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2 2 Preliminaries

For t> 0, let (ϕ) denote the reproducing kernel Hilbert space with kernel Ht 1 ϕ(λ)ϕ(z) tkϕ(z,λ)= t − (ϕ ), 1 λz ∈ S − and we will use notations tkϕ(z)= tkϕ(z,λ) and (ϕ)= (ϕ). Then, since λ H H1 ϕ ϕ ϕ 1 ϕ ϕ tkλ , tkz t(ϕ) = tk (z,λ)= t− tkλ , tkz (ϕ), h iH h iH the trivial linear mapping f f from (ϕ) onto (ϕ) is bounded and invertible. Par- 7→ H Ht ticularly, t(ϕ) = (ϕ) as vector spaces. In this section, we construct the exponential of (ϕ)H and giveH its basic properties. The contents of this section are well known to Ht specialists. For example, see Exercise (k) in p. 320 of Nikolski [11] and Chapter 7 in Paulsen-Raghupathi [13]. However, we give the details for the sake of readers.

2.1 Construction of exp Ht(ϕ) Let (ϕ)n be the reproducing kernel Hilbert space obtained by the pull-back construction Ht with the n-fold tensor product space

n (ϕ)⊗ = (ϕ) (ϕ) Ht Ht ⊗···⊗Ht and the n-dimensional diagonal map

∆ : D Dn, λ (λ,...,λ) n → → ϕ n (for the pull-back construction, see Theorem 5.7 in [13]). We note that (tkλ )⊗ ∆n = ϕ n n n ◦ (tkλ ) is the reproducing kernel of t(ϕ) . Let n∞=0 t(ϕ) denote the Hilbert space with the inner product H ⊕ H

∞ 1 ∞ n n (f0, f1,...)⊤, (g0,g1,...)⊤ t(ϕ) = fn,gn t(ϕ) , h i⊕n=0H n!h iH n=0 X where we set (ϕ)0 = C. Moreover, we define linear map Γ as follows: Ht

f0 f0 ∞ 1 n Γ: f1 fn f1 ∞ t(ϕ) .  .  7→ n!  .  ∈ ⊕n=0H  . n=0 .   X         Proposition 2.1. The following statements hold:

n (i) Γ is a map from ∞ (ϕ) to Hol(D). ⊕n=0Ht (ii) ker Γ is closed.

3 n Proof. For any F =(f , f ,...)⊤ in ∞ (ϕ) , we have 0 1 ⊕n=0Ht m 1 m 1 f (λ) f (λ) ℓ! ℓ ≤ ℓ! ℓ ℓ=n+1 ℓ=n+1 X X m 1 ϕ ℓ ℓ ℓ fℓ t(ϕ) (tkλ ) t(ϕ) ≤ ℓ!k kH k kH ℓ=Xn+1 m 1/2 m 1/2 1 2 1 ϕ ℓ 2 ℓ ℓ fℓ t(ϕ) (tkλ ) t(ϕ) ≤ ℓ!k kH ! ℓ!k kH ! ℓ=Xn+1 ℓ=Xn+1 m 1/2 m 1/2 1 2 1 ϕ 2ℓ ℓ = fk t(ϕ) tkλ t(ϕ) . ℓ!k kH ! ℓ!k kH ! ℓ=Xn+1 ℓ=Xn+1 Hence, ∞ 1 f (λ) n! n n=0 X converges uniformly on any compact subset of D. This concludes (1). Next, suppose that (ℓ) (ℓ) n F = (f , f ,...)⊤ belongs to ker Γ and F converges to F in ∞ (ϕ) . Then, for ℓ 0 1 ℓ ⊕n=0Ht sufficiently large L, we have

∞ 1 (ℓ) 2 2 n ∞ n fn fn t(ϕ) = Fℓ F t(ϕ) < 1 (ℓ L). n!k − kH k − k⊕n=0H ≥ n=0 X (ℓ) n Hence we have fn fn t(ϕ) < √n! for every n 0 if ℓ L. It follows from this H inequality that,k for any− ℓ k L, ≥ ≥ ≥ 1 (ℓ) 1 (ℓ) ϕ n n n fn (λ) fn t(ϕ) (tkλ ) t(ϕ) n! ≤ n!k kH k kH

