THE REGGE PROBLEM FOR STRINGS,
UNCONDITIONALLY CONVERGENT EIGENFUNCTION EXPANSIONS, AND UNCONDITIONAL BASES
OF EXPONENTIALS IN L*(—T, T)
S V HRUSCEV
A string is the interval [0, +00) carrying a non-negative measure dm The x
function m(x) = \ dm evaluates the mass of the string supported by [0, x] The point x = 01s assumed to be a point of growth of m, i.e m(x) > 0 for x > 0 It is supposed also that the string is obtained from the classical homogeneous string (corresponding to Lebesgue measure dx) by a finite perturbation The latter means that dm = dx on (a, +00) for some a < -++oo In what follows
a,, & inf{a : dm = dx on (a, -+oo)}
Given a > 0 let L7({0, a], dm) denote the Hilbert space of all m-measurable functions f with
WIR, = \ ICO dm(x) < -beo
9
Every string determines the formal differential operator af -
dmdx fo
defined on-the class Dy of functions fon R = (—o0, +00) such that
fO)+/-O)x for x <0
ƒ#œ)=]} xg ì
Trang 268 S V HRUSCEV
2
with g satisfying g € £70, a],dm) for every a> 0 Clearly Se = g for mdx
such an / The symbols f+ (x) and f-(x) denote the right-hand and left-hand deri- vatives of f respectively
Fix a > @,, and let o(7) be the set of all complex numbers k such that the
equation
d3
= —k*y, y-(0)=0, y*@ + iky(a) =0
dmdx
(1)
has a non-zero solution ),(x, k) In fact the set o(m) does not depend on the parameter a when a 2 a,, and coincides with the zero-set of an entire function
It can be shown (and we will do it later) that o(m) is disposed in the open upper
half-plane C, The spectrum o(m) is always symmetric with respect to the imagi- nary axis because }',(x, kg) is a non-zero solution of (1) corresponding to k == —k, provided k, € øữn)
The spectra which occur in the eigenfunction problem (1) are described by Arov’s theorem [1]:
THEOREM I A closed countable subset o of C, symmetric with respect to the imaginary axis coincides with the spectrum of a problem (1) if and only if o is the zero-set for an entire function F of exponential type with
X + x?9)-1:|Ƒ(x)~3dx < +co, Ạo + x#)-1!log+| F(x)| dx < +-œ Given a 2 a,, the Regge problem [2], [3] is to determine whether the family
{7% A) eeacmy is complete in L?(0, a], dm) or not Let T(x) = (90565
i.e T(x) is the time required for a point perturbation of the end x == 0 to reach the point x
The following result solves the completeness problem which, of course, is of most interest for the critical value a = a,, + T(a,,) It is assumed that the spec- trum o(m) is simple, i.e the associated function F has only simple zeros
THEOREM 2 The family {y,(x, k)}xeocm is complete ¡in L^(O, a],dm) for
Qm <a < a, Tla,) and is not complete ifa > a, + Tan)
Trang 3alone does not permit us, of course, to expand any given function fe L7({0, a], dim) in an unconditionally convergent series
fxy= Yo uy y), % EC k€a(m)
Recall that a family of non-zero vectors {e,} in a Hilbert space H is called an unconditional basis in H if every element x e¢ H can uniquely be decomposed
in an unconditionally convergent series x = Ơ)4,-e,, %,ÂC The classical
n
G Kéthe—O Teoplitz theorem says that a complete family {e,} in a Hilbert space
forms an unconditional basis iff the following ‘approximate Parseval identity’’ holds no?
