Master Thesis Canonical Systems

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Master Thesis

Canonical Systems

Institute of

Analysis and Scientific Computing Technische Universit¨at Wien

Supervisor

Assoz. Prof. Dr. Harald WORACEK

Author JohannesKaiser

18 M¨arz 2018

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1 Introduction

In this section we introduce an initial value problem and show that it always has a unique solution and that this solution is an entire function ofz. Further we establish some basic properties including a rough growth estimate.

We consider an initial value problem of the form

(y⁰(x) =−zA(x)y(x), x∈[a, b],

y(a) =y₀, (1.1)

wherez∈C,y₀ ∈C² and A∈L¹([a, b],C^2×2).

We call y : [a, b] → C² a solution of (1.1), if y is defined on [a, b], takes values in C², is componentwise absolutely continuous, its derivative fulfills (1.1) for almost all x ∈ [a, b] and y(a) =y₀.

We will mainly focus on canonical systems, which are initial value problems of the form (1.1), with A(x) = J H(x), where H ∈ L¹([a, b],R^2×2) is real and positive semidefinite, and J =

0 −1

1 0

.

Remark 1.1. Note that in some books (1.1) is given without the ”−” on the right hand side.

This results in minor changes to some theorems.

This section will follow lecture notes of Professor Michael Kaltenb¨ack from the Technische Universit¨at Wien.

1.1 Existence and Uniqueness of Solutions

Theorem 1.2 (Existence and Uniqueness). Let z ∈C, A∈L¹([a, b],C^2×2) and y₀ ∈C², then there exists a unique function y: [a, b]→C², such thaty is a solution of (1.1).

Proof. Let B^M [a, b],C²

be the space of all bounded and Borel measurable functions on [a, b]

with values in C² equipped with kfk∞ := sup_x∈[a,b]kf(x)k₂. This is a closed subspace of the Banach spaceB [a, b],C²

of all bounded functions on [a, b] with values inC², because uniform convergence preserves measurability. HenceB^M [a, b],C²

is a Banach space as well. We define Λ(x) :=

Z x a

kA(t)kdt, x∈[a, b],

wherek · k denotes the operator norm on C². LetG(x) := exp − |z|Λ(x) , then kfk_G := sup

x∈[a,b]

kf(x)k₂G(x) gives a norm equivalent tok · k∞onB^M [a, b],C²

. Now letT :B^M [a, b],C²

→ B^M [a, b],C² be defined by

T(f)(x) :=y₀− Z x

a

zA(t)f(t)dt.

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As an integral of an L¹-function T(f) is absolutely continuous, and T indeed maps into B^M [a, b],C²

. Further forf, g∈ B^M [a, b],C²

we obtain G(x)kT(f)(x)−T(g)(x)k₂=G(x)

Z x a

zA(t) g(t)−f(t) dt

2

≤ G(x)

Z x a

|z|kA(t)kexp |z|Λ(t)

exp − |z|Λ(t)

kf(t)−g(t)k₂dt≤ G(x)kf−gk_G

Z x a

|z|kA(t)kexp |z|Λ(t)

dt=G(x)kf −gk_G Z x

a

d dt

exp |z|Λ(t) dt≤ kf−gk_Gexp − |z|Λ(x)

exp |z|Λ(x)

−1

. Taking the supremum overx∈[a, b], we get

kT(f)−T(g)k_G≤

1−exp − |z|Λ(b)

· kf −gk_G. By the Banach Fixed Point Theorem there exists a uniquey ∈ B^M [a, b],C²

, satisfyingy =T y.

In particularyis absolutely continuous and obviouslyy(a) =y0holds as well. Taking the almost everywhere defined derivative, we obtain that y satisfies (1.1).

If, on the other hand, an absolutely continuous u satisfies u(a) = y₀ and the differential equation (1.1) almost everywhere, we get u ∈ B^M [a, b],C²

and by the fundamental theorem of calculus, we obtainT u=u. Hence y=u.

In the next theorem we show that the solution depends holomorphically on z.

Theorem 1.3. For each y₀ ∈C², there exists a unique function y: [a, b]×C→C² such that (i) for eachz∈C the function y(·, z) is the solution of (1.1),

(ii) y is continuous on[a, b]×C,

(iii) for eachx∈[a, b]the function y(x,·) is holomorphic.

Remark 1.4. Note that Theorem 1.3 is a slight modification of Theorem 1.2 and its proof, to obtain holomorphy in the second argument.

To prove this theorem, we need the following lemma.

Lemma 1.5. Let us now regardG(x, z) := exp −k|z|Λ(x)

, for some fixedk >1, as a function on[a, b]×C→(0,∞). For a function f : [a, b]×C→C² we define

kfk_G:= sup

z∈C

sup

x∈[a,b]

kf(x, z)k₂·G(x, z).

Let X denote the vector space of all f : [a, b]×C→C², such that (i) kfk_G<∞,

(ii) for eachz∈C the function x7→(x, z) is continuous.

(iii) for eachx∈[a, b]the function z7→f(x, z) is holomorphic.

Then X equipped with k · k_G is a Banach space.

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Proof. First it is a well known fact that the space B [a, b]×C,C²

of all bounded functions on [a, b]×C with values in C² equipped with k · k_∞ is a Banach space. Hence the space B_G [a, b]×C,C²

of all functionsf on [a, b]×Cwith values inC²such thatkfk_G <∞, equipped withk · k_G is a Banach space as well, because it is isometrically isomorphic toB [a, b]×C,C² via f 7→f·G.

Obviously X is a linear subspace ofB_G [a, b]×C,C²

. Further it is closed: Assumefn∈X and f_n → f in B_G [a, b]×C,C²

. Then for every x ∈ [a, b] and for every compact K ⊆ C, the uniform convergence of f_n|_[a,b]×K to f|_[a,b]×K, as well as the uniform convergence of z 7→

fn(x, z) to z7→f(x, z), for z∈K follows. Hencef|_[a,b]×K and therefore f, are continuous and z7→f(x, z),z∈Cis holomorphic for everyx. Hencef ∈X.

Proof of Theorem 1.3. Let X and G(x, z) be defined as in Lemma 1.5, and let T :X →X be defined by

T(f)(x, z) :=y₀− Z x

a

zA(t)f(t, z)dt.

