj(x)1 :::: I/(x)l/n i ::::
I/(x)1 ::::
(2
+ Ixl)N
for some N. This concludes the proof that the family (f gj )j is bounded in Ap(C) while (gj)j is not. 0 The last lemma shows that condition (8) implies concludes the proof of Theorem 6.1.23.
I is slowly decreasing. This 0
6.1.26. Remark. Using the same type of argument the reader can find in [Eh3] a few other important equivalences of the slowly decreasing condition. For instance, I is slowly decreasing if and only if: (9) {t
* V' (~) = V' (~);
(IO) {t*: V(~) -+ V(~) is an isomorphism onto a closed subspace of (11) for any v E £' (~), if {t * v E V(~), then v E V(~). These equivalences are also valid in [H02] for details.
~n.
V(~);
The reader should consult [Eh3] or
For {t slowly decreasing such that cv(supp {t) = [a,,8] let us consider the Cauchy Problem {
{t*1
=g,
II[a, ,8] = h,
where g E £(~), h E £([a, ,8]). From Corollary 6.1.15 we know that if V(F{t) is an interpolation variety for Ap(C) then (*) has at most one solution I E £(~). (Remark 6.1.16 implies that the uniqueness is true for any {t f. 0.) The obstruction to find a solution is only to be able to solve the simpler problem { {t
* 10 = 0,
101[a,,8]
= ho,
for ho E £([a, ,8]). Namely, to solve the Cauchy problem (*) we choose an arbitrary solution II of {t * 11 = g, which exists by Proposition 6.1.21. If f is
6. Hannonic Analysis
374
a solution of the Cauchy problem (*), then 10 = I - It is a solution of (**) with ho = h - (fll[a, .8]). Conversely, if (**) is solvable we take I = 10 + II to solve (*). It is not true that (*) and (**) are always solvable when /.t is slowly decreasing. The first difficulty is that (**) requires obvious compatibility conditions for ho and its derivatives: for any integer n ::: 0,
0= (/.t * lrin»(O) = (/.t, (fo)(n») = (/.t, h~n»). The corresponding compatibility conditions for (*) are g(n)(o) = (/.t
* I(n»)(o) =
(/.t, (h(n») = (/.t, hen»).
The above procedure of reducing (*) to (**) when F/.t is slowly decreasing preserves the compatibility conditions. In any case, it is not generally true that (**) can be solved. If every h satisfying 0 (/.t, hbn») can be extended to a solution of (**) one says that /.t is hyperbolic. The necessary and sufficient condition for hyperbolicity is the following:
=
6.1.27. Theorem. A slowly decreasing distribution /.t such that V is an interpolation variety is hyperbolic if and only if there exists a constant C > 0 such that:
lor every zero
Zk
01 F /.t.
6.1.28. Remark. The theorem is valid without the assumption that V(F/.t) is an interpolation variety [Eh3]. The proof is similar to the one given below except for the grouping of terms. Proof. Let 001 = Ker /.t* and E = {4> E [([a, .8]): (/.t, 4>(n») = 0, Vn EN}. Since the map I ~ fl[a,.8] is an injective continuous map from 001 into E, it is surjective if and only if it is a topological isomorphism. Hence, if /.t is hyperbolic and e > 0 is fixed, there are C I > 0 and N EN such that for any f E 001 I/(x)1 ~ C I
sup xE(a-e,/l+eJ
Let
Zk
be a zero of Fp, and I(x) sup a-e:",x:",/l+e
sup 1/(j)(y)l. yE(a.IlJ O:",j:",N
= e- iXZk . We obtain immediately
I/(X)I=eXlmZk~CI(2+lzki)Nexp(
sup YlmZ k). yE[a,IlJ
From this inequality one can easily conclude that eel Imzkl ~
C 1(2 + IZk I)N ,
which is the condition we wanted to obtain. Note that we only need to use the extension to an open interval of length> L fJ - a. We continue the proof with the following lemma:
=
6.1. Convolution Equations in R
375
6.1.29. Lemma. Let be 0"# IL E £'(lR), cv(supp IL) = [a, ,8], a < ,8,:FIL slowly decreasing, V (:FIL) = V = {(Zk, md} an interpolation variety satisfying 11m Zk I ::: C log(2 + IZk I)
(k:::: l).
If .rIL(O) = 0 we denote Zo = 0, if not the index k runs only through k :::: 1. Then,for every S E [' OR), there exist two distributions So and SI, such that SI is of the form Lk>O S I.k, the series is convergent in [' (JR). Each S I,k is a linear combination of th; derivatives ILk:] for a convenient q E N (q = 0 for k = 0), so that S = IL
* So + SI
cv(supp SI) ~ [a,,8] and the ILk,1 are the distributions defined in Lemma 6.1.14. Furthermore, for any B > 0, q can be chosen to depend only on B so that the map S ~ SI is linear and continuous from F-I(Ap,B(C» into £(",lll(JR) (the distributions with support in [a, ,8].) Finally,for R > 0 and any x E JR, Ixl ::: R, the coefficients of ILk,/ for the distributions S/,k which correspond to S = Ox are linear combinations of x j e- iXZI , 0 ::: j ::: mk - 1.
Proof of Lemma 6.1.29. From the proof of Theorem 2.2.10 we obtain disjoint disks B(Zko rk), 0 < rk ::: such that on aB(zk, rd we have
!,
IFIL(~)I
The condition on the zeros
Zk
:::: £1 exp(-C1P(Zk». implies that this inequality can be replaced by
IFIL(~)I ::::
£
(1
+ IZkl)N
for some £ > 0 and N E N. Let q be a nonnegative integer to be chosen later. For those o ¢ B(zko rk) we define a function
Zk
such that
0 such that the weight function w(z) = H(Imz) + (q + r) log(2 + Izl) satisfies Iw(z) - w(w)1 ~ K
if Iz - wi ~ I. Therefore, by the maximum principle, the above estimate for 1/11. holds inside the disk B(Zko 2rk) up to a small modification. Since the series
f;
1
(2
+ IZk 1)2
< +00.
it follows that: which proves the claim. We will show now that g interpolates the values of FS on the variety V. If we consider a zero Zk of F f..L, then all the functions 1/Ij. j =I k. are multiples of Ffl. in a neighborhood of Zk. Therefore. it is enough to show that Ffl. + 1/11. vanishes at Zk with multiplicity at least mk. Consider a point w E B(Zko rk)\{zd. Let 0 < e < Iw - zkl, then letting y = a(B(Zko rk)\B(zk. e» we have FS(w) wqFfl.(w)
=
-1-1 2:rri
FS(~)
y ~q.Ffl.(~)(~ - w)
d~.
6.1. Convolution Equations in IR
377
hence
For Iz - zkl > rk one has
therefore, for those z, l/1k(Z)
r
= zqFIL(z) 2rci
1a8(Zk.e)
FS(~) d~. ~qFIL(~)(~ - z)
Since the two sides in this expression are holomorphic in C\B(Zb £) they coincide at the point w. Hence, the above expression (t) for FS(w) can be rewritten as with hk(w)
=
wq 2rci
r 188(2ko")
FS(~)
~qFIL(~)(~ - w)
d~ ,
which is holomorphic in B(zk, rk). This proves that FS - g is a multiple of FIL in Jt'(C). Since FIL is slowly decreasing we conclude there is a function h E Ap(C) such that FS = g + FIL' h. So
Let SI E E'(lR) with cV(SUPPSl) 5; [a,,8) be such that FS 1 = g and define E £'(lR) by FSo = h, then we have S=IL*SO+SI.
To conclude the proof we first observe that, from the expression of 4Jk as a linear combination of 1/[(z - zd j ], 1 ~ j ~ rnk, it follows that each l/1k is a linear combination of F«d/dx)q ILk.!), 0 ~ I ~ rnk - 1, for k ::: 1, and similarly, of F(ILo.,), 0 ~ I ~ rna - 1, if Zo = 0 is a zero of FIL. Here, the ILk,/ are the distributions introduced in Lemma 6.1.14. Furthermore, if S = 8x for some x E JR., the computation of the residues that define 4Jk shows that the coefficients of l/[(z - zk)i] are linear combinations of x'e- iXZk with 0 ~ I ~ rnk - 1. We note that the coefficients of these linear combinations depend on q, which can be taken to be the same for all x, Ixl ~ R. This concludes the proof of Lemma 6.1.29. 0 Now let us return to the proof of Theorem 6.1.27. We need to show that the inequalities on the zeros of F IL imply the hyperbolicity. To see how to proceed,
6. Hannonic Analysis
378
= 11- * So + Sl. then f(x) = (ox * nCO) = (f * 11- * So)(O) + (Sl * ncO) = (Sl * ncO) = (SI. f)· Therefore. if we start with h E E. we define an extension f of h to the whole suppose
f
E
!m and let 8x
real line by f(x) = (SI. h).
* So + SI. First. we need to show that this definition of f Ox = 11-
is independent of the choice
of decomposition of Ox. In fact. let Ox = 11-
* So + SI = 11- * To + TI
with cv(supp Sl) !:; [cr. p] and cv(supp TI ) 5; [cr. ,B]. Then TI - SI = 11-
* (So -
To)
and, from the support theorem. [cr. p]
+ cv(supp(So -
To»
= cv(supp(TI -
SI» 5; [cr.
Therefore. cv(supp(So - To» = {O}, which indicates that So - To for some polynomial P. Hence (TI-SI.h) = \11-*P
Pl.
= P(d/dx)(8o)
(~) (oo),h)
= \11-. P ( -
:x) (h»)
= 0
by the compatibility conditions defining E. Note that. in particular, if x E [cr.,B] it follows that f(x) = hex)
since we can take So = 0 and SI = Ox in this case. Second. we are going to show that the function f is of class Coo. If we fix R > 0, then the decomposition of Ox can be made so that R: S ~ SI given by Lemma 6.1.29 is linear and continuous. It is a general fact from functional analysis that f is Coo since the map x ~ Ox is Coo for Ixl < R. For the sake of completeness let us show the existence of the first derivative. For x fixed and o "" It I small we have 1 v v /1 ) t(f(x + t) - f(x» = \ t[R(ox+t) - R(ox)]. h
The limit
. OX+I - Ox 11m
1-+00
t
= .'
Ox
6.1. Convolution Equations in R
379
is valid in the sense of distributions, hence
1 v lim -(f(x t
1..... 00
+ t) -
v
I
f(x» = ("R.(ox)' h).
Repeating this procedure we see that j 0, N > 0 depend only on 4>. In order to obtain the Fourier expansion of the T -mean-periodic function f we need to identify the strong dual A~(V) of the space A*(V). It is easy to see that any element b E A~(V) can be written as a sequence b = (bn)n?) , bn E A'(Vn ), with the dual norm IIbnll~ =max{(an,bn): IIanli n S I},
where the duality bracket is (an' bn ) =
L
XjYj,
!~j~dn
when we identify A(Vn ) with Cd. by enumeration of the values g(l)(ak)/ I!, an = Pn (g). Hence, for every D > 0 we have
"bn"~ :=
L
IIbn"~eDRn < +00
n?!
and
(a, b)
= L(an , bn }. n?;l
From these considerations it follows that to every f E Jf"(C) such that = 0 corresponds to a unique element bU) E A~(V) in such a way that for S E .Jf1(C) T
*f
(~)
(S, f) =
(p(~(S»,
b).
(The verification of this last identity is done the same way as in Theorem 6.1.11.) Moreover, the continuity of f as a linear functional on .Jfl(e) tells us that the convergence of the series that are implicit on the right-hand side of (~) are uniform for any bounded family of analytic functionals S. In particular, we can take S = 8z , z belonging to a compact subset of C. Then zj eakZ ) p(~(8z» = ( k?~
-y-
O~J:5mk
6. Hannonic Analysis
390
and
where
o It is clear that the choice of grouping is not unique, though one can construct examples showing that they are necessary for the convergence of the series [Leo]. One natural way to group terms is to use the connected components of S(II, s, A), this observation and basically the same proof as that of Theorem 6.2.6 can be used to prove the expansion of Coo JL-mean-periodic functions in R when JL E £' (R) is just slowly decreasing. In order to find an explicit formula for the coefficients bk,t//! we need to construct analytic functionals Tk •j such that: p(:J(Tk,l»j.j =
(~(Tk,I»(j)
.I
1
(aj) = -/I81,j . 8k,i.
J. . Let fJ be a holomorphic function near w = 0 which vanishes at w = 0 with order exactly m E N*. Then 1 A-m+I A-I - =A-m -+- + ... +-+ .... O(w) wm wm- I W
Denote the principal part of wi /fJ(w) at w = 0 by
[(J7~)L
'
then, for 0 ::::: 1 < m, Wi] [--
(J(w)
0
_ I -w
=
I W
(A_ A-+ --+ ... +-w
I- I )
m
wl
m
(1 {A_ (J(w) -
I
~
I
A_I }) + A-I+l Wl-I + ... + --;:;;- + ...
WI
= O(w)
+ HI (w),
where HI is holomorphic near w fJ(w)
= O. Hence,
[~] = (f(w) 0
wi
+ fJ(w)HI(w),
6.2. Convolution Equations in
rc
391
for w near 0, and the order of vanishing of the term () HI at the origin is bigger or equal to m. Therefore, when O::s j < m,
We define Tk,l by the formula
~(Tk,l)(~) := (~) [(~!;~kt] a, ' where ['la, represents the principal part, as done above. Clearly the right-hand side of this expression is an entire function of exponential type, namely it coincides with times a rational function. Besides, it is obvious that is vanishes at every a;, for i "# k, with multiplicity at least mi. Therefore this analytic functional Tk,l has the properties we were looking for. As we did in the previous section we can write more explicitly the functional Tk,l. First we define a functional Tk by the formula
(Tko f) = ( T"
ea,z (mk - I)!
