                                                A CANONICAL SYSTEM OF DIFFERENTIAL EQUATIONS
                                                   ARISING FROM THE RIEMANN ZETA-FUNCTION

                                                                                  MASATOSHI SUZUKI


                                                  Abstract. This paper has two main results, which relate to a criteria for the Riemann
                                                  hypothesis via the family of functions Θω (z) = ξ( 21 −ω−iz)/ξ( 12 +ω−iz), where ω > 0 is
                                                  a real parameter and ξ(s) is the Riemann xi-function. The first main result is necessary
                                                  and sufficient conditions for Θω to be a meromorphic inner function in the upper half-
                                                  plane. It is related to the Riemann hypothesis directly whether Θω is a meromorphic




arXiv:1204.1827v2 [math.NT] 23 Sep 2016
                                                  inner function. In comparison with this, a relation of the Riemann hypothesis and
                                                  the second main result is indirect. It relates to the theory of de Branges, which
                                                  associates a meromorphic inner function and a canonical system of linear differential
                                                  equations (in the sense of de Branges). As the second main result, the canonical system
                                                  associated with Θω is constructed explicitly and unconditionally under the restriction
                                                  of the parameter ω > 1 by applying a method of J.-F. Burnol in his recent work on
                                                  the gamma function to the Riemann xi-function. If such construction is extended to
                                                  all ω > 0 unconditionally, we get a criterion for the Riemann hypothesis in terms of
                                                  a family of canonical systems parametrized by ω > 0, which explains the validity of
                                                  the Riemann hypothesis as positive semidefiniteness of the corresponding family of
                                                  Hamiltonian matrices.




                                                                                   1. Introduction
                                             Let ζ(s) be the Riemann zeta function. The set of all non-trivial zeros of the Riemann
                                          zeta function coincides with the set of all zeros of the Riemann xi-function
                                                                               1                 s
                                                                        ξ(s) = s(s − 1) π −s/2 Γ      ζ(s).
                                                                               2                   2
                                          The Riemann hypothesis, which is often abbreviated to RH, assert that all zeros of ξ(s)
                                          lie on the critical line ℜ(s) = 1/2. We attempt to understand the nontrivial zeros of the
                                          Riemann zeta function via the family of functions
                                                             1                                     i
                                                  Aω (z) := (ξ(s + ω) + ξ(s − ω)), B ω (z) := (ξ(s + ω) − ξ(s − ω)),             (1.1)
                                                             2                                     2
                                          where s = 1/2 − iz and ω is a positive real parameter. Functions Aω (z) and B ω (z) take
                                          real values on the real line and satisfy the functional equations Aω (z) = Aω (−z) and
                                          B ω (z) = −B ω (−z) by the functional equations ξ(s) = ξ(1 − s) and ξ(s) = ξ(s̄).
                                             If all zeros of Aω (z) lie on the real line for every ω > 0, it implies RH by Hurwitz’s
                                          theorem in complex analysis. Conversely, all zeros of Aω (z) lie on the real line for
                                          ω ⩾ 1/2 unconditionally and for 0 < ω < 1/2 under RH by a result of Lagarias [15]
                                          (see also Li [21] for an unconditional result for 0 < ω < 1/2). We abbreviate to RH(Aω )
                                          (resp. RH(B ω )) the assertion that all zeros of Aω (z) (resp. B ω (z)) lie on the real line,
                                          and abbreviate RH(Aω ) and RH(B ω ) as RH(Aω , B ω ) . Then the above things are stated
                                          as follows:
                                          Proposition 1.1 RH holds if and only if RH(Aω ) holds for all ω > 0.
                                          The latter condition is easier to study in that it is currently known to hold for all ω ⩾ 1/2.
                                          Also it is known to be related to some operators. We will study the latter problem of
                                          finding linear differential equation systems with boundary conditions for which the zeros
                                          of Aω (z) are eigenvalues, for a suitable range of ω.
                                            Date: Version of September 13, 2018.
                                                                                             1
2                                        M. SUZUKI

   It is believed that a promising way to prove RH is the Hilbert-Pólya conjecture which
asserts that the non-trivial zeros of the Riemann zeta function correspond to eigenvalues
of some positive operator if RH is true. Therefore, if we refer to Proposition 1.1, it
is an interesting problem to find a canonical way realizing the zeros of Aω (z) as the
eigenvalues of some positive operator. Fortunately, as shown in [15] (see also [16]), it
is possible for ω ⩾ 1/2 unconditionally and for 0 < ω < 1/2 under RH if we use the
theory of de Branges spaces that are kind of reproducing kernel Hilbert spaces consisting
of entire functions. However, unfortunately, RH is used essentially in [15] to construct
corresponding de Branges spaces for 0 < ω < 1/2.
   According to a general theory of de Branges spaces, there exists a unique canoni-
cal system of linear differential equations associated with a given de Branges space up
to a normalization. And also, it is known that a special class of canonical system is
transformed into a pair of Schrödinger equations endowed with a pair of (distributional)
potentials. At this stage, the validity of RH(Aω ) is encoded in analytic properties of
potentials (see [16], and also [17]). Hence, a possible way to avoid assuming RH in the
construction of the de Branges space arising from Aω (z) for 0 < ω < 1/2 is a direct
construction of a pair of potentials without RH. However, in general, it is difficult to
determine a pair of potentials corresponding to a given de Branges space, and it is so for
the de Branges space arising from Aω (z) even if ω ⩾ 1/2.
   A goal of the present paper is to describe unconditionally for ω > 1, a canonical
system and corresponding pair of potentials associated with a de Branges space arising
from Aω (z) in terms of Fredholm determinants of certain compact integral operators
(Theorem 2.3). The restriction ω > 1 is expected to be relaxed to ω > 0 if RH is true
(see comments after Theorem 2.3 and Section 5 for details).
   In order to explain the above things more precisely, we review results on de Branges
spaces, canonical systems and model subspaces.
1.1. de Branges spaces and canonical system. At first, we review the theory of de
Branges spaces according to de Branges [9] and Lagarias [16, 17] (see also Remling [24]).
Let E be an entire function satisfying the Hermite-Biehler condition
                             |E(z)| > |E ♯ (z)| for ℑ(z) > 0,                           (1.2)
where E ♯ (z) = E(z̄). Then entire function E generates the de Branges space
                     B(E) := {f | f is entire, f /E and f ♯ /E ∈ H 2 }
endowed with norm kf kB(E) := kf /EkL2 (R) , where H 2 = H 2 (C+ ) is the Hardy space
in the upper half-plane C+ which is defined Rto be the space of all analytic functions f
in C+ endowed with norm kf k2H 2 := supv>0 R |f (u + iv)|2 du < ∞. An entire function
F (z) is called a real entire function if F (z) = F ♯ (z) (:= F (z̄)). Condition (1.2) implies
that real entire functions
                           1                                 i
                 A(z) := (E(z) + E ♯ (z)),        B(z) := (E(z) − E ♯ (z))
                           2                                 2
have real zeros only, and these zeros interlace. Moreover, if E(z) 6= 0 on the real line, all
zeros are simple ( [8, Lemma 5]). A de Branges space B(E) has an unbounded operator
(M, D(M)), multiplication by the independent variable (Mf )(z) = zf (z) with the domain
D(M) = {f ∈ B(E) | zf (z) ∈ B(E)}. The multiplication operator M is symmetric and
closed, and if D(M) is dense in B(E), it has deficiency indices (1, 1), and hence has a
family of self-adjoint extensions Mθ parametrized by θ ∈ [0, π). In particular, Mπ/2 and
M0 have pure discrete spectrum located at zeros of A(z) and B(z) respectively.
   We put the normalization E(0) = 1 for entire functions E satisfying (1.2) for a con-
venience. Then, for a given de Branges space B(E), there exists a chain of de Branges
spaces B(Ea ) ⊂ B(E), 0 < a ⩽ c (⩽ ∞), endowed with a family of entire functions
Ea (z) satisfying (1.2) and Ea (0) = 1 such that B(Ea ) ⊂ B(Ea′ ) for a < a′ , and the
                    A CANONICAL SYSTEM OF DIFFERENTIAL EQUATIONS                                   3


parametrized pair of real entire functions (Aa , Ba ) := ( 21 (Ea + Ea♯ ), 2i (Ea − Ea♯ )) satisfies
the canonical system
                                                                             
            ∂ Aa (z)          0 −1            Aa (z)                  α(a) β(a)
                        =z            H(a)             , H(a) =
           ∂a Ba (z)          1 0             Ba (z)                  β(a) γ(a)
of linear differential equations with the initial condition
                                   lim (Aa (z), Ba (z)) = (1, 0)
                                  a→0+
for each z ∈ C, and Ec (z) = E(z) (see [9, Theorem 40], but note that it is formulated
in terms of integral equations). Here the matrix H(a) is a measurable and real positive
semidefinite symmetric matrix for almost all 0 < a ⩽ c, and which is integrable over
the interval. The matrix H(a) is often called a Hamiltonian of a canonical system.
These properties of H(a) are crucial, because the initial function E can be recovered
from H(a) by solving the canonical system with the above initial condition ([9, Theorem
41]). On the other hand, the spectrum of the extended multiplication operator Mθ
coincides with the spectrum of the above canonical system with the boundary condition
lima→0+ (Aa (z), Ba (z)) = (1, 0) and Ac (z) sin θ − Bc (z) cos θ = 0.
   If H(a) is diagonal (β(a) = 0) and α(a)γ(a) = 1 almost everywhere in (0, c], the
corresponding canonical system is transformed into a pair of Schrödinger equations
                                                                              
          d2       ±                  2            ±        1 α′ (a) 2 1 α′ (a) ′
        − 2 + V (a) ψ(a, z) = z ψ(a, z), V (a) =                         ±           ,
         da                                                 4 α(a)         2 α(a)
and the initial E is recovered by solving the pair of Schrödinger equations under the
corresponding initial conditions.
   Eventually, condition (1.2) of E is encoded in analytic properties of H(a) or V ± (a).
In general, it is difficult to determine H(a) or V ± (a) for given E except for few special
examples (see Chapter 3 of [9], and also [6, 17]).
1.2. Spectral realization of zeros of Aω and B ω . Suppose that the condition
                                                                 1
                           |ξ(s + ω)| > |ξ(s − ω)| for ℜ(s) >                          (1.3)
                                                                 2
holds. Then we find that E(z) = E ω (z) := ξ( 12 + ω − iz) satisfies (1.2) by using the
functional equations ξ(s) = ξ(1 − s) and ξ(s) = ξ(s̄). Thus the de Branges space B(E ω )
is defined, and RH(Aω , B ω ) holds. By a result of [15], condition (1.3) holds for ω ⩾ 1/2
unconditionally and for 0 < ω < 1/2 under RH. This is the reason why RH implies
RH(Aω ) for all ω > 0. However, for fixed ω > 0, condition (1.3) is only a sufficient
condition to RH(Aω , B ω ), that is, RH(Aω ) or RH(B ω ) may be true even if condition
(1.3) does not hold.
   Anyway, we can regard the zeros of Aω (z) and B ω (z) as discrete spectrum of self-
adjoint extensions of (M, D(M)) on B(E ω ) for ω ⩾ 1/2 unconditionally and for 0 < ω <
1/2 under RH. Therefore, a natural problem on RH(Aω ) and a spectral realization of
the zeros of Aω (z) is to find a way avoiding RH for 0 < ω < 1/2. A possible approach is
to construct H(a) or V ± (a) associated with B(E ω ) without assuming RH, and recover
E ω , Aω and B ω from the canonical system attached to H(a) or the pair of Schrödinger
equations attached to V ± (a). We attempt to follow this way by using the theory of
model subspaces.
1.3. Model subspaces. For further discussions, we review a theory of model spaces
according to Havin–Mashreghi [12, 13] (see also Baranov [1], Makarov–Poltoratski [22]).
A function Θ is called an inner function in C+ if it is a bounded analytic function in
C+ such that limv→0+ |Θ(u + iv)| = 1 for almost all u ∈ R with respect to Lebesgue
measure. If an inner function Θ in C+ is extended to a meromorphic function in C, it is
called a meromorphic inner function in C+ . It is known that every meromorphic inner
function is expressed as Θ = E ♯ /E by using an entire function E satisfying (1.2). For
4                                       M. SUZUKI

an inner function Θ, a model subspace (or coinvariant subspace) K(Θ) is defined by the
orthogonal complement
                                 K(Θ) = H 2 ⊖ ΘH 2 ,                             (1.4)
where ΘH 2 = {Θ(z)F (z) | F ∈ H 2 }. It has the alternative representation
                                   K(Θ) = H 2 ∩ ΘH̄ 2 ,                                (1.5)
where H̄ 2 = H 2 (C− ) is the Hardy space in the lower half-plane C− . If Θ is a meromorphic
inner function such that Θ = E ♯ /E, the model subspace K(Θ) is isomorphic to the de
Branges space B(E) as a Hilbert space by K(Θ) → B(E) : f 7→ f E. In particular,
K(Θ) is a reproducing kernel Hilbert space. The reproducing kernel of K(Θ) is given by
                                   1 1 − Θ(z)Θ(w)
                      K(z, w) =                            (z, w ∈ C+ ),               (1.6)
                                  2πi    z̄ − w
and the reproducing formula f (z) = hf, K(z, ·)iL2 (R) (f ∈ K(Θ), z ∈ C+ ) remains true
                                                                       R
for z ∈ R if Θ is analytic in a neighborhood of u, where hf, giL2 (R) = R f (u)g(u)du.