First we want to show thatT(f) actually lies inX. Let (x₁, z₁),(x₂, z₂)∈[a, b]×Cand assume without loss of generality that x1 ≤x2. Then

kT()(x1, z1)−T(f)(x2, z2)k₂ ≤ kT(f)(x1, z1)−T()(x2, z1)k₂+kT(f)(x2, z1)−T(f)(x2, z2)k₂≤

Z x2

x1

z₁A(t)(t, z₁)dt 2

+

Z x2

a

A(t) z₂)f(t, z₂)−z₁f(t, z₁) dt

2

≤ Z x2

x1

kA(t)kkz₁f(t, z1)k₂dt+ Z b

a

kA(t)k

z1f(t, z1)−z2f(t, z2)

2dt. (1.2)

If we fix (x1, z1), then we get, for |z₁−z2| ≤1 and ρ:=|z₁|+ 1, z₁f(t, z₁)−z₂f(t, z₂)

2≤2 |z₁|+ 1

sup

x∈[a,b],|z|≤ρ

kf(x, z)k₂ ≤2 |z₁|+ 1 kfk_G G(b, ρ).

Hence, by the Dominated Convergence Theorem, the second integral in (1.2) converges to zero for (x₂, z₂) → (x₁, z₁). Because kz₁f(t, z₁)k₂ ≤ ρ_G(r,ρ)^kf^k^G, the same holds for the first integral.

Hence T(f) is continuous on [a, b]×C.

Let us regard H : z 7→ T(f)(x, z) = y₀ −Rx

a zA(t)f(z, t)dt as a parameter integral, with holomorphic integrand z 7→ zA(t)f(z, t) for every t and integrable integrand t7→ zA(t)f(z, t) for everyz. We have the estimate

zA(t)f(t, z)

2≤ kA(t)k max

(τ,ζ)∈[a,b]×K|ζ|kf(τ, ζ)k₂, K ⊆Ccompact, z ∈K, t∈[a, b], where the right hand side is integrable, and conclude thatH is holomorphic. Further we obtain

kT(f)(x, z)k₂≤ ky₀k₂+ Z x

a

|z|

A(t)

exp k|z|Λ(t)

exp −k|z|Λ(t)

f(t, z) 2dt≤ ky₀k₂+1

k

exp k|z|Λ(x)

−1

kfk_G.

Multiplying withG(x, z) and taking the supremum over (x, z)∈[a, b]×C, we obtainkT(f)k_G≤ ky₀k_G+¹_kkfk_G<∞. Hence we getT(f)∈X.

Finally T is a strict contraction: For f, g∈X we have T(f)(x, z)−T(g)(x, z)

2G(x, z)≤exp −k|z|Λ(x) Z _x

a

|z|kA(t)k

f(t, z)−g(t, z) 2dt≤

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exp(−k|z|Λ(x)) Z x

a

|z|kA(t)kexp(k|z|Λ(t))kf−gk_Gdt= 1

kexp(−k|z|Λ(x))(exp(k|z|Λ(x))−1)kf−gk_G,

and hence kT(f)−T(g)k_G ≤ ¹_kkf −gk_G. By the Banach Fixed Point theorem, there exists a unique f ∈ X, which satisfies T f = f. In particular x 7→ f(x, z) is absolutely continuous.

Differentiating almost everywhere, shows that x 7→ f(x, z) satisfies the differential equation (1.1) and obviouslyf(a, z) =y₀ holds as well.

1.2 Properties of the Fundamental Solution

In the proof of Theorem 1.3 we saw that the solutiony(x, z) lies in the spaceX. Thus, for each k >1, it satisfies an estimate of the form (x∈[a, b])

ky(x, z)k₂ ≤C_kexp(k|z|Λ(x)), t∈C,

with some constant C_k >0. The Gronwall Lemma says that such an estimate holds for k= 1 as well.

Lemma 1.6 (Gronwall). Let y(x, z) be a solution of (1.1). Then ky(x, z)k₂ ≤ ky₀k₂exp

|z|

Z x a

kA(t)kdt

, (x, z)∈[a, b]×C. (1.3) Proof. Obviously it holds that

y(x, z) =y0−z Z x

a

A(s)y(s, z)ds, (1.4)

which implies

ky(x, z)k₂ ≤ ky₀k₂+

|z|

Z x a

kA(t)kky(t, z)kdt

. Fora≤s≤x≤b we further evaluate

d dsln

ky₀k₂+|z|

Z _s

a

kA(t)kky(t, z)k₂dt

= |z|kA(s)kky(s, z)k₂ ky₀k₂+|z|Rs

akA(t)kky(t, z)k₂dt.

The right hand side is bounded from above by |z|kA(s)k, hence integrating from a tox with respect tosyields

ln

ky₀k₂+|z|

Z x a

kA(t)kky(t, z)k₂dt

−ln ky₀k₂

≤ Z x

a

|z|kA(s)kds.

Applying the exponential function on this inequality completes the proof.

Definition 1.7. Let A ∈ L¹([a, b],C^2×2). Let y1 and y2 be the solutions of (1.1) with initial value

1 0

and

0 1

, respectively. Then the matrix functionM : [a, b]×C→C^2×2 defined as M(x, z) := y₁(x, z)|y₂(x, z)

,

is called fundamental solution of (1.1). It is the solution of the matrix-valued initial value problem

( d

dxY(x, z) =−zA(x)Y(x, z), x∈[a, b],

Y(a, z) =I, (1.5)

Its value at the right endpoint is called themonodromy matrix of (1.1).

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Remark 1.8. Note that callingM a fundamental solution is not very precise, since the dimen- sions for the initial value do not match.

Proposition 1.9. The fundamental solution M(x, z) of (1.5) has the following properties.

(i) Leta≤y≤x≤b, and let My,x be the fundamental solution for A|_[y,x]. Then M(c, z) =My,x(c, z)M(y, z), c∈[y, x], z∈C.

(ii) M is continuous on [a, b]×C → C^2×2, absolutely continuous in the first argument and holomorphic in the second argument.

(iii) IfA is real a.e., then the entries M_ij(x, z) of M(x, z) satisfy

Mij(x, z) =Mij(x, z), x∈[a, b], z∈C. (1.6) (iv) LetQ∈GL(2,C) and a≤y ≤x≤b. Then

kQM(x, z)k ≤ kQM(y, z)kexp

|z|

Z x y

kQA(t)Q⁻¹kdt

. (1.7)

Further, the fundamental solution is estimated as kM(x, z)k ≤exp

|z|

Z x a

kA(t)kdt

, (x, z)∈[a, b]×C. (1.8) (v) The fundamental solution satisfies detM(a, z) = 1 and the differential equation

d

dx(detM(x, z)) =ztrA(x) detM(x, z). (1.9) Hence,

detM(x, z) = exp

z Z x

a

trA(t)dt

. (1.10)

Proof. Let a≤y≤x≤b. We haveMy,x(y, z) =I by definition and hence M(y, z) =M_y,x(y, z)M(y, z).

Further M(c, z) and M_y,x(c, z) both satisfy the differential equation (1.5) forc ∈ [x, y]. Since M(y, z) is a constant factor on the right, alsoM_y,x(c, z)M(y, z) satisfies the differential equation (1.5). By uniqueness of solutions, we obtain (i).

