Bachelorarbeit

De Branges’ Theorem on Weighted PolynomialApproximation

Daniel Hainschink

2016S

Contents

Introduction 2

1 Approximation in Regular Function Spaces 41.1 Regular Function Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Hall Majorant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.4 Analytic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 De Branges’ Lemma 92.1 Dual of C0(W ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 De Branges’ Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3 De Branges’ Proof 113.1 Consequences of de Branges’ Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.2 Analytic Extensions for L1(µ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.3 Representation of L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.4 Construction of a Krein Class Function . . . . . . . . . . . . . . . . . . . . . . . . . . 14

4 Proof of Sodin and Yuditskii 144.1 Chebyshev Sets and ω-Extremality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154.2 Properties of R(C+) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164.3 Construction of a Krein Class Function . . . . . . . . . . . . . . . . . . . . . . . . . . 17

A Some Results on Entire Functions 18

References 20

1

Introduction

The subject of this thesis is de Branges’ theorem on the classical Bernstein problem on weightedpolynomial approximation. If, for a lower semicontinuous function W : R → (0,∞], referred to asthe weight function, we let C0(W ) be the space of all f ∈ C(R) with f(x)/W (x) → 0 as x → ±∞,with the norm ‖f‖ = ‖f/W‖∞, the simplest instance of the Bernstein problem may be stated as:

For which weight functions W are the polynomials P dense in C0(W )?

Classical results on the Bernstein problem usually rely on the Hall majorant

m(z) = sup|f(z)| : f ∈ P, ‖f‖W ≤ 1, (I.1)

i.e. the norm of the point evaluation at z on P. Characterizations of denseness include (cf. [5,pp. 147,158,164]):

1) m(z) =∞ for some (all) z ∈ Cr R. (Mergelyan)2)∫ log m(x)

1+x2 dx =∞. (Akhiezer)3) sup

∫ log |f(x)|1+x2 dx : f ∈ P, ‖f‖W ≤ 1 =∞. (Pollard)

The key idea here is that if P 6= C0(W ), then m is everywhere finite, and the elements of P actuallyextend to entire functions, with the point evaluation on P still being bounded by m.Before stating de Branges’ theorem, let us remark that only a handful of properties of the class ofpolynomials are relevant to most of the theory of the Bernstein problem, so that it is natural to workwith the following generalization:I.1 Definition. A linear space L of entire functions is called an algebraic de Branges space if

1) If F ∈ L, then so is F ](z) = F (z).2) If F ∈ L and F (w) = 0, then F (z)/(z − w) is in L.

This definition includes, in particular, the space Ea of all entire function of exponential type ≤ a.While the above extension property is also relevant to the proof of de Branges’ theorem, the moti-vation behind de Branges’ theorem comes from the Hahn-Banach characterization of non-densenessin C0(W ): If we identify the dual of C0(W ) with the space M(R) of complex measures on R via(Tµ)f :=

∫fW−1 dµ (as is done in Theorem 2.1), then the theorem may be interpreted as a de-

scription of the annihilating measures of a given algebraic de Branges space L ⊂ C0(W ) in terms ofparticular entire functions:I.2 Definition. The weighted Krein class K(L,W ) consist of all entire functions B such that:

1) B = B]. B has at least one zero, and all zeros of B are real and simple.2) For F ∈ L, F/B is of bounded type in the half planes C+,C−, and yF (iy) = o(B(iy)) as

y → ±∞.3)∑

B(x)=0

∣∣∣W (x)B′(x)

∣∣∣ <∞.

Given B ∈ K(L,W ), condition 3) ensures that

µ =∑

B(x)=0

1B′(x)δx (I.2)

defines a complex measure on R with L ⊂ L1(µ). Condition 2) implies that for F ∈ L, the interpola-tion formula

(z − x0)F (z)B(z) =

∑B(x)=0

(x− x0)F (x)B′(x)(z − x) , x0 ∈ R, B(z) 6= 0, (I.3)

holds. To see this, note that both functions are meromorphic in C with the same simple poles andresidues. The left hand side is of bounded type in both half planes and tends to 0 along both

2

imaginary axes by condition 2), and the right hand side has the same properties by Lemma A.6(being the Cauchy integral of the measure

∑(x − x0)F (x)/B′(x)δx) and dominated convergence.

It follows that the difference is an entire function in N(C±) tending to 0 along the imaginary axes,which must vanish identically by Lemma A.10. If we choose x0 ∈ R with B(x0) 6= 0 and let z = x0,(I.3) turns into 0 =

∫F (x) dµ(x), so that µ ∈ L⊥.

We have just shown that K(L,W ) 6= ∅ implies the non-denseness of L, which is the simple part ofde Branges’ theorem:I.3 De Branges’ Theorem. Let W : R → (0,∞] be lower semicontinuous. An algebraic de Brangesspace L ⊂ C0(W ) is dense in C0(W ) if and only if K(L,W ) 6= ∅.

We present two proofs of the necessity of K(L,W ) 6= ∅: The original proof of de Branges [3], whichconstructs the function B from an extremal annihilating measure, using a lemma on extremal mea-sures from [4], and a more recent proof by Sodin and Yuditskii [11], based on a normal family argu-ment, which makes no reference to annihilating measures. Both proofs were orginally formulated forthe case of polynomials; the generalizations to algebraic de Branges spaces we present are containedin [1] and [12], respectively.Before we actually give the proofs in Sections 3 and 4, respectively, we develop some of the theory ofapproximation by analytic functions in regular function spaces according to Pitt [7] in Section 1, andgive Summers’ [13] description of the dual space of C0(W ) as a space of complex measures and deBranges’ lemma [4] on extremal annihilating measures in Section 2. For completeness, a few resultson entire functions of exponenial and bounded type are given in Appendix A.

Notation

1Ω . . . . . . . . . . . . . . . indicator function of Ω, 1Ω(x) = 1 for x ∈ Ω, 0 otherwise‖f‖Ω . . . . . . . . . . . . for f : X → C and Ω ⊂ X, ‖f‖Ω = supx∈Ω |f(x)|C0(W ) . . . . . . . . . . weighted C0-space, page 2Cn(R) [Cn

c (R)] . . . n times differentiable functions [with compact support]C+,C− . . . . . . . . . . upper and lower half planes, C± = z ∈ C : ± Im z > 0δx . . . . . . . . . . . . . . . Dirac measure δx(A) = 1 if x ∈ A, 0 otherwiseD,D(w, r) . . . . . . . complex unit disc; disc with radius r about wD . . . . . . . . . . . . . . . test functions in the sense of distribution theory, cf. e.g. [9, p. 151]H(Ω) . . . . . . . . . . . . holomorphic functions in an open set Ω ⊂ Riff . . . . . . . . . . . . . . . if and only ifL . . . . . . . . . . . . . . . . algebraic de Branges space, page 2lsc . . . . . . . . . . . . . . . lower semicontinuousM(X) . . . . . . . . . . . finite complex Borel measures on X, page 10m(z) . . . . . . . . . . . . . Hall majorant in a regular function space, page 8N(Ω) . . . . . . . . . . . . holomorphic functions of bounded type in Ω, page 19R(Ω) . . . . . . . . . . . . holomorphic functions with positive imaginary part in Ω, page 16Rw . . . . . . . . . . . . . . resolvent operator Rwf(z) = (z − w)−1f(z), page 7Rw,R1

w . . . . . . . . . . difference quotient operator Rw[F,G](z) = (z −w)−1(F (z)G(w)− F (w)G(z)) onL; corresponding operator on L1(µ), page 13

W . . . . . . . . . . . . . . . lower semicontinuous weight function, page 2Z(F ) . . . . . . . . . . . . zero set of an entire function, Z(F ) = z ∈ C : F (z) = 0

3

1 Approximation in Regular Function Spaces

We first give results by Pitt [7] implying that for algebraic de Branges subspaces of a large class offunction spaces (including C0(W ), but also L1(µ)), non-denseness implies boundedness of the Hallmajorant and the extendability of functions in the closure to entire functions, whose quotients are ofbounded type in both half-planes.