1.4. Model subspaces related to Aω and B ω . Now we apply the theory of model
subspaces to the spaces B(E ω ) of Section 1.2. For positive real ω, we define the mero-
morphic function Θω (z) in C by
                                             ξ( 21 − ω − iz)
                                 Θω (z) :=                   .                         (1.7)
                                             ξ( 21 + ω − iz)
Then we have
                             Θω (z)Θω (−z) = 1 for z ∈ C,                              (1.8)
                                |Θω (u)| = 1 for u ∈ R,                                (1.9)
                                        Θω (0) = 1,                                  (1.10)
by functional equations ξ(s) = ξ(1 − s) and ξ(s̄) = ξ(s).
  The inequality (1.3) can now be reinterpreted as the condition
                               |Θω (z)| < 1 for ℑ(z) > 0                             (1.11)
and vice versa. Recall that condition (1.3) is known to hold for ω ⩾ 1/2 unconditionally
and for 0 < ω < 1/2 under RH. By (1.9), when condition (1.11) holds, it implies that
Θω (z) is a meromorphic inner function in C+ . Therefore, whenever (1.11) holds, we
obtain a model subspace K(Θω ) which is isomorphic to the de Branges space B(E ω )
generated by E ω (z) = ξ( 21 + ω − iz). Here we mention the following equivalence relation.
Proposition 1.2 Let ω0 ⩾ 0. Then the following are equivalent:
    (1) ζ(s) 6= 0 for ℜ(s) > 12 + ω0 ,
    (2) Θω (z) is a meromorphic inner function in C+ for every ω > ω0 .
Proof. Assume that 0 ⩽ ω0 < 1/2 since we have nothing to say for ω ⩾ 1/2. By applying
Theorem 4 of [18], we find that (1) implies that (1.11) holds for every ω > ω0 . Thus
we obtain (1)⇒(2). The converse implication (2)⇒(1) is proved by a way similar to the
proof of Theorem 2.3 (1) in [26].                                                   

   The changing of consideration from B(E ω ) to K(Θω ) has the advantage that spaces
Θω H 2 , Θω H̄ 2 , H 2 ⊖ (H 2 ∩ Θω H 2 ) and H 2 ∩ Θω H̄ 2 are defined even if Θω (z) is not
necessarily a meromorphic inner function in C+ (see (1.4) and (1.5)), and it allows us
to study these spaces for the range 0 < ω < 1/2 without assuming RH. (Note that
ΘH 2 6⊂ H 2 in general if Θ is not necessary a inner function in C+ .) To make a further
discussion, we use Fourier analysis.
                    A CANONICAL SYSTEM OF DIFFERENTIAL EQUATIONS                                 5

1.5. An operator related to K(Θω ). As usual we identify H 2 and H̄ 2 with subspaces
of L2 (R) = L2 ((−∞, ∞), du) via nontangential boundary values on the real line such
that L2 (R) = H 2 ⊕ H̄ 2 . Then the shifted Fourier transform
                                                             Z ∞
                                                                         1     dx
                 2                    2
         F1/2 : L ((0, ∞), dx) → L (R) : (F1/2 f )(z) =          f (x) x 2 +iz    ,
                                                              0                x
                                                                Z ∞
                                                 −1           1                1
         F−1
          1/2 : L 2
                    (R) →  L 2
                               ((0, ∞), dx) : (F 1/2 g)(z) =         g(u) x− 2 −iu du
                                                             2π −∞
provides an isometry of L2 -spaces up to a constant such that H 2 = F1/2 L2 ((1, ∞), dx)
and H̄ 2 = F1/2 L2 ((0, 1), dx) by the Paley-Wiener theorem.
  Fourier analysis on K(Θω ) and Θω H 2 enables us to state equivalent or sufficient
conditions that Θω (z) is a meromorphic inner function in C+ (Theorem 2.2).
  On the other hand, condition (1.9) allows us to define the Hankel type operator
                                             Z ∞
                                  ∗
                                (Hω f )(x) =     h∗ω (xy) f (y) dy
                                        0
    2
on L ((0, ∞), dx) endowed with the kernel given by
                                               Z ∞
                                           1                   1
                              h∗ω (x) =              Θω (u) x− 2 −iu du.                    (1.12)
                                          2π   −∞
Of course the definition of H∗ω has only a formal sense because of the problem of the
convergence of integral in (1.12). However h∗ω (x) is going to be identified with the
function hω (x) in Section 2, and then H∗ω is going to be justified as the operator Hω
obtained by replacing the kernel h∗ω (x) by hω (x). Moreover the operator Hω is extended
to an isometry from L2 ((0, ∞), dx) to L2 ((0, ∞), dx) for ω ⩾ 1/2 unconditionally, and
for 0 < ω < 1/2 under RH (see Lemma 4.1).
   As developed in Burnol [6] (and his other related works [3–5]), the Hankel type op-
erator Hω and its kernel hω (xy) is quite useful to study a structure of subspaces of
F−1
  1/2
      K(Θω ) corresponding to de Branges subspaces of B(E ω ) ≃ K(Θω ). By applying
Burnol’s theory to Hω and hω (x), we derive a canonical system of B(E ω ) under the
restriction ω > 1 (Theorem 2.3 and studying in Section 4). Recall that the structure of
subspaces of a de Branges space is controlled by its canonical system.
1.6. Summary of issues. Briefly, we have two issues. The first is to state a (nice)
criterion for the innerness of Θω (z). It is directly related to the zero-free region of ζ(s)
(Proposition 1.2). The second is to describe the Hamiltonian Hω (a) of the canonical
system of B(E ω ) explicitly by assuming that Θω (z) is a meromorphic inner function
in C+ if 0 < ω < 1/2. If it is done, we can state that Θω (z) is a meromorphic inner
function in C+ if and only if (Aω , B ω ) = (Ac , Bc ) for the solution (Aa , Ba ) of the canonical
system for Hω (a) on a ∈ (0, c] satisfying lima→0+ (Aa , Ba ) = (E ω (0), 0). This description
explains the innerness of Θω (z) as a consequence of properties of Hω (a), and it provides
a criterion for a zero-free region of ζ(s) in terms of a family of canonical systems attached
to {Hω (a)}ω>ω0 via Proposition 1.2.
   However the second problem is not trivial even if ω ⩾ 1/2. In this paper, we deal with
the case ω > 1 for the second problem as the first attempt.
1.7. Organization of the paper. The paper is organized as follows. In Section 2, we
state main results Theorem 2.2 and Theorem 2.3 after a small preparation of notation.
The first one is equivalent conditions on the Hermite-Biehler condition (1.11) in terms
of the function hω (x) for fixed ω > 0. This is proved in Section 3. The second one is
a result on the canonical system of B(E ω ) ≃ K(Θω ) under the restriction ω > 1. It
is proved in Section 4 together with related studies and auxiliary results. In addition,
we present more sufficient or equivalent conditions that Θω (z) is a meromorphic inner
function in C+ in Appendix A (Theorem A.1).
6                                        M. SUZUKI

   Here we mention that this paper, particularly Appendix A, is a sequel to [26], though
it is independent and can be read separately. The operator H∗ω of Section 1.5 is also
justified as the Watson transform:
                       Z ∞                                  Z ∞
         ∗∗          d      ∗∗          dy       ∗∗       1      Θω (u) 1 −iu
       (Hω f )(x) =        hω (xy) f (y) ,      hω (x) =                 x2    du,
                    dx 0                y                2π −∞ 12 − iu
which gives a linear involution on L2 ((0, ∞), dx) under (1.8) (only for real z) and (1.9)
(see Titchmarsh [27, §8.5], Bochner–Chandrasekharan [2, Chap.V, §2]). Moreover, H∗∗   ω =
Hω if Θω (z) is inner in C+ . The Watson transform has the advantage that h∗∗ ω (x) always
exists in L2 -sense by (1.9), and belongs to L2 ((0, ∞), dx). While the modified function
                                          Z ∞
                               h1i      1      Θω (u) 1 −iu
                             hω (x) =                 x2      du
                                       2π −∞ −iu
does not belong to L2 ((0, ∞), dx) although it is justified as a function (Appendix A).
However it is also useful to study the space K(Θω ) and the operator Hω because of
formula (1.6) for the reproducing kernel. In fact, several sufficient or equivalent condi-
                                                          h1i
tions that Θω (z) is inner in C+ are stated in terms of hω (x) (Theorem A.1) as well as
Theorem 2.2. Moreover, if Θω (z) is inner in C+ , we obtain
                         Z ∞                             Z ∞
                                                √ d√
            (Hω f )(x) =      hω (xy) f (y) dy = x     x      hh1i
                                                               ω (xy) f (y) dy
                          0                        dx      0

for compactly supported smooth functions f , and it is extended to L2 ((0, ∞), dx) (The-
                            h1i
orem A.2). The function hω (x) was introduced and studied in [26] for more general
L-functions, but a relation with spaces B(E ω ) ≃ K(Θω ) and operators Hω were not
mentioned there. In this sense, this paper is a sequel to [26].

1.8. De Branges’ works. Finally, we comment on de Branges’ works on B(E ω ). The de
Branges space B(E ω ) was considered first for the special value ω = 1/2 in de Branges [10,
pp.10–14], motivating to generalize the Lax-Phillips scattering theory to the Laplace-
Beltrami operator, and for ω ⩾ 1/2 in the subsequent paper [11, pp.205–210]. (Precisely,
we need to replace ζ(s) by a Dirichlet L-function L(s, χ) attached to an even primitive
Dirichlet character χ in [10]). De Branges gave a sufficient condition on B(E) attached
to general entire function E satisfying (1.2) such that the zeros of E(z) lie on the line
ℑ(z) = −1/2, which implies the (generalized) RH when E = E ω for ω = 1/2. However
Conrey and Li [7] showed that B(E ω ) (ω = 1/2) does not satisfy de Branges’ condition.
For ω ⩾ 1/2 de Branges studied the space B(E ω ) by associating it with the weighted
Hardy space F(W ) = W H 2 for the weight function W (z) = 14 (s + ω)(s + ω − 1)Γ( s+ω  2 )
with s = 21 − iz, but we omit the details of this topic (see [11], and also [7]).
   In any case, de Branges directly related RH with a condition on B(E ω ) for fixed
ω ⩾ 1/2. On the other hand, we reduced RH to the family of spaces {B(E ω )}ω>0 , and
study each space B(E ω ) depending on a level of difficulty, which is determined by the
value ω. This is a major difference with de Branges’ approach and ours.
Acknowlegements I heartily thank the reviewer for many detailed and helpful com-
ments and corrections. In particular, the readability of the paper was quite improved,
and an error of the proof of Lemma 4.3 in the initial version was corrected by com-
ments of the reviewer. This work was supported by KAKENHI (Grant-in-Aid for Young
Scientists (B)) No. 21740004.

                                    2. Main Results
   Our first result is to derive an expression for Θω (z) as a Mellin transform of a function
hω (x) defined for 0 < x < ∞, which is valid for all real ω > 0 (Proposition 2.1). To
                   A CANONICAL SYSTEM OF DIFFERENTIAL EQUATIONS                             7

define this function we first define the numbers
                                      X µ(d)       Y     1
                                                             
                                    ω            ω
                        cω (n) := n           =n      1 − 2ω                            (2.1)
                                          d2ω            p
                                     d|n              p|n

for natural numbers n, where µ(n) is the Möbius function, that is, µ(n) = 0 if n is not
a square free number, and µ(n) = (−1)k if n is a product of k distinct primes. The
arithmetic function n 7→ J2ω (n) := nω cω (n) is called Jordan’s totient function, which
gives Euler’s totient function ϕ(n) for ω = 1/2.
   Next we introduce a function gω (x) defined on (0, ∞) by
                                                       Z 1                   
                      2π ω     2−ω       2 ω−1      ω−1       1
                                                                −ω     ω−1
            gω (x) =         x     (1 − x )    − ωx         t 2 (1 − t)    dt
                      Γ(ω)                               x2
for 0 < x < 1, and gω (x) = 0 for x > 1. It is continuous on (0, 1) and (1, ∞). The
behavior of gω near x = 1 and x = 0 is as follows. We have
                             (2π)ω
                    gω (x) =       (1 − x)ω−1 + o(1) as x → 1− .                (2.2)
                             Γ(ω)
Therefore gω is continuous at x = 1 if and only if ω > 1, and it is L1 (resp. L2 ) at x = 1
if ω > 0 (resp. ω > 1/2). On the other hand, we have
              
                      ω−1/2 Γ(3/2 − ω) xω−1 + o(1),
              −4 ωπ
                   √
                                                       0 < ω < 3/2,
     gω (x) = 4 π x (3 log x + 4 − 3 log 2) + o(1), ω = 3/2,             as x → 0+ .
              