By Theorem 1.3 y₁, y₂ : [a, b]×C → C² are continuous, absolutely continuous in the first argument and holomorphic in the second. Hence M : [a, b]×C → C^2×2 satisfies the same properties and therefore it satisfies (ii) as well.

Since

d

dx(M(x, z)) = d

dxM(x, z) =−zA(x)M(x, z) almost everywhere on [a, b], the matrixM(t, z) is the solution of

( _d

dxY(x, z) =−zA(x)Y(x, z), x∈[a, b], Y(a, z) =I.

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In particular, if A(x)∈R^2×2 for x∈[a, b], uniqueness of solutions implies M(x, z) =M(x, z), (x, z)∈[a, b]×C.

For each a≤y≤x≤ba fundamental solution satisfies the integral equation M(x, z) =M(y, z)−z

Z x y

A(t)M(t, z)dt. (1.11)

Now let Q∈GL(2,C). Then we can rewrite this equation as follows QM(x, z) =QM(y, z)−z

Z x y

QA(t)Q⁻¹

QM(t, z)dt.

From now on we follow the lines of the proof of Lemma 1.6: The upper equality implies kQM(x, z)k₂≤ kQM(y, z)k₂+

|z|

Z x a

kQA(t)Q⁻¹kkQM(t, z)kdt

. Fora≤s≤x≤b we further evaluate

d dsln

kQM(y, z)k₂+|z|

Z s a

kQA(t)Q¹⁻kkQM(t, z)k₂dt

=

|z|kQA(s)Q⁻¹kkQM(s, z)k₂ kQM(y, z)k₂+|z|Rs

akQA(t)Q⁻¹kkQM(t, z)k₂dt.

The right hand side is bounded from above by |z|kQA(s)Q⁻¹k, hence integrating from a to x with respect tosyields

ln

kQM(y, z)k₂+|z|

Z x a

kQA(t)Q⁻¹kkQM(t, z)k₂dt

−ln kQM(y, z)k₂

≤

≤ Z x

a

|z|kQA(s)Q⁻¹kds.

Applying the exponential function on this inequality yields assertion (v).

For the last property we start by evaluating the left hand side, (omitting the arguments of our fundamental solution to simplify notation)

detM⁰ =(M11M22−M12M21)⁰=M₁₁⁰ M22−M₁₂⁰ M21+M₂₂⁰ M11−M₂₁⁰ M12=

=z

A11(x)M11M22+A12(x)M21M22−A11(x)M12M21−A12(x)M22M21+ +A₂₁(x)M₁₂M₁₁+A₂₂(x)M₂₂M₁₁−A₂₁(x)M₁₁M₁₂−A₂₂(x)M₂₁M₁₂

=

=zA₁₁(x)M₁₁M₂₂−zA₁₁(x)M₁₂M₂₁+zA₂₂(x)M₂₂M₁₁−zA₂₂(x)M₂₁M₁₂=

=ztrA(t) detM

and we obtain the desired equality. Plugging the right hand side of (1.10) in (1.9) completes the proof.

The next result is easy to prove, but a very important fact.

Theorem 1.10. Let J ∈C^2×2, with J^∗ =−J and assume that J A is hermitian a.e., then M(x, w)^∗J M(x, z)−J = ( ¯w−z)

Z _x

a

M(t, w)^∗J A(t)M(t, z)dt z, w∈C. (1.12)

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Proof. We evaluate d

dx[M(x, w)^∗J M(x, z)] = d

dxM(x, w) ∗

J M(x, z) +M(x, w)^∗J d

dxM(x, z)

=

−J d

dxM(x, w) ∗

M(x, z) +M(x, w)^∗J d

dxM(x, z)

. Because J Ais hermitian the right hand side further equals

¯

wM(x, w)^∗J A(x)M(x, z)−zM(x, w)^∗J A(x)M(x, z) = ( ¯w−z)M(x, w)^∗J A(x)M(x, z), and the equality follows by integrating with respect tox.

Corollary 1.11. Let J ∈C^2×2, with J^∗=−J and assume that−J Ais hermitian and positive semidefinite a.e., then

M(x, z)^∗J M(x, z)−J

i ≥0, x∈[a, b], Im(z)>0. (1.13)

Proof. We can put w=z in (1.12), because if −J A is hermitian, J Ais hermitian as well, and we obtain

M(x, z)^∗J M(x, z)−J =−2iIm(z) Z x

a

M(t, z)^∗J A(t)M(t, z)dt.

After we divide byi, the right hand side is greater or equal to 0, by positive semidefiniteness of

−J A.

Corollary 1.12. Let J ∈C^2×2, with J^∗=−J and assume that−J Ais hermitian and positive semidefinite a.e.. Then the entry M11(x, z) has no nonreal zeroes.

Proof. EvaluatingM(x, z)^∗J M(x, z)−J, yields (omitting the arguments (x, z) from the matrix entries)

M11M21−M21M11 M12M21−M11M22+ 1 M₁₁M₂₂−M₂₁M₁₂−1 M₁₂M₂₂−M₂₂M₁₂

.

Since, by Corollary 1.11, this is positive semidefinite for Im(z) > 0, after dividing by i, we conclude that

M₁₁M₂₁−M₂₁M₁₁

i ≥0.

Note that, due to which, for M21(x, z)6= 0, is equivalent to

M11

M21 −

M11

M21

i ≥0.

Hence Im M11

M21

≥0. Now, if Im M11

M21(x, z0)

= 0, for a z0 ∈C⁺, by the maximum principle we obtain that Im

M11

M21(x, z)

≡ 0, for z ∈ C⁺ and hence Im(M11(x, z))≡ 0 for z ∈ C holds as well. Therefore Re(M₁₁(x, z)) has to be constant as well, and hence has to be equal to the initial value, 1.

Corollary 1.13. Let J :=

0 −1

1 0

, and assume that J A is real and hermitian a.e.. Then detM(x, z) = 1 for x∈[a, b], z∈C.

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Proof. We have that

J A(x) =

−A₂₁(x) −A₂₂(x) A₁₁(x) A₁₂(x)

is real and hermitian, or in other words, A11(x) =−A₂₂(x). Therefore trA(x) = 0 and hence (1.10) yields the assertion.

The results above motivate the following definition.

Definition 1.14. A (2×2)-matrix-valued function W(z) belongs to the classM₀ if its entries w_ij(z) are real entire functions,W(0) =I, detW(z) = 1 and if

H_W(w, z) := W^∗(w)J W(z)−J w−z is a positive semidefinite kernel.

The following lemma shows that M₀ is closed with respect to products.