cy lon? = [en ll? Š I>; Oey ||? Š c1 lczlP -|JzlÌP
Ht H "„
for some c, 0 < c < 1, and for every finite sequence of complex numbers {a,}
The unconditional basis problem for {y,(x, k)}, eam is intimately connected with the same problem for exponentials {e!*}, ¢4¢m in L(0, a) In few words the
relationship between the problems looks as follows Given a string m and a 2 a,,
one can associate with (1) a semigroup {Z,},,9 of contractions in an auxiliary
Hilbert space K* so that the characteristic function of {Z,},,9 is S = 0-B, where 0: : &@-4w? and B is the Blaschke product in C, with the roots placed at the points of o(m) The eigenfunctions %, of {Z,},,), and &% of the conjugate semigroup
{Z*},>9 can easily be expressed in terms of {y,(x, k)},eacm (see (10) below) On the
other hand the semigroup {Z,},5» is unitarily equivalent to the so-called model semi-
group {I,},,9 whose eigenvectors are related to the exponentials via the usual
Fourier transform
This new approach to the problem based on the Lax-Phillips scattering theory for unitary groups and originating in earlier papers by B S Pavlov has been developed in [4] (see Part 1V) to investigate the basis problem for a special class of strings
In the present paper we exploit the connection indicated above in the direction inverse to one considered in [4] This yields the following result
THEOREM 3 Let A be a subset of C, invariant under z— —z such that inf{Im2 : Äe A} > 0 Suppose that {e*},¢4 is an unconditional basis in L*(0, 2d) for some d > 0 Then there exists a string m with o(m) = A such that fora =a, +d
the family {y(x, k)}xeocmy) forms an unconditional basis in L*((0, a], dm)
Notice that for strings obtained T(a,,) > 0 because d = T(a,,)
Trang 4wv
70 S V HRÙSCEV
The paper is organized as follows Section | contains preliminaries It deals
mainly with the construction of the corresponding functional model For reader’s convenience we present the proof of Theorem 1 in §2 This section also deals with the completeness problem, i.e with the proof of Theorem 2 The most interesting case here is the case a = a,, + T(a,,) with T(a,,) > 0 In §3 the proof of Theorem 3
is given (see also Theorem 3.1 below)
Theorem 2 is closely related to similar results obtained by M G Krein and A A Nudel’man in [6], [7], [8] The papers [6], [8], besides other things, deal with
the completeness problem of root elements of the dissipative operator associated
with a string which is constrained to satisfy slightly different boundary conditions The main technical tool used in [6], [8] to prove the corresponding completeness
theorem is the well-known criterion of M.S Livgic, while in the present paper the proof of Theorem 2 is based on the theory of entire functions
Acknowledgements 1 am grateful to the Institute INCREST in Bucharest, Romania for the support of this investigation I am indebted to L de Branges for valuable discussions concerning Hilbert spaces of entire functions during the fall of 1982 It is my pleasure to express a gratitude to M G Krein and A A Nudel’man who turned my attention to some inaccuracies of the preliminary manuscript
1 THE CONSTRUCTION OF THE FUNCTIONAL MODEL
1.1 THE OPERATOR G The string defines a self-adjoint operator in the Hilbert space M == L4([0, -+co), dm) which can be specified as the restriction of d?/dimdx to the domain