              
                −6π ω (2ω − 3)−1 Γ(ω)−1 x2−ω + o(1), ω > 3/2,
Thus gω is L1 (resp. L2 ) at x = 0 if 0 < ω < 3 (resp. 1/2 < ω < 5/2). The size of
the singularity at x = 1 will be important in the sequel because it influences the type
of operators Hω,a below, while there is no need to be careful about the behavior around
x = 0 in this paper.
   Finally, we define the real-valued function hω on (0, ∞) by
                                             ⌊x⌋           n
                                           1X
                                hω (x) =         cω (n) gω                              (2.3)
                                           x n=1            x
for x > 1, and hω (x) = 0 for 0 < x < 1. The value hω (1) may be undefined, since
cω (1) = 1 and gω (1− ) = +∞ for 0 < ω < 1 by (2.2). By definition, hω has a support in
[1, ∞), and is L1 (resp. L2 ) on every finite interval [1, b] if ω > 0 (resp. ω > 1/2). On
the other hand, the behavior of hω at x = +∞ is not obvious from its definition (see
(4.22) below). Now the first result is stated as follows.
Proposition 2.1 For ω > 0 and ℑ(z) > 1/2 + ω, we have
                             Z ∞
                                           1     dx
                                  hω (x) x 2 +iz    = Θω (z),                           (2.4)
                               0                  x
where the integral converges absolutely.
  We introduce more notation in order to sate the main results mentioned in the intro-
duction. By (1.9), F (z) 7→ Θω (z)F (z) defines a map L2 (R) → L2 (R). We denote it also
by Θω if no confusion arises, and define
                 b ω = F−1 Θω F1/2 : L2 ((0, ∞), dx) → L2 ((0, ∞), dx).
                 Θ        1/2

If Θω (z) is an inner function in C+ , images Θω H 2 and Θ b ω L2 ((1, ∞), ds) are subspaces
    2         2
of H and L ((1, ∞), dx), respectively. Obviously the map Θ     b ω is related to the function
hω by (2.4). In fact the innerness of Θω (z) is described in terms of hω as follows.
Theorem 2.2 Let ω > 0. The function Θω (z) is a meromorphic inner function in C+
if and only if one of the following conditions holds:
8                                        M. SUZUKI

         b ω f = hω ∗ f for every f ∈ L2 ((1, ∞), dx), where
     (1) Θ
                                            Z ∞
                                                             dy
                             (hω ∗ f )(x) =     hω (x/y)f (y) .
                                             0                y
     (2) b
         Θω f vanishes on (0, 1) for every f ∈ L2 ((1, ∞), dx).
     (3) hω ∗ f belongs to L2 ((0, ∞), dx) for every f ∈ L2 ((1, ∞), dx).
    Suppose that Θω (z) is an inner function in C+ . Then
                                          Z ∞
                             (Hω f )(x) =      hω (xy) f (y) dy                         (2.5)
                                              0
defines a bounded operator from L2 ((0, ∞), dx) to L2 ((0, ∞), dx) (Lemma 4.1). For
a > 0, we denote by Pa the orthogonal projection from L2 ((0, ∞), dx) to L2 ((0, a), dx),
and define
                  Hω,a := Pa Hω Pa : L2 ((0, a), dx) → L2 ((0, a), dx).             (2.6)
A study of Hω and Hω,a yields a canonical system as follows:
Theorem 2.3 Suppose that ω > 1. (It implies automatically that Θω is inner in C+ .)
Then the operator Hω,a is a Hilbert-Schmidt type self-adjoint operator with a continuous
kernel for every a > 1, and Hω,a = 0 for 0 < a ⩽ 1. Moreover 1 ± Hω,a are invertible for
every a > 0. Define
                                              det(1 + Hω,a )
                            m(a) := mω (a) =
                                              det(1 − Hω,a )
by using Fredholm determinants. Then m(a) is real-valued continuous function on (0, ∞),
and the canonical system
                                                          
             ∂ Xa (z)         0 −1 m(a)−2         0       Xa (z)
         −a             =z                                         (0 < a < ∞)
            ∂a Ya (z)         1 0         0     m(a)2 Ya (z)
has the explicit solution (Xa , Ya ) = (Aa , Ba ) given by (4.26) in Section 4 such that
    (1) Aa (z) and Ba (z) are real entire functions as a function of z for every fixed a > 0,
    (2) Aa (−z) = Aa (z) and Ba (−z) = −Ba (z) as a function of z for every fixed a > 0,
    (3) (A1 (z), B1 (z)) = (Aω (z), B ω (z)) and
                lim (Aa (z), Ba (z)) = lim (Aa (z), Ba (z)) = (Aω (z), B ω (z))
               a→1+                    a→1−

        hold uniformly on every compact subset in C, where Aω (z) and B ω (z) are real
        entire functions defined in (1.1).
Furthermore, the canonical system can be transformed into the pair of Schödinger equa-
tions                                     
                            ∂ ∂
                        −a a       + V (a) ψ ± (a, z) = z 2 ψ ± (a, z)
                                        ±
                            ∂a ∂a
with the pair of potentials
                                          2                         
                   ±           1     ∂            ∂      1       ∂
                 V (a) =           a m(a) ∓ a                 a m(a)
                              m(a) ∂a            ∂a m(a) ∂a
by taking ψ + (a, z) = m−1 (a)Aa (z) and ψ − (a, z) = m(a)Ba (z).
   The assumption ω > 1 in Theorem 2.3 is required to obtain a continuity of the kernel
hω (xy) in the proof in Section 4.3 and 4.4, since hω (x) has a singularity at x = n ∈ Z>0
for 0 < ω ⩽ 1. However, we observed that singularities at x = n ∈ Z > 0 are in L2
exactly for ω > 1/2, and are in L1 for all ω > 0. This implies that that the function
hω (x) on any interval [x0 , x1 ] with 0 < x0 < x1 < ∞ lies in the same function spaces,
and it affects the behavior of associated Hω,a . In fact, Hω,a is a Hilbert-Schmidt type
self-adjoint operator such that 1 ± Hω,a are invertible for every a > 0 if ω > 1/2 (Lemma
4.2 and 4.4 below), and is a compact self-adjoint operator for all ω > 0. In addition, the
                  A CANONICAL SYSTEM OF DIFFERENTIAL EQUATIONS                             9

type of singularities at x = n ∈ Z > 0 presumably affects the canonical system since it
is given by determinants of 1 ± Hω,a if ω > 1.
   On the other hand, as mentioned in Section 1.2 and 1.4, Θω (z) is an inner function in
C+ for all ω ⩾ 1/2 unconditionally, and for all ω > 0 under RH.
   Therefore, it is plausible that all results of Theorem 2.3 can be extended to ω > 1/2
unconditionally without essential difficulties. Moreover, it is expected that Theorem 2.3
is generalized to ω > 0 if we assume RH for ζ(s). See Section 5 for further comments on
the validity of Theorem 2.3.
   Finally, we emphasize that the limit behavior lima→+∞ (Aa (z), Ba (z)) is still open
even if ω > 1. The expected result is lima→+∞ (E ω (0), 0) = (ξ( 21 + ω), 0) if we note that
E is normalized as E(0) = 1 in Section 1.1. Provably, this limit behavior is related to
the arithmetic properties of ζ(s) in more deep level, because we need information for all
{cω (n)}n⩾1 to understand it differ from the situation that we need only finitely many
cω (n)’s to understand Hω,a for a finite range of a. However, we do not touch this problem
further in this paper.

                 3. Proof of Proposition 2.1 and Theorems 2.2
3.1. Proof of Proposition 2.1. For convenience, we use variable s = 1/2 − iz. Put
γ(s) = 12 s(s − 1)π −s/2 Γ(s/2) so that ξ(s) = γ(s)ζ(s). Then
                                      s−ω
                                                                   
                   γ(s − ω)       ωΓ    2 + 1      2ωπ ω Γ s−ω2 + 1
                             =π                 −                     .
                   γ(s + ω)        Γ s+ω2 +1
                                                  s + ω − 1 Γ s+ω
                                                               2 +1
We have                                           Z 1
                     Γ( s−ω
                         2 + 1)    2                                           dx
                        s+ω     =                       x2−ω (1 − x2 )ω−1 xs           (3.1)
                     Γ( 2 + 1)    Γ(ω)             0                            x
for ℜ(s + 2) > ω > 0 by [23, (5.35) of p.195], and
                                           Z 1
                                  1                     dx
                                         =      xω−1 xs                                (3.2)
                              s+ω−1          0           x
for ℜ(s) > 1 − ω. Applying Theorem 44 of [27] to (3.1) and (3.2) together with
                 Z 1                                                    
             2         2−ω       2 ω−1      ω−1 dx   y ω−1    2 3
                     x     (1 − x )    (y/x)       =       β y , − ω, ω ,
           Γ(ω) y                               x    Γ(ω)        2
we obtain            Z ∞                         Z 1
                                      s dx         dx   γ(s − ω)
                           gω (x) x          =        =gω (x) xs
                    0            x      0          x    γ(s + ω)
for ℜ(s) > max(ω − 2, 1 − ω). On the other hand, we have
                                                1 X µ(d)  n ω X cω (n)
                   ∞
                   X              ∞         X∞                      ∞
        ζ(s − ω)        µ(m)m−ω X nω
                 =                        =                      =
        ζ(s + ω) m=1       ms    n=1
                                      ns    n=1
                                                ns     dω   d      n=1
                                                                       ns
                                                                   d|n

by definition (2.1), where the series converges absolutely for ℜ(s) > 1 + ω. By definition
(2.3), we have formally
       Z ∞                   X∞         Z ∞
                    1−s dx                                    dx
           hω (x) x        =     cω (n)     x−1 gω (n/x) x1−s
        0                x   n=1         0                     x
                             X∞         Z
                                 cω (n) ∞                    dx    γ(s − ω) ζ(s − ω)
                           =         s
                                            gω (n/x) (n/x)s      =                   ,
                                   n      0                   x    γ(s + ω) ζ(s + ω)
                             n=1

and it is justified by Fubini’s theorem for ℜ(s) > 1 + ω. Replacing s by 1/2 − iz, we
obtain (2.4).
10                                       M. SUZUKI

3.2. Proof of Theorem 2.2.          It is sufficient to prove the following three assertions:
i) condition (1) is equivalent that Θω is inner in C+ , ii) condition (2) implies that Θω is
inner in C+ , and iii) condition (3) implies that Θω is inner in C+ , since (1) implies (2)
and (3) by definition of Θb ω and hω . We prove them after the following lemma.

Lemma 3.1 Assume that Θω H 2 ⊂ H 2 . Then Θω is inner in C+ .
Proof. Let δ > 0. We find that Θω (z) is uniformly bounded on the upper half-plane
ℑ(z) ⩾ 1/2 + ω + δ by using a usual estimate for the Dirichlet series ζ(s − ω)/ζ(s + ω)
and the Stirling formula for the gamma-function. On the other hand, we know (1.9), and
the assumption implies that Θω has no poles in C+ . Hence, by applying the Phragmén-
Lindelöf convexity principle to Θω in the strip 0 ⩽ ℑ(z) ⩽ 1/2 + ω + δ, we find that Θω
is bounded on 0 ⩽ ℑ(z) ⩽ 1/2 + ω + δ. Therefore Θω is a bounded analytic function is
C+ satisfying (1.9). This is the definition of an inner function in C+ .              

   i) Suppose that Θω is inner in C+ . Then Θω F ∈ H 2 for every F ∈ H 2 . Thus the
inverse (shifted) Fourier transform along the line ℑ(z) = c
                                      Z
                        b           1                      1
                        Θω f (x) =           Θω (z)F (z)x− 2 −iz dz
                                   2π ℑ(z)=c

is independent of c > 0, and belongs to L2 ((1, ∞), dx), where f = F−11/2 F and the integral
                           2
converges in the sense of L . On the other hand
                                       Z
                                     1                        1
                     (hω ∗ f )(x) =             Θω (z)F (z)x− 2 −iz dz
                                    2π ℑ(z)=c′
for c′ > 1/2 + ω by Proposition 2.1 and [27, Theorem 65], where the integral converges
also in the sense of L2 . Comparing these two formula for large c, we obtain (1).
    Conversely, suppose that (1) holds. Write g = b   Θω f = hω ∗ f for arbitrary fixed
       2                                  2
f ∈ L ((1, ∞), dx). Then g belongs to L ((0, ∞), dx), since Θ  b ω maps L2 ((0, ∞), dx) to
  2
L ((0, ∞), dx) by definition. In addition, g has a support in [1, ∞), since both hω and f
have support in [1, ∞). Therefore g belongs to L2 ((1, ∞), dx). Because f was arbitrary,
we have Θω H 2 ⊂ H 2 . Hence Θω is inner in C+ by Lemma 3.1.                           
   ii) Suppose that (2) holds. Then it implies Θb ω L2 ((1, ∞), dx) ⊂ L2 ((1, ∞), dx), since
b ω maps L2 ((0, ∞), dx) to L2 ((0, ∞), dx) by its definition. It means Θω H 2 ⊂ H 2 by
Θ
              b ω . Hence Θω is inner in C+ by Lemma 3.1
definition of Θ                                                                           
  iii) Suppose that (3) holds. Then hω ∗ f belongs to L2 ((1, ∞), dx) for every f in
L2 ((1, ∞), dx), since hω ∗ f has a support in [1, ∞) by its definition. Therefore
                               Z ∞
                                                1     dx
                                   hω ∗ f (x) x 2 +iz    ∈ H 2.
                                0                      x
Additionally, we suppose that f belongs to the dense subset L1 ((1, ∞), dx)∩L2 ((1, ∞), dx).
Then                     Z ∞
                                           1     dx
                              hω ∗ f (x) x 2 +iz    = Θω (z)F (z)
                           0                     x
for ℑ(z) > 1/2 + ω by [27, Theorem 44]. Therefore
                                                   
                b
                Θω L1 ((1, ∞), dx) ∩ L2 ((1, ∞), dx) ⊂ L2 ((1, ∞), dx),

                  b ω L2 ((1, ∞), dx) ⊂ L2 ((1, ∞), dx), since Θ
This implies that Θ                                             b ω is continuous by its def-
inition. Therefore Θω H 2 ⊂ H 2 by definition of Θ   b ω , and hence Θω is inner in C+ by
Lemma 3.1.                                                                                 
                   A CANONICAL SYSTEM OF DIFFERENTIAL EQUATIONS                                11

                               4. Proof of Theorem 2.3
   In this section, we study operators (2.5), (2.6), and their kernels toward Theorem 2.3
referring to Burnol [6]. However here we use classical arguments rather than the theory
of distributions used in [6].