Lemma 1.15. Let W1, W2 ∈ M₀. Then,

H_W₁_W₂ =H_W₂+W₂^∗(w)H_W₁(w, z)W₂(z) Proof. We compute

HW1W2 = W₂^∗(w)W₁^∗(w)J W₁(z)W₂(z)−W₂^∗(w)J W₂(z) +W₂^∗(w)J W₂(z)−J

w−z =

=W₂^∗(w)W₁^∗(w)J W₁(z)−J

w−z W₂(z) +W₂^∗(w)J W₂(z)−J

w−z =

=W₂^∗(w)H_W₁(w, z)W₂(z) +H_W₂(w, z), and have shown the assertion.

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2 Exponential Type

We recall the notion of exponential type, and present a first result connected with the initial value problem (1.5). Namely we present Theorem 2.4, which provides a much better estimate for the exponential type of the solutions of (1.5) than the Gronwall Lemma (see Lemma 1.6). In fact we show that for canonical systems this upper estimate coincides with the type (Corollary 2.9).

Definition 2.1. Let X be a Banach space, and f :C→ X. We say f is of finite exponential type, if there exist α∈[0,∞) and Cα ≥0 such thatkf(z)k ≤Cαexp α|z|

for all z∈C.

We call τ(f) := inf{α >0 :∃C_α >0 :kf(z)k ≤Cαexp α|z|

} theexponential type of f. Remark 2.2. Assuming thatf andgare elements of a Banach algebra, we immediately obtain the following two properties:

(i) τ(f g)≤τ(f) +τ(g),

(ii) τ(f +g)≤max{τ(f), τ(g)}.

Lemma 2.3. Let X=C^2×2, andA= (Ai,j)i,j∈{1,2} :C→X. Then τ(A) = max

i,j∈{1,2}{τ(A_ij)}.

Proof. Since all norms are equivalent onC^2×2 we can use the maximum entry norm to compute τ(A).

Theorem 2.4. Let A ∈L¹([a, b],C^2×2), and let M(x, z) be the fundamental solution of (1.5).

Set

φ(x) := inf

Q∈GL(2,C)

QA(x)Q⁻¹

, Φ(x) :=

Z x a

φ(t)dt, x∈[a, b].

Then τ M(x, z)

≤Φ(x) for x∈[a, b].

Proof. First we note that the assertion is trivial if we consider x=a. Next we want to verify that x7→ τ M(x, z)

is absolutely continuous. To this end let a≤y < x≤b. By Proposition 1.9 (i), we have

M(x, z) =M_y,x(x, z)M(y, z),

which is equivalent to M(y, z) =M_y,x(x, z)⁻¹M(x, z). Hence, by Remark 2.2 and Lemma 2.3, we obtain

τ M(x, z)

≤τ My,x(x, z)

+τ M(y, z) , τ M(y, z)

≤τ M_y,x(x, z)⁻¹

+τ M(x, z) . Combining these two inequalities yields

τ M(x, z)

−τ M(y, z)

≤max

τ My,x(x, z)

, τ My,x(x, z)⁻¹ . (2.1) The first term in the maximum can be estimated from above byRx

y kA(t)kdt, due to the Gronwall Lemma. To estimate the second term, let cofM_y,x(x, z) denote the cofactor matrix ofM_y,x(x, z).

By Remark 2.2 and Lemma 2.3, τ cofMy,x(x, z)

=τ My,x(x, z)

. Cramers rule says My,x(x, z)⁻¹ = 1

detM_y,x(x, z) cofMy,x(x, z)T

,

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and applying Proposition 1.9 (v), we obtain τ M_y,x(x, z)⁻¹

≤τ

1 detM_y,x(x, z)

+τ (cofM_y,x(x, z))^T

≤

Z x y

trA(t)dt

+τ My,x(x, z)

≤ Z x

y

|trA(t)|+kA(t)k dt.

Hence we can estimate the left hand side of (2.1) by Rx

y |trA(t)|+kA(t)k

dt, and conclude thatx7→τ M(x, z)

is absolutely continuous.

Let Q∈GL(2,C) and let again a≤y < x≤b. By (1.7) the fundamental solution satisfies kQM(x, z)k ≤ kQM(y, z)kexp

|z|

Z x y

QA(t)Q⁻¹ dt

.

Since multiplying M(x, z) by Q ∈ GL(2,C) does not change the exponential type, i.e., τ QM(x, z)

=τ M(x, z)

, we get τ M(x, z)

−τ M(y, z)

≤ Z x

y

QA(t)Q⁻¹ dt.

This is equivalent to

τ M(x, z)

−τ M(y, z)

x−y ≤

Rx y

QA(t)Q⁻¹ dt

x−y .

For almost everyx both sides have a limit wheny→x, and we obtain τ M(x, z)0

≤

QA(x)Q⁻¹

, x∈[a, b] a.e..

Taking the infimum over Q∈GL(2,C), and integrating yields our assertion.

Lemma 2.5. Let A∈C^2×2 with trA= 0 and let k · k denote the spectral norm. Then

Q∈GL(2,inf C)

QAQ⁻¹ =p

|detA|.

Proof. Taking an appropriate basis of C^2×2 and composing it to a matrix C ∈ GL(2,C), we obtain that either

CAC⁻¹ = a 0

1 a

, orCAC⁻¹= a 0

0 b

. By assumption and because tr CAC⁻¹

= tr CC⁻¹A

= tr(A), we havea= 0 in the first case and b=−ain the second.

In the first case we get p

|detA|= 0. Forr >0 and B = diag ¹_r, r

C, we haveBAB⁻¹ = 0 0

r 0

and hence inf_Q∈GL(2,C)

QAQ⁻¹ = 0.

In the second case we have p

|detA|=|a|=

CAC⁻¹

and hence inf

Q∈GL(2,C)

QAQ⁻¹ ≤a.

For arbitrary Q ∈ GL(2,C), the norm

QAQ⁻¹

is greater or equal to the maximum of the absolute values of the eigenvalues ofQAQ⁻¹, and hence inf_Q∈GL(2,C)

QAQ⁻¹ ≥ |a|.

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Corollary 2.6. If trA(x) = 0 a.e., then τ M(x, z)

≤ Z x

a

pdetA(t)dt. (2.2)

Proof. This result follows immediately from Lemma 2.5 and Theorem 2.4.

While the estimate (2.2) of the type from above holds under an assumption on the trace of A, the reverse inequality will follow from a definiteness assumption.

Theorem 2.7. Let J =

0 −1

1 0

and assume that −J A is a.e. positive semidefinite and real.

Then

τ M(x, z)

≥ Z x

a

pdetA(t)dt. (2.3)

To prove this result we need the following general lemma.

