Duality theorems for coinvariant subspaces of $H^1$

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by  R. V. Bessonov

DUALITY THEOREMS FOR COINVARIANT SUBSPACES OF H1
R.V.BESSONOV
4
1
0 Abstract. Letθ beaninnerfunctionsatisfyingtheconnected levelsetcon-
2 dition ofB.Cohn, andletKθ1 bethe shift-coinvariant subspace of theHardy
n space H1 generated by θ. We describe the dual space to Kθ1 in terms of a
boundedmeanoscillationwithrespecttotheClarkmeasureσα ofθ. Namely,
Ja we prove that (Kθ1 ∩zH1)∗ = BMO(σα). The result implies a two-sided
estimate for the operator norm of a finite Hankel matrix of size n×n via

2 BMO(µ2n)-norm of its standard symbol, where µ2n is the Haar measure on
thegroup{ξ∈C:ξ2n=1}.

]
V
C
.
h
t 1. Introduction
a
m A bounded analytic function θ in the open unit disk D = z C : z < 1 is
[ called inner if θ(z) =1 for almost all points z on the unit c{ircl∈e T in| t|he se}nse
of angular bou|ndary| values. With every inner function θ we associate the shift-
1 coinvariant[19] subspace Kp of the Hardy space Hp,
v θ
2
Kp =Hp z¯θHp, 16p6 . (1)
5 θ ∩ ∞
4
As usual, functions in Hp are identified with their angular boundary values on the
0
1. unit circle T; formula (1) means that f ∈ Kθp if f ∈ Hp and there is g ∈ Hp such
that f(z) = z¯θ(z)g(z) for almost all points z T. An inner function θ is said to
0
4 be one-component if its sublevel set Ωδ = z ∈D : θ(z) < δ is connected for a
{ ∈ | | }
1 positivenumberδ <1. ThisclassofinnerfunctionswasintroducedbyB.Cohn[12]
: in 1982. It is very useful in studying Carleson-type embeddings Kp ֒ Lp(µ) and
iv Riesz bases of reproducing kernels in Kp, see [3–5,7,12,13,17,26]θfo→r results and
X θ
further references.
r
a In this paper we describe the dual space to the space K1 generated by a one-
θ
component inner function θ. Our main result is the following formula:
(K1 zH1) =BMO(σ ), (2)
θ ∩ ∗ α
where σ denotes the Clark measure of the inner function θ. Below we state this
α
result formally and apply it to the boundedness problem for truncated Hankel
operators.
2010 Mathematics Subject Classification. Primary30J05.
Key words and phrases. Inner function, Clark measure, discrete Hilbert transform, bounded
meanoscillation,atomicHardyspace,truncated Hankeloperators.
ThisworkispartiallysupportedbyRFBRgrants12-01-31492,14-01-00748,byISFgrant94/11,
by JSC “Gazprom Neft”, and by the Chebyshev Laboratory (Department of Mathematics and
Mechanics,St. Petersburg StateUniversity)underRFGovernment grant11.G34.31.0026.
1
DUALITY THEOREMS 2
1.1. Clark measures of one-component inner functions. Let θ be a non-
constant inner function in the open unit disk D. For each complex number α of
unit modulus the function Re α+θ is positive and harmonic in D. Hence there
α θ
exists the unique positive Borel−measure σ supported on the unit circle T such
α
(cid:0) (cid:1)
that
α+θ(z) 1 z 2
Re = −| | dσ (ξ), z D. (3)
α θ(z) T 1 ξ¯z 2 α ∈
− Z | − |
The measures σ are usually referred to as Clark measures of the inner
α α=1
{ }| |
function θ due to seminal work [11] of D. N. Clark where their close connection to
rank-oneperturbationsofsingularunitaryoperatorswasdiscovered. Foramodern
exposition of this topic and subsequent results see survey [21].
Each Clark measure σ of an inner function θ is singular with respect to the
α
Lebesgue measure on the unit circle T. Conversely if µ is a finite positive Borel
singular measure supported on T and α = 1, then there exists the unique inner
| |
functionθ satisfying(3)withσ =µ. Thus,thereisone-to-onecorrespondencebe-
α
tweeninnerfunctionsintheunitdiskD andsingularmeasuresontheunitcircleT.
It was unknown which singular measures on T correspond to the Clark measures
of one-component inner functions. We fill this gap in Theorem 1 below.
ForeveryBorelmeasureµonthe unitcircleTdenote bya(µ)the setofisolated
atoms of µ. Then the set ρ(µ) = suppµ a(µ) consists of accumulating points in
\
the support suppµ of µ. We will say that an atom ξ a(µ) has two neighbours
ina(µ)ifthereisanopenarc(ξ ,ξ )oftheunitcircleT∈withendpointsξ a(µ)
+
− ± ∈
suchthatξ istheonlypointin(ξ ,ξ ) suppµ. BymwewilldenotetheLebesgue
+
measure on T normalized so that−m(T)∩=1.