ϕ n n √ ( fn t(ϕ) + n!) tkλ t(ϕ) k kH k kH ≤ n! ϕ n (M + 1) tkλ t(ϕ) k kH , ≤ √n! 2 1/2 where we set M = supn 0( fn t(ϕ)/n!) . Then, by the ratio test, ≥ k kH ϕ n ∞ (M + 1) tkλ t(ϕ) k kH √ n=0 n! X is finite. Hence, by the Lebesgue dominated convergence theorem and the assumption that (ℓ) (ℓ) Fℓ =(f0 , f1 ,...)⊤ belongs to ker Γ, we have

∞ ∞ ∞ 1 1 (ℓ) 1 (ℓ) fn(λ)= lim fn (λ) = lim fn (λ)=0. n! ℓ n! ℓ n! n=0 n=0 →∞ →∞ n=0 X X X 4 This concludes (2). By Proposition 2.1, the pull-back construction can be applied to Γ.

Definition. We define exp (ϕ) as the reproducing kernel Hilbert space obtained by the Ht pull-back construction with the linear map

n Γ: ∞ (ϕ) Hol(D). ⊕n=0Ht →

2.2 Basic properties of exp Ht(ϕ) We summarize basic properties of exp (ϕ). Ht Proposition 2.2. exp (ϕ) is a reproducing kernel Hilbert space consisting of holo- Ht morphic functions on D. More precisely, for any f in exp t(ϕ), there exists a vector n H (f , f ,..., )⊤ in ∞ (ϕ) such that 0 1 ⊕n=0Ht ∞ 1 f = f n! n n=0 X converges uniformly on any compact subset of D. Moreover,

(i) the following estimate holds:

∞ 2 1 2 n f exp t(ϕ) fn t(ϕ) , k k H ≤ n!k kH n=0 X (ii) the reproducing kernel of exp (ϕ) is Ht ∞ 1 (tkϕ)n = exp tkϕ, n! λ λ n=0 X that is, ϕ f(λ)= f, exp tkλ exp t(ϕ) h i H for any λ in D,

(iii) the following growth condition holds:

2 2 2 1 ϕ(λ) f(λ) f exp t(ϕ) exp t −| | | | ≤k k H 1 λ 2  −| |  for any λ in D.

5 Proof. By the definition of the norm and the inner product of exp t(ϕ), we have (1) and (2). We shall show (3). By (2) and the Cauchy-Schwarz inequality,Hwe have

2 ϕ 2 f(λ) = f, exp tkλ exp t(ϕ) H | | |h 2 i | ϕ 2 f exp t(ϕ) exp tkλ exp t(ϕ) ≤k k H ·k k H 2 ϕ = f exp t(ϕ) exp tkλ (λ) k k H · 2 2 1 ϕ(λ) = f exp t(ϕ) exp t −| | . k k H 1 λ 2  −| |  Thus we have (3).

3 Unbounded multipliers

We shall investigate into unbounded multipliers of exp (ϕ). Ht Lemma 3.1. Let ψ be a function in (ϕ). Then, for any function f in (ϕ)n, ψf Ht Ht belongs to (ϕ)n+1. Ht Proof. We define bounded linear operator τψ as follows: n n+1 τ : (ϕ)⊗ (ϕ)⊗ , F ψ F. ψ Ht → Ht 7→ ⊗ Then, the following diagram commutes:

n τψ n+1 (ϕ)⊗ (ϕ)⊗ Ht −−−→ Ht

∆n ∆n+1

 n  n+1 t(ϕ) t(ϕ) , H −−−−−−→Mψ H (ϕ)n H y | t y

where Mψ denotes the multiplication operator with symbol ψ. This concludes the proof.