Lemma 2.8. Let M(x, z) be the fundamental solution corresponding to A ∈ L¹([a, b],C^2×2), and let Ψ : [a, b]→Rbe some absolutely continuous function with Ψ(a) = 0. Then

MΨ(x, z) := exp izΨ(x)

M(x, z) is the fundamental solution of the system

( d

dxY(x, z) =−z A(x)−Ψ⁰(x)iI

Y(x, z), x∈[a, b], Y(a, z) =I.

Proof. First we note that MΨ(a, z) =M(a, z) =I. Further we obtain d

dxM_Ψ(x, z) =izΨ⁰(x) exp iΨ(x)z

M(x, z) + exp izΨ(x) d

dxM(x, z) =

=izΨ⁰(x)M_Ψ(x, z)−exp izΨ(x)

zA(x)M(x, z)

=−z A(x)−iΨ⁰(x)I

M_Ψ(x,z), and by uniqueness of solutions the proof is complete.

Proof of Theorem 2.7. Let −J A ≥ 0 and write −J A(x) =

h1(x) h3(x) h3(x) h2(x)

, where hi(x), i ∈ {1,2,3}, are real andh1, h2 ≥0. Let Ψ(x) :=Rx

a

pdetA(t)dt, then

−J A(x)−Ψ⁰(x)iI

=

h₁(x) h₃(x)−ip

detA(x) h3(x) +ip

detA(x) h2(x)

. We have

det

−J A(x)−Ψ⁰(x)iI

=h₁(x)h₂(x)−h²₃(x)−detA(x) = 0, and hence−J A(x)−Ψ⁰(x)iI

≥0. Fory >0, this implies by Corollary 1.11 det

MΨ(x, iy)^∗J MΨ(x, iy)−J i

≥0. (2.4)

Now assume thatτ M(x, z)

<R_x

a

pdetA(t)dt= Ψ(x). Choose >0 such thatτ(M(x, z))+

<Ψ(x) and C >0 such that

|M_ij(x, z)| ≤Cexp (τ(M(x, z)) +)|z|

, z∈C.

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Then

y→∞lim M_Ψ(x, iy) = lim

y→∞exp −yΨ(x)

M(x, iy) = 0.

and hence

y→∞lim

MΨ(x, iy)^∗J MΨ(x, iy)−J

i =iJ.

Since detiJ =−1, that contradicts (2.4).

Putting together Theorem 2.7 and Corollary 2.6 for a canonical system, equality holds in (2.2).

Corollary 2.9. If trA(x) = 0 a.e. and −J Ais positive semidefinite and real a.e., then τ M(x, z)

= Z x

a

pdetA(t).

Proof. This follows immediately from Corollary 2.6 and Theorem 2.7.

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3 Growth Functions and λ-type

The question arises to which extent Theorem 2.4, or the method leading to its proof, can be used to also capture different growth behaviour, for example with respect to an order smaller than 1.

This chapter is of a preparatory nature. We introduce the notion of growth functions and give some auxiliary results.

Throughout this chapter the order of an entire function will often be of importance or interest. Therefore we start with its definition.

Definition 3.1. Letf be an entire function. Then, we define theorder off as ordf := lim sup

z→∞

log log |f(z)|

log |z| ∈[0,∞].

3.1 Proximate Orders and Growth Functions

This section follows the section of the same name of [5], but a similar chapter can be found in [6] as well

Definition 3.2. We call a functionλ:R⁺→R⁺ a growth function, if it satisfies the following properties

(i) The limit γ := limr→∞ log(λ(r))

log(r) exists and is a finite nonnegative number, (ii) λis differentiable for all sufficiently large values of r and limr→∞rλ⁰(r)

λ(r) =γ, (iii) log(r) =o λ(r)

.

Let either α >0 and βi ∈R fori∈ {1, . . . n}, or α = 0,β1 = 1,βj ∈R and forj < i ≤n.

Then an examples for growth functions are given by functions of the form λ(r) =r^α log(r)β1

·. . .·(log_n(r))^βⁿ, where log_i(r) is defined as the itimes iterated logarithm,

The following definition gives us information about the relative growth of f compared toλ.

Definition 3.3. Letf be an entire function andλa growth function. Theλ-type of f is given by

σ_f^λ:= lim sup

z→∞

log⁺ |f(z)|

λ |z| ∈[0,∞]. (3.1)

Further we introduce the following notion, which is closely related to growth functions. We will elaborate their connection in Lemma 3.6.

Definition 3.4. A proximate order ρ(r) for the orderρ ≥0, is a function ρ :R⁺ → R⁺ such that

(i) limr→∞ρ(r) =ρ, (ii) limr→∞ρ⁰(r)rlogr = 0.

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For the entire function f(z) we set

σ_f = lim sup

r→∞

log |f(r)|

r^ρ(r)

and call σ_f the type of f with respect to ρ. If σ_f ∈(0,∞), ρ(r) is called a proximate order of the functionf(z).

Remark 3.5. We note that for any positive continuous increasing function f of finite order, there exists a proximate order with respect to whichf is of finite and nonzero type. For a proof see [5], Appendix II.

Clearly the proximate order and its corresponding type of a given function are not uniquely determined. For example, if we add log(c)/log(r) to the proximate order, then we obtain a new proximate order for the function, and now the type has been divided byc.

Lemma 3.6. The following assertions hold:

(i) Let ρ(r) be a proximate order with respect to the order ρ > 0. Then λ(r) := r^ρ(r) is a growth function.

(ii) Letρ(r) be a proximate order with respect to the order ρ= 0 andlog(r) =o r^ρ(r) . Then λ(r) :=r^ρ(r) is a growth function.

(iii) Conversely, if λ(r) is a growth function, then ρ(r) := ^log(λ(r))_log(r) is a proximate order with respect to the order γ, where γ := limr→∞ log(λ(r))

log(r) as indicated in Definition 3.2.

Proof. First we prove (i). To this end let ρ(r) be a proximate order, with respect to the order ρ and λ(r) := r^ρ(r). Then property (i) of Definition 3.2 follows, because we get log λ(r)

= ρ(r) log(r) and apply Definition 3.4 (i). To validate (ii), we evaluate

λ⁰(r) =r^ρ(r)

log(r^ρ⁰^(r)) +ρ(r) r

.

Multiplying this by _λ(r)^r and applying Definition 3.4 (i) and (ii) yields the desired assertion.

Further, Definition 3.2 (iii) is obvious, because ρ(x)>0 by definition, and we obtain assertion (i).

If we consider the situation in assertion (ii), we observe that Definition 3.2 (i) and (ii) follow analogously. The last point of this definition is fullfilled by the assumption in (ii).

To prove (iii) letλbe a growth function. For ρ(r) := ^log(λ(r))_log(r) Definition 3.4 (i) is satisfied, by Definition 3.2 (i), and we observe thatρ=γ. To verify Definition 3.4 (ii), we evaluate

ρ⁰(r) =

λ⁰(r) log(r)

λ(r) − ^log(λ(r))_r log²(r) ,

and hence we obtain limr→∞ρ⁰(r)rlogr= 0, by applying Definition 3.2 (i) and (ii).