Theorem 1. Let α =1. The following conditions are necessary and sufficient for
| |
a Borel measure µ to be the Clark measure σ of a one-component inner function:
α
(a) µ is a discrete measure on T with isolated atoms, m(suppµ) = 0, every
atom ξ a(µ) has two neighbours ξ in a(µ), and every connected compo-
nent of∈T ρ(µ) contains atoms of µ±;
\
(b) A ξ ξ 6µ ξ 6B ξ ξ forallξ a(µ)andsomeA >0,B < ;
µ µ µ µ
(c) the|d−iscr±e|te Hi{lbe}rt tran|sfo−rm±|(H 1)(z)∈= dµ(ξ) is bounded on a(∞µ):
µ T z 1 ξ¯z
we have (H 1)(z) 6C for all z a(µ). \{ } −
| µ | µ ∈ R
The necessity of conditions (a) and (b) in Theorem 1 is well-known. I would
like to thank A. D. Baranov who tell me the fact that condition (c) is necessary
as well. The proof of sufficiency part in Theorem 1 relies on a characterization
of one-component inner functions in terms of their derivatives which is due to
A. B. Aleksandrov [3].
1.2. Themainresult. HavingadescriptionoftheClarkmeasuresofone-component
inner functions, we now turn back to formula (2). For a measure µ with proper-
ties (a) (c) define the space BMO(µ) by
−
1
BMO(µ)= b L1(µ): b µ∗ =sup b b ∆,µ dµ< ,
∈ k k µ(∆) | −h i | ∞
(cid:26) ∆ Z∆ (cid:27)
where∆runsoverallarcsofTwithnon-zeromassµ(∆) and b = 1 bdµ
h i∆,µ µ(∆) ∆
is the standard integral mean of b on ∆. The following theorem is the main result
R
of the paper.
DUALITY THEOREMS 3
Theorem 2. Let θ be a one-component inner function and let σ be its Clark
α
measure. We have (K1 zH1) =BMO(σ ). That is, for every continuous linear
θ ∩ ∗ α
functional Φ on K1 zH1 there exists a function b BMO(σ ) such that Φ=Φ ,
θ ∩ ∈ α b
where
Φ :F Fbdσ , F K1 zH . (4)
b 7→ T α ∈ θ ∩ ∞
Z
Conversely,foreveryfunctionb BMO(σ )thefunctionalΦ isthedenselydefined
α b
∈
continuous linear functional on Kθ1∩zH1 with norm comparable to kbkσα∗.
Everymeasureµwithproperties(a),(b)fromTheorem1generatesthedoubling
metric space suppµ, , µ in the sense of R. Coifman and G. Weiss [14]. For
|·|
such measures µ we have H1(µ) = BMO(µ), where H1(µ) is the corresponding
(cid:0) a(cid:1)t ∗ at
atomic Hardy space,
H1(µ)= λ a : a are µ-atoms, λ < . (5)
at k k k k k| k| ∞
nX X o
Byaµ-atomwemeanacomplex-valuedfunctiona L (µ)supportedonanarc∆
∞
of T, with kakL∞(µ) 61/µ(∆), and such that hai∆,∈µ =0. The norm of f ∈Ha1t(µ)
is the infinum of λ over all possible representations f = λ a of f as
k| k| k k k
a sum of µ-atoms. We see from Theorem 1 that Theorem 2 admits the following
P P
equivalent reformulation.
Theorem 2. Let µ be a measure with properties (a) (c). Then f H1(µ) if
and only if f′admits the analytic continuation to the uni−t open disk D as∈a fuatnction
F K1 zH1, where θ is the inner function with the Clark measure σ = µ.
∈ θ ∩ α
Moreover, such a function F is unique and the norms kfkHa1t(µ), kFkL1(T) are
comparable.
Forthecountingmeasureµ=δZ onthesetofintegersZTheorem2′followsfrom
theresultsbyC.Eoff[15],S.BozaandM.Carro[8]. Theyprovedthatf H1(Z)if
andonlyiff admitstheanalyticcontinuationtothecomplexplaneCas∈afuantction
from the Paley-Wiener space PW1 . It seems difficult to adapt the technique
[0,2π]
of [8] (where convolution operators were used to relate H1(Z) and ReH1(R)) for
at
thegeneralmeasuresµwithproperties(a) (c). Insteadwegiveacomplex-analytic
−
proof based on the Cauchy-type formula
F(z)/z
F(ξ)dσ (ξ)= dz, (6)
α
1 α¯θ(z)
Z∆ IΓ −
where ∆ is an arc of T, Γ is a simple closed contour in C which intersects T at the
endpointsof∆, andF K1 zH1. Oncewehaveagoodestimateforthe function
∈ θ∩
F(z)/z on Γ, formula (6) gives us an upper bound for the mean F on the
1 α¯θ(z) h i∆,σα
ar−c∆. ThenwecanuseastandardCalder´on-Zigmunddecompositiontoobtainthe
representationofF as a sumofatoms with respectto the measureσ . The idea of
α
usingacontourintegrationistakenfromtheclassicalproofofatomicdecomposition
of Re(zH1), where the contour Γ comes from the Lusin-Privalov construction. In
our situation we have to modify this construction so that the contour Γ does not
approachthe subsets of the unit disk D where the function α θ is small.
| − |
DUALITY THEOREMS 4
1.3. Truncated Hankel operators. One of important applications of the classi-
cal Fefferman duality theorem is the boundedness criterium for Hankel operators
on the Hardy space H2. Theorem 1 yields a similar criterium for truncations of
Hankel operators to coinvariantsubspaces of H2.