D(6) = {/e Dạ :/-(0) =0, [flim + IISflim < +00}
¬ d?A 5
Given k € C denote by A(x, k) the unique solution in Dy of Gndx —k°A
satisfying A-(0, k) = 0, A(O,k) = 1 It can be obtained as a solution of the following integral equation
x t
(2) A(x, kK) = 1 — “| | \ A(s, k) amG) dt
0 —_
A*(x, k) which implies that both k + A(x, k) and k > B(x, k) =— cm are entire functions, in fact of exponential type Let
Trang 5Clearly, (see (2)), E(«, 0) = 1, x e[0, +co) and the set of zeros of E*(a, k)®:
#° E(a, k) coincides with the spectrum o(m) of (1)
The functions 4, Ư, E, (the last is called a de Branges function), play an essential role for the spectral representation of G
Let 4 be a principal spectral function of S which is an increasing odd function on R completely determined by © [5] Consider the Hilbert space Z(A4) consisting of all functions on R with
fiz = + (umes < +00
R
and two orthogonal subspaces Zeyen(4) and Zoga(A) there, formed by even and odd functions respectively
The “even’’ transform
+00
(Aeren() = \ A(x, 1ƒ) dmQ)
0—
defines a unitary mapping of M onto Zeyen(4) Accordingly the ‘‘odd”’ transform
f )oda(y) = B(x, y)f(x) dx
0
is a unitary mapping of L2((0, +00), dx) onto Zogg(A)
1.2 DE BRANGES FUNCTIONS Any entire function E satisfying
|E(z)| > |E*()l, ze Cy
is called a de Branges function We assume that E satisfies the reality condition
E*(z) = E(—z), zeEcC
and that E(O) = 1 Let us notice that this class of functions appeared for the first time in 1938 in a paper by M G Krein (see the English translation [9], p 214—260)
A de Branges function of exponential type is called short if
\c -+- y?)~1|#(yJW~?dy < +œ
Trang 672 S V HRUSCEV Clearly, any short de Branges function is root free on C, UR and the trivial inequality log-x < x-? implies
\c + 3)=!log~¡E():dy < +eo
R
which together with the assumption of finite exponential type of E yields by the Carleman formula that
\a +)" *logti£Q)i dy < +00
R
The class of all entire functions of exponential type satisfying the last condition is called Cartwright’s class @ The basic facts concerning @ can be found
in [10], [LH]
Let A be the set of zeros of a short de Branges function £ Because of the inclusion Ee @ the function E admits the following factorization
(4) E(Œ) =: e~*Z.v.p TỊ ( — 7] ›
LEA 2
where c € R and v.p He lim Jf Let Biz) = EĨ(I — z/Ä)\( — z/2)-1 be the
AEA ĐR#=+oolA|<R 4
Blaschke product corresponding to 21 The function £# : £~1! being bounded in C, , It follows that c > 0:
t*Œ
(5) 6), =e?}z B(z)
E()
There exists a nice correspondence between the class of strings under consideration and the class of short de Branges functions The proof of the following result can be found in [5], Sections 6.3, 6.12
THEOREM 1.1 Given a string m and a> QO the function E(a, z) is a short de Branges function of exponential type
T= \ [m'(s)}/? ds
7?
The function A(y) = \IZ0)7*4y is the principal spectral function of the string
0
with mass function
n(x) = Í m(x) for x <a
Trang 7The converse is also true Given a short de Branges function E, E(0) = 1,
satisfying the reality condition there exist precisely one number a > Q and precisely
one mass function m with a,, < a such that E(y) = EG, y) LEMMA 1.2 Given a string m and a > a,, we have
Ea, z) — ead? v.p H ( _— 7) ^€4