1 z
f(w)(z - w)m,-le-a,w dw ) .
Zo
One can see that this formula is independent of the choice of base point zoo Changing Zo amounts to adding (Tz, ea,z P(z)} where P is a polynomial of
degree less than or equal to mk - 1, but this quantity is zero because ak is a zero of multiplicity mk of . Using the same observation one obtains by integration by parts ~(Tk)(O = (r., - ak )m k
It is also easy to see that if K is the minimal convex compact carrier of T then K is a convex compact carrier of Tko for instance, by taking Zo E K on the explicit integral formula. Now one can consider the polynomial Pk,l of degree 1 P (r).- (r -a )mk k.l.,
.-.,
k
[(tI!(O -ad] a,'
then
Moreover,
6.2.7. Proposition. Any mean-periodic function f Fourier representation of the form
"# 0 in
Je(C) has a unique
392
6. Hannonic Analysis
where the series (in n) is uniformly and absolutely convergent on every compact subset of C, the CXk are among the zeros of some function
8}. Then It:;.m f(zo • ... , zm)1
:s
2m 8m
Q,
8 > 0, and
IIflloo.
Proof. It is clear that it is enough to prove the statement when the points are distinct. We proceed by induction on m. For m = 0 it is evident. For m :::: 1 let us define g(z) = fez) - f(zo) , z of. zoo { Z - Zo g(zo) = f'(zo),
in other words, g(z) = t:;.l /(zo. z). Clearly g is holomorphic in Q and if we have Iz - zol :::: 8 then Ig(z)1
:s 211~lIoo.
By the maximum principle, this is true everywhere in m-I
It:;.
g(z\' ...• zm)l:s
Q.
By induction
2m- I llglloo 8m- 1 •
Since t"m /(zo, ZI .... , zm) = t:;.m-lg(ZI, ... , zm) the induction step is correct.
o
the number of zeros counted with multiplicity of 4> = Denote by z8, .... Z~._I these points, each of them repeated according to their multiplicity. It is clear that the divided differences t:;. j (g) for Pn (g) = an E A (Vn ) are independent of the choice of g. For this reason, we denote t:;.j (an) := t:;.j (g)(zo' ... , zj), Let us denote by
Vn
J(T) in the annulus
en.
395
6.2. Convolution Equations in IC
o~ j
~ Vn - 1. The diameter 8n of C n is 2R n , using this notation, we introduce a new norm in A(Vn )
Let Pn be the Newton polynomial of degree
L
Pn(z) =
Vn -
1 given by
L~)(an)(z - zo)'" (z -
Z}_I)'
O::sj:Sv,,-l
then Pn(Pn) = an and, for z
E
Cn, we have
L
IPn(z)1 ~ Illanlll n
o,;-jlz - zol" 'Iz - z}_11
O::sj::Svn - l
~ 1110.111. (,,~_, 1) ~ v.llla.III., Since
Vn ~
KoRn for a constant KO > 0, which depends only on , we have lIanll n ~ IIPnil oo ~ KoRnlllanllln.
To obtain an inequality in the opposite direction, given an E A (Vn) we consider the sequence a = (a m)mo::l E A.(V) with am = 0 if m i= n. Then, for D > 0 fixed Iiallo = lIanline-ORn. From the proof of Theorem 6.2.7 we know there is h p(h) =
Ih(nl We apply Lemma 6.2.10 with
~
n=
E
Exp(
t. = (t.(O)(a) • ...• t.(v.-I)(a».
Its inverse can be computed very simply, we just observe that
L
ak = Pn(zZ) =
t.j(a)(z~ - z3)'" (z~ -
z1-1)
O:U:svn-1
L
=
t. j
(a)(z~
- z3)'"
(z~
- z1-1)'
O 0, b > 0, and set z = -i, then Ib/,oe-Cl/,oi I = e2nm // a ,
m,
which is unbounded for a certain sequence -+ +00. This example which shows that groupings may be necessary is due to Leont'ev.
400
6. Hannonic Analysis
6.2.11. Remark. It follows from the proof of Theorem 6.2.6 that a sufficient condition on the analytic functional T for the convergence of the Fourier series representations f (z) = L:k> I Pk (z )eUkZ without groupings is that for some c > 0, A > 0, each component-of S(l4>I, c, A) contains only one zero of 4>. This happens precisely when V is an interpolation variety so that also the series
'""' ck,).zj eak ' ~ k;o>:l
O:Sj<mk
is absolutely and uniformly convergent over compact sets. In the previous formulation we are really grouping all the terms (L:o:sj<mk Ck,jzj)eUk = Pk(z)e UkZ together. We know from Section 2.4thatifT E Jf"'(C)\{O}, then T * f = g,g E Jf"(C), is always solvable in Jf"(C). It is clear how to solve explicitly an equation of the form T * f = zme uz but it is not so clear how to find an explicit solution for an arbitrary entire function g. Nevertheless, there is a method, due to Gelfond and Leont'ev, which we will explain in detail in the following section for the very simple equation fez + 1) - fez) = g(z). It consists of the following: Given g(z) = L:n;o>:o(an/n!)zn, there is an increasing sequence Rn > such that the function
°
fez) = P(z)
+ LanBn(z, Rn) n;o>:O
is an entire function solving the equation T * f = g, where P is a polynomial of degree strictly less than m, m is the order of vanishing of 4> = J(T) at z = 0, and Bn(z, Rn) = - 1
2lri
1
lul=Rn
e zu - du -u n 4>(u)
(see [Ge], Chapter 5, Section 7, p. 358). Observe that the map g ~ f we have just defined is not in general a linear operator since the choice of Rn depends on g. Another application of Theorem 6.2.7 is the analysis of overdetermined systems of convolution equations. For instance, let us consider the system (*)
{S
*f = g T*f =h
where g, h E Jf"(C) are given and we are looking for an entire solution is clearly a compatibility conditions, namely,
f. There
T *g = S *h. Is this condition sufficient for solvability of the system (*)? To simplify we consider only the case where 'J(T) and 'J(S) have no common zeros. We leave the general case to the exercises.
6.2. Convolution Equations in C
401
6.2.12. Lemma. Assume that ~(T) and ~(S) have no common zeros. The necessary and sufficient condition for the previous system to have a solution for every pair satisfying the compatibility condition is the existence of S I. TI E J't' (C) such that SI * S + TI * T = 8. Proof. Consider R := {(g. h) E (J't(C»2: S * h - T closed subspace of (-*'(C)) 2 . Let
*g =
O}. Then R is a
a: J't(C)""""* R. aU) := (S
* J. T * f).
This map is injective as a consequence of the hypothesis that J(T) and ~(S) have no common zeros (see Proposition 6.2.7). Therefore a is surjective if and only if a is an isomorphism. and this is equivalent to the fact that la is also an isomorphism. By the Hahn-Banach theorem every element of the dual space R' can be represented as a pair (SI. T I ). SI. TI E J't'(C). Therefore there exist SI. TI such that On the other hand. (la(S], Td. f)
= (SI. Tl), a(f)} = (SI. TI). (S * f. T * f)} = (SI. S * f) + (Tl • T * f) = (SI * S + TI * T. f).
i.e .• la(SI. TI) = SI
* S + TI * T = 8.
Conversely, if this convolution equation has a solution then we can choose
f = SI * g + TI * h. which one can verify is a solution of the system because g and h satisfy the compatibility condition. 0
*
The equation S1 * S + TI T = 8 is sometimes called the Dezout equation. It is equivalent to finding functions 1. 1/11 E Exp(C) such that 1 ~(S)
+ 1/11 ~(T)
= 1.
We have already considered this last problem in Lemma 2.6.15 and we can state the following corollary:
6.2.13. Corollary. Under the assumptions of Lemma 6.2.12, the necessary and sufficient condition for the existence of a solution f of the previous overdetermined system for every pair satisfying the compatibility condition is the existence of constants e > 0, C > 0, such that IJ(S)(~)I
+ 13'(T)(~)1
(strongly coprime condition)
::: ee-Cl~1
(~ E IC)
402
6. Hannonic Analysis
We would like to consider next the case in which J(T) and J(S) do not have common zeros but they do not satisfy the strongly coprime condition. The question is: What extra conditions should g and h satisfy in order that the system is solvable? To simplify the analysis let us assume all the zeros of J(S) and J(T) are simple. Since the convolution operator S* is surjective, we can find a function G E Jf'(C) such that S * G = g. Consider F =
f -
G, then letting H := h - T * G we have
* F = 0, T * F = H. S
The compatibility condition becomes S * H = S * h - S * T * G = S * h - T * g = O. Hence F and H are S-mean-periodic. Let V J(S), then F(z)
= {ad
be the variety of zeros of
= Lake,,"kz, k;::1
H(z) = Lbke,,"Az, k;::1
where these series may require groupings to be convergent. Therefore, H(z)
= T * F(z) = LakJ(T)(ak)e,,"kz k;::!
is the Fourier-expansion of the S-mean-periodic function H = T uniqueness of the coefficients in the expansion we conclude that
* F.
By the
bk
ak = J(T)(ak).
(0)
The extra necessary conditions on the functions g and h appear now as a condition on the sequence (bd so that the sequence of the coefficients (ak) defined by (0) must satisfy the growth conditions discussed before Remark 6.2.11. For instance, if V is an interpolation variety for Exp(C), then the condition for solvability is lim log lak I = lakl
-00.
k-H'O
That is, lim
log Ibkl -log IJ(T)(ak)1
k~oo
= -00.
lakl This condition is definitely stronger than the minimal condition lim log lakl = lakl
k~oo
-00
6.2. Convolution Equations in C
because we have assumed that particular we have
403 ~(T)
and
~(S)
are not strongly coprime, and in
. -log 1~(T)(adl hmsup k-+oo
lakl
= +00.
If V is not an interpolation variety and we need groupings we can make a similar analysis using the norms III . 11111' EXERCISES
6.2.
I. Verify the details of the example of Leont'ev given in the text about groupings for any solution I E Jt'(C) of the equation J1. * f = 0, J1. = 8o+ b - 8a - 8b + 80 • Consider also the case alb is not real.
2. Study the system S * I = g, T * 1= h, as given in the text, in the cases (a) there are no common zeros but some zeros of Ij'(S), Ij'(T) are multiple; (b) there are common zeros to Ij'(S) and 1j'(T); and (c) there are no common zeros but the variety of zeros of Ij'(S) is not an interpolation variety. 3. This exercise is a self-contained treatment of results from the recent manuscript [BezGr]. (a) Show that Jt'(C) is a reflexive Frechet space and that the transposed map ir: (Exp(C»' ~ Jt'(C) is a topological isomorphism such that irlJt"(C) Ij'. Deduce that one can introduce a convolution product in (Exp(C)' by means of the formulas:
=
* /)(z) := (R~, I(z + n) (Exp(C)' and I E Exp(C), which satisfies R * IE Exp(C), and (RI * R2, f) := (RI, R2 * f) (R
for R
E
for R I , R2
E
(Exp(C)', f
E
Exp(C). Show that Ij'I(RI
* R2 ) = \j'(R I)\j'(R2 ).
=
=
Moreover, if g ir(R) and c E C, show that ir(eC~ R)(z) g(z + c). (b) Show that I E X(C) is T -mean periodic (T E X'(C) if and only if Ij'(T)R = 0 in the sense of (Exp(C)', where R E (Exp(C)' is such that Ij't(R) I. Show also that for R E (Exp(C»" g E Exp(C), one has R * g = 0 in Exp(C) if and only if the product pT = 0 in X'(C), where p = ir(R) and Ij'(T) = g. For pEN and R E (Exp(C)', define the derivative R(p) by (R(p), g) := (-l)P(R, g(p») for any g E Exp(C). Show that R(p) = 8&P) * R, where 80 is the Dirac delta at the origin. (c) Show that for any am E C, a E C, R E (Exp(C»', 1= ir(R) we have
=
Show also that if Qrn(z) := 1l'kC,):=
:L05 O. R > O. with the property that IP(e"')1
+ IQ(eP:)1
2: y
Izl 2: R.
whenever ~ E
Deduce the existence of a nonzero polynomial P(e":)
C[z) such that the two functions
Q(e P:) h(z):= - - .
I,(z):= - - . ~(z)
~(z)
are entire functions of exponential type without common zeros and show there are C > O. satisfying
f.
> 0,
=E IC.
for all
Prove that under these conditions any entire function rp satisfying the system of equations M
L
N
amrp(z
+ rna)
= L
bnrp(z
+ nfJ} == 0
m=O
must be an exponential polynomial. (e) Let a, fJ E iC be JR.-linearly independent, ao • .... am E C, ao I: 0, Qo • ...• QN be polynomials in C[z), QN ::/= O. The purpose of the following question is to show that if I E Jt" (iC) satisfies the system of equations M
N
Laml(z
L
+ rna) == 0,
m=O
Qn(z)/(z
+ nfJ) == O.
n=O
then I must be an exponential polynomial. Denote P(z) = E:=oamz n • F(l;} = P(e"'). R E (Exp(IC)', such that ~(R) = I. and let Jrk(l;) be the sums of exponentials introduced in part (c). (ed Show that R verifies the equations (in (Exp(lC)') N
(e2) Let L
E
FR = LJrkR(k) = o. k=O N* be arbitrary. p the column vector (R(N+L-I). R(N+L-2) •... , R'. R)
with entries in (Exp(IC)'. By differentiating (N - I) times the functional F Rand (L - 1) times E~=OJrkR(k) find a (N + L) x (N + L) upper triangular matrix A with entries in Exp(1C) of the fonn L
A=
o F
o such that Ap = O.