1.1 Regular Function Spaces

The following concept of function space from [7, Sec. 2] allows us to prove results that will be neededfor both C0(W ) and L1(µ) simultaneously. Let Cn(R) be endowed with the topology of locally uniformconvergence of the first n derivatives.1.1 Definition. A Banach space B together with a subspace B0 of CN0(R) and a linear map B0 → B(in general neither continuous nor injective) is called a regular function space if

1) B0 is dense in B.2) B0 is preclosed in B, i.e. if fn ∈ B0 converge to 0 in CN0(R) and fn form a convergent sequence

in B, then they actually converge to 0 in B.3) The multiplication operators (Ttf)(x) = eitxf(x), t ∈ R form a strongly continuous operator

group of polynomial growth on B0, i.e.

‖Ttf − f‖ → 0, t→ 0, and ‖Tt‖ ≤ (C + |t|)N1 . (1.1)

Notation for the map B0 → B will be suppressed throughout. We can interpret the elements of B0as representants for equivalence classes of (not necessarily continuous) functions, of which B consistsin many important cases. If fn ∈ B0 converge to some f ∈ CN0(R) as well as to some element f ofB, then condition 2) says that f is uniquely determined by f , so that the mapping B0 → B and ourinterpretation extend to the projection B1 of the closure of B0 as a subset of CN0(R)×B to CN0(R).The most interesting condition 3) is meant to formalize the requirement for ‖f‖ to depend only onthe first N1 derivatives of f .1.2 Proposition. C0(W ) for lsc weights W and Lp(µ) for Borel measures µ on R are regular functionspaces.Proof. For C0(W ), we may let N0 = 0, B0 = C0(W ), so that 1) is clear and the embedding is actuallyinjective. To see 2), note that the limit of a sequence in both CN0(R) and C0(W ), if existent, mustbe equal to the pointwise limit. For 3), choose M > 0, t > 0 with |f/W | < ε outside [−M,M ] and|eitx − 1| < ε on [−M,M ], so that

‖Ttf − f‖W = ‖(eitx − 1)f/W‖∞ ≤ ε(‖f/W‖[−M,M ] + ‖eitx − 1‖Rr[−M,M ]) ≤ ε(‖f‖W + 2). (1.2)

For Lp(µ), use B0 = Cc(R), the existence of pointwise convergent subsequences of Lp-convergentsequences and dominated convergence.

1.2 Multipliers

Next, we introduce a space of multipliers for B1 (which will be used to formalize division of zeroson analytic subspaces of regular function spaces). Write N = maxN0, N1, N(t) = (C + |t|)N , andnote N(t+ s) ≤ N(t)N(s).1.3 Definition. M is the space of all complex measures µ on R such that

∫N(t) d|µ|. M is the

space of all Fourier transforms φ = µ of measures in M , with ‖φ‖ =∫N(t) d|µ|. M+ is the subspace

of all Fourier transforms of measures in M supported in (−∞, 0].

4

1.4 Lemma. M forms a Banach algebra under pointwise multiplication. For φ ∈ M and f ∈ B1,φf ∈ B1 with ‖φf‖ ≤ ‖φ‖‖f‖ and

φf(x) =∫Ttf(x) dµ(t), φ = µ. (1.3)

Note that the right hand side of (1.3) makes sense for all f ∈ B, so that we can use it definemultiplication with φ for arbitrary f ∈ B.Proof. To see that M is a Banach algebra, note that

‖µν‖ =∫N(t) d|µ ∗ ν| ≤

∫N(t) d|µ| ∗ |ν| (1.4)

=∫ ∫

N(t+ s) d|µ| d|ν| ≤∫ ∫

N(t)N(s) d|µ| d|ν| = ‖µ‖‖ν‖. (1.5)

Since ‖Ttf‖ ≤ N(t)‖f‖, the B-valued (Riemann) integral in (1.3) has ‖∫Ttf dµ‖ ≤

∫N(t)‖f‖ d|µ| =

‖φ‖‖f‖. It remains to show (1.3). For µ = δt and φ = eitx, this is trivial. For general µ, get atight sequence of discrete measures νn converging to N(t)µ weakly1, and let µn = N(t)−1νn. Thenφn = µn → φ follows immediately from weak convergence, and∥∥∥∥∫ Ttf dµn −

∫Ttf dµ

∥∥∥∥ ≤ ‖f‖ ∫|t|≤t0

N(t) d|µn − µ|+ ‖f‖∫

|t|>t0(d|νn|+N(t)d|µ|). (1.6)

Now choose t0 > 0 large enough to make the second term < ε for all n, and note that the first termtends to 0 as n→∞ if we additionally take t0 to have µ(t0) = 0. Hence the lim sup of the left sideis smaller than ε.

The obvious identity

(x− z)−1 = i

∫ 0

−∞ei(x−z)t dt = i

∫ 0

−∞eitx dµ(t), Im z > 0 (1.7)

where dµ = e−iztdt, allows us to interpret multiplication by (x− z)−1 for z ∈ C+ as the action of anelement of M+, whose norm we can now bound by a simple analysis of µ.1.5 Lemma. If B is a regular function space, f(x) ∈ B and z ∈ C with y = Im z 6= 0, then(x− z)−1f(x) ∈ B with

‖(x− z)−1f(x)‖ ≤ C‖f‖(1/|y|+ 1/|y|N+1

), (1.8)

where C > 0 does not depend on y or f .

Proof. With µ as above,∫N(t) d|µ| =

∫N(t)e−yt dt = y−1

∫N(y−1t)e−t dt (1.9)

≤ y−1N(y−1)∫N(t)e−t dt ≤ Cy−1(1 + y−N ). (1.10)

1Both concepts are taken to have their probabilistic meaning, although we do share the opinion that weak* convergence

would be the more appropriate term. One easily verfies that the measures

νn =∞∑

k=−∞

(Nµ)(n−1(k − 1, k])δn−1k

will do.

5

1.6 Definition. For Im z 6= 0, Rz : B → B denotes multiplication by (x− z)−1.

Rz is a bounded operator on B by (1.8).1.7 Lemma. If A is a linear subspace of B and RwA ⊂ A for some w ∈ C+, then M+A ⊂ A.Proof. The continuity of Rw immediately implies RwA ⊂ A. Rz satisfies the resolvent equation

Rz −Rw = (w − z)RzRw. (1.11)

For |z − w| < ‖Rw‖−1, it follows that

∞∑n=0

(z − w)nRn+1w = Rz

∞∑n=0

(z − w)nRnw − (z − w)n+1Rn+1

w = Rz, (1.12)

which gives the invariance of A under Rz for z close to w. It also follows that for f ∈ B, l ∈ B∗, thefunction z → (Rzf, l) is analytic on C+ and vanishes in a neigborhood of w, hence on all of C+.For arbitrary φ = µ ∈ M+, f ∈ A and l ∈ A⊥, we have by (1.3)

(φf, l) =∫ 0

−∞(Ttf, l) dµ, (1.13)

but we already know that also l ∈ (RzA)⊥, so that the Laplace transform of the integrand vanishes:

0 = (Riyf, l) = −i∫ 0

−∞eyt(Ttf, l) dt. (1.14)

It follows by the invertibility of the Laplace transform that (φf, l) = 0, and hence l ∈ (M+A)⊥.