Proof Straightforward computations show that E(a, z) = e7 €@~*“z”* E(a,,, z)
So we need to prove the equality only for a=a,, Clearly (4) holds for E(z)=E(a,,, 2): with c > 0 If c > 0 we can consider an auxiliary short de Branges function E*(z) == == el F(q,,,z) which by Theorem 1.1 generates the same string m® because |E| =
=:|E"|, and there exists 2 > 2„ such that E*(z) == E(a,z) =e \“~™*E(a,,, z)
which obviously contradicts the assumption c > 0 Z 1.3 THE WAVE EQUATION Let N denote the space of all functions on [0, 4-00) with
ifs = \ If" Pax < $00,
0
Being factored by the subspace of constant functions and endowed with the corres- ponding factor-norm, the space N becomes a Hilbert space
The Cauchy problem for the wave equation is defined by
Ox d3⁄
= > ~ 0,7 =a 0
or? dmdx Ont)
w(x, 0) == g(x), (x, 0) = 4, (x) and the space E =: N @ M supplied with the norm
+00 +00
(¿) =2 la4#dx +-<L Iza(9IÊ đm() Ley \E 2 2
0ˆ lI⁄lle —= “yg
Trang 874 S V HRUSCEV
THEOREM 1.3 The operator
e=i(' W p(⁄) = |[(°Ì:z¿£ Đ(G), « < Ma NỈ ay
ds self-adjoint in E
The proof of the theorem is essentially the proof of self-adjointness of S which can be found for example in [2], see also [1] for a partial case
The operator # being self-adjoint generates the strongly continuous unitary
group U,=expi¥t Given any data #0 =[Ìs E this group defines
tly
U(t) = Mà 2) = ,2(0) and za(x, ?) 1s the solution of the Cauchy problem
a(x, t
The spectral representation for U, can be obtained with the help of even and odd
transforms Define an operator ¥ :E > Z(4) by
FU) =F ("
ty Jo =-i '
⁄ạB(zx, y)dx + A(x, y)dm(x)
0—
‘Then
I2l$ = \!Z2*44 2m
R
and ,% = #-!el'zZ22, 4c E
1.4 THE SEMIGROUP OF CONTRACTIONS {Z,},„ạ Fix ø > đ„ and consider the
subspace K“ (or briefly K) of all elements of E such that |z¿(3)! + |s4(x)| = Ư for
x> a
Let #x denote the orthogonal projection onto K It is easy to check that
K=EO(@_ @@,), where Q_ = lC) tạ = 19, a(x) = 0 for x < 2} and
⁄
ae :
Gi= |( | : —y =#g,q(X) = Ư or x < 2} are the spaces of incoming and
tly
outgoing waves correspondingly The family
Z, 2 PU |K, 1 >0
: : : ⁄ vo
is a strongly continuous semigroup of contractions Notice that Px ) =( ) ay Vy
Trang 9THEOREM 1.4 The generator T of the semigroup z, = eT is a maximal com-
pletely dissipative operator in K with
D(T) = | 2v := () e D(Z), z‡(2) + z¡() = of ,
T(Px%) = PerL4U, for PxU € D(T) The adjoint operator T* is defined on
D(T*) = {ox :4/= Đ e D(Z), wé(@) — (a) = of , ty
T*(Pe®) = PLU for PeU € D(T*)
The proof can be found in [4] (see Theorem 2.1 of Part IV) Denote by UM, and W* the eigenvectors of T and T* : TW, = k&,, T*U® = kU An easy
computation (using Theorem 1.4) shows that
29) ae<( 09)
Iky„(x, &) 1ky„(x, k)
=~ for 0 < x < a, where & ranges over the spectrum o(m)
(6) %6 =|
1.5 THE MODEL SEMIGROUP {SIl,},.9 ¥ transforms K* onto the class J4) of entire functions fe Z(A) of type <T The relationship between T and a is the
following: a is the biggest root of the equation
7 =\ m'conds
0
which clearly has only one root aif a > a,,
The entire function (a, z), being a short de Branges function, generates a de Branges space of entire functions This space B(E) consists precisely of entire functions f satisfying
ini? = | (V@IEGJPdy < +ec
R
(7) |/Œ)I < C;-(mz)-1⁄#|E(z)| if Imz > 0
(8) If(2| < C,-(Imz)-"|E*(z)| if Imz <0