N
o
F
6.3. The Equation I(z (e~)
+ 1) - I 0 independent of n such that for It I = (2n + I)Jr let - 11 ~ c,
as one can reduce this inequality to the Lojasiewicz inequality for sin z. Hence, using Stirling's fonnula, we can prove the existence of A > 0 such that for
410
6. Harmonic Analysis
all z, n IB
(z)1 < n+1.n+l
-
(n
+ I)!
c«2n + 1)Jl')n+l
< Ae(2n+l)rr(lzl+1).
e(2n+l)rrlzl
-
This bound, together with the Hadamard estimate for gn, shows that the series defining I converges absolutely and uniformly on any disk of C. 0 6.3.6. Remarks. (1) If g is a function of infraexponential type, i.e.,
lim 10gM(g, r) = 0, r
r~OO
where M(g, r) = max{lg(z)l: Izl :s r}, then one can prove that the series
is locally uniformly convergent (see Chapter 4). (2) In general, one can choose kn in the last proof so that for any given s > 0, e > 1 there are constants c > 0, ro > 0, 8 = (1 + s) flog e, so that the function I defined in Proposition 6.3.5 satisfies I/(z)l:s CM(g,er)~,
Izl :s r.
In particular, I is of the same order as g, and of finite (resp., minimal) type if g is of finite (resp., minimal) type. (See [Ge, Chapter 5].) (3) In case g is a merom orphic function, the case we shall consider next, one can also find a merom orphic sum I with good estimates on the Nevalinna characteristic T(r, f). (See [Whi, Theorem 4).) We want \!O find now the sum of a meromorphic function. We observe that the difference equation (*) can be used to find two formal solutions as follows: I(x) = -g(x)
+ I(x + 1) =
-g(x) - g(x
+ 1) + I(x + 2)
= .. "
so that one is lead to consider the "right formal sum" Ir(x) = -
L g(x + n). n~O
If this series converges in a certain region, it is a solution of (*) in that region.
Similarly, we have I(x) = g(x)
+ I(x
- 1) = g(x)
+ g(x -
which leads to the "left formal sum" hex) =
L g(x n~O
as a candidate for a solution.
n)
1)
+ I(x + 2)
= "',
6.3. The Equation f(z
+ 1) -
f (z)
= g(z)
411
Returning to the equation (*) with g meromorphic, let us assume that there is real number p so that all the poles of g lie in the half-plane Re z < p. The idea is to try to find polynomials Yo, YI, ... so that the series !PI(Z):= (Yo(z) - g(z)}
+ (YI(Z) -
g(z
+ I)} + ...
converges to a meromorphic function in the whole plane. (Note that this function is obtained by applying the Mittag-Leffler procedure to the right formal sum of g.) In order to choose the Clj, let kEN be the smallest value such that p - k < O. Choose Yo = ... = Yk-I = O. The poles of g(z + n) lie in the halfplane Re z < p - n, hence for any n ::: k we have that g(z + n) is holomorphic in a neighborhood of the disk Izl : : : n - p and n - p > O. Choose Yn as the partial sum of the Taylor series of g(z + n) about z = 0 such that Ig(z
+ n) -
for
Yn(z)1 :::::: Z-n
Izl:::::: n -
p.
For any R > 0 we have that the series L
(Yn(z) - g(z
+ n)}
converges absolutely and uniformly on Izl : : : R, hence it defines a holomorphic function in Izl < R. It follows that the series !PI converges to a meromorphic function in C. Moreover, outside a discrete set, !PI (z
+ 1) -
fIJI (z) = g(z) - Yo(z)
+ L{Yn(Z + 1) -
Yn+1 (z)},
n;:,O
and the series converges locally uniformly. Thus, we have
where 1/11 is an entire function. From Proposition 6.3.5 we know we can choose an entire function fh = 61/11, then
is a merom orphic sum of g. If g had all its poles to the right of a vertical line we could have modified in the same way the left formal sum of g to obtain a meromorphic sum of g. In general, choose any p E lR. such that the line Re z = p contains no poles of g. The Mittag-Leffler theorem (see [BG]) allows us to find a meromorphic function g 1, whose poles lie to the right of that vertical line and whose principal parts at their poles coincide with those of g. Then g = gl + g2, where the poles of g2 lie to the left of the same line. Thus, from the previous construction we have two meromorphic functions II, h such that 11 = 6g l , h = 6g2 • Clearly, we can take 6g = I := 11 + fz. Let us remark that this construction of a sum of the meromorphic function g has a special feature, if all the poles of g are simple, then the same is true for this particular sum I. Moreover, if the residues of g at those poles are integral,
6. Hannonic Analysis
412
the same is true for f. The importance of this feature lies in the following application. Recall that for a meromorphic rp, its logarithmic derivative rp'lrp has only simple poles with integral residues. Conversely, for any merom orphic function cf> with this property we can find rp merom orphic such that cf> = rp' / rp. These two remarks together tell us that given a meromorphic function 1/1 there is another merom orphic rp such that rp'lrp = 6(1/1'/1/1).
That is, rp'(z + 1) rp'(z) 1/I'(z) -'------'-- =rp(z
+ 1)
rp(z)
1/I(z)
By integration we obtain that rp(z + 1) = 1/I(z). rp(z)
Retracing the steps we have shown that given a merom orphic function 1/1 one can find merom orphic function rp such that (**) is satisfied. Clearly we can multiply rp by an arbitrary periodic meromorphic function and obtain a new solution of (**). All solutions are obtained this way. Being meromorphic, rp is of the form e f PI I P2 , where f is entire and PI, P2 are Weierstrass canonical products, but following the technique of the right and left formal sums used to solve (**) one can anticipate the form of PI and P2 • We shall illustrate this below in the case of the r -function. We can now state the main theorem of this section. 6.3.7. Theorem. Let ao, al be a pair of meromorphic functions in the plane such that neither is identically zero. For any meromorphic function g in C there is a meromorphic solution f of the linear difference equation aICz)f(z
+ 1) + ao(z)f(z)
= g(z)
(z E C).
Proof. From the previous remarks we know there exists a nontrivial meromorphic function fl such that
/I(z
+ 1)
h (z)
=
al (z) ao(z)
The equation (t) now becomes h(z
+ l)f(z + 1) -
fl (z) fl(z)f(z) = ---g(z). ao(z)
We have also shown there is a meromorphic
h
such that
/I (z) h(z + 1) - h(z) = ---gCz). aoCz)
Letting fCz)
= h(z)//ICz)
we obtain a solution to Ct).
o
6.3. The Equation f(z
+ 1) -
413
I(z) = g(z)
As a title of example let us consider the equation f(z
(!)
+ 1) =
zf(z).
If we assume f(1) = I, then f(n) = (n - 1)! for n EN so that f interpolates the factorials. One of the solutions to this equation is Euler's [,-function. Following the previous method we let q;(z) := l'(z)lf(z) so that: q;(z
1
+ I) -
q;(z) = -.
z
The right formal sum of this equation is 1 L:n~O + n
q;r(z) = -
Z
which has its poles at the points z = 0, -I, -2, .... From here we conclude that a solution of the equation (!) is given by et(z)
I(z)=
00
z
IT
n=l
z
(I + -) en
' z/ n
where 1jr is entire and the denominator is the Weierstrass canonical product with zeros at z 0, -1, -2, .... Letting
=
fm (z)
m! exp (1jr(Z)
et(z)
= Z
IT
=
Z
m
+ _) e-z/n
(I
+ n~ zln)
---:--~---:-:':"""":,---:""':"""
z(z
+ 1) ... (z + m)
n
n=l
we have f(z) = limm ... oo Im(z) exists, locally uniformly in C\{O, -1, -2, ... }, and zf(z) = lim zfm(z) f(z + 1) m ... OO fm(z + I)
oo + m + I) exp [1jr(Z) -1jr(z + I) -
= lim (z m...
=
(I
Ji.moo + m:
n=l
n
1)exp [1jr(Z) -1jr(z + +~og(m + ~ ~
The limit y = lim
t !.]
m"""""""oo
1)
[~!. ~n
-
log(m
+
1)-
) ].
1)] ~ 0.557
n=l
is the Euler-Mascheroni constant. Inserting this value and using equation (!) we obtain zf(z) 1 = f(z + 1) = exp[1/t(z) - y,(z
+ 1) -
y).
6. Hannonic Analysis
414
In other words,
1/1 satisfies a difference equation of the fonn
+ 1) -
1/I(z
for some k we have
E
= -y + 2rrik,
1/I(z)
Z. The simplest solution is 1/I(z)
= -yz. With this choice of 1/1
Hence, f(l)
= =
lim fm(l)
m-->oo
lim exp
m ..... oo
[1/1(1) + ~ .!. -log(m + 1)] ~ n n=l
= m-->oo lim exp [-y +
t .!. n=l
n
-log(m
+ 1)] = 1.
This f satisfies the extra condition fO) = 1, so that it is the correct choice. This solution is exactly the r -function. r(z) =
g(1
[eYZ z
+ ~) e- z / n
]-1
We conclude this section studying a sort of unexpected relation between difference equations and analytic continuation, which is hinted at in the proof of Lemma 6.3.1. Consider a function F which is holomorphic in the simply connected region C\] - 00, 0] and assume it admits an analytic continuation along any path in C*. Thus, we can talk about F(~) for any ~ E C* as a multivalued holomorphic function, or, what is the same, a holomorphic function on the Riemann surface of log ~. Let us denote for the rest of this section .Ij the vector space fonned by these multivalued functions F. It follows that for; < 0 the values F(; ± iO) = lim~ ..... o+ F(; ± I rJ) are well-defined but the jump is not necessarily equal to zero. Let ~(n
:= F(;
+ iO) -
F(; - iO)
(; =1= 0).
This function is identically zero for; > 0, continuous for; < 0, and it represents a distribution on lR if and only if there is k ::: 0 such that ITJlk F(; + iTJ) remains bounded in a punctured neighborhood of ~ = 0 (see [BO, Proposition 3.6.12]). In general ~ is a hyperfunction, as explained in Chapter 1. Introducing a change of variables ~ = e 21riz (or unifonnizing parameter z), let us define f by F(O = fez), in other words,
f
(~IOg~) = F(n 2m
Thus, f is an entire function of z which satisfies, for x = Re z the difference equation fez + 1) - fez) = ~(e21fiz) = ~(_e-2"'Y).
= - t, y = 1m z,
6.3. The Equation f(z
+ 1) -
f(z)
= K(Z)
415
Note that with this notation it is natural to rewrite the original equation as F(e2rri~) - F(n = rp(~)
(~ < 0).
(Equations of the fonn F(q~) - F(~) = rp(~), 0 < q, often arise in Number Theory and are related to the theta functions mentioned at the beginning of this section.) Let us consider a special example, rp(~) = Log I~I for ~ < O. Then we have rp(_e- 2rry ) = -21fY for Y E JR.. By an easy inspection we see that f(z) = 1fiz 2 satisfies f(z + 1) - f(z) = -2rry (Rez =
-!).
Hence,
F(~) =
-4 1 . Log2~, 1f1
in C\] - 00,0]. Since rp E V'(JR.), this answer could also have obtained with the help of the Cauchy transfonn, as suggested by the general theory in Chapter 1 and in [BG, §3.6]. After these preliminaries, let us consider a type of question raised by Hurwitz [Hu2, Vol. 2, p. 752]. The homotopy group 1f, (C*) of C* is isomorphic to Z and it is generated by the unit circle IS I = 1, oriented counterclockwise. This group acts on Sj by sending a function F to the function obtained by analytic continuation along the unit circle. This is precisely what we have denoted by F(e2rri~). In this case F(e2rrin~), n EN, means the new function in Sj obtained from F by analytic continuation along the unit circle traversed n times counterclockwise, and for n < 0, traversed Inl times clockwise. Since the space Sj is very large it is natural to consider subspaces invariant under the action of rr,(C*). Typically they are defined by functional equations. For instance, the equation F(e2rri~) = F(~)
describes all the functions in Sj that are single valued, that is, the collection of all Laurent series
convergent in C*. Observe that if F E Sj, then its derivative F' also belongs to Sj. So that in Sj we have three types of operators acting: D: F(~) ~ F'(~); A: F(~) ~ A(~)F(~), multiplication by A E Jf'(C*); and r: F(~) ~ F(e2rri~), "monodromy" operator. It is not hard to see that r commutes with D and with A. For this reason Hurwitz
considered the following simple-looking equation in 1918: DF = r F, that is, (H) F'(~) = F(e 2rri The solutions of such an equation, if there are any, also fonn an invariant subspace for the action of 1f( (C*). It is clear that if we assume F is single
n.