The reason why we consider the above lemma is that for A an analytic subspace, its assumptions willfollow from infiniteness of the Hall majorant (while having little to do with the division of zeros inalgebraic de Branges spaces) and its conclusion will often imply densenss by the criterion formulatedbelow.1.8 Definition. A subspace A of a regular function space is called analytic if its members extend toentire functions.ZR(A) is the set of common real zeros of A.

Note that according to our definition of algebraic de Branges spaces (which includes division of realzeros), ZR(A) = ∅ for A an algebraic de Branges space. For this reason, we need only a special caseof [7, Prop. 2.4], which also shows denseness in some cases with ZR(A) 6= ∅.1.9 Lemma. If A 6= 0 is an analytic subspace of a regular function space with MA ⊂ A andZR(A) = ∅, then A = B.Proof. Assume A 6= B, so that A ( B0, and get l ∈ A⊥, f ∈ B0 with K = supp f compact‖l‖, ‖f‖ = 1 and (f, l) 6= ∅. Then u(φ) = (φf, l), φ ∈ D defines a nontrivial distribution of order Nsupported in K.2 Since ZR(A) = ∅ and f has compact support, we may represent f as

f =∑

φiaif =∑

(φif)ai, φi ∈ D , ai ∈ A, (1.15)

where φif is compactly supported and in CN , hence φif ∈ M by [9, Thm. 7.15]. But then f ∈ A,so that l ∈ A⊥ implies u = 0.

2As usual, D is the class of test functions, contained in M by the invariance of the Schwartz class under Fouriertransforms and the Inversion theorem [9, Thm. 7.7]. u being of order N means that u is bounded in the norm ofCN

c (K). While 1 6∈ D , it is still clear that u is nontrivial since D is dense in the norm of M .

6

1.10 Remark. In the case ZR(A) 6= ∅, we could still have applied the construction in (1.15) tog = ψf , where ψ ∈ D with ψ(x) = 1 for d(x,ZR(A)∩K) < ε and φ(x) = 0 for d(x,ZR(A)∩K) ≥ 2εto obtain that u is supported on ZR(A). By discreteness of ZR(A), multiplying f by a function ψ ∈ Dsupported near a single point x ∈ ZR(A) reduces u to a distribution with one-point support, whichmust be of the form u(φ) =

∑ni=0 ciφ

(i)(x) by [9, Thm. 6.25]. Again multiplying f by ψ ∈ D withψ(i)(x) = 0, i < n and ψ(n)(x) = c−1

n , we get a representation of the point evaluation at x on M inthe form (φf, l) = φ(x).Such points x are said to be in the point spectrum σp(B) of B, and we have just seen that it wouldsuffice to assume ZR(A) ∩ σp(B) 6= ∅. In the case B = C0(W ), this is of no great use since takingf ∈ C0(W ) with f(x) = 1, and l the point evaluation at x on C0(W ) shows that σp(C0(W )) = R.For B = L1(µ), σp(B) is the set of atoms of µ, since for µ(x) = 0 and f ∈ L1(µ), there are φ ∈ Mof arbitrarily small L1(µ)-norm with φ(x) = 1.1.11 Remark. At this point, the lack of injectivity assumptions on the embedding of A into B willprobably have made the reader somewhat suspicious. Therefore, let us remark that the above proofreally takes place in B0, first showing denseness in B0 and only then using the denseness of B0 in B.This rules out pathologies such as A 6= 0 as a space of functions and MA ⊂ A, but A = 0 as asubspace of B, which appear to be possible from the statement of the lemma.

1.3 Hall Majorant

We are now able to give the key statements on finiteness of the Hall majorant and denseness ofalgebraic de Branges subspaces, as contained in [7, Prop. 3.1,3.2, Thm. 3.1]. Fix an algebraic deBranges subspace H of a regular function space B, and let Hz = h ∈ H : h(z) = 0.1.12 Definition. For f ∈ B, ‖f‖+ := ‖Rif‖.

Since Rz is ‖ · ‖-bounded, ‖ · ‖+ is weaker than ‖ · ‖. By the resolvent equation (1.11),

‖Rzf‖ ≤ ‖Rwf‖+ |w − z|‖RzRwf‖ ≤ (1 + |w − z|‖Rz‖)‖Rwf‖, f ∈ B, z, w ∈ C (1.16)

so that the norms ‖Rzf‖ are all equivalent.1.13 Definition. For z ∈ C, the Hall majorants of the algebraic de Branges subspace H are definedby

m(z) = sup|h(z)| : h ∈ H, ‖h‖ ≤ 1, m+(z) = sup|h(z)| : h ∈ H, ‖h‖+ ≤ 1 (1.17)

Clearly, m(z) ≤ ‖Ri‖m+(z). By the symmetry requirement H] = H, m(z) = m(z),m+(z) = m+(z).1.14 Lemma. If m+(z) = ∞ for some z ∈ C r R, then RzH ⊂ H. Conversely, if RzH = H, thenm+(z) =∞.Proof. m+(z) = ∞ occurs iff the point evaluation at z is unbounded on H, which happens iff itskernel Hz is dense in H, ie

d‖·‖+(f,Hz) = 0, f ∈ H. (1.18)

By the equivalence of the norms ‖Rzf‖, this is the same as

0 = d‖·‖(Rzf,RzHz) ≥ d‖·‖(Rzf,H), (1.19)

where the latter inequality is by RzHz ⊂ H. This shows that Rzf ∈ H.If RzH is dense in H, the inequality in (1.19) becomes an equality, so that we can reverse theargument.

Together with Lemma 1.7, Lemma 1.9, the symmetry of algebraic de Branges spaces and m(z) ≤‖Ri‖m+(z), we obtain

7

1.15 Theorem. If an algebraic de Branges space H is not dense in a regular function space B, thenits Hall majorant m(z) is finite on Cr R.

To obtain analytic extensions for the functions in H, we actually need local boundedness, which iscontained in the next theorem:1.16 Theorem. If H is not dense in B, then m(z) is continuous and finite on C, and logm(z) issubharmonic and thus locally L1.

The statement that m(z) is either infinite on all of C or bounded in the sense of the theorem iscommonly referred to as the Riesz-Mergelyan alternative. Recall that every subharmonic function6≡ −∞ is locally integrable by the sub-mean value property [8, Thm. 2.3,2.4 (ii)], and that log |f | issubharmonic for any analytic function f [8, Thm. 2.12].Proof. First, we show local integrability. Note that log− m(z) is bounded below by each of thesubharmonic functions log− |g|, g ∈ H, ‖g‖ = 1, so that it suffices to show log+ m(z) ∈ L1

loc. SinceH is not dense, Lemma 1.9 shows that H cannot be invariant under all operators Rz. Hence fixl ∈ H⊥ and f ∈ H with (Rzf, l) 6= 0 for some z ∈ C r R. For arbitrary g ∈ H and z fixed, we have(w − z)−1(f(z)g(w)− f(w)g(z)) ∈ H as a function of w, so

0 =((w − z)−1(f(z)g(w)− f(w)g(z)), l

)(1.20)