Trang 10vv
76 S V HRUSCEV
Let now @;,-@,f=f/E denote the map which maps Z(A) isometrically onto L°(R, dx) It follows from (7) that.@,f¢ H? for any f € B(E) and (8) implies
Mf € SH? on R for such an f with S = £*/E inner Therefore @,B(E) < K,, where K, is a “model space’ defined as K, = H? © SH? In fact @,B(E): = K,
because every function from K, has a pszudo-analytical (in our case usual analytical) continuation in the lower half-plane C_ to a meromorphic function with simple poles at zeros of E Multiplied by £ this function clearly turns into an entire function belonging to B(E)
Let J .&@,oF:E—-> LR) and define the norm in L? by file
= = {fPdx Then obviously
“TL R
1#|=|Z#ll:, U„# ~ Z-!e'zZ%
with Ø K2 = K,and Š = E*/E Summarizing we get
THEOREM I.5 The mapping J defines a unitary equivalence of the semigroup
(Z,),;>9 to the model semigroup
M f= Perf, feKk,,t>0,
where P, denotes the orthogonal projection of H¥, onto K,
S(z)-/2Imk
1.6 EIGENVECTORS It is well-known that — 1s the family
~~ L€ơ(m})
J2imk - is the family of
z—k Jxrean)
of eigenvectors of the generator A of {I,\,.9 and |
eigenvectors of A* Theorem 1.5 implies that
S(y)-V2Imk > Z2 = — ———-› ke alin)
9 TU), =
(9) pe ay
with appropriate coefficients c,, y.(x, kK) = ¢,- A(x, k), where k € o(m) U om), in (6) The following formula borrowed from [5], p 234
\ A(x, K)A(x, y) din(x) -+ [Bes k)B(x, y) dx = ES (EQ) — EE*)
—21@ — k)
0~
yields = |ĐImk-(kE(â)~!, e =§Imk(K-E(K))~! for k € o(m) Therefore
Trang 11as k € o(m), and we get
—k
„@) =( %AG &) } 4/*_ = tớ 0Ì ik-c, A(x, k) Nike, A(x, k)
Finally
1 0 xa Vax, &)\
d0) 4 2 { fm, $k = k tr} ky (x, k) =1, — UF =( a 2 { k tr} Ộ 0
for k e ơ(m)
LEMMA 1.6 L) Let k # —k e o(m) Then %, 1 %¿ and theref0re
l2, + *„I = 2
2) Let k = ib € ơ(m) with b > 0 Then
Ne te UE eI? = LF WiS'Gd)}
Proof lf k # —k € o(m) then
(Uy U8 de (TU, TURD) yy = — | Tem 2k ay 0
, an \}(y—k) ytk
R
because g(z) = S(z)-(z — k)-1 € H% and g(—k) =: 0 Let now k = ib € o(m) with b > 0 Then
Gin + WHIP = 2(1 + Re(2„, 27)}
But
(Win, US) = _°U) _ dy == —2biS’(id) 2ni \(y — ib)?
R
The spectrum o(m) being invariant under z + —Z, we have S(z) =: S(—z) which
implies —2biS’(ib) € R Z
Set 6,, = O if S does not have roots on the imaginary axis and let 6,, ==
= sup{2b|S’(ib)| : ib € o(m), b > 0} Clearly 0 < 6,, < 1 In case S = ei” B(z),
where d > 0 and B is a Blaschke product we always have 6,,< 1 Indeed,
5, =: sup{2be~”- |B’(ib)| : 1b 6 ơ(m), b > 0} < exp{—dinf 5} < 1
ibeo
IONAL BASES OF EXPONENTIALS A subset A = {2„} of C, is
called a Carleson-Newman set if
Trang 1278 S V HRUỀCEV
and the last holds iff {4,} is an interpolating sequence for the Hardy algebra H™ (see for details [12], [13] or {14))
Given a > 0 consider the class %, of all entire functions of exponential type with indicator diagram [0, —ia] and satisfying the Muckenhoupt condition (A,) on the line:
(Ag) sop (\ rnds)(ÀIIrse‡ < +œo
€7 \ỊI| i Z| ĩ
Here J stands for the family of all intervals It is well-known that any function F satisfying (A) satisfies also
Ạo + 19-!:1FQ)J*dy < +©o, Ạ + 3)=1-|F)|~*dy < -+œ
which implies „Z„ c #
Let ø be a subset of C; S 'fz C„ :Imz > ơ} for some ổ > 0, let B be the
Blaschke product corresponding to o with understanding that B, = 0 if o isa
uniqueness set for H™, and let S=0-B=e'”-B with d>0 Then Ky :=