Let θ be an inner function and let K2 be the corresponding coinvariant sub-
θ
space(1)ofthe Hardy spaceH2. Denote by Pθ¯the orthogonalprojectionin L2(T)
to the subspace zK2 = f L2(T) : f = zg, g K2 . The truncated Hankel
θ { ∈ ∈ θ}
operator with symbol ϕ L2(T) is the densely defined operator Γ :K2 zK2,
∈ ϕ θ → θ
Γϕ :f 7→Pθ¯(ϕf), f ∈Kθ∞. (7)
ThesymbolϕofΓ isnotunique. However,itiseasytocheckthateverytruncated
ϕ
Hankel operator on K2 has the unique “standard” symbol ϕ K2 zH2, which
θ ∈ θ2 ∩
plays the same role as the antianalytic symbol of a Hankel operator on H2.
Two special cases of truncated Hankel operators are of traditional interest in
the operator theory. If θ = zn, then the operators defined by (7) are classical
Hankel matrices of size n n. Indeed, in this situation the space K2 consists of
× θ
analytic polynomials of degree at most n 1 and the entries of the matrix of Γ
ϕ
−
in the standardbases of K2 and zK2 depend only on the difference k l: we have
θ θ −
(Γ zk,zl+1)=ϕˆ( k l 1) for 06k,l 6n 1. Similarly, for the inner function
ϕ
θ : z eiaz in th−e u−ppe−r half-plane C = z− C : Imz > 0 the corresponding
a +
coinva7→riant subspace K2 of the Hardy spac{e H∈2(C ) can be}identified with the
θa +
Paley-Wiener space PW2 ; truncated Hankel operators on PW2 are unitarily
[0,a] [0,a]
equivalenttotheWiener-Hopfconvolutionoperatorsontheinterval[0,a],see[9,23].
The question for which symbols ϕ L2(T) the truncated Hankel operator Γ is
ϕ
∈
boundedonK2(andhowtoestimateitsoperatornormintermsofϕ)admitsseveral
θ
equivalentreformulations. Ithasbeenstudiedin[5,6,9,18,23,24],seethediscussion
in Section 4. Most of known results are Nehary-type theorems: under certain
restrictionstheyaffirmthe existenceofaboundedsymbolforaboundedtruncated
Hankel/Toeplitz operator with control of the norms. Until now, the only BMO-
typecriteriumfortruncatedHankeloperatorswasknown. In2011,M.Carlsson[9]
provedthataHankeloperatorΓ onPW2 withstandardsymbolϕisboundedif
ϕ [0,π]
andonly if the sequence ϕ(n) n Z lies in the space BMO(Z). Recall thatwe have
PW2 = 2 for the sp{ecial}on∈e-component inner function θ : z eiπz in the
uppe[0r,πh]alf-pKlθaπne C+. The counting measure δZ on Z can be regπarded7→as the Clark
measureν forthe innerfunction θ2 (foreveryinnerfunctionθ the Clarkmeasures
1 π
of θ2 will be denoted by ν ; from (3) we see that ν = (σ +σ )/2, α = 1).
α α α α¯
− | |
Therefore the folowing result is a generalization of the criterium by M. Carlsson.
Theorem 3. Let θ be a one-component inner function, and let ν be the Clark
α
measure of the inner function θ2. The truncated Hankel operator Γ : K2 zK2
ϕ θ → θ
with standard symbol ϕ is bounded if and only if ϕ BMO(ν ). Moreover, we have
α
∈
c1 ϕ ν∗ 6 Γϕ 6c2 ϕ ν∗, (8)
k k α k k k k α
for some constants c , c depending only on the inner function θ.
1 2
Similarly,onecandescribecompacttruncatedHankeloperatorsintermsoftheir
standard symbols: we have Γ S if and only if ϕ VMO(ν ), see Section 4.
ϕ α
∈ ∞ ∈
DUALITY THEOREMS 5
Theorem3fortheinnerfunctionθ =zn yieldsthefollowinginterestingcorollary
for finite Hankel matrices.
Corollary 1. Let Γ=(γ ) be a Hankel matrix of size n n; consider
j+k 06k,j6n 1
its standard symbol ϕ=γ z¯+γ z¯2+−...γ z¯2n 1. We have ×
0 1 2n 2 −
−
c1 ϕ µ∗ 6 Γ 6c2 ϕ µ∗ , (9)
k k 2n k k k k 2n
where the constants c ,c do not depend on n and µ = 1 δ is the Haar
1 2 2n 2n 2√n1
measure on the group ξ C:ξ2n =1 .