Proposition 3.7. If ρ(r) is a proximate order with ρ > 0, then r^ρ(r) is strictly increasing for r sufficiently large.

If ρ(r) is a proximate order with ρ= 0, then r^ρ(r) is increasing for r sufficiently large, if

r→∞lim

rρ⁰(r) log(r)

ρ(r) = 0. (3.2)

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Proof. We have _dr^d r^ρ(r)

=ρ(r)r^ρ(r)−1+r^ρ(r)ρ⁰(r) logr.

First letρ >0, then, by Definition 3.4 (i) and (ii), we getρ(r)> ρ/2 and|ρ⁰(r)rlogr|< ρ/4 for all r sufficiently large. For such r, we conclude

r^ρ(r)−1 ρ(r) +rρ⁰(r) logr

> r^ρ(r)−1ρ/4, and becauser^ρ(r)−1>0 we obtain our assertion.

Let ρ= 0, then ^rρ⁰^{(r) log}_ρ(r) ^r ≥ −¹₂ forr sufficienly large, by (3.2). Hence d

dr

r^ρ(r)

=r^ρ(r)−1ρ(r)

1 +rρ⁰(r) logr ρ(r)

>0, forr large enough, which completes the proof.

Remark 3.8. Since in the study of asymptotic properties of entire functions we are only interested in their properties for r sufficiently large, we can always change ρ(r) on a bounded set, without affecting the asymptotic properties we study. Thus forρ >0, we can always assume thatr^ρ(r) is strictly increasing forr >0.

3.2 Slowly varying functions

Definition 3.9. Let Ψ : [X,∞) → R⁺ be a measurable function. We say that Ψ is slowly varying, if

x→∞lim Ψ(kx)

Ψ(x) →1, k >0. (3.3)

Remark 3.10. Note that limx→∞ Ψ(kx)

Ψ(x) = 1 if and only if limx→∞ Ψ(¹_kx)

Ψ(x) = 1. Hence, in order to have Ψ slowly varying, it is enough to check (3.3) on any setM ⊆(0.∞) withM∪M¹ = (0,∞).

The following theorem provides a useful result, in the context of slowly varying functions.

For the sake of completeness, we present a proof taken from [3], section 1.2..

Theorem 3.11. [Uniform Convergence] LetI be a compact interval in(0,∞) and Ψbe slowly varying, then

x→∞lim Ψ(kx)

Ψ(x) →1, uniformly for k∈I. (3.4)

Proof. Write h(x) = log(Ψ(e^x)). Then our assumption is equivalent to

x→∞lim (h(x+u)−h(x)) = 0, (3.5) for all u ∈ R and we have to prove uniform convergence for u ∈ I, where I is any compact interval inR. It suffices to prove uniform convergence on [0, A], for by translation this gives us uniform convergence on every compact interval.

Now choose ∈(0, A). For x >0 let I_x:= [x, x+ 2A],

E_x:={t∈I_x :|h(t)−h(x)| ≥/2},

E_x^∗:={t∈[0,2A] :|h(x+t)−h(x)| ≥/2}=Ex−x.

Then E_x, E_x^∗ are measureable, as Ψ is, and |E_x| = |E_x^∗|, where | · | denotes the Lebesgue measure. By (3.5) the indicator function ofE_x^∗tends pointwise to 0 asx→ ∞. So by dominated convergence its integral |E^∗_x|tends to 0. Thus|E_x| ≤/2 for x large enough, say, forx≥x.

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Now, for c∈[0, A],I_x+c∩I_x= [x+c, x+ 2A] has length 2A−c≥A, while forx≥x

|E_x+c∪E_x| ≤ |E_x+c|+|E_x|< < A.

So for c∈[0, A] and x≥x

(I_x+c∩I_x)\(E_x+c∪E_x)

has positive measure, and therefore is non-empty. Lett be a point of this set, then

|h(t)−h(x)|< /2,

|h(t)−h(x+c)|< /2.

So, by the triangular inequality, for all c∈[0, A] and x≥x,

|h(x+c)−h(x)|< , proving the desired uniformity on [0, A], and hence the theorem.

The following theorem is very important in the study of proximate orders.

Theorem 3.12. If ρ(r) is a proximate order, thenΨ(r) =r^ρ(r)−ρ is a slowly varying function.

Proof. By definition of Ψ, we obtain log

Ψ(kr) Ψ(r)

= log

(rk)^ρ(kr)−ρ

−log

r^ρ(r)−ρ

=

= (ρ(kr)−ρ) (log(k) + log(r))−(ρ(r)−ρ) log(r) = (3.6)

= (ρ(kr)−ρ) log(k) + (ρ(kr)−ρ(r)) log(r).

Let >0 and chooser₀ such that

|ρ⁰(x)xlog(x)| ≤, x≥r₀ (3.7)

log(r)

log(r) + log(k) ≤2 r≥r0 (3.8)

|ρ(x)−ρ| ≤ x≥r₀. (3.9) Now assume k∈(0,1). Then, by the Mean Value Theorem,|ρ(r)−ρ(rk)|=|ρ⁰(ξ)|(r−rk)|

for someξ ∈(rk, r). Hence, for r≥ ^r_k⁰, we have

log(r)|ρ(rk)−ρ(r)| ≤ |ρ⁰(ξ)ξlog(ξ)|(r−rk) log(r)

ξlog(ξ) ≤

≤ |ρ⁰(ξ)ξlog(ξ)|r(1−k) log(r)

krlog(kr) ≤

≤1−k k 2, by (3.7) and (3.8). Further by (3.9)

|(ρ(rk)−ρ) log(k)| ≤|log(k)|

and hence, by (3.6),

log_Ψ(kr)

Ψ(r)

< 2^1−k_k +|log(k)|

. By Remark 3.10, we conclude that

log_Ψ(kr)

Ψ(r)

< for all r > r₀ andk∈(0,∞).

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Corollary 3.13. Let >0. Then, forr sufficiently large

(1−)k^ρr^ρ(r)<(kr)^ρ(kr)<(1 +)k^ρr^ρ(r), r > r0().

Proof. By Theorem 3.12, we obtain

(1−)≤ (kr)^ρ(kr)r^ρ

(kr)^ρr^ρ(r) ≤(1 +), forr sufficiently large, and the assertion follows immediately.

Further we will need the following result, and prove it following the lines of [3], section 1.3.

Theorem 3.14. Let I be a compact interval. Then the following two assterions hold.

(i) If the function Ψis slowly varying, it may be written in the form Ψ(x) =c(x) exp

Z x a

(u) u du

, x≥a, (3.10)

for some a >0, where c(x) and(x) are measurable functions with c(x)→c∈(0,∞) and (x)→0 for x→ ∞, respectively.