== clos(Kg + Kg) It is easy to see that Ky; © K, = 0-K, THEOREM 1.7 The following are equivalent:
1) {e'4*},., is an unconditional basis in L*(0, đ);
2) ao € (CN), the angle between Kx and 0K, Is positive and Ks = Kg + OKg;
3) o@ € (CN) and there exists a function in 4, with simple zeros whose zero-set is 6
See the proof in [4] (Theorem 2, Corollary on p 231, Theorem 7, Theorem 9, Theorem 1.2 of Part III)
2, THE COMPLETENESS PROBLEM
2.1 PROOF OF THEOREM 1 The first implication is clear because o(m) is the set of roots of F(a, z) satisfying |E*(y)| = |E(y)| on R, and E is a short de Branges function
The converse is a direct corollary of Theorem 1.1 provided it is known that F is a short de Branges function The function F being of exponential type and satisfying
Trang 131tfollows Ƒ'e #, We can assume therefore that
F@) =v.p Il ('=‡}
AEo(m A
Denote by B the Blaschke product corresponding to o(m) Clearly F/F* = B
in C, This implies that F* is a de Branges function which in fact is short because
of (11) ZB
2.2 PROOF OF THEOREM 2 The following lemma is borrowed from [4] (see Part IV, Theorem 2.3 there) The proof is given for reader’s convenience
LEMMA 2.1 The family {y.(x, k)}, etm) is complete in L*([0, a), dm), a > ay if and only if the joint family (%,, UE}, coin) Of eigenvectors of T and T* is
complete in K?,
Proof Identify K* with N* + M*, where
N?= {fe N :f'(x) =0, x> 2}, M2 = 720, a], dm)
Suppose span{2,, 2# : k e ơứn)} = K“ and let g e M“* with gø 1 y„(x, k), k € ơ(m)
in M“ The function Œ = () being orthogonal to N”, we get from (10) that §
GL%&,, GLU for k € o(m) and therefore g = 0
Suppose now that span{y,(x, k) : k € o(m)} = MY’, and pick any function g € N’, g(a) = 0 with g 1 y,(x, 4) for k € o(m) in N* Then
a a
0=: he EtG) dx = [psc k)de(x) = —K? \ valor, Kets) d(x), o~
which implies g = 0 It follows that span{y,(x, k) : k € o(m)} = N* and finally
span {%, , Uf :k € ao(m)} = K* (see (10)) ZB
Since semigroups {Z,},,9 and {U,},,, are unitarily equivalent by Theorem 1.5,
the question of completeness of {%, , %*} in K? reduces to that for the model semi- group {M,},,5, ie to the joint completeness problem in K, = Hi â SHƠ of
I
(12) L ca t n
z—k k€z(m} Z—K Jyeotm)
S(z) = e'"%m B(z) = OB being the characteristic function of {9,},59
In case a = a,, we have S = B and it is well-known that eaeh of the above
families is complete in Kg (see, for example, [12])
Trang 1480 S V HRUSCEV Lemma 2.2 The families (12) span Ks,S= eB = 6B if and only if
fe} ca(m) is complete in L(0, 2d)
Proof We have S(z) 8G) i tke ain)| z—k
Ks © K, 8K, < 0:span| ‘ke atl = span]
Z—
"Therefore it is clear that the families (12) span K; iff the family of orthogonal
projections Pf ! z) k € o(m) spans Ky The Fourier transform
2 zc —
fx) = \ na
R
maps ? isometrically onto £°(0, +00) The space K, is mapped onto the subspace £20, 2d), while P, turns into the multiplication by the indicator X¢o.2d) of (0, 2d),
Finally
I Po-
Z—
=> —2rie "Xo2a)s ^ ._—R k € o(m)
The last lemma reduces the joint completeness problem to a special case of «completeness problem of exponentials in L2(0, 2d)
LEMMA 2.3 Let F be an entire function of exponential type with the width of indicator diagram equal to 2d > 0 Suppose that all roots of F are simple and that
\ cơ —.— 1 dy < +00
[+ y iF@)?