416
6. Harmonic Analysis
valued, then there is essentially only one solution for (H), namely F(l;) = eel;, for any e E C*. Hurwitz seemed to believe this was the only solution. Note that when using the uniformizing parameter l; = e 21riz this equation becomes essentially more complicated than those in Theorem 6.3.7. Namely, d 2 . - fez) = 271:ie 1rlZ fez
(H*)
+ 1),
dz where fez) = F(e 21riz ) as earlier. In 1975, Naftalevich [Na] used the method of the formal sums to obtain many nontrivial solutions of (H*), i.e., different from e exp(e Z1riZ ). Earlier Hans Lewy had found the explicit solution
t)
(_1_.
Fo(O = ('" e-I;t exp Log2 dt, Jo 47rl which is well defined and holomorphic for Re l; > O. To show that this function belongs to the class n, let us consider for 0 E IR the auxiliary functions Fo(l;):=
l
1
ooe ;9
o
exp(-l;t
+ -.logZt)dt. 471:1
In the integral defining Fo we have t = re iO , 0 < r < 00. If l; = peiO/, p > 0, then the integrand is absolutely convergent when cos(O + a) > 0, in particular, if -71:/2 - 0 < a < 71:/2 - O. It follows that Fe is holomorphic in the half-plane -71: /2 - 0 < arg l; < 71: /2 - (). Note that if 0 < ()2 - 0, < 71:, then the functions Fe] and F~ have a common domain of definition, and applying Cauchy's theorem they can be seen to coincide there. It follows that the collection (Fe)eEIR defines a single function FEn. Moreover, for Re l; > 0 we have F(e Z1ri l;) = F21r (l;). This last integral can be computed from the observation that for r > 0 _1_. log2(reio) =
471:1 For
e=
~(Logr + iO)2 47rl
=
~ LogZ 471:1
() 02 r+ - L o g r - - . 271: 471:i
271:, we get
so that F21r(l;) = -
roo re-I;r exp (_1_. Log2 r)
Jo
dr.
471:1
On the other hand, F6(i;) can be computed differentiating under the integral sign F6(l;)
=
roo ~(e-I;r) exp (_1_. Log 2 r) dr, dl; 471:1
Jo
re-I;r exp (~ Log2 r) dr, Jo 471:1 which shows that Hans Lewy's function Fo is indeed a solution to the Hurwitz equation (H). = _
roo
6.3. The Equation I(z
+ I) -
I(z) = g(z)
417
It is not clear that this solution is obtainable by Naftalevich's method. It has been shown recently [BSe] that Fo generates all the solutions to (H). Namely, any such solution F E 5) can be written in one and only one way in the form
F(~) = ce{
+ LCnF(e2Jrin~), nEZ
where the series converges locally uniformly in the Riemann surface of log ~ . This example indicates that it is natural to consider a wider class of equations for multivalued holomorphic equations. Their natural name should be monodromic differential equations, e.g., an equation of the form
L
An.k(~)Dkrn F(~) = D
n.k
for F E 5), where the (finite) sum has coefficients A n.k E .)t"(IC*). More generally, we could consider infinite order differential operators or consider other domains like IC\{D, l},IC\{al, ... , ad, IC\Z, etc. We remind the reader that the case IC\ {D, I} is intimately related to the hypergeometric functions. Here the group Jrl (IC\ {D, I}) has two generators, and Riemann showed that if a multivalued function F has the property that at any point ~ =1= D, I, the number of linearly independent branches of F is exactly 2, then F satisfies a hypergeometric differential equation (see [BG, Exercise 5.15.2], [Tr] [Pool]). The domain IC\Z appears in the closely related theory of resurgence [Ec] , [Ram], [Mal]. EXERCISES
6.3.
I. Prove the converse of Lemma 6.3.2. Show also that if I is periodic merom orphic and F(?;) = 1((1/2rri) log ?;), the order of the corresponding zeros for I and F coincide. When is I entire? When is the function F entire?
2. Construct a periodic function I holomorphic in C except for a discrete set of points, where it has essential singularities. 3. Recall that a trigonometric polynomial of degree m is a function of the form E:-m cne Z1ri n= with either C- m or em is :j:- O. Show that it has the form I(z) = ao/2+ E:=I(akcosmkz +bk sin2rrkz). Show that an entire periodic function I is a trigonometric polynomial if and only if I is a function of exponential type (Le., lim sup, ~oo (log M (r, f) / r) < 00). Could it be of infraexponential type if I '" constant? Could it be of type < 2rr if I '" constant? I(z) =
4. (a) Let I be a periodic meromorphic function such that limx~oo I(x + iy) = a for every y E R. Show that I(z) == a. (b) Let I be a periodic meromorphic function such that the limits limy->±oo I(x + iy) a± (possibly infinite) exist and are independent of x E [0, 1], then I is a rational function of e 2rriz •
=
5. Verify that the following sums are correct: (a)
(5
) ( nZ) = (nZ +l '
where
(Z)=Z(Z-I)"'(Z-n+l) n
-
n!
•
n
E
No
418
6. Hannonic Analysis
(b) 6e 2rr ;n,
= ze2rr;n,.
n E N. cos(z - 1/2) (c)6sinz=- 2sin(1/2) . (d) u(z)(6v)(z
+ I) + v(z)(6u)(z) = v(z)(6u)(z + I) + u(z)(6v)(z).
6. Compute the first five Bernoulli polynomials B.(z). 7. Use right formal sums to find the following sums: 1
(b)6 z (z+I); What happens with the left formal solutions in these cases? 8. Does Theorem 6.3.7 hold when either
ao
or al are identically zero (but not both)?
*9. Let g be an entire function of infraexponential type, g(z) = ~n>O gnzn its Taylor series about z = O. Show that the series I(z) = ~n>O[gn/(n + 1»)Bn+;(z) converges to a sum of g and it is also of infraexponential type. 10. Study the solvability of the equation I(z done in the proof of Proposition 6.3.5.
+ 1) -
al(z)
= g(z), a E iC\(O, Ij. as
11. (a) Show that a solution of (z - 1)/(z + 1) = 2z(z - 2)/(z) is the function I(z) = 2'[r(z)/(z - 2»). Find all the solutions of this equation. (b) Let r(z) be a rational function. Find all the meromorphic solutions of the equation I(z + 1) = r(z)/(z) with the help of the r -function. 12. Let 1/I(z) = r'(z)j rez). Show that 1_ = (_l)m 1/I<m)(z), 6_ zm+1 m!
mEN.
Determine 6(z - a)-m. Verify that 610g z = Log r (z) (at least for z E )0,00[.) 13. Show that any function F E 5) admits the expansion ~.>o an logn S', absolutely and locally uniformly convergent in the Riemann surface of log f 14. For a
E
C solve the equation F(e 2rr ;S')
= aF(S),
FE 5).
15. For IE .1t'(IC), D = djdz, a E C, one defines eOv I(z) := ~n>o(an /n!)Dn I(z). (al Show that eOv I(z) = I(z + a). (b) Fix P E .1t'(1C) and define operators Land K in the space .1t'(iC) by the formulas LI := e- v 1- e P DI. KI := e-v(e v D/). For an entire function ((J define U((J to be the formal series U((J = ((J + K((J + K2rp + .... Show that formally
LUrp(z)
= ((J(z + I).
16. (a) Let I(z) = ~n>O ane 2rr ;nz be such that an 2: 0 for all n E N. Show that if I is not a trigonometric polynomial, then I is a function of infinite order. (b) Let I be an entire function, WE IC* is called an asymptotic period if the order of growth of the function z ~ I(z + W) - I(z) is strictly less than that of I. For I of infinite order, this difference is assumed to be of finite order ([WhiJ). Let I(z) = ~n=1 [e 2rr ;.!, j(n!)!]. Show that I is of infinite order but that for every W E Q*, f (z + w) - f (z) is a function of order 1. (That is, every nonzero rational is an asymptotic period.)
6.4. Differential Operators of Infinite Order
419
(c) Use Exercises 9 and 3 of this section to show that if f is a function of infraexponential type it cannot have any asymptotic period.
6.4. Differential Operators of Infinite Order Let T
E )If' (C)
be an analytic functional, if
~(T)({)
=L
an {",
n~O
then (T
* f)(z) = L
anf(n)(z).
n~O
In other words, one could write T = L(-l)n ano (n). n~O
We are interested in a particular class of operators of this kind, those for which ~(T) is an entire function of exponential type zero, that is, T is carried by {O}. As we shall see, these operators have a number of interesting properties, not shared by general convolution operators. 6.4.1. Definition. An operator of the fonn (T, f) = Lanf(n)(O), n~O
where the coefficients an satisfy lim sup n yTa;;T = 0 n-+OO
is called a differential operator of infinite order. The function (O = l:n>O an{n is in this case an entire function of exponential type zero (see [BG, Chapter 4]) and we usually denote T * f by
(!!...) (f)(z) = L anf(n)(z). dz n~O
Recall that we denote Exp({O}) the space of entire functions of exponential type zero. 6.4.2. Proposition. If n is an open set of C and ct>(djdz) an infinite order differential operator, then ct>
(:z ):~(n) ~ ~(n)
420
6. Harmonic Analysis
is a continuous operator. The same is true for this operator acting on the space Oa = Je({a)), a E C. Proof. The proposition is clearly correct once we know it holds for disks. On the other hand. Propositions 2.4.2 and 2.4.3 have already shown this for any 0 open convex set Q. If L is a linear continuous operator in 0 0 • it is automatically a continuous operator from Je(C) into Je(B(O. r» for some r > O. by the definition of the topology of 00. Therefore. it makes sense to say L commutes with translations 1:a • at least for small lal. In fact. Je(C) is dense in 0 0 so that L is completely detennined by its action on Je(C). then. if lal < rand f E Je(IC). the function 1:a (L(f)(z) = L(f)(z - a) is holomorphic for Izl < r - lal, that L commutes with 1:a means that L(1:a (f»(z)
= 1:a (L(f)(z).
Izl
< r
-101.
Let us consider
L: 0 0 --+ 00 linear continuous and such that L commutes with translations. We can define it also as an operator from Oa --+ Oa by
Now, f ~ L(f)(O) defines an analytic functional T. We want to show that F(T) = CI> is an entire function of exponential type zero and L(f) = T * f. The last identity is an immediate consequence of the definitions. On the other hand. the continuity of L: 0 0 --+ 0 0 implies that for every e > 0 there exists a constant Ce > 0 such that I(T,
f}1
= IL(f)(O)1 :::: Ce sup If(z)l. 1,I:::e
In particular, ICI>(OI
= I(T., eZ{}1 :::: C e exp(sup Iz~1) = Ceeel~l. 1.I:::e
This proves the following proposition:
6.4.3. Proposition. Any linear continuous operator L: 00 --+ 0 0 which commutes with translations is an infinite order differential operator. We now proceed to give a detailed proof of a result already mentioned in Chapter 2, §4.
6.4.4. Theorem. Let Q be a convex open set in C (resp. K convex compact) and CI>(dJdz) =I- 0 an infinite order differential operator. Then: (i) CI>
(:z)
(AP(O» = AP(O).
6.4. Differential Operators of Infinite Order
(ii) 4>
(:z)
421
(JIf(K» = JIf(K).
(iii) If M = Ker{(d/dz): JIf(S1) ~ JIf(S1)} and Mo = span{zkell!Z EM}, then Mo = M. (iv) Similar to (iii) for JIf(K). Proof. (i) The transpose of the map h E JIf(Q) ~ (d/dz)(h) E JIf(S1) is given by S 1-+ T * S from JIf'(S1) into JIf'(S1), where :F(T) = . To prove the surjectivity of (d/dz) it is necessary and sufficient to show that its transpose is injective and has closed image. Since :F(T S) = :F(S), the injectivity is obvious and we only have to prove that the ideal Exp(S1) is
*
closed in Exp(S1). Due to the properties of the space Exp(S1), it is enough to prove that, for every f E Exp(Q), which is of the form f
= lim
n-->oo
fn,
fn E Exp(S1) (the limit in the topology of Exp(Q», then f = g for some function g E Exp(Q) (see Corollary 2.3.13). Since convergence in Exp(S1) implies convergence in JIf(C) we know there is a unique entire function g such that f
= g.
From [BG, §4.5.5) (or the following lemmas), g is an entire function of exponential type. We need to show that the indicator function of g is of the form He, the supporting function of C, for some compact convex subset C of S1. This depends on the following lemmas: 6.4.5. Lemma. Let C be a convex compact subset of C, let S E JIf' (C) be an analytic functional carried by C, f = :F(S). If is an entire function of exponential type zero and g = fI is an entire function, then there is an analytic functional R carried by C such that g = :F(R). We shall prove first a lemma due to Martineau, which is a variation on a result of V. Avanissian (see [BG, §4.5.5.).) Let us recall the notation for the area average A(, z, r) = (l/Jrr2) i8(z.r) dm. 6.4.6. Lemma. Let be an entire function of exponential type zero such that 4>(0) '" 0, then,for every A > 0 and every s > 0, there exists a constant DE E lR such that for z '" 0 A(log 11, z, Alz!)
~
(
1 +1..)2 -1..log 14>(0)1
+
(1 - C~ r) A
(s(1
+ A)lzl + DE)·
Proof of Lemma 6.4.6. Let R = (1 + A)lzl. Since is of exponential type zero, given s > 0, there exist DE E lR such that log 1<J>(u)1 :s slul + DE' Hence, the
422
6. Hannonic Analysis
subhannonic function U t-+
log leIl(u)1 - sR - De
is negative in lui::: R. On the other hand, B(z, >"Izl) ~ B(O, R);
therefore, R2 A (log leIll - 8R - De, 0, R) ::: (>"lzI)2 A (log leIll- sR - De, z, >"Izl).