4.1. Fredholm integral equations.

Lemma 4.1 Suppose that Θω is inner in C+ . Define Hω f by integral (2.5) for compactly
supported smooth functions f . Then Hω f belongs to L2 ((0, ∞), dx), and the linear map
f 7→ Hω f is extended to the isometry Hω : L2 ((0, ∞), dx) → L2 ((0, ∞), dx) satisfying

                            (F1/2 Hω f )(z) = Θω (z) (F1/2 f )(−z)                      (4.1)

for z ∈ R. Moreover, (4.1) holds for ℑ(z) ⩾ 0, if f ∈ L2 ((0, ∞), dx) has a support in
[0, b] for some b > 0.

Remark This is applied unconditionally to ω ⩾ 1/2, and also to 0 < ω < 1/2 under
RH by discussion in Section 1.2 and 1.4.

Proof. If f is a compactly supported smooth function, we have
                    Z ∞Z ∞
                                        1    dx
  (F1/2 Hω f )(z) =          hω (xy) x 2 +iz     f (y) dy
                     0   0                    x
                    Z ∞                   Z
                                 1     dx ∞            1     dy
                  =     hω (y) x 2 +iz         f (y) y 2 −iz    = Θ(z)F (−z)   (F = F1/2 f )
                     0                  x 0                  y
for ℑ(z) > 1/2 + ω by Proposition 2.1, and F (−z) is an entire function satisfying
F (−z) = O(|z|−n ) as |z| → ∞ in any horizontal strip c1 ⩽ ℑ(z) ⩽ c2 for arbitrary fixed
n > 0. Therefore, we find that Hω f belongs to L2 ((0, ∞), dx) by applying the Fourier
inversion formula to Θω (z)F (−z) along a line ℑ(z) = c > 1/2 + ω and then moving the
path of integration to the real line ℑ(z) = 0, since Θω is inner in C+ by assumption.
Moreover
                          kHω f k = kΘω (·)F (−·)k = kF k = kf k
by (1.9). Recall that the set of all compactly supported smooth function in L2 ((0, ∞), dx)
is dense in L2 ((0, ∞), dx). Therefore f 7→ Hω f is extended to all f ∈ L2 ((0, ∞), dx) by
continuity, and the extended operator is obviously isometric.
   Equality (4.1) holds for real z by the continuity. Suppose that f ∈ L2 ((0, ∞), dx) has
a support in [0, b] for some b > 0. Then Hω f belongs to L2 ((0, ∞), dx) and has a support
in [1/b, ∞). Therefore the left-hand side of (4.1) is defined by the shifted Fourier integral
and analytic in C+ . On the other hand, (F1/2 f )(−z) in the right-hand side of (4.1) is
also defined by the shifted Fourier integral and analytic in C+ , since f has a support
in [0, b] by the assumption. Hence both sides of (4.1) are analytic functions in C+ , and
they are equal on the real line. Thus equality (4.1) holds for ℑ(z) ⩾ 0.                   

Lemma 4.2 Suppose that ω > 1/2. (It implies automatically that Θω is inner in C+ .)
Then the operator Hω,a = Pa Hω Pa defined in (2.6) is a self-adjoint Hilbert-Schmidt type
operator if a > 1, and Hω,a = 0 if 0 < a ⩽ 1.

Proof. If 0 < a ⩽ 1 and 0 < x < a, we have
                                        Z a
                          Hω Pa f (x) =     hω (xy)f (y) dy = 0,
                                           0

since hω (x) = 0 for 0 < x < 1, and 0 ⩽ xy < a2 ⩽ 1. Hence Hω,a = 0 if 0 < a ⩽ 1.
12                                           M. SUZUKI


   Denote by K(x, y) = hω (xy) the kernel of Hω,a . We have K(x, y) = K(y, x), since
hω (xy) is real-valued. Thus Hω,a is self-adjoint. For a > 1, we have
      Z aZ a                    Z a Z a
                        2
               |K(x, y)| dxdy =            |hω (xy)|2 dxdy
        0   0                         1/a   1/a
                                     Z a      Z a2                                   Z a2
                                       dy                              2
                                 ⩽                           |hω (x)| dx = 2 log a          |hω (x)|2 dx.
                                   1/a y              1/a2                            1
      R a2
Here 1 |hω (x)|2 dx < ∞ if ω > 1/2, since hω (x) has only finitely many singularities at
x = n (1 ⩽ n ⩽ ⌊a2 ⌋) in [1, a2 ], and hω (x) ≪ |x − n|ω−1 around x = n by (2.2) and
(2.3). Hence, K(x, y) = hω (xy) is a Hilbert-Schmidt kernel if a > 1 and ω > 1/2.     
Lemma 4.3 Let a > 0. Suppose that ω > 1/2. Then the support of Hω Pa f is not
compact for f ∈ L2 ((0, ∞), dx) unless Hω Pa f = 0.
Proof. We prove this by contradiction. Suppose that Hω Pa f 6= 0 and has a compact
support. Then F1/2 Hω Pa f is an entire function of exponential type by the Paley-Wiener
theorem. On the other hand, we have
                           F1/2 Hω Pa f (z) = Θω (z) · F1/2 Pa f (−z).
This implies that G(z) := Pa f (−z)/ξ( 12 + ω − iz) is entire, because (1.2) holds uncondi-
tionally for the denominator E(z) = ξ( 12 + ω − iz) of Θω defined in (1.7), and E(z) 6= 0
on ℑ(z) = 0. Thus, we have
                                                         
                                               1
                        F1/2 Hω Pa f (z) = ξ     − ω − iz · G(z),
                                               2
where the right-hand side is a product of entire functions. The point is that the zeros in
the numerator of Θω can not kill the poles of the denominator, which therefore must be
killed by zeros of Pa f (−z). This allows ξ( 21 − ω − iz) to be factored out.
   The entire function on the right-hand side has at least π1 T log T zeros in the disk of
radius T around the origin, as T → ∞ ( [28, Theorem 9.4]). However all entire functions
of exponential type have at most O(T ) zeros in the disk of radius T around the origin, as
T → ∞, because of the Jensen formula ( [20, §2.5 (15)]). This is a contradiction. (The
proof contained an error in the first version, but it was revised by the reviewer.)     
Lemma 4.4 Let ω > 1/2 and a > 1. We have i) Hω,a f = 0 for every f ∈ L2 ((0, 1/a), dx),
ii) kHω,a f k < kf k for every 0 6= f ∈ L2 ((0, a), dx), and iii) kHω,a k < 1. In particular,
1 ± Hω,a are invertible operator on L2 ((0, a), dx).
Proof. If 0 < x < 1/a, we have
                                             Z a
                             Hω Pa f (x) =             hω (xy)f (y) dy = 0,
                                                  0
since hω (x) = 0 for 0 < x < 1, and 0 ⩽ xy < 1. Hence i) is proved.
   To prove ii), it is sufficient to show kHω,a f k 6= kf k unless f = 0, because kHω k = 1,
kHω,a k ⩽ kPa k · kHω k · kPa k = 1, and kHω,a f k ⩽ kHω,a k · kf k ⩽ kf k. Here kHω,a f k 6= kf k
is equivalent to kPa Hω f k 6= kf k, since Pa f = f for f ∈ L2 ((0, a), dx). Suppose that
kPa Hω f k = kf k for some 0 6= f ∈ L2 ((0, a), dx). Then it implies kPa Hω f k = kHω f k by
kHω f k = kf k. Therefore
                             Z a                  Z ∞
                                           2
                                 |Hω f (x)| dx =      |Hω f (x)|2 dx.
                             0                               0
Thus Hω f (x) = 0 for almost every x > a. On the other hand, we have
                           Z a                   Z a
                Hω f (x) =     hω (xy)f (y) dy =     hω (xy)f (y) dy = 0
                                 0                               1/x
                   A CANONICAL SYSTEM OF DIFFERENTIAL EQUATIONS                              13

for 0 < x < 1/a by f ∈ L2 ((0, a), dx). Hence Hω f has a compact support contained in
[1/a, a]. However, it is impossible for any f 6= 0 by Lemma 4.3. As the consequence
kHω,a f k < kf k for 0 6= f ∈ L2 ((1/a, a), dx).
   Finally, we prove iii). By Lemma 4.2, Hω,a is a self-adjoint compact operator. There-
fore, Hω,a has purely discrete spectrum which has no accumulation points except for 0,
and one of ±kHω,a k is an eigenvalue of Hω,a . However, by ii), every eigenvalue of Hω,a
has an absolute value less than 1. Hence kHω,a k < 1.                                 
Lemma 4.5 Let ω > 1/2, a > 1 and ε ∈ {±1}. Then the integral equation
                              Z a
                    X(x) + ε      hω (xy)X(y) dy = hω (ax)                               (4.2)
                                      0

has unique solution X = φaε in L2 ((0, a), dx), which is real-valued almost everywhere in
[0, a] and vanishes almost everywhere in [0, 1/a].
   Moreover, if ω > 1, the solution φεa is a real-valued continuous function on [0, a]
vanishing on [0, 1/a].
Proof. By Lemmas 4.2 and 4.4, Hω,a is a compact operator such that ±1 belong to its
resolvent set. Therefore, integral equation (4.2) has unique
                                                         R a solution    φεa in L2 ((0, a), dx)
                                                                      ε
by the Fredholm alternative. We have hω (ax) = 0 and 0 hω (xy)φa (y) dy = 0 for almost
every 0 < x < 1/a, since 0 < xy < 1 for 0 < y ⩽ Ra, and hω (x) = 0 for 0 < x < 1.
                                                    a
    On the other hand, if ω > 1, the integral 0 hω (xy)f (y) dy defines a continuous
function on [0, a] which vanishes on [0, 1/a] for every f ∈ L2 ((0, a), dx), since the kernel
hω (xy) is continuous on [0, a] × [0, a] by (2.2). Hence φεa is continuous on [0, a] and
φεa (x) = 0 for 0 ⩽ x ⩽ 1/a. Obviously φεa is real-valued, since hω (x) is real-valued,      
Lemma 4.6 Let ω > 1/2, a > 1 and ε ∈ {±1}. Then the integral equation (4.2) has
unique extended solution X = φ̃aε in L2 ((0, b), dx) for arbitrary b > a, which is real-valued
almost everywhere in [0, b], and φ̃aε (x) = φaε (x) for almost every 0 < x < a.
   Moreover, if ω > 1, the integral equation (4.2) has unique extended solution X = φ̃aε
in C 0 (0, ∞), which is real-valued on [0, ∞) and satisfies φ̃aε (x) = φaε (x) for 0 < x < a.

Proof. The solution φaε of Lemma 4.5 is extended to the solution φ̃aε on (0, b) by
                                               Z a
                        φ̃aε (x) = hω (ax) − ε     hω (xy)φaε (y) dy.                    (4.3)
                                                 0

The right-hand side belongs to L2 ((0, b), dx) by the Cauchy-Schwartz inequality, since
hω (x) belongs to L2 ((0, b′ ), dx) for every 0 < b′ < ∞ when ω > 1/2 and the integral on
the right-hand side vanishes for almost every 0 < x < 1/a. Clearly, φ̃aε (x) = φaε (x) for
almost every 0 < x < a. Conversely, equality (4.3) shows that every solution of (4.2) on
(0, b) is determined by its restriction on (0, a). Hence the uniqueness of solutions follows
from Lemma 4.5. By the way of the extension, φ̃aε is real-valued almost everywhere.
   If ω > 1, we obtain unique extended continuous solution φ̃aε on (0, ∞) by (4.3), since
hω (x) is continuous on (0, ∞) and C 0 (0, a) ⊂ L2 ((0, a), dx).                          

   In what follows, we denote by φaε the extended solution φ̃aε for a > 1 if no confusion
arise. For 0 < a ⩽ 1, we take the convention that
                          φ+        −
                           a (x) = φa (x) = hω (ax) x ∈ (0, ∞).