{ ∈ } P
Corollary1impliestheboundednesscriteriumforthestandardHankeloperators
onH2. Recallthatthe HankeloperatorH :H2 zH2 with symbolϕ L2(T) is
ϕ
→ ∈
densely defined by
H :f P (ϕf), f H ,
ϕ ∞
7→ − ∈
where P denotes the orthogonal projection in L2(T) to zH2. It follows from the
−
classicalFeffermandualitytheoremthatH isboundedifandonlyifitsantianalytic
ϕ
symbolP ϕlies inBMO(T). Moreover,the operatornormof H is comparableto
ϕ
P ϕ , −the norm of P ϕ in BMO(T). Taking the limit in (9) as n one can
k − k∗ − →∞
prove the estimate c ϕ 6 H 6 c ϕ for every antianalytic polynomial ϕ.
1 ϕ 2
k k∗ k k k k∗
Thisisalreadysufficienttoobtainthegeneralversionoftheboundednesscriterium
for Hankel operators on H2, see details in Section 4.
2. Proof of Theorem 1
2.1. Preliminaries. Given aninner function θ, denote by ρ(θ) its boundary spec-
trum, that is, the set of points ζ T such that liminfz ζ,z D θ(z) = 0. In this
paper we always assume that ρ(θ)∈=T, because this is so→for∈on|e-com|ponent inner
6
functions and for functions satisfying condition (a) in Theorem 1 (see Lemma 2.1
below). As is well-known,the function θ admits the analyticcontinuationfromthe
open unit disk D to the open domain D G , where G = T ρ(θ) z : z >
θ θ
∪ \ ∪{ | |
1, θ(1/z¯)=0 . The analytic continuation is given by
6 } (cid:0) (cid:1)
1
θ(z)= , z G . (10)
θ
θ(1/z¯) ∈
Moreover,D G is the maximal domain to which θ can be extended analytically.
θ
∪
We need the following known lemma.
Lemma 2.1. Let θ be an inner function with the Clark measure σ , α =1. Then
α
ρ(θ) = ρ(σ ). A point z T ρ(θ) belongs to suppσ if and only| i|f θ(z) = α.
α α
∈ \
Moreover, in the latter case we have z a(σ ) and σ z = θ (z) 1.
α α ′ −
∈ { } | |
Proof. As is easy to see from formula (3), we have
α+θ(z) 1+ξ¯z α+θ(0)
= dσ (ξ)+iIm , z D G . (11)
α θ(z) T 1 ξ¯z α α θ(0) ∈ ∪ θ
− Z − −
Since θ is analytic on D G , a point z T ρ(θ) belongs to suppσ if and only
θ α
∪ ∈ \
if θ(z) = α, and in the latter case there is no other points of suppσ in a small
α
neighbourhood of z. Hence z a(σ ) and we see from (11) that
α
∈
σ z =(α¯zθ (z)) 1 = θ (z) 1.
α ′ − ′ −
{ } | |
It follows that T ρ(θ) T ρ(σ ). For every z T ρ(σ ) either z is an isolated
α α
\ ⊂ \ ∈ \
atom of σ or z / suppσ . In both cases formula (11) shows that the function
α α
∈
DUALITY THEOREMS 6
θ admits the analytic continuation from D to a small neighbourhood of z. Hence
z T ρ(θ) and we have ρ(θ)=ρ(σ ). (cid:3)
α
∈ \
The following result is in [3], see Theorem 1.11 and Remark 2 after its proof.
Theorem (A. B.Aleksandrov). An inner function θ is one-component if and only
if it satisfies the following conditions:
(A1) m(ρ(θ)) = 0 and θ is unbounded on every open arc ∆ T ρ(θ) such
′
| | ⊂ \
that ∆ ρ(θ)= ;
(A2) θ satisfi∩es the6es∅timate θ (ξ) 6C θ (ξ)2 for all ξ T ρ(θ).
′′ ′
| | | | ∈ \
2.2. Proof of Theorem 1. Essentially, we will show that conditions (a) (c) in
−
Theorem 1 are equivalent to conditions (A1), (A2) above.
Necessity. Let θ be a one-componentinner function andlet σ be its Clark mea-
α
sure. By Lemma 2.1 we have ρ(θ) = ρ(σ ). It was proved in [3] that m(ρ(θ)) = 0
α
andσ (ρ(θ))=0. Hence σ is adiscrete measurewith isolatedatoms andwe have
α α
m(suppσ )=0. Let ∆ be a connected component of the set T ρ(σ )=T ρ(θ).