R
Then the family {e*} q 9 is complete in L*(0, 2d)
Proof The function F belongs to the Cartwright class @ and we can assume ‘without | oss of generality that the set o of all roots of F is contained in C, Other- ‘wise we can multiply F by a corresponding Blaschke product to transform all roots in C, Multiplying F by an appropriate exponential we may assume that
F(z) = el -v.p TT (: -=} bì
À€ơ
Trang 15Suppose that the family {e*},¢, is not complete in L7(0, 2d) Then there exists
a non-zero function g € L2(0, 2d) such that
G(z) = \ e*g(x) dx
Cer
BP
vanishes on o Multiplying G by an appropriate Blaschke product we can reflect all the zeros of G in C_ to C, keeping |G|? invariant on R So we can assume that _ G does not have zeros in C_ Denote
a = inf{?:/ supp(g)}, B == sup{t : 1 € supp(g)} Since G € H? , we have
C,e7B,(z)G(z) for z, Im
G(z) =
C_ e“G,(z) for z, Im
where C, € C, |C,| == |C_| = 1, Bg stands for the Blaschke product and
ef 1! Jj |
foe) Hemp | VE EE al Ti t—2 1+ 7
denotes the outer factor of f Similarly,
F@) = CLBAz)F(z) for z,Imz 20
, C!e?4? F(z) for z, Imz < 0
, " i(2d— § G(z)
Consider now an auxiliary entire function /(z) = e' oy which can be F(z factored as follows Ce ei0d- 01-4), Be Ge > Imz>0 Cy B, F, (13) h(2z) = C- Ge Imz <0 CL OF,
Notice that /(z) = expi(2đ — (8 — «))z-B,/B, is an inner function in C, h(z)
Z—
|H(@)i dx = _— JŒŒl ig, < (\ Sal  ote) ax)"
R R |x — ki -|FQ)| Đ |x — kj*\f@)l? 2
Case I I(k) = 0 for some ke C, Let H(z) = i Then
which implies that He Hin Ht = 0 (see (13), [14])
Trang 16v
82 S V HRUSCEV
Case 2 Bạẹ = Bp but Œ(xạ) = 0 for some real xạ Put again H(z): ~
A = — '&) Then Z— Xo d 1/2 1,8 LHWf(x)j dx <( \ a iG(x)? dx _ +00 1X — Xoi?| F(x)? |x-xgi>1 [x= Xp| PL R
and therefore H © Hin Ht 0
In both cases we get G = 0 which contradicts to the assumption g # 0 We have to consider the last possibility
Case 3 Bg = B, and G is root free on the real line In this case every root
of G is simple and G and F have the same set of roots Therefore G = e"1"- F-c,
for some c, € R, c; € C, c, # 0 We have
'F(y)| dy 12.1 Lẻ
:©oO =: Tran dy < (\ ao) _— { S0)*4;]
q +?) r0), Œ- ?),Ƒ0)* ico, \
R R _ R
and ở £ L*{R) which clearly is a contradiction Đỳ
I (=J(2)) is said to be the completeness interval of 2={4,} if {e'5“} is com-
plete in L? on every interval of length less than J and on no larger interval See the proof of the following result in [15] (Theorem 28)
THEOREM 2.4 Let {1,} be the zeros of an entire function F of finite type with the width of indicator diagram 2d > 0 and such that
lost FOO! g
I+ x? x < +00 R
Then (0, 2đ) =- 12)
1 follows from Theorem 1.1 and from the well-known fact that the indicator diagram of the canonical product
v.p TT (: — =|
AEA *
is [—Ti, Ti] (see [10] or [1I]), that (0, 27) with d= | [m'(s)}/2ds is the completeness
0
interval for {k}.eotn)- Moreover {e"*},¢o(m 18 complete in 1340, 2đ) in view of
Lemma 2.3 The proof is finished by application of Lemma 2.1 and Lemma 2.2 HH
Trang 173 UNCONDITIONAL BASES
3.1 The main ingredient of the proof of Theorem 3 is the following THEOREM 3.1 The following are equivalent:
1) A is a subset of C, (for some 6 > 0) invariant under the transform z+ —Z and such that {e'**},¢4 forms an unconditional basis in L(0, 2d) for some d > 0
2) There exists a string m with a = a,,-+ d and o(m) = A such that the joint
family {%,, UE :k € o(m)} of the eigenvectors of T and T* forms an unconditional
basis in K*
First we show how the proof of Theorem 3 may be finished now
The part 1) = 2) clearly implies the existence of a string m with ơữn) = A Lemma 1.6 and (10) would imply the desired conclusion if we could prove that
Sm = sup{26|S’(ib)| : ib € o(m)} < 1 But S = e”-B(z) and d> 0 It has been
noticed in 1.6 that this implies 6,, < 1 G
3.2 PROOF OF THEOREM 3.1 1) => 2) Let A satisfy the hypotheses of 1) By Theorem 1.7 there exists an entire function F in %,,, F(O) = 1 with simple
zeros whose zero-set is precisely A Since F satisfies the Muckenhoupt condition (A,) on R it follows that
\ 1 dy
et —_—— <
I-+-y? (FO)?