= f(z)(Rzg, l)− g(z)(Rzf, l) = f(z)G(z)− g(z)F (z), (1.21)

where G(z) = (Rzg, l) and F (z) = (Rzf, l) are entire by (1.12). From (1.8), we obtain that

|G(z)| ≤ ‖g‖C(1/|y|+ 1/|y|N+1

), (1.22)

where by calculus∫ 1

0log

∣∣∣1/|y|+ 1/|y|N+1∣∣∣ ≤ −C1

∫ 1

0log y dy = −C1 [y log y − y]10 <∞. (1.23)

Since log+ |g| = log+ |f | + log+ |G| − log+ |F |, the functions log+ |g| for ‖g‖ ≤ 1 can be uniformlyestimated by a locally integrable function, so that m(z) is indeed locally integrable.By the area sub-mean value property3 applied to log+ |g|, it follows that the family g ∈ H : ‖g‖ ≤ 1is locally bounded, hence normal. This gives the continuity of m(z) and the upper semicontinuity oflogm(z). Since

logm(z) = sup‖g‖≤1

log |g(z)| ≤ sup‖g‖≤1

12π

∫ 2π

0log |g(z + reiθ)| dθ (1.25)

≤ 12π

∫ 2π

0sup

‖g‖≤1log |g(z + reiθ)| dθ = 1

2π

∫logm(z + reiθ) dθ, (1.26)

logm(z) also has the sub-mean value property and is thus subharmonic [8, Thm. 2.3], which completesthe proof.

3The area sub-mean value property reads

u(w) ≤ 1r2π

∫∫|z−w|<r

u(z) dxdy, u subharmonic, (1.24)

and follows from the sub-mean value property by integration over r.

8

1.4 Analytic Extensions

With the results of the preceding section at hand, we can easily prove the properties of the functionsin H we need (cf. [7, Prop. 3.3,3.4, Thm. 3.2]).1.17 Theorem. If H is not dense in B, then every function f ∈ H has an extension h ∈ H(C)satisfying

|h(z)| ≤ ‖f‖m(z). (1.27)

Proof. By local boundedness of m(z), the unit ball of H is a normal family, so that every ‖ · ‖-convergent sequence in H has a locally unformly convergent subsequence. Since each member of thesequence satisfies (1.27), so does the limit.

1.18 Theorem. If H is not dense in B and f, g ∈ H with g 6= 0, then f/g is a meromorphic functionof bounded type in C±.Proof. Since the quotient of funtions of bounded type is again of bounded type, it clearly suffices toprove this for a fixed g ∈ H and all f ∈ H, so that we may assume Rzg 6∈ H. As in the proof oftheorem 1.16, we get a representation

f(z)/g(z) = F (z)/G(z), F (z) = (Rzf, l), G(z) = (Rzg, l). (1.28)

Estimate (1.22) shows (together with lemma A.5) that f/g is of bounded type in the half planesy < −1, y > 1. We now prove∫ 1

−1

∫ ∞

−∞

log+ |f(x+ iy)/g(x+ iy)|1 + x2 dxdy <∞, (1.29)

which will imply f/g ∈ N(C±) by the generalization of Krein’s theorem given in [7, Thm. A.1].This amounts to estimating the same integral with log+ |F | and log− |G| in place of log+ |f/g|. Forlog+ |F |, use (1.23). By subharmonicity of G(z) and the Poisson formula for the half plane [8, Sec.5.2], we get that

1π

∫ ∞

−∞

log |G(x+ ih)|1 + (x− x0)2 dx ≥ log |G(x0 + i+ ih)|, h > 0, x0 ∈ R. (1.30)

Since G 6≡ 0, the integral of the right hand side over h ∈ (−1, 1) must be finite for some x0 by localintegrability of log |G(z)|. Since (1 + x2)/(1 + (x− x0)2) is bounded, this completes the proof.

2 De Branges’ Lemma

We proceed to prove Summers’ description of the dual space of C0(W ) from [13], and de Branges’Lemma [4] on extreme points of the annihilator of a subspace of C0(W ).

2.1 Dual of C0(W )

In this section, we may as well consider a locally compact space X in place on R, and define C0(X)to be the space of all f ∈ C(X) such that for every ε > 0, there is K ⊂ X compact such that |f | isbounded by ε on X rK. For W : X → (0,∞] lower semicontinuous, C0(W ) is defined accordingly.From the Riesz Representation Theorem for C0(X), it seems natural to identify the dual C0(W )∗ ofC0(W ) with the space M(X) of (finite) complex Radon measures4 on X via Tµ(f) :=

∫fW−1 dµ.

4Radon measures are locally finite inner regular measures on the Borel sets of X; by σ-compactness, the requirementof inner regularity is void for X = R.

9

If W is continuous and everywhere finite, then f → fW−1 is an isomorphism between C0(W ) andC0(X), so that the statement indeed reduces to the Riesz Theorem. However, by the following resultof Summers [13, Thm. 3.1], lower semicontinuity is already sufficient for the existence of representingmeasures, although they need not be unique if W is infinite:2.1 Theorem. If W : X → (0,∞] is a lower semicontinuous function on a locally compact space X,T maps M(X) onto C0(W )∗.2.2 Lemma. If f : X → [0,∞] is lsc on a Tychonoff space X, then f = supg : f ≥ g ∈ Cc(X)pointwise.Proof. Given x ∈ X and c < f(x), f > c is a neighborhood of x, so get g : X → [0, 1] withg(x) = 1, g|f≤c = 0, and consider cg.

The key to the proof is the following lemma [13, Lemma 3.3]:2.3 Lemma. If an inner regular positive measure ν on X satisfies

∫f dν ≤ ‖f‖W , f ∈ Cc(X), then

ν = W−1µ for some finite positive measure µ.Proof. From inner regularity and W > a =

⋃f > a : W ≥ f ∈ Cc(X)

it follows easily that∫

W dν = sup∫

f dν : W ≥ f ∈ Cc(X)≤ 1, (2.1)

since f ≤W iff ‖f‖W ≤ 1, i.e. W ∈ L1(ν). Hence µ = Wν is a finite measure.

2.4 Lemma. The topology of locally uniform convergence on Cc(X) is finer than the topology inducedby C0(W ) on Cc(X).Proof. The set U = f ∈ Cc(X) : ‖f‖W < 1 is a neighborhood of 0 in Cc(X), since W is boundedfrom below by some c > 0 on each compact K ⊂ X and U ∩ C(K) ⊃ f ∈ C(K) : ‖f‖∞ < c/2.

Proof (Theorem 2.1). If φ ∈ C0(W )∗, and ‖φ‖ = 1, then φ|Cc(X) is again continuous, so that by theRiesz Theorem for Cc(X), there is a measure ν with φ(f) =

∫f dν, f ∈ Cc(X). Applying Lemma 2.3

to the parts of the Jordan decomposition of ν, we obtain µ ∈ M(X) with ν = W−1µ. Since Cc(X)is dense in C0(W ), φ = Tµ follows by continuity.

2.2 De Branges’ Lemma

As a last preparation for de Branges’ proof, we need a property of extremal annihilating measuresfirst stated in [4] in a non-weighted context (and used there for a short proof of the Stone-WeierstrassTheorem). We use the version [1, Lemma 2.2], adapted to our specific situation:2.5 De Branges’ Lemma. Given an lsc weight function W and a non-dense linear subspace L ofC0(W ), there exists a positive measure µ on R such that

a)∫W dµ <∞ (so that C0(W ) ⊂ L1(µ)), and

b) the annihilator of L in the duality (L1(µ),L∞(µ)) is spanned by a real function g of constantmodulus 1.