By subhannonicity, log leIl(O)1 - sR - D. ::: A (log leIll- sR - De, 0, R). Thus.
In other words. >..2 A (log leIll. z, >"Izl) ~ (1 + >..)2 log Iell (0) I + (>..2 - (1 + >..)2)(sR + De),
o
which implies the stated inequality.
Proof of Lemma 6.4.5. Let 8 > 0 be given. We are going to show that Ig(z)1 ::: A. exp(HcCz)
+ 81zl)
for some A. > O. Up to translation we can assume that eIl(O) =I- O. The hypothesis implies that for every Sl > 0 there exists Ee , > D such that log If(w)1 ::: HcCw) + sllwl + E.,. Let M := max{HcCu): lui::: I}, z =I- 0, and>.. > 0, then max
WEB(z,Alzl)
HcCw)::: HcCz)
+ >..Mlzi.
Since log Ig I is subhannonic we have log Ig(z)1 ::: A (log Igl, z, >"Izl) = A (log If I. z. >"Izl) - A (log leIll. z. >"Izl). For 81 > D and S2 > 0, to be fixed later, there exist constants E., and Dez given by Lemma 6.4.6 such that: log Ig(z)1 ::: HcCz) + >..Mlzi
+ 81(1 + >")Izl + E e,
1+>..)2 log Iell (D) I + ((1+>..)2) - ( ->..->..- 1 (82(1 + >")Izl + D.z )· Let us choose A > 0, 81 > 0,
S2
> 0, such that
>"M < s/3, 81(l +),,) < 8/3.
6.4. Differential Operators of Infinite Order
423
(C: rA
1) (1 + A)82 < 8/3.
Let 1+A)2 log 1ct>(0)1 + ((1+A)2) Ne = - ( -A-A- 1
DB2
+ Eel'
then log Ig(z)1
~
Hc(z) + 81z1 + N e ·
In other words, there is R an analytic functional carried by C such that the function g = F(R). 0 From Lemma 6.4.5 we can now conclude that part (i) of the theorem is correct. (ii) Since (Jf(K»' is the space of analytic functionals carried by K the proof of part (i) works verbatim. (iii) Let JJ. E Jf'(Q) be orthogonal to Mo, this implies that J(JJ.) is divisible by ct>, J(JJ.) = ct> . h, h is an entire function. By Lemma 6.4.5 h E Exp(Q). Hence there is R E Jf'(Q) with J(R) = h, thus JJ. = T *R.
If
f
EM, then (JJ., f) = (T
* R, f) =
(R, T
therefore Mo = M. (iv) The same proof as (iii).
* f) =
0,
o
We shall now prove that any solution f E Jf(Q) of a homogeneous differential equation of infinite order has a Fourier-expansion convergent in Jf(Q) of the same kind as that found in the previous sections. It is clear that once we have proved this result for an arbitrary open convex set Q it also holds in the space Va. This Fourier representation goes back to the work of Valiron [Val], Gelfond [Ge], Dickson [Di2]. On the other hand, the novelty of our proof is that it obtains estimates for the coefficients of the expansion and it is essentially the same as Theorem 6.2.7 up to technical details. 6.4.7. Lemma. Let ct> i: 0 be an entire function of exponential type zero and let 0< 8 ~ ~. There exists R = Re > 0 such that for any z with Izl ~ R there are constants 0 < a, 8/8 < A < 8/4 so thatfor any t; such that Aizi - a ~ It; - zl ~ Aizi one has while exp(-28Izl)
~ a ~
1.
Proof. Let us recall a property of entire functions of exponential type zero (see [Bo, Corollary 3.7.3, p. 52]), given l3 > 0 and TJ > 0 there exists ro > 0 and a measurable set Ea ~ [0, +oo[ such that if r ~ ro m(Ea
n [0, rD
~ TJr
424
6. Harmonic Analysis
and if r ::: ro, r
~
£8, then
log 1(01 ::: -8r,
I~I
= r.
We shall choose convenient 8 > 0, 1'} > 0, later. The first condition on 1'} is that 41'}(1 +.oo
k
lim -
(Zn)n~I'
with
=+00.
Moreover, taking the constant Re from Lemma 6.4.7 bigger if necessary we can assume nCr) S er Assume has infinitely many zeros, otherwise is a polynomial.
6. Harmonic Analysis
426
Let us denote V = {Cab mk), k::: I}, latl ::: la21 ::: "', with distinct ak. Let en = 2- n and Rn = max(Re., 2R n_ l ) as given by Lemma 6.4.7. We denote by kl.l the first index k such that lakl ::: R I . Let Au > 0 associated by Lemma 6.4.7 to au = akl,1 and e = el. Denote Bl,I:= B(al,l,Al,Ilal.tI).
We let hi be the collection of indices such that ak E Bu. If there is any k > k l • 1 so that ak is in the annulus RI ::: Izi < R2 and it does not belong to Bl,I, we let k1,2 be the first such index and consider the corresponding disk with Ak'2 chosen with respect al.2 := akl.2 and e = el. We let J1.2 be the collection of indices of the ak E B1,2 \ B 1,1. We continue in this fashion until we exhaust all the zeros of in this annulus. We obtain in this way disks BI,}. 1 ::: j ::: N], and corresponding disjoint index sets II,}, 1 ::: j ::: NI, with II.} the collection of indices of the ak belonging to the set BI,} \ (
U Bl,l). I::'OI::'O}-I
Observe that if ak does not belong to U} BI,} and R3 > lakl ::: R2, then the index k > kl,N 1 ' We let k2,1 the first such index (if there is any such zero). Denote with A2,1 chosen with respect to a2,1 := ak2,1 and e2 according to Lemma 6.4.7. One can construct in this way a double indexed sequence of zeros (al,}), 1 ::: j ::: NI, disjoint index sets II,} and closed disks B I ,}. Let 10 denote the collection of indices such that lakl < RI and also ak ¢ UI::'O}::'ON 1 B I ,}.
This construction exhausts all the zeros of and defines the groupings that we need in the definition of III . III. Let us also point out that BI,} never intersects Iz I > RI +2. Let us now denote by VI,} the set of zeros aj of counted according to their multiplicity when i E .ft.}, and let VI,} be the total number of points in Vt,}. Remark that we have VI.} :::
n(la/.}I(l + AI,}»
~ 2sdal,jl.
Hoping that there is no confusion with the previous notation in Section 6.3, we denote a E A (V) as
We recall that
6.4. Differential Operators of Infinite Order
427
where Ol,} is the diameter of BI,} Ol,j
= 2AI,jlal,jl.
Let (Kp)p':.l be an exhaustion of ()) by convex compact sets, and let Hp = H Kp ' then we denote
With the last definition in mind we define A*(V) := {a E A(V): 3p ~ 1: Illalllp < +oo}
considered with the natural inductive limit topology. For further reference, we let Mp = sUPlul::"l Hp(u) and let f/p > 0 be such that (z E
q,
that is, Kp+l contains the f/p-neighborhood of Kp.
6.4.8. Theorem. Let be a nonzero entire function of exponential type zero, then Exp(Q) p Exp(Q) ~A*(V) is an isomorphism. Moreover, every f a Fourier expansion fez)
=
E .K(Q)
L(L I,j
aEl,
such that (d/dz)(f) = 0 has
Pa(Z)e(1Z) , J
where Pa is a polynomial of degree < ma = multiplicity of a as a zero of . The series on the indices (I, J) converges absolutely and uniformly on compact subsets of Q. The frequencies for which Pa '" 0 and the corresponding coefficients of the polynomial Pa depend only on the function f, and the coefficients satisfy estimates of the same kind as in Theorem 6.2.7. Proof. Given g E Exp(Q) there exists p is a constant Ce ~ 0 such that
~
1 so that for every 0 < s < 1 there (z
Consider a then
= peg) = (al,}) l~i(al,})I8f,j .::::
Eq.
and apply Lemma 6.2.10 to the set B(ak"j' Ol,}),
i
max{lg(z)l: Iz - al,jl
:s Ot,j}'
For Iz - al,} I .:::: Ol,j we have Hp(z)
+ slzl
+ sla/,JI + (Mp + s)o/,J .:::: Hp(a/,J) + la/,J I(e + 2A./,J (Mp + 1». .:::: Hp(al,j)
428
6. Harmonic Analysis
On the other hand, we can find l(p) 2: 1 such that for I 2: l(p), we have 2A l ,j(Mp
Choose
S
+ 1) ::::
Sl 2(Mp
Yip
+ 1) :::: 2'
= Yip /2 then
IIlal.jllll,j = max , I~i (al,j)8i) :::: C, exp(Hp(al.j)
+ Yiplal,ji)
:::: C, exp(Hp+l(al,j». Since there are only finitely many indices l, j with I < I (p) there is a constant Bp > 0 such that
with
This implies the continuity of the restriction map p. To prove the surjectivity and the openness of p, we observe first that given a E A*(V) such that Illalllp < +00, if Pl,j denotes the Newton interpolation polynomial such that Pl,j(Pl,j) = al,j. then IP/.j(z)1 :::: vl,jeHp(U'J)lllalllp
for
Z E Bl,j'
We recall that Vl,j :::: 2SJ\al,jl,
It is also clear that for z E Bl,j' one has lal,jl ::::
Izi + Al,jlal,jl
::::
Izi + klal.jl,
Hence For the same reason, Therefore, for z in Bl,j. IPI,j(z)1 :::: IIlalllpeHp(z)+,/(3+Mp/2)lzl.
We need to construct disjoint open sets dinate to them, so that the Coo function
Ul,j
and cut-off functions
()I,j
subor-
o = L Ol,j PI,j I,j
coincides with PI,j on a neighborhood of any a E VI,j and there are good estimates for iJ() jiJz. to be able to apply the idea of the semilocal interpolation Theorem 2.6.4. The difficulty here is the need to estimate, very precisely, the
429
6.4. Differential Operators of Infinite Order
constants so that at the end of the process we obtain a function of Exp(Q). We introduce the disjoint open sets
BI.1' U1.2 := B1,2\BI.1' UI.I :=
and an open set Uo which is the union of small disjoint disks surrounding each with k E Jo and which is also disjoint from the U/. j . We need to estimate the size of a region inside the aU/. j where we have good lower estimates for (~)1
::::
e-Etlaul,
whenever Aulaul - au :::: I~ - al.Ii :::: Au/aul· For U2 , I we have the subregion union of the two open subsets: (i) AI.liau/ < /~ - al.Ii < Aulal,) 1+ al,)
(ii) Al.2ial,2/ - au
(~)/
2:
e-E/1a/,jl
2:
e- E /(4 Ia k ,M\)Ict>(.;)/
2: e-SB.lak,de-I/2.
6. Hannonic Analysis
430
For the same reason the width of that portion of the collar is the inequality
al.j
which satisfies
Therefore we can find functions B/,j E V(VI.j) which are identically 1 except on the collar, 0 ~ fll.j ~ 1, and 1
a:~j (Z)I ~ CeI6etla',JI
for some constant C independent of I and j. Furthermore, the Coo function of compact support (l/ef>)(afll,j/(Jz) satisfies 1_ ael,j ef>(z) az
_ 1
(Z)I -< eI/2Ce24etlal,jl
belongs to Exp(Q). For this reason, we choose \II as a solution of the equation 1 ae
(J\II az
(*)
= -q;- az = -
" 1 ael,j L..J I• j ' PI,jq;- az .
In VI,j, one has the estimate 1
~Z) :~ (z) 1 ~ e I / 2er IlalllpeHp(z)+e,(51+Mp/2)lzl.
For a convenient choice for
I:::: i(p).
Thus, up to a multiplicative constant Dp , which depends only on p, we have everywhere 1
ef>~Z) :~ (Z)I ~ DplllalllpeHp(z)+'/p/2IZI.
From Theorem 2.6.3, with the subharrnonic weight qJ(z) = 2Hp(z)
+ I1plzl + 2 log (1 + IzI 2 ),
we use the fact that
[I ~Z) :~
2
(Z)1 e-q;(z)dm
~ D~lllalll; [
(1
~ D~lllalll~ to choose \II so that it solves (*) and
J
11/I(z)12 e-q;(z)dm < lD' Illall1 2 . + Iz12)2 - 2 P P
(1
C
:~j2)2
6.4. Differential Operators of Infinite Order
431
Therefore, the holomorphic entire function g defined above, satisfies peg) = a and Ig(z)1 2 ::: 2(161(z)1 2 + It(z)1 2 Iq,(z)1 2 ) ::: 2(Dp II lall l;e2Hp(z)+~plzl
+ It (z) 12C;eelzl).
Now, we introduce CPo(z) = cp(z)
and we obtain
+ 210g(l + Iz12) + slzl
J
Ig(z)1 2e-'Po(z)dm ::: 2(D;
+ C;D~)IIlalll;.
IC
Let epl(Z)
= max lepo(u + z)1 ::: epo(z) + E p , luI:::: I
by the mean value property, we obtain Ig(z)1 2
:::
Bp.ee'Po(Z)IIlalll;
for some convenient constant Bp.e which depends on p and
E.
Hence,
Ig(z)1 ::: B~:;IIlalilpeHp(z)+[(e+~p)/2Jlzl+2Iog(l+IZI2). We can choose
E
= '1p/2 and find a new constant Bp > 0 such that Ig(z)1 ::: BpillalilpeHp(z)+~plzl ::: BpillalilpeHp+,(z).