α α
\ \
Byproperty(A1) the argumentofθ on∆is amonotonic functionunboundednear
both endpoints of ∆. It follows that the arc ∆ contains infinitely many points ξ
k
suchthatθ(ξ )=α. Enumeratethesepointsclockwisebyintegernumbers. Wesee
k
fromLemma2.1thatξ a(σ )forallk Zandeveryatomξ hastwoneighbours
k α k
∈ ∈
ξ , ξ . This shows that the measure σ satisfies condition (a). The fact that
k 1 k+1 α
−
σ satisfies condition (b) follows from Lemma 5.1 of [7]. Now check condition (c).
α
Fix an atom ξ a(σ ). From (11) we see that
0 α
∈
1 dσ (ξ)
= α +c , z D G , (12)
1 α¯θ(z) T 1 ξ¯z α ∈ ∪ θ
− Z −
where c =αθ(0)/(1 αθ(0)). Hence,
α
−
1 σ ξ
α 0
(H 1)(ξ )+c = lim { } .
σα 0 α z→ξ0(cid:18)1−α¯θ(z) − 1−ξ¯0z(cid:19)
Consider the analytic function kξ0 :z 7→ 11−−α¯ξ¯θ0(zz) on the domain D∪Gθ. We have
1 1 1
(H 1)(ξ )+c = lim
σα 0 α z→ξ0 1−ξ¯0z (cid:18)kξ0(z) − kξ0(ξ0)(cid:19) (13)
= ξ0kξ′0(ξ0) = αθ′′(ξ0).
− k2 (ξ ) −2θ(ξ )2
ξ0 0 ′ 0
From here and the estimate in (A2) we see that H 1 is bounded on a(σ ). Sur-
σα α
prisinglysimple relation(13) betweenthe discreteHilbert transformH 1 andthe
σα
inner function θ is the key observation in the proof.
Sufficiency. Let µ be a measure with properties (a) (c). Construct the inner
−
function θ with the Clark measure σ = µ. To prove that θ is a one-component
α
inner function we will check conditions (A1) and (A2).
By Lemma 2.1 we haveρ(θ)=ρ(σ ). Hence m(ρ(θ))=0 by property (a) of the
α
measure σ . Let ∆ be an open arc of T such that ∆ T ρ(θ) and ∆¯ ρ(θ)= .
α
⊂ \ ∩ 6 ∅
Then it follows from property (a) of the measure σ that ∆ contains infinitely
α
many atoms of σ . Since σ is finite and σ ξ = θ (ξ) 1 for every ξ a(σ ),
α α α ′ − α
{ } | | ∈
the function θ cannot be bounded on ∆. This gives us condition (A1).
′
| |
DUALITY THEOREMS 7
Condition (A2) is more delicate. To check it we need the following lemma.
Lemma 2.2. Assume that the Clark measure σ of an inner function θ has prop-
α
erties (a) (c). Then there exists a number κ > 0 such that for every ξ a(σ )
α
the set D −(κ)= z C: ξ z 6κσ ξ is contained in D G and we∈have
ξ α θ
{ ∈ | − | { }} ∪
1 α θ(z) 2
6 − 6 (14)
2σ ξ ξ z σ ξ
α{ } (cid:12) − (cid:12) α{ }
(cid:12) (cid:12)
for all z Dξ(κ). (cid:12) (cid:12)
∈ (cid:12) (cid:12)
Proof. Pick an atom ξ a(σ ) and rewrite formula (12) in the following form:
0 α
∈
1 dσ (ξ) σ ξ
= α + α{ 0} +c , z D G . (15)
1−α¯θ(z) ZT\{ξ0} 1−ξ¯z 1−ξ¯0z α ∈ ∪ θ
We have
dσ (ξ) ξ z dσ (ξ)
α 6 (H 1)(ξ ) + | 0− | α .
(cid:12)(cid:12)ZT\{ξ0} 1−ξ¯z(cid:12)(cid:12) | σα 0 | ZT\{ξ0} |ξ−z|·|ξ−ξ0|
(cid:12) (cid:12)
By property(cid:12)(cid:12)(c), |(Hσα1)(ξ0(cid:12)(cid:12))|6Cσα. Put κ∗ =(2Bσα)−1. For for ξ ∈a(σα)\{ξ0}
and z D (κ ) we have ξ z 6 κ σ ξ 6 ξ ξ /2 by property (b) of the
∈ ξ0 ∗ | 0− | ∗ α{ 0} | 0− |
measure σ , which gives us the inequality ξ z > ξ ξ ξ z > 1 ξ ξ .
α | − | | − 0|−| 0− | 2| − 0|
It follows that for z D (κ ) we have
∈ ξ0 ∗
z ξ dσ (ξ) dσ (ξ)
ZT\{ξ0} ||ξ−−z|0·||ξ−α ξ0| 62κ∗σα{ξ0}ZT\{ξ0} |ξ−αξ0|2.