Explicitly, b) says that if g ∈ L∞(µ) satisfies∫fg dµ = 0 for all f ∈ L, then g is a multiple of g. We

denote the annihilators with respect to the dualities (C0(W ),M(R)) and (L1(µ),L∞(µ)) by ⊥0 and⊥1, respectively.Proof. The closed convex subset Σ = σ ∈ L⊥0 : σ real, ‖σ‖ ≤ 1 of M(R) is nonempty, since, Lbeing a space of real functions, for any nonzero σ ∈ L⊥0 , Reσ and Imσ are in the annihilator aswell, and one of them must be nonzero.Let σ be an extreme point of Σ, put µ = W−1|σ| and g = d|σ|/dσ ∈ L⊥1 . If L⊥1 is not spanned byg, then there is a nonconstant f ∈ g−1L⊥1 . Let h be the function f + 2‖f‖∞ ≥ 0, normalized to

10

have∫h d|σ| = ‖σ‖. Then ‖h‖∞ > 1 since otherwise∫

|1− h| d|σ| =∫

1− h d|σ| = ‖σ‖ −∫h d|σ| = 0, (2.2)

meaning that h would have to be a constant σ-a.e. Letting t = ‖h‖−1∞ < 1, so that 1 − th ≥ 0, we

obtain convex combinations

1 = th+ (1− t)1− th1− t , σ = t(hσ) + (1− t)

(1− th1− t σ

). (2.3)

As is easily verified from the choice of h and t, this actually exhibits σ as a strict convex combinationof elements of Σ, a contradiction.

3 De Branges’ Proof

We are now in the position to give de Branges’ proof of the necessity of K(L,W ) 6= ∅, in the versionfrom [1].

Assumptions. For the rest of this section, fix a lower semicontinuous weight W , an algebraic deBranges space L ⊂ C0(W ) with L 6= C0(W ), and a measure µ ∈ M(R) with a function g ∈ L∞(µ)satisfying the conclusion of de Branges’ lemma. Define m(z) to be the (everywhere finite andlocally bounded) Hall majorant with respect to the norm of L1(µ). Let ρ0 : H(C)→ C(R) be therestriction operator ρ0F = F |R, and let ρ be the corresponding operator C0(W )→ L1(µ).

Note that the results of sections 1.3 and 1.4 apply to H = ρL and B = L1(µ). In particular, theorems1.17 and 1.18 give3.1 Lemma. For every f ∈ ρL, there is F ∈ H(C) with ρF = f and

|F (z)| ≤ ‖f‖1m(z), z ∈ C. (3.1)

The quotient of any two such functions F and G is of bounded type in C±.

3.1 Consequences of de Branges’ Lemma

Since (ρL)⊥ = spang implies ρL = g⊥, we get the following characterization of ρL:3.2 Lemma. f ∈ ρL iff

∫fg dµ = 0. In particular, ρL has codimension 1 in L1(µ).

We deduce two important properties of the measure µ as proved in [1, p. 891]:3.3 Lemma. µ is discrete, and ρ maps L injectively into L1(µ).Proof. Assume that x0 is an accumulation point of suppµ, and take an interval (a, b) 63 x0 that hasat least two points in common with suppµ.5 It follows that L1(µ|(a,b)) is at least two-dimensional,so that there must be f ∈ ρLr 0 with f |supp µr(a,b) = 0. If F is any entire function with ρF = f ,it follows that ρ0F must vanish on a set of points accumulating at x0.6 The identity theorem givesF = 0, and hence f = 0, a contradiction.Now let F ∈ L with ρF = 0. By discreteness, this means that F (x) = 0 for each x ∈ suppµ. If l isthe multiplicity of some x0 ∈ suppµ as zero of F , G(z) = (z − x0)−lF (z) ∈ L is nonzero at exactlyone point of suppµ, which contradicts

∫Gg dµ = 0.

5Recall that the support supp µ of µ is the intersection of all closed sets whose complement is a µ-null set.6Argue, for example, as follows: For each k, the set Ak = |ρ0F | < 1/k ∩ supp µ has complement a µ-null set, so that

it must be dense in supp µ. Ak is also clearly open, so that by the Baire category theorem applied to the completemetric space supp µ, the intersection

⋂Ak = ρ0F = 0 ∩ supp µ is dense in supp µ.

11

3.2 Analytic Extensions for L1(µ)

From the preceding lemma, it follows that ρ has an inverse ι : ρL → L, which is continuous in thetopology of locally uniform convergence by (3.1). Extend ι to a map ι : ρL → H(C) by continuity,and let L = ι(ρL). We show that the properties of ι and L carry over to ι and L (cf. [1, pp. 888-890]).3.4 Proposition. ρ|L = ι−1

This is nontrivial since ρ is in general discontinuous. However, 1Kρ is continuous for any compactK since locally uniform convergence of Fn implies uniform convergence, and hence L1-convergence ofρFn on K.Proof. Since L = ran ι, it suffices to show ρι = IdρL. Let ρFn → f ∈ ρL in L1(µ). Then ιρFn → ιflocally uniformly, so that

ρ0ιf ← (ρ0ι)ρFn = ρ0(ιρ)Fn = ρ0Fn pointwise on R. (3.2)

In particular, ρFn → ριf µ-a.e. If we pass to an µ-a.e. convergent sequence ρFn → f , it follows thatριf = f µ-a.e.

3.5 Proposition. |ιf(z)| ≤ ‖f‖1m(z), f ∈ ρL.

Proof. We already know that this holds for f ∈ ρL, and since the composition ρL ι−→ H(C) χz−→ C,where χz is the point evaluation at z, is continuous, the statement extends to the closure of ρL.

Next, we consider division of zeros in L. To this end, define the difference quotient operatorRw[F,G](z) = (z − w)−1(F (z)G(w) − G(z)F (w)), F,G ∈ L, and the corresponding operator R1

w =ρRw(ι× ι) on ρL. By Proposition 3.4, Rw = ιRw(ρ× ρ) on L.It is clear that invariance under Rw is equivalent to invariance under RwH(z) = (z − w)−1H(z):Rw[F,G] = Rw(G(w)F −F (w)G) and conversely, if F (w) = 0 and we take G0 with G0(w) = 1, thenRwF = Rw[F,G0].3.6 Proposition. Rw and R1

w are continuous.Proof. The Schwarz lemma (applied to the function F (z)G(w)−G(z)F (w) on the unit disc D(w, 1)around w) together with the trivial estimate

|(z − w)−1F (z)G(w)−G(z)F (w)| ≤ 2‖F‖K‖G‖K , z, w ∈ K, |z − w| > 1 (3.3)

shows that‖Rw[F,G]‖K ≤ ‖F (z)G(w)−G(z)F (w)‖K ≤ 2‖F‖K‖G‖K (3.4)

for any compact K ⊃ D(w, 1). (Here ‖F‖K := supz∈K |F (z)|.)The estimate |(x− w)−1(f(x)ιg(w)− g(x)ιf(w))| ≤ m(w)(|f(x)|+ |g(x)|), |x− w| > 1 shows that

‖1RrD(w,1)R1w[f, g]‖1 ≤ 2m(w)‖f‖1‖g‖1, (3.5)

which gives the boundedness of 1CrD(w,1)R1w. Since 1D(w,1) is continuous, so is 1D(w,1)R1

w.

3.7 Lemma. L is an algebraic de Branges space.Proof. The proof consists in carrying invariance statements from H(C) to L1(µ) and back usingintertwining relations with ι and ρ.For invariance under ], note that ρF ] = ρF , F ∈ L. By definition of ι, it follows that (ιf)] = ιf , f ∈ρL. Since and ] are involutory automorphisms of L1(µ) and H(C), respectively, (ιf)] = ιf , f ∈ ρLfollows by continuity. Since leaves ρL invariant by continuity, this gives invariance under ].