Obviously, these are continuous on (0, ∞) if ω > 1. This convention is compatible with
Lemma 4.5 and 4.6, since integral equation (4.2) for 0 < a ⩽ 1 should be X(x) = hω (ax)
by Lemma 4.2, and hω (ax) = 0 on (0, a) for 0 < a ⩽ 1. Then its extension φ̃aε (x) to
(0, ∞) should be hω (ax) by (4.3).
14                                           M. SUZUKI

4.2. Differentiability of the solution. In this part, we handle the differentiability
of the extended solution φεa (x) with respect to x and a under the restriction to the
parameter ω > 1. This restriction is required in order to obtain the continuity of the
kernel K(x, y) = hω (xy).
   Let a > 1. The solution φaε of (4.2) is related to the kernel of the resolvent (1−λHω,a )−1
as follows. The kernel K(x, y) = hω (xy) of Hω,a is continuous on [0, a] × [0, a] by the
assumption ω > 1. Then there exists a continuous function R(x, y; λ; a) for (x, y, λ) ∈
[0, a] × [0, a] × C satisfying integral equations
                                      Z a
                    R(x, y; λ; a) − λ     K(x, z)R(z, y; λ; a) dz = K(x, y),
                                       0
                                      Z a                                               (4.4)
                    R(x, y; λ; a) − λ     K(z, y)R(x, z; λ; a) dz = K(x, y)
                                        0
(see Smithies [25, Chap. V], Lax [19, Chap. 24], for example). By taking y = a and
λ = −ε in the first equation of (4.4), we have
                                    Z a
                 R(x, a; −ε; a) + ε     hω (xz)R(z, a; −ε; a) dz = hω (ax).
                                        0
Therefore, we obtain
                                      φεa (x) = R(x, a; −ε; a)                                   (4.5)
for 0 < x < a by the uniqueness of solutions of (4.2). In particular, we obtain the
continuity of φεa (x) for x again, and
                              lim φεa (a) = lim R(a, a; −ε; a) = 0                               (4.6)
                             a→1+            a→1+

by Lemma 4.2. We investigate the differentiability of φεa (x) by using the resolvent kernel
R(x, y; λ; a). The following inequality is going to be used often.
Hadamard’s inequality (see [25, Theorem 5.2.1], for example). Let A = (aij ) be a
n × n complex matrix. If |aij | ⩽ M (1 ⩽ i, j ⩽ n), then | det A|2 ⩽ nn M 2n .
     We introduce the notation
                                                                                            
                                              K(x1 , y1 ) K(x1 , y2 )      · · · K(x1 , yn )
                                           K(x2 , y1 ) K(x2 , y2 )
               x1 , x2 , · · · , xn                                       · · · K(x2 , yn ) 
                                                                                             
           K                          = det     ..          ..            ..       ..       
               y1 , y2 , · · · , yn              .           .                .     .       
                                              K(xn , y1 ) K(xn , y2 )      · · · K(xn , yn )
as usual. The Fredholm determinant d(λ; a) and the first Fredholm minor D(x, y; λ; a)
of the continuous kernel K(x, y) on Ωa = [0, a] × [0, a] are defined as follows:
                                                  ∞
                                                  X
                                      d(λ; a) =         dn (a)λn ,                               (4.7)
                                                  n=0
                                                  X∞
                               D(x, y; λ; a) =          Dn (x, y; a)λn ,                         (4.8)
                                                  n=0

where d0 (a) = 1, D0 (x, y; a) = K(x, y) and
                            Z       Z a                       
                     (−1)n a              x1 , x2 , · · · , xn
           dn (a) =             ···    K                         dx1 . . . dxn (n ⩾ 1),  (4.9)
                       n!    0       0    x1 , x2 , · · · , xn
                            Z       Z a                     
                     (−1)n a              x, x1 , · · · , xn
      Dn (x, y; a) =            ···    K                        dx1 . . . dxn (n ⩾ 1).  (4.10)
                       n!     0      0    y, x1 , · · · , xn
The kernel Dn (x, y; a) are clearly continuous in (x, y). It is well-known that the series
(4.8) converges uniformly and absolutely in (x, y, λ) when λ is confined in a compact
                     A CANONICAL SYSTEM OF DIFFERENTIAL EQUATIONS                                        15

subset of C, and D(x, y; λ; a) is a continuous function on Ωa for every λ ∈ C (see [25,
Theorem 5.3.1], for example). If d(λ; a) 6= 0, the resolvent kernel R(x, y; λ; a) is given by

                                                      D(x, y; λ; a)
                                    R(x, y; λ; a) =                 .                              (4.11)
                                                        d(λ; a)

Note that d(±1; a) 6= 0 for every a > 1 when K(x, y) = hω (xy) and ω > 1/2 by Lemma
4.4 and Theorem 5.6.1 of [25].

Lemma 4.7 Let ω > 1, a > 1 and ε ∈ {±1}. Then the extended solution φεa (x) is
continuously differentiable on x ∈ [0, ∞) \ {n/a | n ∈ N}.

Proof. By (4.3), (4.5) and (4.11), it is sufficient to prove that the Fredholm minor
D(x, y; λ; a) for the kernel K(x, y) = hω (xy) on Ωa = [0, a] × [0, a] is continuously
differentiable on x ∈ Da := [0, a] \ {n/a | n ∈ N} for every fixed y ∈ [0, a], since hω (ax)
                                                                                       ∂
is continuously differentiable on x ∈ [0, ∞) \ {n/a | n ∈ N}, and | ∂x                   hω (xy)φεa (y)| is
integrable on [0, a]. To prove it, we modify the proof of Theorem 5.3.1 in [25]. We have
                                                                                      
                                              K(x, y) K(x, x1 ) · · · K(x, xn )
                                          K(x1 , y) K(x1 , x1 ) · · · K(x1 , xn ) 
          x, x1 , · · · , xn                                                          
      K                          = det          ..           ..    ..        ..       
           y, x1 , · · · , xn                    .            .        .      .       
                                              K(xn , y) K(xn , x1 ) · · · K(xn , xn )
                                                                                                
                                                           0       K(x, x1 ) · · · K(x, xn )
                                                     K(x1 , y) K(x1 , x1 ) · · · K(x1 , xn ) 
                              x1 , · · · , xn                                                   
         = K(x, y)K                             + det     ..             ..     . .       ..    .
                              x1 , · · · , xn              .              .         .      .    
                                                        K(xn , y) K(xn , x1 ) · · · K(xn , xn )

Therefore, by (4.9) and (4.10), we have
      ∂                        ∂
         Dn (x, y; a) = dn (a) K(x, y)
      ∂x                      ∂x                                                      
                                                  ∂                      ∂
                Z a     Z a
                                    0             ∂x K(x, x1 )    ···    ∂x K(x, xn )
         (−1) n                 
                                 K(x  1 , y)     K(x1 , x1 )     ···    K(x1 , xn ) 
                                                                                     
      +             ···     det     ..                 ..        ..          ..        dx1 . . . dxn
           n!     0      0           .                  .           .         .       
                                 K(xn , y)         K(xn , x1 )    ···    K(xn , xn )
                ∂
      = dn (a) K(x, y) + Dn† (x, y; a),
               ∂x
say. Then, we obtain
                                                         X         ∞
                    ∂                          ∂
                       D(x, y; λ; a) = d(λ; a) K(x, y) +   Dn† (x, y; a)λn
                    ∂x                        ∂x
                                                                  n=0

by (4.8). The first term on the right-hand side is continuous on x ∈ Da . Therefore,
                                                          ∂
in order to prove the existence and the continuity of ∂x    D(x, y; λ; a) on x ∈ Da , it is
sufficient to prove that the series on the right-hand side converges uniformly on every
compact subset in Da . Put
                                                            Z a
                                                                 ∂
           M1 (a) = a sup |K(x, y)|,        M2 (a) = sup            K(x, y) dy.
                      (x,y)∈Ωa                       x∈[0,a] 0  ∂x
                                                 Ra ∂               Ra
The second constant M2 (a) is well-defined, since 0 ∂x K(x, y) dy = 0 |h′ω (xy)| ydy
is continuous on [0, a] (by ω > 1). Using the row expansion of the determinant and
16                                                 M. SUZUKI

Hadamard’s inequality, we have
                ∂                              ∂
                                                                    
          0      ∂x K(x, x1 ) · · ·             ∂x K(x, xn )                                    n X
      K(x1 , y) K(x1 , x1 ) · · ·
                                                K(x1 , xn ) 
                                                                             1
                                                                                n       M1 (a)
                                                                                                    n
                                                                                                         ∂
 det     ..          ..      ..                        ..           ⩽n      2                             K(x, xj ) .
          .           .          .                      .                              a               ∂x
                                                                                                   j=1
       K(xn , y) K(xn , x1 ) · · ·               K(xn , xn )
Therefore, we obtain
                             1
                               n               n Z a             Z aX
                                                                      n
                         n   2         M1 (a)                                 ∂
       |Dn† (x, y; a)| ⩽                                     ···                 K(x, xj ) dx1 . . . dxn
                          n!            a           0              0          ∂x
                                                                        j=1
                             1
                               n            n Z a
                                            X                                                       1
                          n  2                               ∂                  M2 (a) n 2 n M1 (a)n
                      =       M1 (a)n                           K(x, xj ) dxj ⩽                      .
                          an!                      0         ∂x                  a       (n − 1)!
                                            j=1
                      P       †
Therefore, the series ∞                  n
                        n=0 Dn (x, y; a)λ converges uniformly and absolutely in (x, y, λ) ∈
Ωa × C, when λ is contained in a compact subset of C. Hence, for fixed y ∈ [0, a],
D(x, y; λ; a) − d(λ; a)K(x, y) is a continuously differentiable function on [0, a] such that
                   ∂                                X ∞
                      D(x, y; λ; a) − d(λ; a)K(x, y) =     Dn† (x, y)λn .
                   ∂x                                  n=0

We complete the proof of the lemma.                                                                                  
Lemma 4.8 Let ω > 1 and ε ∈ {±1}. Then the extended solution φεa (x) is continuous
in a ∈ (1, ∞) for every fixed √
                              x > 0. In addition, it is continuously differentiable with
respect to a in (1, ∞) \ {n/x, n | n ∈ N}.
Proof. The continuity in a follows from (4.3) and (4.5). Before the proof of the differen-
tiability, we note that d(λ, a) is continuous in a. In fact, we have
                              1
                                                                      !
                            n 2 n M1 (a)n
                 |dn (a)| ⩽                 M1 (a) = a sup |K(x, y)|
                                  n!                    (x,y)∈Ωa

by definition (4.9) and Hadamard’s inequality, and hence the series of (4.7) converges
absolutely and uniformly on a compact subset of (λ, a) ∈ C × (1, ∞).
  Let λ ∈ C such that d(λ; a) 6= 0 for every a > 1. We have
                                ∂                                        ∂
             ∂                     D(x, a; λ; a)d(λ; a) − D(x, a; λ; a) ∂a d(λ; a)
                R(x, a; λ; a) = ∂a                         2
             ∂a                                     d(λ; a)
by (4.11). Therefore, in order to prove the lemma, we need (i) the existence and the
                 ∂
continuity of ∂a   d(λ; a) and (ii) the existence, the continuity and the integrability of
 ∂
∂a D(x, a; λ; a) by (4.3), (4.5) and
                  Z a                                      Z a
               ∂                                                      ∂
                      hω (xy)φεa (y) dy = hω (ax)φεa (a) +     hω (xy) φεa (y) dy.
              ∂a 0                                          0         ∂a
We prove (i). By definition (4.7) and (4.9), we have
                          X∞         Z a      Z a                     
 ∂                             (−λ)n                 a, x2 , · · · , xn
   d(λ; a) = −λK(a, a) +                   ···    K                       dx2 . . . dxn
∂a                               n!     0       0    a, x2 , · · · , xn
                                   n=2
              XZ a
              n−1          Z a                                            
                                  x1 , · · · , xk−1 , a, xk+1 , · · · , xn
           +          ···    K                                               dx1 . . . dxk−1 dxk+1 . . . dxn
                           0      x1 , · · · , xk−1 , a, xk+1 , · · · , xn
             k=2 0
             Z a     Z a                                                  
                             x1 , x2 , · · · , xn−1 , a
           +     ···      K                               dx1 . . . dxn−1 .
              0        0     x1 , x2 , · · · , xn−1 , a
                      A CANONICAL SYSTEM OF DIFFERENTIAL EQUATIONS                                           17