R
which implies by Theorem | that A is the spectrum of (1) for some string m and a > đ„ To calculate a consider the indicator diagram of F which is [0, —i2d], and
the canonical product
Fans 2) = v0 (I~)
AEA
whose indicator diagram is equal to [—i7(a,,), iT(a,,)] The zero-sets of F* and E being identical, the lengths of the indicator diagrams 2d and 27(a,,) of F and B must coincide Therefore 7(a,,) = d > 0,
E*(q, 2) = ew" E*(a,, 2)= F(@), and the characteristic function of {Z,},50 is
E*(4,z) —
SO = aD 2B — 0.B,
Trang 18v
84 S V HRUSCEV
By Theorem 1.7 K, == Kg + 0K, and the angle between K, and @-K, is positive The families (12) form unconditional bases in their closed linear spans 0K, and Kg, because Ae (CN) [12] This implies (see (9)) that {2„, 2È :k 4} forms an unconditional basis in K*
2) = 1) Let m be a string satisfying the hypotheses of 1) By Theorem 1.5 the semigroup {Z,},, is unitarily equivalent to the model semigroup {S,} with the characteristic function S = e'#-B Clearly the angle between the spaces
0- Kp = span{2Z 2, : & € o(m)}, Ky = span{ 7 UF :k € o(m)}
(see (9)) must be positive which implies, in particular, that d > 0 On the other hand K, +0-K,z = Ks which implies a =a,,+ T(a,,) and therefore d= 7(a,,) (see Theorem 2) The spectrum o(m) lies in C;, 6>0 Indeed, the orthogonal projection
50K
of | ae kea(m) onto K, has the norm (I — |Ø(k)j?}*2, The space 0-Kp
Zz —
is the orthogonal complement of K, in K,; Therefore by Pythagora theorem the
k
projection of pmE onto Ø- K; has the norm ,60(&)| It follows
Zz —
\O(k)| < cose < |
where « denotes the angle between K, and 0-Kg It remains to apply Theorem 1.7 3.3 Examples of unconditional bases of exponentials {eo} ez in L7(0, a) with lim Im/, = -co given in [1] lead to interesting examples of spectra o(#1)
|2]+ +00
In particular, it can be proved (see Theorem 3.4 (V I Vasjunin), part FE of [1)) that any sequence of points a, in C, satisfying {a,} € (CN) and limIma, = 0o can be complemented up to such a family {/,},ez of points in C,; (for some 6 > 0) for a given a > 0 that {en ez forms an unconditional basis in L°(0, a)
In terms of problem (1) this means that such a family {a,} can be complement- ed to be the spectrum of (1) The obtained string m may be considered as a “small” perturbation of the homogeneous string in the sense that
0 < inf |E(a,,, x)| < sup|£(a,,, x)| < +00
xER xER
(see [4]), which implies that {y,(x,k)},¢o(m) forms an unconditional basis in L°({0, a], dm), a = a,, + T(a,,), but at the same time
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