(ii) If the functionΨcan be written in the form given above, it satisfies (3.4).

Remark 3.15. Note that this will not yield an alternative proof of the Uniform Convergence Theorem 3.11, since we apply it to show assertion (i).

Remark 3.16. Before we start the proof we make some observations.

(i) Since c, Ψ, may be altered on finite intervals, the value ofa is unimportant, and one may also take cbounded. We may rewrite (3.10) as

Ψ(x) = exp

c1(x) + Z x

a

(u) u du

, (3.11)

wherec₁(x), (x) are bounded and measurable, limx→∞c₁(x) =d∈R, limx→∞(x) = 0.

(ii) We will write h(x) := log (Ψ(e^x)). Thus we can prove Theorem 3.14 by showing that h satisfies limx→∞(h(x+y)−h(x)) = 0, y∈Rif and only if h may be written as

h(x) =d(x) + Z x

b

g(y)dy, (3.12)

with x ≥ b = log(a), d(x) = c₁(e^x), g(x) = (e^x), where limx→∞d(x) → d ∈ R, limx→∞g(x) → 0. This follows by applying the definition of h and by substitution in the integral.

Lemma 3.17. If Ψ is positive, measurable, defined on some interval[A,∞) and

x→∞lim Ψ(kx)

Ψ(x) →1,

for every k > 0, then Ψ is bounded on all finite intervals far enough to the right. If h(x) :=

log (Ψ(e^x)),h is also bounded on finite intervals far enough to the right.

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Proof. By the Uniform Convergence Theorem 3.11 and Remark 3.16 (ii), we get for x₀ sufficiently large,

|h(x+u)−h(x)|<1, x≥x0

foru ∈[0,1]. Hence, if we considerx =x₀ and writey :=x₀+u by the triangular inequality,

|h(y)| ≤1+|h(x₀)|fory∈[x₀, x₀+1]. By induction we get|h(y)| ≤h(x₀)+nfory∈[x₀, x₀+n], for n = 1,2, . . ., and we get the conclusion for h. Further this implies the conclusion for Ψ(x) = exp(h(log(x))).

Proof of Theorem 3.14. Assume that (3.10) holds and let k∈I. Then we get Ψ(kx)

Ψ(x) = c(kx) c(x) exp

Z kx x

(u) u du

.

Choose >0. Then, for all sufficiently largex, the right hand side lies between (1±) exp (±|log(k)|),

from which (3.4) follows.

By Lemma 3.17, his integrable on finite intervals far enough to the right, by being bounded and measurable on them. For X large enough, we may therefore write

h(x) = Z x+1

x

h(x)−h(t) dt+

Z x X

h(t+ 1)−h(t) dt+

Z X+1 X

h(t)dt, x≥X.

The last term on the right is a constant, say c. If g(x) := h(x+ 1)−h(x), then g(x) → 0 as x→ ∞by Remark 3.16 (ii). The first term on the right is R1

0(h(x)−h(x+u))du, which tends to 0 as x → ∞, by the Uniform Convergence Theorem 3.11 and again Remark 3.16 (ii). By puttingd(x) :=c+R1

0(h(x)−h(x+u))du we obtain (3.12)

3.3 Indicator functions with relation to slowly varying functions

In this section we investigate more closely the growth of functions of order 0. The main result is the theorem below. There we denote, for an entire function f,

mf(r) := inf

|z|=r|f(z)|and Mf(r) := sup

|z|=r

|f(z)|.

Theorem 3.18. Letf be an entire function andΨbe a positive slowly varying function. Further assume

α:= lim sup

r→∞

log(M_f(r))

Ψ(r) <∞. (3.13)

Then also

lim sup

r→∞

log m_f(r) Ψ(r) =α.

Remark 3.19. This theorem applies to all functions of zero order. For iff has order zero, and ρ(r) is a proximate order forf, then Ψ(r) :=r^ρ(r)is slowly varying and (3.13) holds with some α >0. It does not remain true for functions of larger order. For example consider f(z) :=e^z. Then M_f(r) =e^r, whilem_f(r) =e^−r.

A proof of this (and some stronger) facts, without appealing to slowly varying functions, can be found in [4], chapter 3.

The proof of Theorem 3.25 will closely follow the lines of [2] section 6, but first we present a corollary of it.

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Corollary 3.20. Assume that (3.13) holds. Then the indicator of f w.r.t Ψ, i.e. the function hf,Ψ(θ) := lim sup

r→∞

log |f(re^iθ)|

Ψ(r) , θ∈[0,2π), (3.14)

is constant equal to α.

Proof of Corollary 3.20. Clearly for everyθ∈[0,2π]

m_f(r)≤

f rexp(iθ)

≤M_f(r).

Thus our corollary follows from Theorem 3.18.

We need a few preliminary observations, before we start with the proof of Theorem 3.25.

First, let us observe that validity of (3.13) for some slowly varying function implies thatf is of zero order.

Lemma 3.21. LetΨbe a slowly varying function. Then for every δ >0 we haveΨ(r) =o(r^δ), r→ ∞.

Proof. Applying Theorem 3.14, we obtain Ψ(r)≤c(r) exp

Z r0

a

max_u∈[a,r₀_](u)

s ds+

Z r r0

(s) s ds

For r0 large enough we get (u) ≤⁰ for u ≥ r0 and hence the second integral is bounded by ⁰log(r). Since, we can choose(x) to be bounded, the first one is a constantK, and hence we obtain

Ψ(r)≤c(r) exp K+⁰log(r)

=c(r)e^Kr⁰. If we take ⁰≤δ we obtain

Ψ(r) =o r^δ

, asr→ ∞, forδ >0, since limr→∞c(r) =c, by Theorem 3.14.

Remark 3.22. Combining Lemma 3.21 with (3.13) shows thatf has zero order, because since Ψ(r) =o(r^δ) holds, we obtain lim sup_r→∞ ^log(M_rδ^f^(r)) = 0 for all δ >0.

Next we want to present two well known results, which we will need during this section.

Their proofs can be found for example in [6].

Theorem 3.23 (Jensen). Let f(z) be holomorphic in a disc of radius R with center at the origin and f(0)6= 0. Then

Z R 0

n(t)

t dt= 1 2π

Z 2π 0

log |f(Rexp(iθ))|

dθ−log(|f(0)|), where n(t) is the number of zeroes off in the disk|z| ≤t.

Theorem 3.24 (Hadamard). An entire functionf(z)of orderρ can be represented in the form f(z) =z^mexp(p(z))

∞

Y

0

G z

zν

, q

,

where z_ν are the zeroes of f, p(z) is a polynomial whose degree does not exceed ρ, m is the multiplicity of the zero at the origin, q≤ρ and

G z

zν

, q

:=

1− z

zν

exp

u+u²

2 +· · ·+u^q q

, G

z zν

,0

:= 1− z zν

.