12

The invariance of L under Rw implies that of ρL und R1w by R1

w (ρ|L × ρ|L) = ρ|L Rw, and bycontinuity, we get the invariance of ρL under R1

w. The identity

Rw[L, L] = ιR1w(ρ× ρ)[L, L] = ιR1

w[ρL, ρL] ⊂ ιρL = L (3.6)

completes the proof.

3.3 Representation of L

The construction of a Krein class function will be based on a representation of the elements of Lin terms of the following functions (cf. [1, pp. 892-893]): For t0 ∈ suppµ fixed an t ∈ suppµ r t0,ht ∈ ρL and Ht ∈ L are defined by

ht(x) :=

−((t− t0)g(t0)µ(t0))−1 x = t0,

((t− t0)g(t)µ(t))−1 x = t,

0 x ∈ suppµr t, t0,Ht = ιht (3.7)

3.8 Proposition. The entire function (z − t)Ht(z) does not depend on t ∈ suppµr t0.Proof. To show Ht(z) = z−s

z−tHs(z) = (Id +(t− s)Rt)Hs(z) in H(C), we prove the analogue in L1(µ):

(Id +(t− s)R1t )hs(x) = ht(x), x ∈ suppµ. (3.8)

This is easily shown for x 6= s, and since both functions are in g⊥, they must agree on all of suppµ.3.9 Proposition. Ht = H]

t with real simple zeros exactly at suppµr t, t0.

Proof. Ht = H]t follows from ht = ht. If Ht has a zero w ∈ Cr suppµ, we may define

G(z) = z − tz − w

Ht(z) = (Id +(w − t)Rw)Ht(z). (3.9)

One easily verifies that G is nonzero on exactly one point of suppµ, but this contradicts∫Gg dµ = 0.

Analogously, if Ht has a multiple zero at w ∈ suppµ, then we may add a zero at t in exchange formaking w a simple zero to get the same contradiction.

3.10 Lemma. Each F ∈ L has a representation

F =∑t6=t0

F (t)(t− t0)µ(t)Ht (3.10)

as a locally uniformly convergent series.Proof. It suffices to prove the analogue in L1(µ), with f = ρF in place of F . Clearly, ‖ht‖1 = 2/(t−t0),so fµ ∈ `1(suppµ) gives the convergence of

f =∑t6=t0

f(t)(t− t0)µ(t)ht (3.11)

in L1(µ). For x ∈ suppµr t0, we compute

f(x) = f(x)(x− t0)µ(x)hx(x) = f(x), (3.12)

so that f, f ∈ L agree on suppµr t0, hence on all of suppµ. This gives the representation of f .

13

3.4 Construction of a Krein Class Function

We can now construct a Krein class function inducing the measure µ (cf. [1, p. 894]), namelyB(z) = (z− t)Ht(z). Condition 1) is satisfied by Proposition 3.9. F/B = (z− t)−1F/Ht is in N(C±)by Lemma 3.1, and dominated convergence gives

F (iy)/B(iy) =∑t6=t0

(µ(t)F (t)) t− t0iy − t

→ 0, y →∞. (3.13)

Finally,

B′(x) =

(x− t0)Hx(x) x ∈ suppµr t0

(x− t)Ht(x) x = t0= (g(x)µ(x))−1 (3.14)

so that ∑B(x)=0

∣∣∣∣W (x)B′(x)

∣∣∣∣ =∫W dµ <∞. (3.15)

4 Proof of Sodin and Yuditskii

Finally, we present a second proof of de Branges’ theorem by Sodin and Yuditskii [11, 12] whichrelies on a normal family argument in L in place of an extremal construction in L⊥. Unless weassume W to be finite on a non-discrete set, this approach will only yield a function B satisfyingslight weakenings of conditions 2) and 3) in the definition of Krein class functions.

Assumptions. Again, let W be a lower semicontinuous weight, L an infinite-dimensional non-dense algebraic de Branges subspace of C0(W ). Denote by m(z) the (everywhere finite andcontinuous) Hall majorant of L with respect to C0(W ). For any F : R → C, write ‖F‖ =‖F/W‖∞. Let B be the set of all locally uniform limits of ‖ · ‖-bounded sequences in L, and BRthe real elements of B.

Note that F ∈ B does not necessarily lie in C0(W ), but it is clear that ‖F‖ <∞ and

|F (z)| ≤ ‖F‖m(z), F ∈ B, (4.1)

so that any ‖ · ‖-bounded subset of B is a normal family.We also need an analogue of Theorem 1.18 for B:4.1 Lemma. For F,G ∈ B, the quotient F/G is a meromorphic function of bounded type in C±.Proof. Take real Fn ∈ L converging locally uniformly to F , and, as in the proof of Theorem 1.18,assume G ∈ L with (RzG,φ) 6= 0 for some z ∈ Cr R and φ ∈ L⊥. Again, we may write

Fn(z)/G(z) = Fn(z)/G(z), Fn(z) = (RzFn, φ), G(z) = (RzG,φ). (4.2)

If we represent φ as W−1µ with µ ∈ M(R) as in Theorem 2.1, we may use dominated convergenceto get a corresponding representation of F/G as quotient of F (z) = (RzF, φ) and G. But F and Gare in N(C±) as the Cauchy integrals of FW−1 dµ and GW−1 dµ by Lemma A.6.

B is still an algebraic de Branges space:4.2 Lemma. If F ∈ B and F (w) = 0, then G(z) = F (z)/(z − w) ∈ B.Proof. Assume that F 6≡ 0, so that for Fn ∈ L converging locally uniformly to F , the Hurwitztheorem implies that there is a sequence wn of zeros of Fn converging to w. The entire functionsGn(z) = Fn(z)/(z−wn) clearly converge to G locally uniformly on CrD(w, 1), hence on all of C bythe maximum principle.

14

Now assume W ≥ c > 0 in D(w, 4). Then by the Schwarz lemma,

|Gn(x)/W (x)| ≤ ‖Gn‖∂D(wn,3)/c ≤ ‖Fn‖∂D(wn,3)/c ≤ ‖F‖‖m‖D(wn,3)/c, x ∈ D(wn, 3), (4.3)

and clearly |Gn| ≤ |Fn| outside D(wn, 1). For n so large that D(wn, 1) ⊂ D(w, 2) ⊂ D(wn, 3) ⊂D(w, 4), it follows that ‖Gn‖ ≤ c−1‖m‖D(w,4)‖Fn‖, so that the sequence Gn is also bounded.

4.3 Remark. While not relevant in the sequel, it is worth noting that any F ∈ B∩C0(W ) is actually inL. To see this take real Fn ∈ L with ‖Fn‖ ≤ 1 and Fn → F locally uniformly. From ‖·‖-boundedness,Theorem 2.1 and F ∈ C0(W ), it follows that Fn converge to F weakly in C0(W ). Komlos’ lemma[9, Thm. 3.13] gives us a sequence Gn of convex combinations of the Fn that converges strongly inC0(W ), so that indeed F ∈ L.

4.1 Chebyshev Sets and ω-Extremality

We shall consider the following notion from [12] of extremality among functions in BR:4.4 Definition. For ω ∈ C±, F ∈ BR is called ω-extremal if F (ω) 6= 0 and any function G ∈ BR withG(ω) = F (ω) has ‖G‖ ≥ ‖F‖.