Denote by ∆ the closed arc of T with endpoints ξ a(σ ). Using property (b),
0 α
± ∈
we obtain the estimate
dσ (ξ) dσ (ξ) 2
α 6 α +
ZT\{ξ0} |ξ−ξ0|2 ZT\∆ |ξ−ξ0|2 A2σασα{ξ0}
dm(ξ) 2
62πB + (16)
σαZT\∆ |ξ−ξ0|2 A2σασα{ξ0}
C
6 1 ,
σ ξ
α 0
{ }
where C is a constant depending only on the measure σ . We now see from (15)
1 α
that
1 σ ξ
= α{ 0} +f (z), z D (κ ) (D G ), (17)
1 α¯θ(z) 1 ξ¯z ξ0 ∈ ξ0 1 ∩ ∪ θ
0
− −
wherethe function f isboundedbytheconstantC =2κ C +C + c . Take
| ξ0| 2 ∗ 1 σα | α|
a number κ6κ such that C 6(2κ) 1. We have D (κ) D (κ ) and
∗ 2 − ξ0 ⊂ ξ0 ∗
1 σ ξ
f (z) 6 α{ 0}
| ξ0 | 2 1 ξ¯z
(cid:12) − 0 (cid:12)
for all z ∈Dξ0(κ)∩(D∪Gθ). From here an(cid:12)(cid:12)(cid:12)d (17) w(cid:12)(cid:12)(cid:12)e get on Dξ0(κ)∩(D∪Gθ) the
estimate
1 σ ξ 1 σ ξ
α{ 0} 6 62 α{ 0} ,
2 1 ξ¯z 1 α¯θ(z) 1 ξ¯z
(cid:12) − 0 (cid:12) (cid:12) − (cid:12) (cid:12) − 0 (cid:12)
whichshowsthat θ adm(cid:12)its the a(cid:12)nal(cid:12)ytic continu(cid:12)ation(cid:12) to a nei(cid:12)ghbourhoodofD (κ)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) ξ0
(that is, D (κ) D (cid:12)G ) and(cid:12)pro(cid:12)ves formul(cid:12)a (14(cid:12)) for poi(cid:12)nts z D (κ). Since
ξ0 ⊂ ∪ θ ∈ ξ0
DUALITY THEOREMS 8
our choice of the number κ is uniform with respect to ξ a(σ ), the lemma is
0 α
∈
proved. (cid:3)
Notation. In what follows we write E . E (correspondingly, E & E ) for two
1 2 1 2
expressions E ,E to mean that there is a positive constant c depending only on
1 2 θ
the inner function θ such that E 6 c E (correspondingly, c E > E ). We will
1 θ 2 θ 1 2
write E E if E .E and E &E .
1 2 1 2 1 2
≍
We are ready to complete the proof of Theorem 1. Differentiating (12) we get
α¯θ (z) ξ¯dσ (ξ)
′ α
= ,
(1 α¯θ(z))2 T (1 ξ¯z)2
− Z − (18)
α¯θ (z) 2α¯2θ (z)2 ξ¯2dσ (ξ)
′′ ′ α
+ =2 .
(1 α¯θ(z))2 (1 α¯θ(z))3 T (1 ξ¯z)3
− − Z −
Pickapointξ a(σ ). LetD (κ)bethesetfromLemma2.2. Denoteby∂D (κ)
0 ∈ α ξ0 ξ0
the boundary of D (κ). By formula (14), α θ(z) > κ/2 on ∂D (κ). Arguing
ξ0 | − | ξ0
as in the Lemma 2.2, from (18) we obtain the estimates
σ ξ dσ (ξ) 1
|θ′(z)|. |1−α{ξ¯00z}|2 +ZT\{ξ0} |1−αξ¯z|2 . σα{ξ0}, (19)
σ ξ dσ (ξ) 1
|θ′′(z)|.|θ′(z)|2+ |1−α{ξ¯00z}|3 +ZT\{ξ0} |1−αξ¯z|3 . σα{ξ0}2
forallz ∂D (κ). By the maximumprinciplewe have θ (z) .1/σ ξ 2 forall
points z∈ D ξ(0κ). On the unit circle T we have | ′′ | α{ 0}
∈ ξ0
ξ¯ z¯
= − .
(1 ξ¯z)2 1 ξ¯z 2
− | − |
From here and formula (18) we get for z D (κ) T the estimate
∈ ξ0 ∩
1 α¯θ(z) 2 1 α¯θ(z) 2 1
|θ′(z)|=σα{ξ0}(cid:12)(cid:12) 1−−ξ¯0z (cid:12)(cid:12) +ZT\{ξ0}(cid:12)(cid:12) −1−ξ¯z (cid:12)(cid:12) dσα(ξ)& σα{ξ0}. (20)
Combining (19) and(cid:12)(20) we se(cid:12)e that θ /(cid:12)θ2 . 1 on(cid:12) D (κ) T. It remains to
obtainthe same estim(cid:12) ate for po(cid:12)ints z |T′′ (cid:12)ρ′(θ|) that d(cid:12)o nξo0t bel∩ong to the union of
∈ \
the sets D (κ), ξ a(σ ), from Lemma 2.2. Take such a point z . We claim that
ξ α 0
∈
θ(z ) α >κ/2. Indeed, assume the converse and find the connected component
0
|∆ of t−he s|et ζ T ρ(θ) : θ(ζ) α < κ/2 containing the point z . Since the
0
{ ∈ \ | − | }
argument of the inner function θ is monotonic on ∆, there exists a point ξ ∆
∈
such that θ(ξ) = α. By Lemma 2.1 we have ξ a(σ ). Next, from (14) we see
α
that α θ(z) > κ/2 for both points in T ∂D∈ (κ). Since ∆ is connected this
ξ
| − | ∩
yields the inclusion ∆ D (κ) which gives us the contradiction with z / D (κ).