Clearly, each term in the series is continuous in a, since K(x, y) is continuous. By using
Hadamard’s inequality,
                      ∞                                  ∞     1
        ∂            X   |λ|n 1 n M1 (a) n n−1 1 X n 2 n
          d(λ; a) ⩽          n 2              na      =                (|λ|M1 (a))n .
       ∂a            n=1
                          n!           a                a n=1
                                                              (n − 1)!
The series on the right-hand side converges uniformly on a compact subset of (λ, a) ∈
C × [0, ∞). Hence d(λ; a) is continuously differentiable for a.
   Successively, we prove (ii). we have
                          ∂
                             D(x, a; λ; a) = Dy (x, a; λ; a) + Da (x, a; λ; a),
                         ∂a
where Dy (resp. Da ) means the partial derivative with respect to the second (resp. the
fourth) variable. We find that D(x, y; λ; a) is continuously differentiable with respect
to y ∈ [0, ∞) \ {n/a | n ∈ N}, and Dy (x, y; λ; a) is a continuous function on (x, y) ∈
[0, ∞) × ([0, ∞) \ {n/a | n ∈ N}) by a way√similar to the proof of Lemma 4.7. Thus
Dy (x, a; λ; a) is continuous on a ∈ (1, ∞) \ { n | n ∈ N} for fixed x, and |Dy (x, a; λ; a)| is
integrable on [0, a] with respect to x. On the other hand, by definition (4.8) and (4.10),
    ∂
       D(x, y; λ; a) = −λK(a, a)K(x, y) + λK(x, a)K(a, y)
    ∂a
          X∞            Z a       Z a                              
              (−λ)n                            x, a, x2 , · · · , xn
       +                      ···         K                            dx2 . . . dxn
                  n!       0         0         y, a, x2 , · · · , xn
          n=2
          XZ a
          n−1           Z a                                                 
                                 x, x1 , · · · , xk−1 , a, xk+1 , · · · , xn
       +            ···     K                                                    dx1 . . . dxk−1 dxk+1 . . . dxn
                0        0       y, x1 , · · · , xk−1 , a, xk+1 , · · · , xn
          k=2
          Z a      Z a                                                  
                             x, x1 , · · · , xn−1 , a
       +      ···       K                               dx1 . . . dxn−1 .
           0         0       y, x1 , · · · , xn−1 , a
Clearly, each term in the series is continuous in (x, y, a), since K(x, y) is continuous. By
the row expansion of the determinant and Hadamard’s inequality, we obtain
                         ∞                    Z         Z aX  n
     ∂                  X   |λ|n 1 n M1 (a) n a
        D(x, y; λ; a) ⩽          n 2                  ···         |K(x, xj )| dx1 . . . dxn−1
     ∂a                 n=1
                             n!           a        0       0              j=1
                                   Z a                     ∞
                                                           X       1
                                                                     n
                              1                                  n 2
                          =              |K(x, x1 )| dx1                (|λ|M1 (a))n .
                              a2    0                      n=1
                                                               (n − 1)!
when 0 ⩽ y ⩽ a. The series on the right-hand side converges uniformly on a compact
subset of (λ, a) ∈ C × (1, ∞). Thus Da (x, a; λ; a) is continuous in a. In addition, the
right-hand side shows that |Da (x, a; λ; a)| is integrable on [0, a] with respect to x.
             ∂                                                  √
    Hence ∂a   R(x, a; λ; a) is continuous on a ∈ (1, ∞) \ { n | n ∈ N} for fixed x, and
  ∂
| ∂a R(x, a; λ; a)| is integrable on [0, a] with respect to x. As a consequence we obtain the
lemma by (4.3) and (4.5).                                                                  
4.3. The first order differential system. As in the previous section, we assume that
ω > 1. Then Θω is an inner function in C+ , the kernel hω (xy) of Hω or Hω,a is continuous,
and φaε (x) is continuously differentiable with respect to x and a outside loci ax = k
(k ∈ N). Under this situation, we derive a first order differential system arising from φaε
(a > 1, ε ∈ {±1}) start from
                                   Z a
                          ε
                        φa (x) + ε     hω (xy)φaε (y) dy = hω (ax).                   (4.12)
                                    0
                         ∂
   Firstly, we operate a ∂a on both side of (4.12). Then,
                                     Z a
                    ∂             ∂                           ∂
                  a φaε (x) + εa         hω (xy)φaε (y) dy = a hω (ax);
                    ∂a            ∂a 0                        ∂a
18                                         M. SUZUKI
                                           Z a
            ∂ ε             ε                            ∂              ∂
          a φa (x) + εaφa (a)hω (ax) + ε       hω (xy)a φaε (y) dy = a hω (ax);
           ∂a                               0           ∂a             ∂a
                    Z a
       ∂                        ∂                                    ∂
      a φaε (x) + ε     hω (xy)a φaε (y) dy = −ε a φaε (a)hω (ax) + a hω (ax).            (4.13)
       ∂a            0          ∂a                                   ∂a
                            ∂
     Secondly, we operate x ∂x on both side of (4.12):
                               Z a
               ∂ ε           ∂                          ∂           ∂
            x φa (x) + εx          hω (xy)φaε (y) dy = x hω (ax) = a hω (ax).
              ∂x            ∂x 0                        ∂x          ∂a
                       ∂             ∂
Using the identity x ∂x  hω (xy) = y ∂y hω (xy) and then applying integration by parts to
the integral of the left-hand side, we have
                                          Z a
         ∂ ε              ε                           ∂  ε             ∂
       x φa (x) + ε a φa (a)hω (ax) − ε       hω (xy)    yφa (y) dy = a hω (ax).
         ∂x                                 0         ∂y                ∂a
                 ∂
Putting δx = x ∂x  + 12 = ∂x
                           ∂
                             x − 12 , we obtain
                               Z a                             
                   1                                    1
      δx φaε (x) − φaε (x) − ε      hω (xy) δy φaε (y) + φaε (y) dy
                   2            0                       2
                                                                                          (4.14)
                                                                     ∂
                                            = −ε a φaε (a)hω (ax) + a hω (ax).
                                                                         ∂a
Next, we rewrite the left-hand side of (4.14) as follows by using (4.12) for the second
term of the left-hand side:
                               Z a                        Z a
                1                                                                 1     
   δx φaε (x) −    hω (ax) − ε     hω (xy)φaε (y) dy − ε       hω (xy) δy φaε (y) + φaε (y) dy
                2               0                           0                        2
                                                                Z a
                                                   1
                                    = δx φaε (x) − hω (ax) − ε       hω (xy)δy φaε (y) dy.
                                                   2               0
                                                                                           (4.15)
Substituting the right-hand side of (4.15) for the left-hand side of (4.14) and rearranging,
we obtain
                  Z a                                        
           ε                      ε              1      ε                    ∂
    δx φa (x) − ε      hω (xy)δy φa (y) dy =       − εaφa (a) hω (ax) + a hω (ax).         (4.16)
                    0                            2                          ∂a
Subtracting (4.16) with choice −ε from (4.13) with ε, we obtain
                                  Z a                            
          ∂ ε            −ε                       ∂ ε           −ε
        a φa (x) − δx φa (x) + ε        hω (xy) a φa (y) − δy φa (y) dy
          ∂a                         0           ∂a
                                                                                        (4.17)
                                                        1
                                                 =−       + εµ(a) hω (ax),
                                                        2
where
                                  µ(a) = aφ+         −
                                           a (a) + aφa (a).                               (4.18)
By (4.6), Lemma 4.7 and 4.8, the function µ(a) is continuous on √  (1, ∞), which satisfies
lima→1+ µ(a) = 0, and is continuously differentiable on (1, ∞) \ { n | n ∈ N }.
                                 ∂ ε
   Equality (4.17) shows that a ∂a φa (x) − δx φ−ε
                                                a (x) is a continuous solution of (4.12).
Hence, by comparing (4.12) with (4.17), we obtain
                                                    
                ∂ ε           −ε            1
              a φa (x) − δx φa (x) = −        + εµ(a) φaε (x) (ε ∈ {±1})           (4.19)
                ∂a                          2
by the uniqueness of solutions (Lemmas 4.5 and 4.6). We use (4.19) in the form
                                  
                     ∂     1
                   a    + + εµ(a) φaε (x) = δx φ−ε
                                                 a (x) (ε ∈ {±1}).             (4.20)
                     ∂a 2
                    A CANONICAL SYSTEM OF DIFFERENTIAL EQUATIONS                                 19

  Now we introduce two special functions
                      √ Z ∞                           √
                aiz     a            1
                                       +iz dx   aiz     a
     Ãa (z) :=     +         +
                            φa (x) x 2        =     +     F (1 − Pa )φ+
                                                                      a (z),
                 2     2 a                  x    2     2 1/2
                      √ Z ∞                           √                                      (4.21)
                aiz     a            1
                                       +iz dx   aiz     a
   −iB̃a (z) :=     −         −
                            φa (x) x 2        =     −     F (1 − Pa )φ−
                                                                      a (z)
                 2     2 a                  x    2     2 1/2
for ℑ(z) ≫ 0 and a > 1. These functions are defined as analytic functions for large
ℑ(z) > 0 by integrals, since φ±
                              a are continuous and have at most polynomial growth at
+∞ by (4.12) and the rough estimate
                   Z U +ic
                 1                   1                1
       hω (x) =            Θω (z) x− 2 −iz dz + O(xc− 2 U 1−ω ) (c > 1/2 + ω)
                2π −U +ic                                                      (4.22)
                         1              1
               = O(xc− 2 U ) + O(xc− 2 U 1−ω ) = O(xω+ǫ ).
As shown below, Ãa and B̃a are analytically continuable to meromorphic functions in
C. We put it off a little and derive a differential system satisfied by Ãa and B̃a . Using
(4.20), we have
                                                               √ Z ∞
      ∂                                     aiz       ∂               a               1
                                                                                         +iz dx
    a    + µ(a) Ãa (z) = (iz + µ(a))           + a       + µ(a)             φ+
                                                                              a (x) x
                                                                                       2
      ∂a                                     2       ∂a              2 a                      x
                       iz   √                    √ Z ∞                       
                     a        a +         1        a           ∂     1                      1
                                                                                              +iz dx
     = (iz + µ(a))        −     φ (a) a 2 +iz +              a    + + µ(a) φ+      a (x) x
                                                                                            2
                      2      2 a                  2 a          ∂a 2                               x
                            √                    √   Z   ∞            
                     aiz      a +         1        a            ∂    1            1
                                                                                    +iz dx
     = (iz + µ(a))        −     φa (a) a 2 +iz +             x     +     φ−
                                                                          a (x) x
                                                                                  2
                      2      2                    2 a          ∂x 2                       x
            iz      √ Z ∞
          a           a                1
                                         +iz dx
     = iz      − iz         φ− a (x) x
                                       2        = z B̃a (z)
           2         2 a                      x
for large ℑ(z) > 0, and then it holds for all z ∈ C by meromorphic continuation (below).
                                     ∂
We obtain a similar formula for (a ∂a   + µ(a))B̃a . As a result, we obtain the first order
differential system
          ∂                                                  
           a ∂a + µ(a)      0         Ãa (z)        0 −1 Ãa (z)
       −                 ∂                      =z                    (a > 1).       (4.23)
                0      a ∂a − µ(a) B̃a (z)           1 0    B̃a (z)
We extend the system to a > 0 by taking the convention that
                                             µ(a) = 0                                        (4.24)
and
                          √ Z ∞                                         
                    aiz     a              1     dx   1  iz
          Ãa (z) =     +        hω (ax) x 2 +iz    =    a + Θω (z)a−iz ,
                     2     2 1/a                 x    2
                          √   Z                                                             (4.25)
                    aiz     a ∞            1     dx   1  iz
        −iB̃a (z) =     −        hω (ax) x 2 +iz    =    a − Θω (z)a−iz
                     2     2 1/a                 x    2
for 0 < a ⩽ 1. Actually the convention (4.24) and (4.25) for 0 < a ⩽ 1 is compatible
with Lemma 4.2 and the convention mentioned in the end of Section 4.1.
   For a > 0, we define
                                                    
                                           1
                        Aa (z) = m(a) ξ      + ω − iz Ãa (z),
                                           2
                                                                            (4.26)
                                  1        1
                        Ba (z) =       ξ     + ω − iz B̃a (z)
                                 m(a)      2
with                             Z a                                   
                                            db                    d
                    m(a) = exp         µ(b)              µ(a) = a log m(a)                   (4.27)
                                     1       b                   da
20                                        M. SUZUKI

under (4.24) and (4.25). Note that m(a) is real-valued by its definition. Then we can
verify that system (4.23) implies that (Aa , Ba ) satisfies the canonical system
                                                         
          ∂ Xa (z)         0 −1 m(a)−2            0       Xa (z)
      −a             =z                                            (0 < a < ∞)   (4.28)
         ∂a Ya (z)         1 0         0      m(a)2 Ya (z)
by elementary ways. It is concluded that (4.28) is the canonical system of Theorem 2.3
if formula
                                        det(1 + Hω,a )
                                m(a) =
                                        det(1 − Hω,a )
                            det(1+H   )
is proved for a > 1, since det(1−Hω,a   ) = 1 for 0 < a ⩽ 1 by Lemma 4.2. This will follow
                                    ω,a


from showing
                                           d
                               φ+
                                a (a) =       log det(1 + Hω,a ),
                                          da
                                             d
                               φ−
                                a (a) = −       log det(1 − Hω,a )
                                            da
by definition (4.18) and (4.27). This is a well-known formula for an integral operator
defined on a finite interval with a continuous kernel. In fact, it is proved by a way similar
to the proof of Theorem 12 of Chapter 24 in [19]. (This also holds for 0 < a < 1, since
φ±             2
  a (a) = hω (a ) = 0 by the convention in the end of Section 4.1 and log det(1± Hω,a ) = 0
by Lemma 4.2.)
  For every fixed 0 < a ⩽ 1, Aa and Ba are real entire functions satisfying Aa (−z) =
Aa (z) and Ba (−z) = −Ba (z), respectively, by (4.25), (4.26) and functional equations
ξ(s) = ξ(1 − s) and ξ(s) = ξ(s̄). Successively, we prove that Aa and Ba have these
properties for a > 1.