It is clear that ω-extremal functions exist: It suffices to take F0 ∈ BR with F0(ω) 6= 0, and Fn ∈ BRwith Fn(ω) = F0(ω) and ‖Fn‖ → r = inf‖F‖ : F ∈ BR, F (ω) = F0(ω), and get a locally uniformlyconvergent subsequence Fn → F ∈ BR. Then F (ω) = F0(ω) and

|F (x)| ← |Fn(x)| ≤ ‖Fn‖W (x)→ rW (x), x ∈ R, (4.4)

so that ‖F‖ = r (while we do not have ‖F − Fn‖ → 0).The following criterion [12, Thm. 1] for ω-extremality in terms of Chebyshev sets will show that ω-extremality does not depend on ω.4.5 Definition. For F ∈ BR, a discrete set Λ ⊂ R is called a Chebyshev set of F if F (λ) =±‖F‖W (λ), λ ∈ Λ with alternating signs. Λ is called maximal if it is not contained in any strictlylarger Chebyshev set.We say that two discrete subsets Γ,Λ of R interlace if for γ1 < γ2 ∈ Γ, (γ1, γ2) ∩ Λ 6= ∅, and forλ1 < λ2 ∈ Λ, (λ1, λ2) ∩ Γ 6= ∅.4.6 Definition. For a region Ω ⊂ C, R(Ω) = F ∈ H(Ω) : ImF ≥ 0.

Note that by the open mapping theorem, F ∈ R(C+) is either identically zero or ImF > 0 in C+.For F ∈ H(C), let Z(F ) denote the set of zeros of F .4.7 Theorem. A ∈ BR is ω-extremal for some ω ∈ C+ iff there is B ∈ H(C) real such that

a) Z(B) is a maximal Chebyshev set of A, andb) for any G ∈ BR with ‖G‖ ≤ ‖A‖, (A−G)/B ∈ R(C+).

Sufficiency is easy to show: If G(ω) = A(ω) and ‖G‖ ≤ ‖A‖, then (A−G)/B ∈ R(C+) vanishes atω, hence in all of C. This argument also establishes uniqueness of ω-extremal functions as soon asthe theorem is proved, for which we will need a lemma [12, p. 7]:4.8 Lemma. If A ∈ BR is ω-extremal and G ∈ BR with ‖G‖ < ‖A‖, then A−G has real simple zerosand Z(A−G) interlaces with any maximal Chebyshev set of A.Proof. We use the Markov corrections

Aδ =(

1− δQP

)A+ δ

Q

PG = A+ δ

Q

P(G−A) ∈ BR, Aδ(ω) = A(ω), (4.5)

15

where δ > 0, Q(z) = (z − ω)(z − ω), and P (z) = (z − α)(z − β) with α, β ∈ Z(A − G) either bothreal or conjugate. From the definition of ‖ · ‖W and an easy pointwise estimate (taking into account‖G‖ ≤ ‖A‖!), it follows that for Ω ⊂ R,

‖1ΩAδ‖ ≤ (1− t0)‖1ΩA‖+ t0‖1ΩG‖, t0 = infδ|Q(x)/P (x)| : x ∈ Ω, |Q/P | ≤ δ−1 on Ω. (4.6)

1) If α is any nonreal zero of G− A and β = α, |P/Q| is bounded above and below on R, so thatfor δ < ‖P/Q‖−1

∞ , we get ‖Aδ‖ ≤ (1− t0)‖A‖+ t0‖G‖ with t0 > 0, a contradiction.2) Now assume α is a multiple real zero, and let β = α. Let ‖G‖ < ‖A‖−3ε, and take a neigborhood

U of α such that |A−G| < εW on U . Since Q/P is bounded outside U , (4.6) will clearly holdfor Ω = R r U and all large δ. While Q/P is not bounded on U , (G − A)/P is, so that forδ < ε‖Q(G−A)/P‖−1

W ,

|Aδ| ≤ |A|+ δ|Q(G−A)/P | ≤ |G|+ 2εW on U. (4.7)

Together, these estimates give the same contradiction.3) From ‖G‖ < ‖A‖, it is clear that A − G changes sign and hence has a zero between any two

consecutive points of a Chebyshev set of A. Conversely, assume α < β are consecutive zeros and‖1(α,β)A‖ < ‖A‖ − ε. Since A(α) = G(α), A(β) = G(β) and ‖G‖ < ‖A‖, we may even assume‖1UA‖ < ‖A‖ − 2ε for a neighborhood U of [α, β]. Again Q/P is bounded outside U . Inside U ,choose δ < ε‖A‖‖(G−A)/P‖−1

W , so that

|Aδ| ≤ |A|+ δ|(G−A)/P | < (1− 2ε)‖A‖W + ε‖A‖W = (1− ε)‖A‖W, (4.8)

again giving a contradiction.

4.9 Lemma. A has infinitely many zeros.Proof. If A has N zeros, then any Chebyshev set Λ of A has at most N + 1 points, and from thepreceding lemma it follows that for G ∈ L, the A−G has at most N + 2 zeros. It follows by LemmaA.3 that (A−G)/A ∈ N(C±) is actually a rational function, with the degree of both numerator anddenominator bounded independently of G. Consequently, L must be finite-dimensional, contradictingour assumptions.

4.2 Properties of R(C+)

4.10 Lemma. If Γ,Λ ⊂ R interlace, then there is a function Π meromorphic in C with simple zerosat Λ and simple poles at Γ such that Im Π 6= 0 in C±.Proof (cf. [6, p. 308]). If Λ = (λj),Γ = (γj) with γj < λj < γj+1, we define

Π(z) =∏ γj

λj· z − λj

z − γj, z ∈ Cr Γ. (4.9)

Since λj , γj interlace, the alternating series∑

1/λj − 1/γj converges, which gives locally uniformconvergence of

∑∣∣∣∣∣1− γj

λj

z − λj

z − γj

∣∣∣∣∣ =∑∣∣∣∣∣1− z/γj

1− z/γj− 1− z/λj

1− z/γj

∣∣∣∣∣ = |z|∑ 1|1− z/γj |

(1λj− 1γj

). (4.10)

The locally uniform convergence of Π follows by a standard argument [10, Thm. 15.5].We compute

arg 1− z/λj

1− z/γj= arg(1− z/λj)− arg(1− z/γj) = arg(z − λj)− arg(z − γj) ≥ 0 (4.11)

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to get

0 ≤ arg Π(z) =∑

arg(z − λj)− arg(z − γj) ≤∑

arg(z − λj)− arg(z − λj+1) = π. (4.12)

Proof (Theorem 4.7). For Λ1 a maximal Chebyshev set for A, Γ1 = Z(A) and Π1 as in Lemma4.10, define B = AΠ1, so that a) is clear. For b), let ‖G‖ < ‖F‖ and put Φ = (A − G)/B. SinceΦ = (A/B)(1−G/A) with A/B,G/A ∈ N(C±) (the first function being Π−1

1 , the second a quotient offunctions in B), we have Φ ∈ N(C±). If Π2 has zeros Λ2 = Z(B) = Λ1 and poles Γ2 = Z(A−G), thenΦΠ2 ∈ N(C±). Since ΦΠ2 is zero-free, it follows from Lemma A.8 that Φ = Π−1

2 , so that Im Φ 6= 0in C±.