ξ 0 ξ
⊂ ∈
Thus, we proved the inequality θ(z ) α > κ/2. Let ξ be the nearest point to
| 0 − | z0
z in a(σ ). We have κσ ξ 6 ξ z 6 B σ ξ . These two estimates
0 α α{ z0} | z0 − | σα α{ z0}
imply (19) and (20) for z = z and ξ = ξ . It follows that θ (z )/θ (z )2 . 1
0 0 z0 | ′′ 0 ′ 0 |
and θ satisfies condition (A2). (cid:3)
Remark. Lemma 5.1 in [7] and formula (13) show that for every one-component
inner function θ there exist positive constants A , B , C such that A 6 A ,
θ θ θ θ σα
B 6B ,C 6C forallClarkmeasuresσ ofθ. Also,itfollowsfromLemma5.1
σα θ σα θ α
in [7] that θ (z) 1/σ ξ for all ξ a(σ ) and all z T between the neigh-
′ α α
| | ≍ { } ∈ ∈
bours ξ of ξ in a(σ ). In particular, we have σ (∆) σ (∆) m(∆) for all β
α β α
± ≍ ≍
DUALITY THEOREMS 9
with β =1 andallarcs∆ofthe unitcircle Tcontainingatleasttwoatomsofthe
| |
measure σ .
α
3. Proofs of Theorem 2 and Theorem 2
′
We first prove Theorem 2. The following result is classical, for the proof see [11]
′
or Chapter 9 in [10].
Theorem. (D.N.Clark)Letθbeaninnerfunctionandletσ beitsClarkmeasure.
α
The natural embedding V :K2 ֒ L2(σ ) defined on the reproducing kernels of the
α θ → α
space Kθ2 by Vα 1−1θ(λ¯λz)θ = 1−1θ(λ¯λz)α can be extended to the whole space Kθ2 as the
− −
unitary operator(cid:16)from K(cid:17)2 to L2(σ ). For every f L2(σ ) the function
θ α ∈ α
1 α¯θ(z)
F(z)= f(ξ) − dσ (ξ) (21)
T 1 ξ¯z α
Z −
in the unit disk D belongs to K2 and V F =f as elements in L2(σ ).
θ α α
ItworthbementionedthatA.G.Poltoratski[22]establishedtheexistenceofan-
gularboundaryvaluesσ -almosteverywhereonTforallfunctionsinthespaceK2,
α θ
thusprovingthattheunitaryoperatorV inClarktheoremactsasthenaturalem-
α
bedding on the whole space K2. In our situation this follows from a very simple
θ
argument, see Lemma 3.3 in Section 3.2.
TheembeddingV :Kp ֒ Lp(σ )definedonthelinearspanofthereproducing
kernels of Kp mightαbe uθnb→oundedαfor 1 6 p < 2 and might have the unbounded
θ
inverse V 1 : Lp(σ ) ֒ Kp for 2 < p 6 , see Section 3 in [2]. However, the
α− α → θ ∞
situationisidealfortheone-componentinnerfunctionsθ,asfollowingresultsshow:
V Kp Lp(σ ) for 1<p< – A. L. Volberg, S. R. Treil [26];
• VαKθp =⊂Lp(σα) for 1<p<∞ – A. B. Aleksandrov [2];
• VαKθp Lp(σα) for 0<p61∞– A. B. Aleksandrov [3].
• α θ ⊂ α
Theorem 2 says that V K1 = H1(σ ) for every one-component inner function θ.
′ α θ at α
We are ready to prove its easy part – the inclusion V K1 H1(σ ).
α θ ⊃ at α
3.1. Proof of the part “ ” in Theorem 2. Let µ be a measure on the unit
′
circle T with properties (a)⇒ (c). Take a complex number α of unit modulus and
−
constructthe one-componentinner function θ with the Clark measure σ =µ. We
α
want to show that every function f H1(σ ) admits the analytic continuation to
tfihresto,paessnuumneitthdaistkfDisaas aσfu-antcotmionsuFp∈p∈orKatteθ1d∩αoznHan1 waricth∆kFkTL1w(Ti)th.ckefntkeHra1ξt(σ.αT).hAent
α c
⊂
f L2(σ ) and the function F in formula (21) lies in the space K2 K1 by Clark
∈ α θ ⊂ θ
theorem. Since fdσ = 0, we have F(0) = 0. Moreover, we see from Lemma
T α
2.1 that F(ξ) = f(ξ) for all ξ a(σ ). Let us check that the norm of F in L1(T)
R ∈ α
is bounded by a constant depending only on the inner function θ. By Aleksandrov
desintegration theorem (see [1] or Section 9.4 in [10]), we have
F dm= V F(ξ) dσ (ξ)dm(β). (22)
β β
T| | T T| |
Z Z Z
Fix a complex number β = α of unit modulus. We claim that V F . 1.