This concludes the proof of the isomorphism between Exp(O)/q, Exp(O) and A*(V).
The rest of the proof of Theorem 6.2.7 and subsequent considerations, including estimates of the coefficients of the Fourier expansion hold verbatim. The only thing that requires a minor remark is the fact that if! has two Fourier-expansions converging in $(0) the coefficients are the same because the operators Tk used in the proof of Proposition 6.2.8 are also infinite order differential operators in this case, since F(Tk)(Z) = q,(z)/[(z - ad m,] are entire functions of exponential type zero. 0 As in the case of mean-periodic functions, if the variety V of 41 is an interpolation variety for the space Exp({O}) of entire functions of exponential type zero, then the Fourier representation of an arbitrary solution of equation q,(dldz)! = 0 converges in )f(0) without groupings. One can prove that the condition for V = (ab mk) being an interpolation variety in the space Exp({O}) is that for every s > 0 there is a constant Ae > 0 such that (k
~
1).
6. Harmonic Analysis
432
(See [BLVJ.) The following theorem ofP6lya-Levinson [Levs, Theorem XXXI, p. 92] furnishes a very simple geometrical condition on the ak. when mk = 1 for all k, for V = (akk~1 to be an interpolation variety in Exp({O}).
6.4.9. Theorem. Let that:
(ak)'\~l
be a sequence of nonzero complex numbers such
lim ~ = o. k--+oo ak (ii) There is a constant y > 0 such that for any k, j E N* (i)
lak - ajl ?: ylk - jl. (iii) There exists e E [0, 21l' [ such that limk--+oo adlak I = e iO . Then, the function
(z) =
II (1 _z:) . k~1
ak
is of exponential type zero and satisfies, for every s > 0,
(a) (b)
\
~Z) \ =
\_1_\
O(e s1zl ),
whenever
= O(eSlakl)
'(ad
'
Iz - akl
?:~,
Vk?: 1;
k ?: 1.
We remark that condition (iii) is missing in the statement of this theorem in [Levs], [Bo, p. 146], but it is used in the proofs given there. Letting a2k+1 = -a2k. one can easily see that some condition on arg ak seems necessary to ensure (b) for the function defined above. Note that condition (b) implies that the variety V of zeros of is an interpolation variety in the space of functions of exponential type zero. In particular, the sequence {ad is an interpolation variety. This is not included in Chapter 2 since Exp({O}) is not Hormander space Ap. One can do this directly as follows, let (ak)k:::1 be a sequence with the property that for every s > 0,
lakl = O(ee1aki )
(k?: 1),
then there is a way to choose integers nk ?: 0 so that
g(z) :=
L
~ (z) (~)nk
'(ak) z - ak ak is an entire function of exponential type zero such that g(ad = ak,
k?: 1,
g(-ak) = O. Instead of the proof of Theorem 6.4.9 we shall give an important generalization of this theorem, due to Vidras [ViI], [Vi2]. Namely, that the condition (iii) is not necessary for the conclusion that the sequence {ad is an interpolation variety.
433
6.4. Differential Operators of Infinite Order
=
6.4.10. Theorem. Let Z {ak} k': \ be a sequence of complex numbers satisfying the following two properties: (i) adk -+ 00 as k -+ 00. (ii) There is c > 0 so that for any n, k E N*: Ian - akl =.:: cln - kl. Then, there exists an entire function F of exponential type zero, with simple zeros {Zm };;'=l which include all the ab and such that "Is > 0 one has
(a) l/W(reioz)1
=
O(eer) whenever Ire iO
-
zml =.:: 1/8c for all mEN as
r -+ 00.
(b) l/IF'(zm)1 = O(eBlzml) as m -+ 00.
In particular, not only the sequence {Zk}bl is interpolating, but this property also holds for the larger sequence {zm};;'=l of zeros of F.
The proof of Theorem 6.4.10 consists of a sequence of lemmas, some of which already appeared in [Levs]. The main idea to get around the problem that we have no control on the arguments arg ak is the following grouping critierion:
6.4.11. Definition. Let Z = {an}~l be a sequence of complex numbers satisfying the properties (i) and (ii) of the statement in Theorem 6.4.10. Then we say that am is paired to an, n #- m, if and only if (iii)
We say that am has the property P if there exists n so that {am, an} fonns a pair (i.e., it satisfies condition (iii». An immediate consequence of the definition is that any am is paired to at most one element of the sequence Z. Indeed, assume that am is paired to ak and a p . The property (ii) implies that
c
~
clk - pi
~
lak - apl
~ lak +aml
< c /2
+ lam +apl
+ c /2 =
c,
which is a contradiction. Thus we are able to separate the tenns of the sequence Z into two disjoint groups Z\ = {an: an not satisfying the property P}, Z2 = {am: am has the property Pl. The following lemma is a simple generalization of Levinson's Theorem 6.4.9 given in [LevsJ. The main differences are that we do not assume that ReAn > 0 or that 1m An/An -+ 00 as n -+ 00.
6.4.12. Lemma. Let A \ = {An }~I be a sequence of nonzero complex numbers satisfying the properties (i) and (ii) of Theorem 6.4.9. Assume further that no point
434
6. Hannonic Analysis
of A I has the property 'P and that there is a constant () satisfying 0 ::::: () < rr 12, so that for every n E N either
I arg An I ::::: ()
or
I arg An - rr I ::::: ().
Then there exists an even entire function Fl of exponential type zero, vanishing ±An and satisfying "Ie > 0: only at the points z
=
(a) I/IFI(reia)1 = O(e") whenever Ire ia (b) I/IF{(An)1 = O(eEIAnl) as n ~ 00.
± A.nl
::: c/8 as r ~ 00.
Proof Let us define FI (z) = 0':'1 [1 - (z2/A;)]. The hypothesis (i) is equivalent to the fact that the counting function nA, (r) = o(r), thus FI E Exp({O)) [Lev]. Since no point of A I has the property 'P, every zero of FI is simple. In order to show that FI has the other desired properties, we first note that since FI is an even function it is enough to prove (a) and (b) for Re z ::: 0 and, given o < e < only for r = Izl sufficiently large. Following Levinson, assume that z ¢ A I, and divide the terms of the sequence into three disjoint sets as follows:
!,
A = {An E AI: IAnl ::::: (1 - e)lzl),
B = {An E AI: (1 - o5')lzl < IAnl < (1 - o5')lzlL C
= {An E AI:
(I + o5')lzl ::::: IAnl}.
Let us denote FA(Z) := OAffEA[l - (z2/A~)J and define similarly FB, Fe. Then FI = FAFBFe ,
and we can divide the proof in several steps. Step 1. Assume An E B, then we have (1 - e)lzl < IAnl < (l + e)lzl, and therefore Izl (l+o5')lzl 1+05' < < - - O. Step 2. This time we are going to estimate !FA (z)1 from below. Similarly to the previous step, let a be the value of the index n that minimizes IAn - zl among An EA. We see, as earlier, that when An E A we have
1
----.,,.........~
11-z2/A~1
IAnf (1 + 8)lzl = IZ-Anllz+Anl-lz-Anlcos8 < ----2(1 - 8)lzl 21z1 < ---'--'--- IAa - Ani cose - cia - nl cos 8
1 be defined by the equation (1 - s)fJ + logs = O. Note that sfJ(e) ~ 0 when s ~ O. Then, for 0 :S x :s 1 - s, the inequality 1/(1 - x) :s sf3 x holds. Therefore, since (r/rn)2 :S r/rn :S 1 - s, we have
I]
(1 -
~2/r;) :S exp
(fJ r2
(~ r~) )
This inequality implies that IFc(z)1 :::: e-2/lclzl.
This concludes the third step of the proof. It is clear that, after convenient renormalization of the constants, we have shown that given s > 0 there are constants rc > and y > 0 such that for Izl :::: re we have ylz - ANle- elzl :S !FI(z)1 :S eelzl.
°
Since F{(AN) = limz->AN F1(z)/(z - AN), the two properties (a) and (b) follow.
o
This concludes the proof of Lemma 6.4.12.
6.4.13. Lemma. Let A2 = {An }~I be a sequence of complex numbers satisfying the properties (i) and (ii) of Theorem 6.4.9. If every term An has the property P then limr->oo 2: IAn I9(l/An ) exists. Proof. By the Cauchy criterion we have to show that for 0
0 if r > 0 is sufficiently large then
= oCr),
given
8
~
~ r::oIA n l9+cl2
~ < nA,(r) < A r -
8.
n
This implies that adding the other element of the pair, if it is not already included in the sum (**), does not change the nature of the problem. On the other hand, if Am, An is a pair we have 1
1 1I Am +An
=
lAm + An I IAmAnl
c
1
~2IAmAnl'
Moreover, IAmAnl ~ IAmI2. Hence, there is a constant C > 0 such that
Since the series L:I(l/IAn I2 ) is convergent, the lemma is completely proved.
o
The main point of the next lemma is that we do not make any assumption on the arguments of the sequence.
=
11.2 {An l:'1 be a sequence of nonzero complex numbers satisfying properties (i) and (ii) of Theorem 6.4.9 and also assume that every element has the property P. Then there exists an entire function F2 E Exp({O}) vanishing exactly at the points of 11.2 and so that for every 8 > 0 the following estimates hold:
6.4.14. Lemma. Let
(a)
(b)
1.
iF2(re,B) I
= O(eer),
1 ----,.__ = iF2(A n ) I
whenever
O(esIAnl)
as n -+
Ire iB
-
An I ~
!c
as r -+
00.
00.
Proof. If the genus of the sequence were zero, we could use a simpler construction of F2 , but we prefer to give a unified formula, irrespective of the genus. Using Lemma 6.4.13 define w:= limr--> 00 LIAn I9(l/An ). Let us consider the function F2(Z) =
e- wz
IT (1 - ~) n=1
An
II
e z/An := lim e- wz r-->oo IAn l::or
(1 - ~) e An
zlAn ,
where the infinite product is defined by the limit of the finite products. The reason is that the IAn I are not necessarily increasing. Note that in the genus zero case, this product is not the Weierstrass canonical product [Lev], while this is true when the genus is 1. Nevertheless, the usual estimates show that F2 E Exp({O}) (see [Lev, Theorem 15, p. 26].) Thus, we need to concentrate in
439
6.4. Differential Operators of Infinite Order
proving lower bounds for Fz, in a way altogether similar to Lemma 6.4.12, so that we shall be able to skip the details when they are the same. Given 0 < s < ~, we fix z ¢ Az, with Izl» 1. Divide the terms of the sequence into three disjoint sets as before A = {An E A2:
B
IAnl :5
= {An E A2: (1 -
(1 -
s)lzl),
s)lzl < IAnl < (1 + s)lzil,
C = {An E Az: (1 +s)lzl :5I AnlJ.
Step 1. This is practically identical to Step 1 in the proof of Lemma 6.4.12. We denote by N the index that minimizes Iz - An I, and assume AN E B without loss of generality. We let L be the number of terms in B district from AN. We obtain
II AnEB.n#N
1
< (1
11-z/AIn
II
+ 8)LlzIL2L
AnEB.n#N
As was done in Lemma 6.4.12, we can now find a constant Izl » 1 we have
Iz -
1
IA N -AI' n K
2: 1 such that for
II 11 - : I < eelz l .
ANle-Kv'Elzl
Exp(Q) ----+A.(V)
is a topological isomorphism. The rest of the proof is the same as in Theorem 6.4.8. 0 Henceforth, we will assume that Q and K are such that
r = r9 =
{z: I Argzl
in the set
r; := {z E C*: B + ~ -
I) :::::
Argz : : : 3; - + o}. (J
Then every solution f in ;f(Q + K) of the equation T function.
*f
= 0 is an entire
Proof. We can use the same groupings and the same definition of the norms III· 1111.j to define a norm in A(V) by Nda)
= sup{lllal,jllll,j exp(-Hdal,j»},
where L is an arbitrary compact convex subset of C. The space A •• (V)
= (a E A(V): 3L.
Nda) < +oo}
is isomorphic to Exp(C)/cI> Exp(C) by Theorem 6.2.7. Therefore, to see that any f E ;f(Q + K) for which T * f = 0, is entire, it is enough to see that the two spaces A.(V) (given by the proof of Theorem 6.4.11)
444
6. Hannonic Analysis
Figure 6.1
and A**(V) are isomorphic. In fact, this means the dual spaces coincide, and so the coefficients of the Fourier expansion of f in Q + K are also the coefficients of the Fourier expansion of some entire function f, T * f = O. It is clear, that f is the analytic continuation of f to the whole plane. In order to prove this isomorphism all we need to show is that, for every compact convex set L in C, we can find L I, a compact convex subset of Q + K such that for all a E Z( 0 we can
ICXk.ll:::s
el~le6Ia'I+D•.
O:ol<m, As V (P) is an interpolation variety for Exp({O}), there is an entire function
1/1 of exponential type zero such that
1/F (I) (CXk) I!