We cite a similar result on R(C+) without proof:4.11 Čebotarev Theorem ([6, Thm. VII.2]). If Φ is real meromorphic in C with simple poles in Rand Φ ∈ R(C+), then Φ has a representation

Φ(z) = az + b+∑

bk

( 1z − ak

− 1ak

), with

∑ bk

a2k

<∞, (4.13)

where ak are the poles of Φ, with residues bk.4.12 Corollary. Φ(iy) = o(y2).Proof. The above representation gives

|Φ(iy)| ≤ a|y|+ b∑

bk|y|

|iy − ak||ak|≤ a|y|+ b+ |y|

∑ bk

a2k

= O(y) = o(y2). (4.14)

4.13 Lemma. For Ω ⊂ C, R(Ω) ⊂ N(Ω).Proof. Using the standard isomorphims ϕ : D → C+ : z → i1+z

1−z and ψ = ϕ−1, we can representf ∈ R(Ω) as

f = φ ψ f = i1 + ψ f1− ψ f , (4.15)

which is clearly the quotient of bounded functions.

4.3 Construction of a Krein Class Function

Let A ∈ BR be ω-extremal for some ω ∈ C+ (hence also for ω by symmetry). As above, let Γ = Z(A)and Λ a maximal Chebyshev set for A, take Π ∈ R(C+) with poles Γ and zeros Λ, and let B = AΠ.B is real with real simple zeros by construction, and A/B = Π−1 ∈ R(C±), which gives A/B ∈ N(C±)by the preceding section. For real F ∈ L and G = ‖A‖F/‖F‖, (A − G)/B ∈ R(C+) by Theorem4.7 b), so that F/B ∈ N(C+) and yF (iy)/B(iy) = o(y3) follow from the corresponding properties ofA/B and the results of the preceding section.By Theorem 4.11 (with Φ = A/B, ak ∈ Λ and bk = Res Φ(ak) = A(ak)/B′(ak)),

‖A‖∑λ∈Λ

W (λ)(1 + λ2)|B′(λ)| =

∑λ∈Λ

|A(λ)|(1 + λ2)|B′(λ)| <∞. (4.16)

So far, we have obtained a function B satisfying conditions 2) and 3) in the definition of a Krein classfunction only up to extra factors y3, λ2. If we assume W <∞ to be non-discrete, then there mustbe infinitely many points x ∈ R with B(x) 6= 0 and W (x) <∞. For three such points x1, x2, x3, thefunction (z − x1)(z − x2)(z − x3)B(z) is actually in the Krein class.

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A Some Results on Entire Functions

We state a few basic facts about entire functions of exponential and bounded type. All results withouta reference are taken from [8, Ch. 6].A.1 Definition. The exponential type of F ∈ H(C) is τ(F ) = lim supz→∞ log |F (z)|/|z|. F is said tobe of exponential type if τ(F ) <∞.

Equivalently, τ(F ) is the infimum of all t > 0 such that F (z)/et|z| is bounded.A.2 Example. Polynomials have exponential type 0; etz has exponential type t.A.3 Lemma. Every zero-free entire function of exponential type is of the form cetz for some t > 0.Proof (cf. [5, pp. 18-19]). If F is zero-free, then F = expϕ with ϕ entire, and F having exponentialtype gives an estimate Reϕ(z) ≤ a|z|+ b. If ϕ(z) =

∑γnz

n, then

12π

∫ 2π

0Reϕ(Reiθ)e−inθ dθ =

∞∑k=0

Rkγk1

2π

∫ 2π

0eikθe−inθ dθ −Rkγk

12π

∫ 2π

0e−ikθe−inθ dθ (A.1)

= Rnγn, n > 0 (A.2)

so that |γn| ≤ R−n(aR+ b)→ 0 as R→∞ for n > 1. It follows that ϕ is linear.

A.4 Definition. F ∈ H(Ω) is of bounded type in Ω (write f ∈ N(Ω)) if log+ |F | has a harmonicmajorant in Ω, i.e. there is a harmonic function h in Ω such that log+ |F | ≤ h.A.5 Theorem. F ∈ N(Ω) iff F = f/g with f, g ∈ H∞(Ω) and g(z) 6= 0 in Ω.A.6 Lemma. For any complex measure µ with

∫(1 + |x|)−1 d|µ| <∞, the Cauchy integral

F (z) =∫dµ(x)x− z

, z ∈ Cr R (A.3)

defines an analytic function F ∈ N(C±).Proof. By the Jordan decomposition, we may assume µ to be positive. But then F ∈ R(C+), so thatthe claim follows from Lemma 4.13.

A.7 Krein’s Theorem. F ∈ H(C) is in N(C±) iff F is of exponential type and has finite logarithmicintegral ∫ log |F (x)|

1 + x2 dx <∞. (A.4)

In this case,τ(F ) = maxτ+, τ−, τ± = lim sup

y→±∞|y|−1 log |F (iy)|. (A.5)

A.8 Lemma. Every zero-free entire function in N(C±) is constant.Proof. By Lemma A.3 and Krein’s theorem, we only have to consider functions of the form F (z) = etz.But t > 0 is impossible again by Krein’s theorem, since then the logarithmic integral of F (z) isdivergent.

A.9 Phragmén-Lindelöf Principle. Let F be analytic in the angle A = Aα = | arg z| < π/2α, α ≥ 1such that

1) lim supz→a |F (z)| ≤M,a ∈ ∂A, and2) |F (z)| ≤ C exp(|z|β), z ∈ A for some β < α.

Then F is bounded by M on A.

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Proof. By slightly increasing β < α, we may assume |F (z)|/ exp(η|z|β)→ 0, |z| → ∞ for all η > 0. Onthe simply connected region A, we can choose a branch of G(z) = exp(ηzβ). For z = x+iy = reiθ ∈ A,we have

|G(z)| = exp(ηRe zβ) = exp(ηrβ cos(βθ)), βθ < π/2, (A.6)

so that exp(ηrβ)/|G(z)| is bounded, and |G(z)| ≥ 1 for arg z = π/2α. It follows that H(z) =F (z)/G(z) is bounded by M on ∂A and tends to 0 as z →∞. Taking R large enough for |H(Reiθ)| <M, |θ| < π/2α, the maximum principle gives

M ≥ |H(z)| = exp(ηRe zβ)|F (z)|, |z| < R. (A.7)

Since exp(ηRe zβ)→ 1 as η → 0, this proves the claim.

The following lemma (extracted from [2, Thm. 1.4.3, 6.2.4]) is necessary to complete the interpolationargument in the simple part of the proof of de Branges’ theorem.A.10 Lemma. If an entire function F is of bounded type in C± and tends to 0 along the imaginaryaxis, then F vanishes identically.Proof. First note that by (A.5), the assumption implies that F has exponential type 0, so that fort > 0 fixed, G(z) = e−tzF (z) is a function of exponential type bounded on each line arg z = θ, |θ| <π/2, as well as on the imaginary axis by assumption. Therefore, the Phragmén-Lindelöf principle isapplicable on the angles Aα, α < 1 as well as the first and fourth quadrant, which gives

‖G‖Aα ≤ ‖G‖∂Aα ≤ max‖G‖iR, ‖G‖R+, α < 1. (A.8)

Letting α 1, we see that G is actually bounded on the half plane Re z > 0, so that anotherapplication of the Phragmén-Lindelöf principle gives

|F (z)| ≤ et Re z‖G‖iR = ‖F‖iR, Re z > 0 (A.9)

Letting t 0, it follows that F is bounded by ‖F‖iR on Re z > 0, and the same works for Re z < 0,so that F is actually bounded in C, hence constant.

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References

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McGraw-Hill, 1987.[11] Sodin, M. and Yuditskii, P. Another Approach to the de Branges’ Theorem on Weighted Poly-

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