4.4. Meromorphic continuation and functional equations. Under assumptions
and notations of Section 4.3, we define
Ẽa (z) := Ãa (z) − iB̃a (z)
                 √ Z ∞                                           √
                    a                           1
                                                  +iz dx    iz     a
        = aiz +            (φ+  (x) − φ−
                                         (x)) x 2        = a   +     F (1 − Pa )(φ+    −
                                                                                  a − φa )(z)
                   2 a        a        a
                                                       x          2 1/2
Ẽa∗ (z) := Ãa (z) + iB̃a (z)
           √ Z ∞                                    √
              a                          1
                                           +iz dx     a
         =              +       −
                     (φa (x) + φa (x)) x 2        =     F (1 − Pa )(φ+    −
                                                                     a + φa )(z)
             2 a                                x    2 1/2
                                                                                      (4.29)
for ℑ(z) ≫ 0 and a > 1. We deal with Aa , Ba via Ẽa (z) and Ẽa∗ (z).
Lemma 4.9 Let ω > 1 and a > 1. Define
                             Z ∞
                                                     1
                                                       +iz dx
                    Ψa (z) =     (φ+        −
                                   a (x) − φa (x)) x
                                                     2        .                       (4.30)
                              a                            x
Then integral of (4.30) converges absolutely for ℑ(z) > 0 and converges in the L2 -sense
on ℑ(z) = 0. Moreover Ψa (z) is extended to a meromorphic function in C which is
analytic in C+ .
Proof. By (4.12), we have
                              φ+    −            +    −
                               a − φa = −Hω Pa (φa + φa ),                            (4.31)
where Pa (φ+      −                                                 +     −
            a + φa ) has compact support in [1/a, a]. Therefore φa − φa belongs to
 2
L ((α, ∞), dx) for every α > 0. Hence integral (4.30) converges absolutely for ℑ(z) > 0
                     A CANONICAL SYSTEM OF DIFFERENTIAL EQUATIONS                                    21

and defines a function of H 2 ([29, Chap. II, §10]). Using (4.31), we have
               Ψa (z) = F1/2 (1 − Pa )(φ+    −
                                        a − φa )(z)
                       = −F1/2 (1 − Pa )Hω Pa (φ+    −
                                                a + φa )(z)
                       = −F1/2 Hω Pa (φ+    −                        +    −
                                       a + φa )(z) + F1/2 Pa Hω Pa (φa + φa )(z)

for ℑ(z) ≫ 0. Here Pa (φ+    −                  +    −
                        a + φa ) and Pa Hω Pa (φa + φa ) have compact support in (0, ∞).
Therefore, we obtain
        Ψa (z) = −Θω (z)F1/2 Pa (φ+    −                         +    −
                                  a + φa )(−z) + F1/2 Pa Hω Pa (φa + φa )(z),                   (4.32)
where F1/2 Pa (φ+     −                          +    −
                a + φa )(−z) and F1/2 Pa Hω Pa (φa + φa )(z) are entire functions. Hence
Ψa (z) is extended to a meromorphic function on C, and is analytic in C+ by (4.32), since
Θω (z) is a meromorphic inner function in C+ .                                         
Lemma 4.10 Let ω > 1 and a > 1. Functions Ẽa and Ẽa∗ of (4.29) are analytically
continuable to meromorphic functions in C satisfying Ẽa∗ (z) = Θω (z)Ẽa (−z), and they
are analytic in C+ . Moreover, both ξ( 12 + ω − iz)Ẽa (z) and ξ( 21 + ω − iz)Ẽa∗ (z) are entire
functions.
Proof. We have
                       Hω Pa (φ+    −                    +        −
                               a − φa )(x) = 2hω (ax) − φa (x) − φa (x)                         (4.33)
by (4.12). Using (1.8), (4.31), and (4.32), we have
                                             √
                              −iz               a
 Θω (z)Ẽa (−z) = Θω (z)a          + Θω (z)       Ψa (−z)
                                          √    2                           √
  (4.32)       −iz                          a            +      −            a
    = Θω (z)a       − Θω (z)Θω (−z)           F1/2 Pa (φa + φa )(z) +          Θω (z)F1/2 Pa Hω Pa (φ+      −
                                                                                                      a + φa )(−z)
                       √                   2                   √            2
  (1.8)                  a            +       −                    a
   = Θω (z)a  −iz
                   −       F1/2 Pa (φa + φa )(z) + Θω (z)            F1/2 Pa Hω Pa (φ+       −
                                                                                     a + φa )(−z)
                        2
                        √                                        √2
  (4.31)                  a                                         a
    = Θω (z)a−iz −           F1/2 Pa (φ+  +  φ −
                                                 )(z) −  Θ ω (z)      F Pa (φ+         −
                                                                                a − φa )(−z)
                         2             a       a
                                                                   2 1/2
for z ∈ C, since Pa (φ+           −                      +      −
                          a ± φa ), and Pa Hω Pa (φa + φa ) have compact support. Further,
by Proposition 2.1, Lemma 4.1, and (4.33), we have
                                     √                                        √
                              −iz      a             +     −                    a
 Θω (z)Ẽa (−z) = Θω (z)a         −      F1/2 Pa (φa + φa )(z) − Θω (z)           F1/2 Pa (φ+      −
                                                                                             a − φa )(−z)
                                      2√                                 √     2
                (4.1)                     a            +      −            a
                 = Θω (z)a      −iz
                                    −       F1/2 Pa (φa + φa )(z) −          F1/2 Hω Pa (φ+       −
                                                                                           a − φa )(z)
                                        √2                                2
                (4.33)           −iz       a
                  = Θω (z)a           −      F Pa (φ+ + φ−     a )(z)
                                          2 1/2 √a
                          √                             a
                       − a F1/2 (hω (ax))(z) +            F1/2 (φ+        −
                                                                   a + φa )(z)
                                       √               2                                √
                (2.4)                     a            +      −                  −iz       a
                 = Θω (z)a      −iz
                                    −       F1/2 Pa (φa + φa )(z) − Θω (z)a          +       F1/2 (φ+    −
                                                                                                    a + φa )(z)
                    √                    2                                               2
                       a
                =        F (1 − Pa )(φ+            −
                                           a + φa )(z) = Ẽa (z)
                                                                ∗
                      2 1/2
for ℑ(z) ≫ 0, since φ+            −
                          a + φa is identically zero on (0, 1/a) and has polynomial growth
                             ∗
at x = +∞. Hence Ẽa (z) = Θω (z)Ẽa (−z) for ℑ(z) ≫ 0. By Lemma 4.9, Ẽa (z) is
meromorphic in C, therefore, Ẽa∗ (z) is also analytically continuable to a meromorphic
function in C. Moreover, Ẽa (z) = Θω (z)(entire) + (entire) from the proof of Lemma 4.9.
Thus
                       Ẽa∗ (z) = Θ(z)Ẽa (−z) = (entire) + Θω (z)(entire)
22                                        M. SUZUKI


by (1.8), and hence Ẽa∗ (z) is analytic in C+ . Simultaneously, these equalities show that
ξ( 21 + ω − iz)Ẽa (z) and ξ( 12 + ω − iz)Ẽa∗ (z) are entire by definition of Θω (z).   
     Lemma 4.10 implies the following immediately.
Lemma 4.11 Let ω > 1 and a > 1. Then Ãa (z) and B̃a (z) are analytically continuable
to meromorphic functions in C, and they are analytic in C+ . Also, Aa (z) and Ba (z) are
both entire functions. In addition, we have functional equations
                   Θω (z)Ãa (−z) = Ãa (z),     Θω (z)B̃a (−z) = −B̃a (z),
                         Aa (−z) = Aa (z),       Ba (−z) = −Ba (z).

Proof. We have 2Ãa = Ẽa + Ẽa∗ and −2iB̃a = Ẽa − Ẽa∗ by definition (4.29). Therefore,
they are analytically continuable to meromorphic function in C and satisfy 2Ãa (z) =
Ẽa (z) + Θω (z)Ẽa (−z) and −2iB̃a (z) = Ẽa (z) − Θω (z)Ẽa (−z) by Lemma 4.10. That
imply the functional equations stated in the lemma. Other things are consequences of
Lemma 4.10.                                                                            
Lemma 4.12 Let ω > 1 and a > 1. Then Aa (z) and Ba (z) are real entire functions.
Proof. At first, we note that if F (z) = F1/2 (f (x))(z) for ℑ(z) ≫ 0, then F ♯ (z) =
F1/2 (x−1 f (x−1 ))(z) and F (−z) = F1/2 (x−1 f (x−1 ))(z) for ℑ(z) ≪ 0. Therefore, if f (x)
(resp. if (x)) is real-valued, F (z) is analytically continued to a meromorphic function in
C, and F (−z) = F (z) (resp. F (−z) = −F (z)), then F ♯ (z) = F (z) holds for z ∈ C. Let
                    1 d 2 d               X∞
                                                                     
           φ(x) :=        x         2
                                 θ(x ) = 2       2π 2 n4 x4 − 3πn2 x2 exp(−πn2 x2 ).
                    2 dx      dx            n=1
                                        R∞
Then φ(1/x) = xφ(x) and ξ(s) = 0 φ(x)xs dx          x for every s ∈ C ( [28, §10.1]). Hence
   1                       −ω
ξ( 2 + ω − iz) = F1/2 (x φ(x))(z) for z ∈ C. On the other hand, by (4.21), Ãa (z) =
       √                                              √
                                                       i a
F1/2 ( 2a (δa + (1 − Pa )φ+                                              −
                          a ))(z) and B̃a (z) = F1/2 ( 2 (δa − (1 − Pa )φa ))(z) for ℑ(z) ≫ 0 ,
where δa (x) is the Dirac delta-function at x = a. Therefore Aa (z) = F1/2 (f + (x))(z) and
Ba (z) = F1/2 (f − (x))(z) for
                               √ Z
              +        m(a) a ∞                                                  dy
            f (x) : =                  (x/y)−ω φ(x/y)(δa (y) + (1 − Pa )φ+a (y))
                             2     0                                              y
                                −ω
                                                        Z ∞                         
                       m(a)x           ω− 12         √               +       ω−1
                    =                a       φ(x/a) + a       φ(x/y)φa (y) y     dy ,
                             2                             a
                             √ Z ∞
                            i a                                                  dy
                −
              f (x) : =                (x/y)−ω φ(x/y)(δa (y) − (1 − Pa )φ−a (y))
                          2m(a) 0                                                 y
                               −ω                      Z ∞                         
                           ix         ω− 12          √              −       ω−1
                      =             a       φ(x/a) − a       φ(x/y)φa (y) y     dy
                          2m(a)                          a
if ℑ(z) ≫ 0. Here f + (x) and if − (x) are both real-valued, since m(a) is real, and φ(x),
φ±a (x) are real-valued. In addition, Aa (−z) = Aa (z) and Ba (−z) = −Ba (z) for z ∈ C
by Lemma 4.11. Hence A♯a = Aa and Ba♯ = Ba .                                            
   Now we complete the proof of Theorem 2.3 (1), (2). The remaining assertion is (3).
In order to prove it, we show the following lemma.
Lemma 4.13 Let ω > 1. Then
                       lim Aa (z) = Aω (z),        lim Ba (z) = B ω (z)                 (4.34)
                      a→1+                        a→1+

hold uniformly on every compact subset in C.
                   A CANONICAL SYSTEM OF DIFFERENTIAL EQUATIONS                             23

Proof. By (4.2) and (4.3), φ±                                                +
                              a (x) → hω (x) uniformly on [1/2, 3/2] as a → 1 . Therefore,
by (4.29) and (4.32), Ẽa (z) converges to a meromorphic function in C uniformly on every
compact subset in C as a → 1+ , since Pa (φ+        −                     +    −
                                              a + φa )(−z) and Pa Hω Pa (φa + φa )(z) both
have support in [1/a, a]. On the other hand, we have
                                         Z a
                     +         −
                   φa (x) − φa (x) = −        hω (xz)(φ+        −
                                                       a (z) + φa (z)) dz            (4.35)
                                           1/a

by (4.12), since φa± (x) = 0 for 0 < x < 1/a. Multiplying by x−v on both sides of (4.35),
and then tending a → 1+ , we have
                                               −       −v
                                lim (φ+
                                      a (x) − φa (x))x    =0
                               a→1+
uniformly on (1, ∞) if v > 0 is large, since hω is of polynomial growth at +∞. Hence
lima→1+ Ẽa (z) = 1 uniformly on every compact subset in ℑ(z) > v. As a consequence
lima→1+ Ẽa (z) = 1, and
                             1                              i
                lim Ãa (z) = (1 + Θω (z)),    lim B̃a (z) = (1 − Θω (z))
               a→1+          2                a→1+          2
uniformly on every compact subset in C. Multiplying by ξ( 12 + ω − iz) on both sides of
these equalities, we obtain (4.34) by (4.6) and (4.26).                              
By definition, we have m(a) = 1 and
                                                         
                               1           iz     1
                  Aa (z) = ξ     + ω − iz a + ξ     − ω − iz a−iz ,
                               2                  2
                                                         
                               1           iz     1
                  Ba (z) = ξ     + ω − iz a − ξ     − ω − iz a−iz
                               2                  2
for 0 < a ⩽ 1. This shows (A1 , B1 ) = (Aω , B ω ) and lim (Aa (z), Ba (z)) = (Aω (z), B ω (z))
                                                      a→1−
uniformly on every compact subset in C. Together with Lemma 4.13, we obtain Theorem
2.3 (3), and hence we complete the proof of Theorem 2.3.

                  5. Comments on the validity of Theorem 2.3
    In this section, we comment on a range of ω > 0 where Theorem 2.3 is expected to be
extended. There might be three levels of difficulties at least: (i) ω > 1/2, (ii) ω = 1/2,
(iii) 0 < ω < 1/2.
    It is natural to expect that Theorem 2.3 is proved unconditionally for the range (i)
as mentioned after Theorem 2.3. In fact, all lemmas in Section 4.1 are already proved
for ω > 1/2. Therefore, the remaining problems are a proof of the differentiability of
φεa (x) with respect to x and a, and formula of m(a) by determinants. However, if we
                                 ∂ ε           ∂ ε
understand partial derivatives ∂x  φa (x) and ∂a φa (x) in the sense of distributions as in
Burnol [6], and if we use the theory of Fredholm determinants for L2 -kernels ( [25, Chap.
VI]), then most of Section 4.3 and 4.4 have reasonable meaning, and we may obtain
Theorem 2.3 for ω > 1/2. This way is plausible, and must be carried out after a suitable
preparation for the theory of distributions.
    The case (ii) have more difficulties, because the kernel of Hω,a is no longer Hilbert-
Schmidt type. However, Θω (z) is still inner function in C+ unconditionally. Therefore,
problems may be restricted to the theory of integral operators, its determinants, and the
theory of integral equations only as well as the case (i). See the later half of comments
on (iii) below.
  It is easily predicted that it is very hard to generalize Theorem 2.3 to the range (iii)
unconditionally. A reason of difficulties is that problems of arithmetic and analysis are
mixed in this range. However, if we assume RH, the function Θω is inner in C+ for
every ω > 0, and hence remaining problems may be restricted to the theory of integral
24                                            M. SUZUKI

operators and the theory of integral equations only. Such analytic problems may be
solved without essential difficulties.
    In fact, if Θω (z) is an inner function in C+ , Hω is extended to an isometry on
  2
L ((0, ∞), dx) by Lemma 4.1. On the other hand, Hω,a is a compact operator on
L2 ((0, a), dx) even for (iii) (and (ii)) because its kernel is a sum of finitely many weakly
singular kernels. Therefore, in particular, the Fredholm alternative holds. Hence we may
obtain reasonable generalization of results in Section 4.1 for ω > 0 under RH, and then
throughout distribution theoretic dealing of Section 4.3 and 4.4, we may arrive at the
generalization of Theorem 2.3 for the range (iii) (and (ii)) under RH. In this strategy, it
is necessary to note that φεa (x) have some possible singularities, which affect definition
(4.18) of µ(a) and definition (4.27) of m(a), and that the definition of determinants
det(1 ± Hω,a ) should be changed as in König [14].
     We leave a justification of the above argument for a future study.