= ak./·
447
6.4. Differential Operators of Infinite Order
Let U~ =
S:
{Z E
Z
+ i~e-itp
which is a nonempty open set. Then, for Z E
U~
E
S},
we have
hence fez + i~e-itp) has an analytic continuation from U~ to the whole set S8' Letting ~ vary in lR we obtain the lemma. 0 6.4.21. Theorem. Let T satisfy the hypotheses of Proposition 6.4.17 and Lemma 6.4.19, ro c Q + K C {z E C: Re z > OJ. Assume further that the zeros of
oo
Moreover, assume that the functions
are entire functions of exponential type zero whose varieties V (Pj ) are interpolation varieties for the space Exp({O}). Under these conditions, any function f E .}f'(Q + K) that satisfies the equation T * f = 0 and has an analytic continuation to a disk B(O, e) for some e > 0, has also an analytic continuation to the open set
where S = ro. In particular, f has an analytic continuation to the cone ro - ie. Proof. We can write f = f+
+ f-, f+ =
L
qa(z)e aZ ,
ae(r;)C
f-
= L qa(z)e'l/Z, aero
where both series are convergent in Q + K. From Lemma 6.4.19 we know that f + is an entire function and therefore, the function f _ has an analytic continuation to B(O, e).
448
6. Hannonic Analysis
The function
f- satisfies the infinite order differential equation PI
(:z) (:z) ... P2
Pr
(:J
f - = O.
Let
which is holomorphic in s,=reUB(O,B)
and satisfies the equation PI
(:z)
gl
=0.
Therefore, by Lemma 6.4.20, gl is holomorphic at least in the half-plane S"'I' Due to the surjectivity of an infinite order differential operator, there is a function hIE .1!f(S"'I) such that
Then
P2 The function
f- -
(:z) ... (:z) Pr
(f- - hI) =
o.
hI is still holomorphic in So. Let now
Thus,
so there is h2 holomorphic in
S'P2
satisfying
In other words,
Continuing in this fashion we get functions (h j) I sj 9-1 holomorphic in that
S"" such
6.4. Ditferential Operators of Infinite Order
449
Applying Lemma 6.4.20 once more we obtain that
hi := f- - hi - h2 - ... - h r is holomorphic in
S"".
Hence,
f-
= hi + ... + hr
I
is holomorphic in
o Theorem 6.4.15 appears for the first time in [BSt 1]. The classical Fabry Gap Theorem (see [Levs], [Die)) is a particular case of Theorem 6.4.15. It states that if 0 < AI < A2 < ... is a sequence which satisfies the conditions: (i) limn_oo(An/n) = 00. (ii) For some constant C > 0 one has An+1 - An ~ C, then, every Dirichlet series Anz , f(z)
= Lanen~l
which has abscissa of convergence point on the line Rez = {Ya.
{Ya,
-00
differential equation P(d/dz)f = 0 with
< +00, has no regular
(Ya
of the infinite order
By Theorem 6.4.9, V(P) is an interpolation variety, hence if f has a regular point zo, Rezo = {Ya, f will be holomorphic in Rez > {Ya - B for some 8 > O. It will satisfy the same infinite order differential equation in this bigger domain. By the representation theorem and uniqueness of the Fourier representation, the Dirichlet series will converge absolutely in Re z > {Ya - B, contradicting the definition of {Ya' Lemma 6.4.19 is the crucial ingredient in the proof of the Fabry Gap Theorem. One can find some particular cases of this lemma in the work of P6lya, Bernstein, and Levinson [Levs], [Berns], [Die], and more recently, using microlocal analysis in [Kawl], [Kaw2]. If the set S of the lemma is an arbitrary open convex set our proof provides some sort of enlargement to a set St. A very precise description of this set and a proof of this lemma without any conditions on the infinite order differential operator appears in the work of [Se], [Ao]. We show now that the ideas exposed in this section can also be used to provide an elementary proof of the general version of Lemma 6.4.19, namely, Theorem 6.4.24. 6.4.22. Lemma. Let be a nonzero entire function of exponential type zero and K, L two convex compact sets such that K £ Land (~ E
Z()).
450
6. Hannonic Analysis
Then,for every entire function f E Exp(L) and every 6> 0 one can find entire functions g and h, such that g E Exp(K,), h E Exp(L,), and
f
= cJ>h
+ g,
where K., L. are the open 6-neighhorhoods of K and L, respectively. Proof. One can certainly assume that K #- L and that cJ> is not a polynomial. (In the latter case it is enough to take g a polynomial that interpolates the values of f on V(cJ».) In the general case, when Z (cJ» is infinite, we use the construction of the sets U,.j from the proof of Theorem 6.4.8 and corresponding standard functions e,.j. Introduce now the function
e:= L(O"d). ',j
For
~ E
U"j we have that
IHd~) - HdCX',j) I :::: M L81,j :::: MLT1Icxl,jl,
where ML = max{Hdu): lui:::: l}, With a similar definition for MK we also have IHK(~) - Hdcxl.j) I :::: M K81,j :::: MKT1Icxl,jl. On the other hand, given Y/ > 0, for I :::: lo(Y/)
»
1, we have
Hdcxl.j):::: Hdcxl.j)+Y/lcxl,jl by hypothesis.
We also have that for Y/ > 0 there is a constant log If(~)1 :::: Hd~) Hence, for I :::: lo(Y/) and
~ E
log If(~)1 :::: HK(cx,.j) ::::
+ Y/I~I + F~
F~
:::: 0 so that (~ E
IC).
Ul,j we have
+ Icx/.jl {2Y/ + ~(ML + MK + Y/)} + F~
HK(~) + RI {4Y/ + ~(2ML + 4MK + 2Y/)} + F~.
Moreover, we have that for some constant Co :::: 0
I ael j I 48 log I- - - - . : (n < - I I~I cJ>(n a~ - 2
+ Co
(~ E
IC).
Given 6 > Owecanchoose Y/ = 6/16 and 1\(6) :::: 10 (6/16) so that if 1 :::: 1\(6), then 1 6 21 (2ML + 4ML + 2y/ + 48) :::: 4' Hence, there is a constant C \ :::: 0 for which 10gIB(~)1 :::: HK(~)
E
+ 21~1 + C\
6.4. Differential Operators of Infinite Order
and log
451
1:~) :; (~)I ~ HdO + ~I~I + C
1•
We can conclude that there is a Coo function 1/1 such that the function
= e + 1/1
g
is an entire function satisfying the inequality log Ig(OI ~ HdO
e
+ 21~1 + 4 log (I + R12) + C2
for some C2 ::=: O. Namely, if p(z) := 2HK (z)
+ elzl + 210g(l + Iz12)
there is a solution 1/1 of the equation
oe
01/1
I
oz
OZ
-=---
satisfying
r
r 1_1_ (Z)1
11/I(z)1 2 e-Pdm < 1 ae 2e-Pdm < +00 + Iz12)2 - 2 Jc (z) az . We obtain the desired inequality for log Igl in the same way as it was done Jc (I
in the proof of Theorem 6.4.6. Define the function h by
f
h =
and apply Lemma 6.4.5 to show that h
-g II> E
Exp(Ls).
o
We shaH now consider some constructions of [Ki] that will help us to describe the largest domain Q into which all solutions in n of an infinite order differential operator will have an analytic continuation. Let A be a nonempty set of complex numbers and let U be a convex open set in Co For every convex compact K in U we set YA(K):=
nlz
E
C: Re(z~) ~ HK(~)}
(EA
and KccU
where the union takes place over all compact convex subsets of U. Let us note that r A (U) is a convex open set. In fact, each YA (K) is clearly a closed convex set, and by taking an exhaustion (Kn)n~l of U by compact convex sets we see that r A (U) is an increasing union of convex sets and hence convex itself. Moreover, if e > 0 is sufficiently small so that Ke = K + B(O, e) ~ U, then YACKo) ;2 YA(K)
which is a neighborhood of YA(K).
+ B(O, e) =
(YA(K»e,
452
6. Hannonic Analysis
Let us assume now that A is an unbounded set, and let, for r 8
A •r
n
(K) =
{z E C: Re(zs) :::;: Hds)
~
0,
+ r},
,eA 8
A (K)
= U8
A ,r(K),
fi':O
8 A (U)
U
=
8
A (K).
KccU
These three sets are convex and 8 A (U) is open. In fact, for K C C U let such that Ke CC U. Then, for every r ~ 0,
e = eK >
°
8 A ,r(Ke) ;2 8 A ,r(K)
+ B(O, e),
Hence,
therefore, o
-"-
(where X denotes the interior of X, while X is the interior of the closure) and KcCU
KCCU
Note that we have shown that
where K runs over the family of convex compact subsets of U. Given a sequence of complex numbers A = (Sj)ji':l such that lim ISj I = +00, )-+00 we denote by .B(A) the subset of the unit circle 'IT' of points S for which there is a subsequence (Sj,hi':l with
S = lim
k-+oo
lL, lSi, I
We denote by a(A) the closed cone with vertex at the origin generated by the directions in .B(A). 6.4.23. Lemma. Let A = (Sj)ji':l, limj-+oo ISjl = +00, and let U be an open convex set in C. Then Proof. It is easy to see that
r A(U) ~ E>A(U) ~ ra(A)(U),
6.4. Differential Operators of Infinite Order
453
If we translate the set U keeping the set of directions peA) fixed, then 8 A (U) and r alA) (U) are translated by the same amount. Therefore, we can assume that o E U. Let K be a compact convex subset of U with 0 E k, then there is a constant )...k such that for I < )... < )...K, the set )"'K is a neighborhood of K still contained in U. We fix such a ).... Observe that there is an £ > 0 such that the assumptions (~E
q
because 0 E k. From this inequality, it is easy to see that there exists /3 > 0 such that the assumptions Z
E
K,
I;
and
E a(A),
imply that Hence, if ~ E
a(A),
then,
Denote
-
a(A, 8) := {I) E C: 3~ E a(A) with I~
1)1 :::: 811)1}.
The inequality (*) implies that Ya(A)(K) ~ Ya(A.8)(K).
It is also clear from the definition of peA) that there exists r > 0 such that Ar := A
n (B(O, r)t
~ a(A, 8).
It is then immediate that
Hence, Ya(A)(K) ~
n
)"'8 A(K) = 8
A
(K)
A>l
and
o
We finally have here the theorem about analytic continuation of solutions of a differential operator of infinite order due to Sebbar and Aoki (see also [BGV]).
6.4.24. Theorem. Let be a transcendental entire function of exponential type zero and let r2 be an open convex set in Co Every solution f E Jot'(r2) of the equation
454
6. Hannonic Analysis
has an analytic continuation F to the open convex set
Q=
ra(Z(z( 0, if E8 := (r ::: 0: mer) < -(ir}, for any '1 > 0, there is ro = ro(8, Tj) such that m)(E8 n [0, TIT for any r ::: o. (Here m) denotes the Lebesgue measure on JR.)
rn :::
"2. From the construction of the components v,.j in the text preceding Theorem 6.4.8, conclude the following result holds for any entire function ct> "# 0 of exponential type zero: Let Z = (z: ct>(z) OJ, then for any e > 0 there is a value R. > 0 such that
=
dist(z, Z) ::: elzl
implies
1ct>(z)l::: e-· 1zl •
(Hint: For R; » 1 any connected component of the set S(ct>, e) = (z E IC: 1(z)1 < exp( -e Iz /)} that contains any point with Iz I ::: must have diameter not exceeding
R;
458
6. Hannonic Analysis
Elzl/2. Assume dist(z, Z) ~ eiz I, Izl ~ RE , z E S(, E), to conclude that log 1(~)1 is harmonic in the component that contains z. Use the minimum principle for harmonic functions to conclude the proof.) Note that this result can be used to give a slightly simpler proof of Lemma 6.4.16. 3. Let (z) = cos.fi, Z = (z E CC: (z) from Theorem 6.4.24, in the case
= OJ. Find the asymptotic cone a(z) and f2 (0
~
() < 2JT)
and
n = 8(0. 1).
6.5. Deconvolution The problem of solving an equation of the form
Jl*[=g, where Jl is a measure with compact support in R appears often in engineering and other applications of mathematics. As we know. this equation does not have a unique solution in the spaces & OR), V' (lR), not even if we assume [ E L 00 OR). For instance, the Fourier transform :F1-t could have a real zero a and then 1-t * eiar = O. For that reason, it is customary to restrict the domain of the [ considered, to a space like L 2 (JR.) or L 1 (JR.), so that there is at least uniqueness of the solution. In fact, if [ ELI (JR.) then the Fourier transform :F[ is a continuous function which tends to zero at infinity, thus Jl * [ = 0 implies (~ E
JR.).
Since F1-t = 0 is a discrete set, it follows that :F[(0 is identically zero. Therefore, [ = 0 a.e. Nevertheless, the operator ELI (JR.) ~ Jl
[
*[
ELI (JR.)
fails to have a continuous inverse in general. This is essentially due to the presence of zeros of FJl. Recall that FJl has no zeros only if 1-t = c8a , for some c '# 0, a E R The same kind of reasoning applies to other function spaces like L 2 (JR.). Given that this problem is very important, there is a large literature about how to find approximate inverses. They are usually called regularization methods [Mor]. The interpretation of the equation 1-t * [ = g is that [ represents an unknown signal, Jl a measuring device, and g the output signal, i.e., the data. In some situations, it is possible to multiplex the signal, in other words, to use several measuring devices JlI,"" Jl". We have therefore a system of convolution equations. fJ.1
* [ = glt···, fJ." * [ = g".
If we could find measures or even distributions of compact support such that VI
* JlI
+ ... + V" * Jl" = 8.