Denote by 2∆ the arc of T6 with center ξ such that m(2∆) = 2mk(∆β )k(Lin1(σthβ)e case
c
DUALITY THEOREMS 10
where m(∆)>1/2 put 2∆=T). Break the integral V F dσ into two parts,
T| β | β
R
V F(ξ) dσ (ξ)= V F(ξ) dσ (ξ)+ V F(ξ) dσ (ξ). (23)
β β β β β β
ZT| | Z2∆| | ZT\2∆| |
By Clark theorem we have kVβFkL2(σβ) = kFkL2(T) = kVαFkL2(σα). Moreover,
we have V F 6 1/ σ (∆) because the function V F = f is a σ -atom
k α kL2(σα) α α α
supported on the arc ∆. This yields the inequality
p
V F(ξ) dσ (ξ)6 σ (2∆) V F 6 σ (2∆)/σ (∆). (24)
| β | β β ·k β kL2(σβ) β α
Z2∆ q q
Note that the arc ∆ contains at least two points in a(σ ) because f has zero σ -
α α
mean on ∆. Hence σ (∆) m(∆) and σ (2∆) m(2∆), see remark after the
α β
≍ ≍
proofofTheorem1. Thisshowsthat V F(ξ) dσ (ξ).1. Letusnowestimate
2∆| β | β
the second term in (23). Take a point z a(σ ) 2∆. Using the fact that f is a
β
R ∈ \
σ -atom we obtain the estimate
α
1 α¯β
V F(z) = f(ξ) − dσ (ξ)
| β | 1 ξ¯z α
(cid:12)Z∆ − (cid:12)
(cid:12) 1 α¯β 1(cid:12) α¯β
(cid:12) (cid:12)
=(cid:12) f(ξ) 1− ξ¯z − 1−(cid:12) ξ¯z dσα(ξ)
(cid:12)Z∆ (cid:18) − − c (cid:19) (cid:12)
(cid:12) ξ ξ (cid:12)
62(cid:12)(cid:12) |f(ξ)| (1 ξ¯z−)(1c ξ¯z) dσα(ξ)(cid:12)(cid:12) (25)
Z∆ (cid:12) − − c (cid:12)
2πm(∆) (cid:12) z ξ (cid:12)
6 z ξ 2 ·s(cid:12)(cid:12)up z− ξc · |f(cid:12)(cid:12)(ξ)|dσα(ξ)
| − c| ξ∈∆(cid:12)(cid:12) − (cid:12)(cid:12) Z∆
4πm(∆) (cid:12) (cid:12)
6 . (cid:12) (cid:12)
z ξ 2
c
| − |
From here we get
dσ (z)
V F(z) .m(2∆) β .1. (26)
ZT\2∆| β | ·ZT\2∆ |z−ξc|2
Hencethe normofF inL1(T) isboundedby aconstantdepending onlyon θ. Now
takeanarbitraryfunctionf H1 (σ )andconsideritsrepresentationf = λ f ,
∈ at α k k
where f are σ -atoms and λ 62 f . Let F be the functions in K2
k α k| k| k kHa1t(σα) k P θ
such that V F = f . Then the sum λ F converges absolutely in L1(T) to a
α k k P k k
function F ∈Kθ1 and we have kFkL1(T)P.kfkHa1t(σα). From formula (21) we get
1 α¯θ(z) 1 α¯θ(z)
F(z)= λ f (ξ) − dσ (ξ)= f(ξ) − dσ (ξ), z D.
k T k 1 ξ¯z α T 1 ξ¯z α ∈
k Z − Z −
X
Since f L1(T), this formula determines the analytic continuation of F to the
domain D∈ G . From Lemma 2.1 we see that F(ξ)=f(ξ) for all ξ a(σ ). (cid:3)
θ α
∪ ∈
3.2. Preliminaries for the proof of the part “ ” in Theorem 2. Let θ be
′
⇐
a one-component inner function with the Clark measure σ . Introduce positive
α
constants A˜ , B˜ such that
σα σα
A˜ m[ξ,ξ ]6σ ξ 6B˜ m[ξ,ξ ], ξ a(σ ).
σα ± α{ } σα ± ∈ α
Here [ξ,ξ ], [ξ,ξ ] are the closed arcs of T with endpoints ξ,ξ a(σ ) such
+ α
− ± ∈
that the corresponding open arcs (ξ,ξ ) do not intersect suppσ . Take a positive
α
±