Theorem 3.2. Let ψ be a function in t(ϕ). Then, the multiplication operator Mψ is a densely defined closable linear operator inH exp (ϕ). Ht n Proof. Let F = (f , f ,...,f , 0 ...)⊤ be a vector having finite support in ∞ (ϕ) . 0 1 N ⊕n=0Ht We set ΓF = f. Then,

N 1 N 1 N 1 N+1 1 ψf = ψ fn = ψfn = (n + 1)ψfn = nψfn 1, n! n! (n + 1)! n! − n=0 n=0 n=0 n=1 X X X X n where we note that nψfn 1 belongs to t(ϕ) by Lemma 3.1. Hence, setting − H G = (0, ψf0, 2ψf1,..., (N + 1)ψfN , 0,...)⊤,

n G belongs to n∞=0 t(ϕ) and ΓG = ψf, that is, ψf belongs to exp t(ϕ). Therefore, Mψ is a densely defined⊕ H linear operator in exp (ϕ). Moreover, it is easyH to see that M is Ht ψ closable.

6 Corollary 3.3. Let ψ be a function in t(ϕ). Then the adjoint operator Mψ∗ of Mψ is a densely defined closed linear operator inHexp (ϕ), and every exp tkϕ is an eigenfunction Ht λ of Mψ∗. More precisely, ϕ ϕ Mψ∗ exp tkλ = ψ(λ) exp tkλ .

4 Main results

Let X be a set. A function k on X X is called a strictly positive definite kernel on X if k(x, y)= k(y, x) for any x and y in ×X and

n

cicjk(xj, xi) > 0 i,j=1 X n for any n in N, any (c1,...,cn)⊤ in C 0 and any n distinct points x1,...,xn in X. For example, it is well known that \{ } k(z,λ) = exp(λz) is a strictly positive definite kernel on C. In fact, this is the reproducing kernel of the Segal-Bargmann space. Now, we note that if ϕ = z2 then

1 1 ϕ(λ)ϕ(z) e− exp − = exp(λz). 1 λz ! − Motivated by this observation, we shall give new examples of strictly positive definite kernels. Let ϕ be a function in . For the canonical factorization ϕ = αzN BSF of ϕ (see Section 5 in Chapter II in GarnettS [7]), where α is a unimodular constant, B is a Blaschke product consisting of nonzero zero points, S is a singular inner function and F is an outer function, we consider three conditions (C1) N 2, (C2) B is nontrivial and (C3) S is nontrivial. We need the following lemma. ≥

Lemma 4.1. Let λ1,...,λn be n distinct points in D. Suppose one of (C1), (C2) and (C3). Then there exists a function ψ in (ϕ) such that ψ(λ ) = ψ(λ ) (i = j). Ht i 6 j 6

Proof. Since t(ϕ) = (ϕ) as vector spaces, it suffices to show the statement for (ϕ). First, we assumeH (C1).H Then, since ϕ/z is in by the Schwarz lemma and (ϕ/z)(0)H = 0, S we have (I T T ∗)z = z T T ∗ T ∗z = z T T ∗ 1= z, − ϕ ϕ − ϕ ϕ/z z − ϕ ϕ/z 2 where Tϕ denotes the Toeplitz operator with symbol ϕ on the H over D. Hence z belongs to (ϕ), and we may take ψ = z. H Secondly, we assume (C2). Let µ be a nonzero zero point of ϕ. Then, we have

ϕ k =(I T T ∗)k = k . µ − ϕ ϕ µ µ 7 1 1 Hence (1 µz)− belongs to (ϕ), and we may take ψ = (1 µz)− . Thirdly,− we assume (C3).H Then, observe that there exists− a point θ [0, 2π) such that √ 1θ ∈ ϕ(re − ) 0 as r 1 (see Theorem 6.2 in [7]). Without loss of generality, we may assume that θ = 0.→ Setting ↑

ϕ ϕ δij(r)= kr (λi) kr (λj) and δ(r) = min δij(r) (0

it suffices to show that δ(r) > 0 for any r sufficiently close to 1. For any distinct λi and 1 λ , we suppose that, for any m in N, there exists a real number r (1 m− , 1) such j m ∈ − that δij(rm) = 0. Then, we have

1 1 1 ϕ(rm)ϕ(λi) 1 ϕ(rm)ϕ(λj) = lim − − = lim δij(rm)=0. 1 λi − 1 λj m 1 rmλi − 1 rmλj m − − →∞ − − →∞