VI, .•. , V"
6.5. Deconvolution
459
Then we can solve the system (*) very simply,
f
=
V(
* gl + ... + Vn * gn'
This inverse formula will be continuous in £ (R) and also in some Sobolev spaces [Ad]. The problem, of course, is to find out whether (**) has a solution and, moreover, to find the deconvolvers Vl, ••. , lin explicitly. Let us consider only the case n = 2 for simplicity. From now on, we use indistinctly il or F IL to denote the Fourier transform of a distribution or measure. The equation (**) is equivalent, via the Fourier transform, to the Bezout equation
and we are looking for a solution pair (ill, il2) E F(£'(R» = Ap(C) for the weight p(z) = I Imzl + log(2 + Izl). From Hormander's Theorem 2.8.15 (see also Theorem 2.8.16) we know that the necessary and sufficient condition or the solvability of (* * *) is that there are constants 8 > 0, C > 0, such that: (~ E
C).
In the engineering literature this condition is sometimes called "strong coprimeness." Generally speaking, this is simply a condition of how separated are the zeros of ill from those of il2' We would now like to construct explicitly deconvolvers VI, V2 for distributions IL I, 1i2, which satisfy a number of conditions that are often found in concrete examples. We point out that most of the limitations that we are going to impose in Proposition 6.5.1 can be eliminated by paying the penalty of more complicated formulas for the deconvolvers. 6.5.1. Proposition. Let ILl> IL2 (i)
E
£'(R) be such that
ill (0), il2(0)
=1= 0 and there exists a constant T > 0 so that all zeros of ild12 lie in the logarithmic strip
I Im~1 :5 T log(2 +
I~I)·
(ii) All the zeros of ilj are simple and they are constants
that
ilj(n
=0
implies
lilj(nl:::
(1
+81~I)L
8
> 0, L > 0, such
(j ::; 1,2).
(iii) There are constants 8 > 0, M > 0, such that A
ILI(~) = 0
and
A
implies
IIL2(~)1 ~ (1
8
+ I~I)M
460
6. Hannonic Analysis
(iv) There is an increasing sequence of piecewise C l Jordan curves (Yn)n;::l (i.e., Int(Yn) S; Int(Yn+I)), and a sequence (rn)n;::1 o/positive real numbers, converging to infinity, such that (a) l(Yn) :;; length of Yn = O(rn ). (b) There exist two strictly positive constants CI > C2, so that
(c) There exist two positive constants)... and N, so that (~ E
Yn, n
~
1).
belong to F(£'(lR» and ~q ILl (ngl (~)
+ ~q IL2(ng2(~) + ILl (nIL2(~)p(~) =
1,
where P is a polynomial of degree at most q - 1, P(~)
= Res
+ ... + u ( ~q-l + u~q-2 ~.
q- l
u q ILl (u)/Jdu)
)
,u =0 .
Proof. Let For
Z
E
1=
Int(Yn) we can rewrite the Cauchy formula as follows:
_1_1_1_ = -1-1 2ni
Yn ~ -
z
2ni
0(0 -O(z) d~ - z)
Yn O(n(~
+ O(z) 2ni
1
d~
Yn e(~)(~
. - z)
We observe that for any z fixed, this formula is valid as long as n is sufficiently large (i.e., n ~ n z ). Let us now compute by the residue theorem the first integral:
-1-1 2ni
Yn
e(~) e(~)(~
e(z) d - z)
~
- ~ (z)' (z)P(z) - /1-1 /1-2 q
~
()
+z /1-1 z
'"
~
IL2(Z)
Rq' (R)~I(R)Z- R P-2(fJ)=O I' ILl I' /1-2 I' I' tlElnl(Yn)
6.5. Deconvolution
461
Let A > 0 be such that Il1j(z)1
s
A(l
+ Izl)AeAI'mzl
(j = 1, 2, z E iC).
Then, if Iz - .81 ::: 1 we have
I:~z~ I s
1112(z)1
s
A(1
+ Izl)AeAI'mzl
and, if Iz - .81 < 1, then
I s Iwl=l max 1112(.8 + w)1 s Ae I112(z) z -.8 S A 2 e A (3
+ Izl)AeAI'mzl
A (2
+ 1.8I)AeAI,mPI (z
E
IC).
Therefore, there is a constant B > 0 such that (z E iC)
and (Z E
iC).
On the other hand,
and, similarly,
for a convenient positive constant C. As q was chosen so that q ::: L + M + 2, the series defining gj converge uniformly and absolutely in the whole plane, since
1
.L
(1
+ lal)2
.L
(1
+ 1.81)2
0. The intersection of this circle with the logarithmic strip of hypothesis (i) will have two connected components, we choose the outermost. We do the same for the point -r, and we complete a Jordan curve by adding two semicircles lying outside the logarithmic strip. Let us now see that under the assumption (.co), the crucial condition (iii) is just a separation condition between the sets Z (ill) and Z (il2)' It is the condition they are well separated (see Section 3.1), that is, there are constants s > 0, N > 0, such that for every a E Z (ill) and fJ E Z (il2) I
la - fJ I ~ s -(1-+-la-!+-!fJ-I)"'7:"N . In fact, if ill (a) = 0, then
d(a, Z(il2»
s ~ (2 + 21al)N
and, by (.co), A
1f-L2(a)! ~ (1
C3
+ l(1)N+N2
for a convenient constant C3 > 0. Conversely, if the zero sets are not well separated there must be sequences an -+ 00, fJn -+ 00, of zeros of ill, il2, respectively, such that I
Ian - fJn I ~ (I
+ Ian! + !fJn I)n .
Since il; is also in F(£'(lR», it follows that A
B(2
+ Ian !)A
1f-L2 (an) ! ~ (l + lan!)n
for some A > 0, B > 0, fixed. Hence condition (iii) cannot be satisfied. In fact, in general, the condition (.co) cannot be expected to be true for il, even for exponential polynomials, unless the origin belongs to the interior of cv(supp f-L). For instance, for f-L = 8) - 83 we have il(t)
= e-il; -
e- 3i l;
=
e- 2i l; (eii; _ e-il;)
= 2ie- 2i i; sint· We have
(t
E 0, j = 1,2), then sinaj~
A
I-Lj(~) = -~A
Z(I-Lj)
=
(~ E C),
7t * -Z .
aj
These functions satisfy (.co) and the only condition we need to verify is whether the zeros are well separated. That means there must exist constants E > 0 and N > 0 such that
I:~ - :: I ~ for all p, q
E
(1
+
IP7t/al~+
Iq7t/a2I)N
Z*. Equivalently,
I:~ -~I ~ (Ipl :lql)N' This is exactly the arithmetical condition that the quotient al/a2 is badly approximated by rational numbers. In particular al a2
r/Q.
We remind the reader that the theorem of Roth [BaJ implies that if al/a2 E Q\Q (that is, an irrational algebraic number) then, for every 1/ > 0, there is 8 > 0 such that
I:~ -~I ~ (IPI+~ql?+~'
For a quadratic irrational we can take 1/ = O. In any case, we can take N = 3 in the earlier inequality whenever ada2 E Q\Q. (2) Let us we consider a distribution I-L of the form I-L =
XI-a,a]
+ a,
where a is a C 2 function in IR with supp(a) ~ [-a, aJ. One can see that satisfies the Lojasiewicz condition (.co) because 18(01 < -
C
(1+1~1)2
eallm~l.
il
6.5. Deconvolution In fact, for some
465 K
> 0,
1t1(~)1 ~ d(~, Z(I1»)(1
+ I~I)-I
for I~ I sufficiently large, since zeros of 11 are asymptotic to (rr /a)Z' and they are simple for I~ I » 1. The multiple zeros will only complicate the formulas but their existence does not affect Proposition 6.5.1. When we have two distributions of this kind we only need to worry about their zero sets being disjoint and the separation condition holds outside some big disk (and therefore everywhere). Let
fJ-2 =
X[-a,.a,)
+ U2,
where al > 0, a2 > 0, and there exist constants 8 > 0, N > integers) such that a, pi 8
la2 - q
°(not necessarily
~ (Ipl + Iql)N
for every p, q E Z', and assume that eQjllml;'1
laj(nl ~
K (1
+ 1~I)Mj'
j = 1,2.
If we impose the differentiability condition
Mj > N
+1
then, it follows that the zeros of 111 and 112 are well separated outside some big disk. It is hard to give an analytic condition that implies the remaining zeros are distinct. In practice, one has to explicitly verify this numerically. (3) The last simple example occurs when fJ-I and fJ-2 are difference-differential operators. In this case, all the conditions of the proposition reduce to verify that the zeros of the exponential polynomials 111 and 112 are well separated. This problem is related to the Ehrenpreis conjecture, which was already discussed in Section 3.1. There is another convolution problem of some importance, the situation in which the function 1 in the system of convolution equations (*) is defined in a fixed finite interval. In this case, even the uniqueness of the solution 1 may fail for strongly coprime convolvers fJ-I, fJ-2. To be more specific let us assume that we consider the system fJ-I
*1
=gl,
* 1 = g2, R, RD, R > 0, cv (supp (/Lj » = [-aj, aj], fJ-2
1 E £' (] -
°
< aj < R. The functions R - aj[, and not defined anywhere else. We shall now see that the main condition for the uniqueness of a possible solution 1 is the very simple requirement
gj are COO in] - R
+ aj,
al
+a2 < R.
466
6. Hannonic Analysis
This is just the one-dimensional version of a theorem in [BOl]. We impose a number of simplifying hypotheses to make the proof simple. 6.5.3. Proposition. Let J-LI be a hyperbolic distribution, assume that cV(SUPP(J-LI)) = [-ai, ad, J-L2 is a distribution with cV(SUPP(J-L2» = [-a2, a2], o < ai, 0 < a2, ill and il2 without any common zeros, and
Let f E t' (] - R, R [) satisfy the pair of equations J-Lj then f
= 0 in ] -
*f
= 0
R, R[.
We remark that when R < 00, which we can assume, this proposition is not a consequence of Proposition 6.5.1 (even assuming the extra conditions required there). The reason is that the deconvolvers VJ, V2, in general, will not have support at a point, hence one cannot obtain information about f in the whole interval] - R, R[.
Proof of Proposition 6.5.3. Let V (ill) = {(Ilk, md, k we conclude that f(x) = Pk(x)e iakX
~
1); from Lemma 6.1.24
L
k:::1
in t'(] - R, RD. Let us denote by t'k the distributions given by (J-Llh,O of Lemma 6.1.14, then cV(SUPP(t'k)) = [-ai, ad and
(t'k
* f)(x) =
Pk(x)e iakX
On the other hand, since il2(ak) #; 0,
J-L2
* (Pk(x)e
iakX )
= Qk(x)eic>v,
where Qk is a polynomial with deg(Qk) Ixl < R - al - a2 we have
Qk(x)e iakX = (J-L2 therefore, Qk ] - R, R[.
= 0,
= deg(Pk).
* Tk * f)(x) =
and so Pk
= O.
Tk
Since al
+ a2
< R, for
* (J-L2 * f)(x) = 0;
This shows that
f
is identically zero in 0
It is clear that the proof we have just given cannot work if al + a2 > R, the reason being that the convolution J-L2 * t'k has too large a support. In fact, the result could be false as the following example shows: Let
467
6.5. Deconvolution
and let R be such that
../2n
< R
R. When al + a2 < R one can also reconstruct explicitly the function 1 in ] - R, R[ from the data f-tl * f in ] - R +aj, R - aj[' j = 1,2. The main elements of such a reconstruction procedure have in fact been presented in this chapter and we refer the interested reader to [BGY], where this is done in the case of n variables. We would like to conclude this section with the statement of some theorems about holomorphic and harmonic functions in the plane whose proofs are analog to the theorems about Fourier representation of mean-periodic functions, Theorem 6.1.11 and Proposition 6.5.1. The extra ingredients are the elements of
6. Hannonic Analysis
468
hannonic analysis of functions of several variables and of holomorphic functions of several variables. For the proof, we refer to [BST], [BZ], [BOl], [BG2], and we also recommend the very lively reports on this subject by Zalcman [Za2] and [Za3]. 6.5.4. Examples. (I) Generalization of Morera's Theorem. Let f be a continuous function in the disk B(O, R) and let To be a closed triangle (or a square, or a simple polygon, etc.) such that
To S; B(O, R/2). Assume that
r
Jar
f(z)dz
=0
for every T S; B(O, R) which is congruent to To (i.e., T is obtained from To by a transfonnation of the fonn z 1-* ei 9 z + a). The conclusion is that f is holomorphic in B(O, R). (2) Generalization of the Mean-Value Characterization of Harmonic Functions. Let us recall that if f is a continuous function in B(O, R), then f is hannonic in B(O, R) if and only if for every B(z, r) S; B(O, R)
f(z)=)...(f,z,r)=_l 21t
r
21t
Jo
f(z+reili)dfJ.
For ro > 0 fixed, it is possible to find a continuous function f(z) = A(f, z, ro)
and rl
f in C such that
(z E C)
f is not hannonic anywhere. Surprisingly, it is possible to choose two radii
> 0 and r2 > 0 so that the mean value property with respect to them implies
hannonicity. More precisely, consider the set of positive numbers E given by E =
{~1: .jiiJo(~j) ~
1 = 0 (j = 1,2) and
!!. > o}. ~
(Here Jo is the Bessel function.) We observe that 1 E E, just take ~l = ~2; in general, the numbers ~j are complex and we only consider those whose quotients are real and positive. Delsarte has shown that this set is finite [De2]. Now let rl > 0, r2 > 0, be such that
0.. r2
and
¢ E
469
6.5. Deconvolution
Under these two conditions, if / E C(B(O, R)) satisfies the mean value properties /(z) = A(f, z, rl) for Izl