                                        Appendix A.
   Suppose that Θω (z) is an inner function in C+ . (It holds unconditionally for ω ⩾ 1/2,
and also for 0 < ω < 1/2 under RH.) Then it defines the reproducing kernel Hilbert
space K(Θω ) which is isomorhic to the de Branges space B(E ω ) (see Section 1.3 and
1.4). According to the theory of de Branges [9], the structure of B(E ω ) is determined by
associated canonical system, which was described in terms of the shifted Fourier inversion
hω (x) of Θω (z) under the restriction ω > 1. On the other hand, the structure of B(E ω )
is also determined by the reproducing kernel of K(Θω ):
                                    1 1 − Θω (z)Θω (w)
                      Kω (z, w) =                          (z, w ∈ C+ )
                                   2πi        z̄ − w
(see Section 1.3). We find that Kω (0, ∗) belongs to L2 (R) by (1.9) and (1.10), and thus
its shifted Fourier inversion F−1                        2
                               1/2 Kω (0, ∗) belongs to L ((0, ∞), dx).
                                                                               −1
   However, if we obtain F−1 1/2 Kω (0, ∗) explicitly enough, we may define F1/2 Kω (0, ∗)
regardless whether Θω (z) is an inner function in C+ . In fact, it is carried out by using
                                       h1i
the weighted summatory function hω (x) defined below. Then sufficient or equivalent
                                                                               h1i
conditions for Θω (z) to be an inner function in C+ are given in terms of hω (x) as in
Theorem 2.2. This is the main result in the appendix.
                   h1i
   The function hω (x) is not only directly related to RH via the innerness of Θω (z), but
also directly related to the operator Hω (Theorem A.2). The above discussion clarifies
the meaning of a part of functions studied in [26] (see the remark after Theorem A.1).
A.1. Notation and Results. Let B(z; p, q) be the incomplete beta function defined by
                     Z z
        B(z; p, q) =     xp−1 (1 − x)q−1 dx (0 ⩽ z ⩽ 1, ℜ(p) > 0, ℜ(q) > 0).
                          0
We use the notation
                                                          Z 1
                  β(z; p, q) := B(p, q) − B(z; p, q) =          xp−1 (1 − x)q−1 dx,
                                                           z
and understand that β(z; p, q) is defined by the integral on the right-hand side if ℜ(p) ⩽ 0,
ℜ(q) > 0, and 0 < z < 1. For example, gω of (2.2) can be written as
                                                                        
                     2π ω     2−ω        2 ω−1       ω−1      2 3 − 2ω
            gω (x) =        x     (1 − x )     − ωx      β x ,          ,ω .
                     Γ(ω)                                           2
                                        h1i
We define the real-valued function gω on (0, ∞) by
                                        Z 1r
                               h1i            y        dy
                              gω (x) :=         gω (y)                                 (A.1)
                                         x    x         y
                  A CANONICAL SYSTEM OF DIFFERENTIAL EQUATIONS                          25

                      h1i
for 0 < x < 1, and gω (x) = 0 for x > 1. Then we have

 gωh1i (x) =
                                                                 
 
        4ω   πω      ω−1    2 3 − 2ω      2ω + 1 −1/2    2 5 − 2ω
                   x     β x ,       ,ω −       x     β x ,       ,ω    , ω 6= 1/2,
      2ω − 1 Γ(ω)                 2          4ω                4
        2  p                        p      
 
 
  √ 2 1 − x2 + log x − log(1 + 1 − x2 ) ,                                 ω = 1/2
         x
                                           h1i
for 0 < x < 1 by elementary ways. Using gω and cω (n) of (2.1), we define the real-valued
          h1i
function hω on (0, ∞) by
                                           ⌊x⌋                  n
                                      1    X
                             hh1i
                              ω (x) =            cω (n) gωh1i                       (A.2)
                                       x                         x
                                           n=1

                h1i                                    h1i
for x > 1, and hω (x) = 0 for 0 < x < 1. Then hω is well-defined on (0, ∞) and has a
support in [1, ∞) as well as hω . We also have
                                         Z xr
                                 h1i           y        dy
                               hω (x) =          hω (y)                        (A.3)
                                           1   x         y
                                                 h1i
for x > 1 by definition (A.1). The function hω is related to Kω (z, w) and Θω as follows.

Theorem A.1 Let ω > 0.
   (1) If Θω (z) is an inner function in C+ , we have

                      −1                 1  −1                    
                    F1/2 Kω (0, ∗)(z) =     x 2 1(1,∞) (x) − hh1i
                                                              ω (x) ,
                                        2π
       where 1(1,∞) is the characteristic function of (1, ∞).
                                                                      1    h1i
   (2) Θω (z) is an inner function in C+ if and only if (x− 2 1(1,∞) (x) − hω (x)) belongs
       to L2 ((1, ∞), dx).
                                                      h1i
   (3) Assume that there exists xω ⩾ 1 such that hω (x) has a single sign for every
       x ⩾ xω . Then Θω (z) is an inner function in C+ .
                             √ h1i
   (4) Assume that limx→∞ x hω (x) exists. Then Θω (z) is an inner function in C+ .
   (5) Assume that Θω (z) is an inner function in C+ for all ω > 0. Then we have
                                 √ h1i
                                   xhω (x) = 1 + o(1)
        as x → +∞ for all ω > 0.
                           √ h1i
Remark The case limx→∞ x hω (x) = 0 is allowed in (4), though it does not hold by
(5) if RH holds for ζ(s).
                                            h1i
Remark Functions hω (x) of (2.3) and hω (x) of (A.2) were introduced and studied
in [26] for more general L-functions, but notation is different a little. The function
  h1i                                   1 h1i                1 h0i
hω (x) (resp. hω (x)) was denoted by x− 2 h1,ω (x) (resp. x− 2 h1,ω (x)) in [26].

Theorem A.2 Suppose that Θω (z) is an inner function in C+ . We define
                                         Z ∞
                               √ d√
                  (H̃ω f )(x) = x      x      hh1i
                                               ω (xy) f (y) dy
                                   dx     0

for compactly supported smooth functions f . Then H̃ω f belongs to L2 ((0, ∞), dx), and
f 7→ H̃ω f is extended to the isometry on L2 ((0, ∞), dx) satisfying H̃ω f = Hω f .
26                                          M. SUZUKI

A.2. Proof of Theorem A.1. We prove each statement of Theorem A.1 separately.
At first, we note the following:
Proposition A.3 For ω > 0 and ℑ(z) > 1/2 + ω, we have
                            Z ∞
                                            1     dx  i
                                hωh1i (x) x 2 +iz    = Θω (z),                            (A.4)
                             0                     x  z
where the integral converges absolutely.
Proof. This is proved by a way similar to the proof of Proposition 2.1 in Section 3.1 (see
also Lemma 4.2 of [26]).                                                                
(3): By the assumption and a theorem of Landau (e.g. Widder [29, Chap.II,§5]), the
integral in (A.4) converges for ℑ(z) > v0 , where iv0 is the first pure imaginary singularity
of Θω (z)/z. On the other hand, Θω (z) has no singularities on the imaginary axis, because
it is known that ξ(s) has no real zeros. Hence Θω (z) is regular in C+ . It implies that
Θω (z) is an inner function in C+ by a way similar to the proof of Lemma 3.1.             
(1) and (2): Suppose that Θω (z) is an inner function in C+ . We have
                               Z U +ic
              h1i      1               Θω (z) − 1 −iz
             hω (x) =     lim                 x 2     dz (c > 1/2 + ω)
                      2π U →∞ −U +ic −iz
for x > 1 by the Mellin inversion formula (e.g. [27, Theorem 28]), since the integral in
                                                    h1i
(2.4) converges absolutely for ℑ(z) > 1/2 + ω and hω (x) is in C 1 (1, ∞). By the Stirling
                                     −ω
formula, we have Θω (u + iv) ≪ω,v u for a fixed v > 1/2 + ω. Therefore
                      Z U +ic
                    1          Θω (z) − 1 −iz            1
          h1i
         hω (x) =                    x 2      dz + O(xc− 2 U −ω ) (c > 1/2 + ω).
                   2π −U +ic −iz
Using the well-known formula
                             Z U +ic
                 − 12     1           1 − 1 −iz              1
                x =                     x 2      dz + O(xc− 2 (log x)−1 U −1 )
                         2π −U +ic −iz
for x > 1 and large U > 1, we have
                      Z U +ic
  − 12  h1i       1           1 − Θω (z) − 1 −iz            1                 1
x − hω (x) =                            x 2      dz + O(xc− 2 U −ω ) + O(xc− 2 (log x)−1 U −1 ).
                 2π −U +ic        −iz
Here the integrand (1 − Θω (z))/z is bounded on C+ ∪ R. Thus the residue theorem gives
                       Z U
  − 12    h1i       1        1 − Θω (z) − 1 −iz            1                1
 x − hω (x) =                          x 2      dz + O(xc− 2 U −ω ) + O(xc− 2 (log x)−1 U −1 ),
                   2π −U        −iz
                   R ±U +ic                      1
since integrals on ±U +i0 are bounded by xc− 2 (log x)−1 U −1 . Tending U to +∞ for fixed
x > 1, we have
                                       Z ∞
                  − 12    h1i        1      1 − Θω (u) − 1 −iu
                 x − hω (x) =                          x 2      du (x > 1).              (A.5)
                                    2π −∞       −iu
                      1               h1i
This implies that x− 2 1(1,∞) (x) − hω (x) belongs to L2 ((1, ∞), dx), since (1 − Θω (u))/u
belongs to L2 (R) by (1.9) and (1.10). In addition, (A.5) implies (1).
                                  h1i
  Suppose that x−1/2 1(1,∞) − hω belongs to L2 ((1, ∞), dx). Then the integral
                          Z ∞h                          i 1
                                   1
                                                             +iz dx
                                x− 2 1(1,∞) (x) − hh1i
                                                   ω (x) x
                                                           2                          (A.6)
                           0                                     x
converges on the real line in L2 -sense, and converges absolutely for ℑ(z) > 0 ( [29,
Chap.II, §10]). Hence integral (A.6) defines an analytic function in C+ . By Proposition
A.3, integral (A.6) is equal to (1 − Θω (z))/(iz) for ℑ(z) > 1/2 + ω. Hence we find that
Θω (z) is an analytic function in C+ , and it implies that Θω (z) is an inner function in
C+ as well as the proof of (3).                                                       
                     A CANONICAL SYSTEM OF DIFFERENTIAL EQUATIONS                                      27

(4): By formula (A.3), the assumption implies that the integral of (2.4) converges at
z = 0 in the sense                   Z T
                                                  1     dx
                                lim      hω (x) x 2 +i0    .
                               T →∞ 1                    x
This implies that the integral of (2.4) converges for ℑ(z) > 0, and defines an analytic
function in C+ ([29, Chap.II, §1]). Hence Θω (z) is an inner function in C+ by (1.9). 
(5): If Θω (z) is an inner function in C+ for all ω > 0, RH(Aω ) holds for all ω > 0.
Hence RH holds by Proposition 1.1. Then we obtain (5) by a way similar to the proof
of Theorem 2.3 (2-b) in [26].                                                      
A.3. Proof of Theorem A.2. Suppose that Θω (z) is an inner function in C+ . Let f
be a compactly supported smooth function. Put F = F1/2 f and define
                                Z c+i∞
                              1                        1
                      g(x) =           Θω (z)F (−z) x− 2 −iz dz
                             2π c−i∞
for c ⩾ 0. Then the right-hand side is independent of c ⩾ 0 by the assumption, and
defines an member of L2 ((0, ∞), dx) by (1.9). Moreover, we have
                   Z x                Z c+i∞
                       g(u)         1                      x−iz
                         √ du =               Θω (z)F (−z)      dz.
                     0     u       2π c−i∞                 −iz
for c > 0. On the other hand, by Proposition A.4, we obtain
               Z ∞                          Z c′ +i∞
                                          1          Θω (z)           1
                   hωh1i (xy) f (y) dy =                    F (−z) x− 2 −iz dz
                0                        2π c′ −i∞ −iz
for c′ ≫ 0. Hence                  Z x             Z ∞
                              1          g(u)
                             √           √ du =          hh1i
                                                          ω (xy) f (y) dy.
                               x    0      u         0
This implies that g = H̃ω f . Thus H̃ω f is defined almost everywhere and belongs to
L2 ((0, ∞), dx), since g belongs to L2 ((0, ∞), dx). Moreover we obtain g = Hω f by the
definition of g and the latter half of the proof of Lemma 4.1. Hence H̃ω f = Hω f , and it
implies the extension of H̃ω to L2 ((0, ∞), dx).                                        

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  Department of Mathematics, Tokyo Institute of Technology, 2-12-1 Ookayama, Meguro-
ku, Tokyo 152-8551, Japan
  E-mail address: msuzuki@math.titech.ac.jp