This concludes that λ = λ , however, which contradicts the assumption λ = λ . Hence i j i 6 j there exists rij (0, 1) such that δij(r) > 0 for any r (rij, 1). Setting R = max1 i∈0 for any r (R, 1). Then, we may take∈ ψ = kϕ. ∈ r Theorem 4.2. Let ϕ be a function in . If ϕ satisfies one of (C1), (C2) and (C3), then the kernel S 1 ϕ(λ)ϕ(z) kt(z,λ) = exp t − (t> 0) 1 λz ! − is strictly positive definite. Proof. It suffices to show that exp tkϕ n is linearly independent for any n in N and any { λj }j=1 n distinct points λ1,...,λn in D. Suppose that n ϕ cj exp tkλj =0 j=1 X for some n in N, some n distinct points λ1,...,λn in D, and some c1,...,cn in C. Then, for any function ψ in (ϕ), by Corollary 3.3 and the assumption, we have Ht n ϕ ϕ 1 1 cj exp tkλ c1 exp tkλ j=1 j ··· ϕ1 n ϕ ψ(λ1) ψ(λn) c2 exp tk j=1 ψ(λj)cj exp tkλ  ···   λ2   P j  . . . . = . . . . . P .  n 1 n 1  ϕ   n 1  − − c exp tk  n − ϕ  ψ(λ1) ψ(λn)   n λn  cjψ(λj) exp tk  ···     j=1 λj       n ϕ   c exp tk  P j=1 j λj n ϕ Mψ∗ cj exp tkλ =  P j=1 j  . P .  n 1 n ϕ  (M ∗) − cj exp tk   ψ j=1 λj    = 0. P

8 Further, by Lemma 4.1, there exists a function ψ in (ϕ) such that Ht (ψ(λ ) ψ(λ )) =0. i − j 6 1 i

1 1 ··· ψ(λ1) ψ(λn)  . ···. .  . . .  n 1 n 1 ψ(λ ) − ψ(λ ) −   1 n   ···  is nonsingular. Therefore, we have that

ϕ c1 exp tkλ ϕ1 c2 exp tkλ  . 2  = 0. .  ϕ  cn exp tk   λn    This concludes that c = = c = 0. 1 ··· n The well-known fact mentioned at the beginning of this section is included in Theorem 4.2.

Corollary 4.3. The kernel function

k(z,λ) = exp(λz)

is strictly positive definite on C.

Proof. For any n distinct points λ1,...λn in C, we set R = max1 j n λj + 1. Then ≤ ≤ 2 | | 2 λ1/R,...λn/R are in D. Hence, by Theorem 4.2 in the case where ϕ = z and t = R , we have

n n R2 2 cicj exp(λiλj)= e− cicj exp(R + λiλj) i,j=1 i,j=1 X Xn R2 2 = e− cicj exp(R (1 + (λi/R)(λj/R))) i,j=1 X n R2 2 1 ϕ(λi/R)ϕ(λj/R) = e− cicj exp R − > 0 i,j=1 1 (λi/R)(λj/R) ! X − n for any (c ,...,c )⊤ in C 0 . 1 n \{ }

9 Although the next result is just a simple consequence of Theorem 4.2, from the view- point of the theory of model spaces (see Garcia-Mashreghi-Ross [8]), it will be worth while mentioning it as a theorem. Theorem 4.4. Let ϕ be an inner function. If ϕ is neither a constant nor eiθz, then the kernel 1 ϕ(λ)ϕ(z) kt(z,λ) = exp t − (t> 0) 1 λz ! − is strictly positive definite. Further, with help of the theory of sub-Hardy Hilbert spaces, we have Theorem 4.5. Let ϕ be a function in . If ϕ is a nonextreme point of the closed unit ball S in H∞, then the kernel

1 ϕ(λ)ϕ(z) kt(z,λ) = exp t − (t> 0) 1 λz − ! is strictly positive definite. Proof. By Theorem (IV-3) in [15] (or Theorem 23.13 in [6]), the are dense in (ϕ). Hence z belongs to t(ϕ). Then, setting ψ = z, the proof of Theorem 4.2 applies Hto this case. H

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