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The basic connection between M_f(r), m_f(r), and Ψ(r) is given by the following theorem, which provides a very important step in the proof of Theorem 3.25.

Theorem 3.25. Let Ψ be a slowly varying function, f entire, and assume that (3.13) holds.

Then

Z 2r r

log

M_f(t) mf(t)

1

tdt=o(Ψ(r)), (3.15)

for r → ∞.

Remark 3.26. We start the proof with a few preliminary observations.

(i) We can assume w.l.o.g. that f(0) = 1. This is because passing from f(z) to f(z)/(cz^p), where p is a nonnegative integer and c a non-zero constant, does not affect the relations (3.13), (3.15).

(ii) By Lemma 3.21 f is of order zero, and hence Hadamard’s Theorem 3.24 gives f(z) =

∞

Y

ν=1

1− z

zν

, wherez_ν are the zeroes of f. Hence, for|z|=r,

∞

Y

ν=1

1− r

|z_ν|

≤ |f(z)| ≤

∞

Y

ν=1

1 + r

|z_ν| . Therefore, we get

log

M_f(r) m_f(r)

≤

∞

X

ν=1

log r+|z_ν| r− |z_ν|

!

. 0< r <∞. (3.16)

To prove Theorem 3.25 we need an estimate of the sum in (3.16), which we will obtain by means of the following lemmata.

Lemma 3.27. Assume that f satisfies (3.13), f(0) = 1, and denote by n(r) the number of zeroes of f in the disc{z∈C:|z| ≤r}, then

n(r) =o Ψ(r)

, r→ ∞.

Proof. Let K >1. From Jensen’s Theorem 3.23, (3.13), and the fact that Ψ is slowly varying, we obtainr0(K)>0 such that

Z Kr 0

n(t)

t dt≤log Mf(Kr)

≤(α+ 1)Ψ(Kr)≤2(α+ 1)Ψ(r), r > r0(K).

Hence

n(r) log(K) =n(r) Z Kr

r

1 tdt≤

Z Kr r

n(t)

t dt≤2(α+ 1)Ψ(r), r > r0(K).

Since K can be chosen arbitrarily large, this yields our assertion.

Lemma 3.28. Fors >0 we have Z ∞

0

log

1 +^s_t 1−^s_t

1 tdt= π

2.

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Proof. We first put s/t=x so that our integral becomes Z ∞

0

log

1 +x 1−x

1 xdx=

Z 1 0

log

1 +x 1−x

1 xdx+

Z 1 0

log

1 +y 1−y

1 ydy, where we have putx= 1/y, when x >1. Further

Z 1 0

log

1 +x 1−x

1 xdx=

Z 1 0

−

∞

X

k=0

(−x)^k+1 k+ 1 +

∞

X

k=0

x^k+1 k+ 1

!1 xdx= 2

∞

X

m=1

Z 1 0

x^2m−2 2m−1dx, where we used the power series representation of the logarithm. Evaluating the integral in the last sum yields

2

∞

X

m=1

Z 1 0

x^2m−2

2m−1dx= 2

∞

X

m=1

1 2m−1

2

= π 4 and this proves our lemma.

Lemma 3.29. We have

X

|zν|>2r

log

1 +_|z^r

ν|

1−_|z^r

ν|

!

=o(Ψ(r)). (3.17)

Proof. First we note log

1 +^r_t 1−^r_t

= log

1 + 2r (1−^r_t)t

≤ 2r

(1−^r_t)t ≤ 4r

t , t≥2r.

Further, by Lemma 3.27 and Lemma 3.21 we get n(t) = o(Ψ(t)) = o(t^δ) for arbitrary δ > 0.

Choosingδ <1, we conclude

t→∞lim n(t) log

1 +^r_t 1−^r_t

= 0.

Applying this, we can write the sum from the left hand side of (3.17) as Z ∞

2r

log

1 +^r_t 1−^r_t

dn(t) =

n(t) log

1 +^r_t 1−^r_t

∞ 2r

+ 2r Z ∞

2r

n(t) t²−r²dt=

=−n(2r) log(3) +o

r Z ∞

2r

Ψ(t) t² dt

,

(3.18)

forrsufficiently large, noting that _t2^t−r² ² ≤4/3 fort∈[2r,∞). Now considert∈[2^pr,2^p+1r] for p∈N. Then, forr large enough

Ψ(t)<(1 +)Ψ(r2^p)<2^1/2Ψ(2r2^p−1)<2Ψ 2r2^p−2

<· · ·<2^p/2Ψ(2r), because Ψ is slowly varying. Hence, forr sufficiently large, we obtain

Z r2^p+1 r2^p

Ψ(t) t² dt <

Z r2^p+1 r2^p

Ψ(2r)2^p/2

t² dt= Ψ(2r)2^p/2

r2^p+1 = Ψ(2r) 2r 2^−p/2. We get

r Z ∞

2r

Ψ(t)

t² dt < Ψ(2r) 2

∞

X

p=1

2^−p/2 =O(Ψ(r)).

Thus, (3.18) and Lemma 3.27 yield the assertion.

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Proof of Theorem 3.25. Let r < t <2r. Then, by Lemma 3.29, X

|zν|>4r

log 1 +_|z^t

ν|

1−_|z^t

ν|

!

≤ X

|zν|>4r

log 1 +_|z^2r

ν|

1−_|z^2r

ν|

!

=o(Ψ(r)), and hence

X

|zν|>4r

log

|z_ν|+t

|z_ν| −t

=o(Ψ(r)).

Thus

Z 2r r

X

|z_ν|>4r

log

|z_ν|+t

|z_ν| −t 1

tdt=o Ψ(r) Z 2r

r

1

tdt=o(Ψ(r)). (3.19) Further, by Lemma 3.28

Z 2r r

X

|zν|≤4r

log

|z_ν|+t

|z_ν| −t

1

tdt≤ X

|zν|≤4r

Z ∞ 0

log

1 +t

|1−t|

1 t ≤ π²

2 n(4r) =o Ψ(r)

. (3.20) Putting together (3.16), (3.19) and (3.20) yields our assertion.

Proof of Theorem 3.18. Let >0 and choose a larger, such that log Mf(r)

>(α−)Ψ(r).

Letk∈[1,2]. Then, because Ψ is slowly varying, we get Ψ(kr)≤(1 +₁)Ψ(r) and hence (α−)Ψ(r)≥ α−

1 +₁Ψ(t)≥(α−2)Ψ(t), with r ≤t≤2r, for r sufficiently large. Further, since log M_f(r)

increases withr we obtain that ifr is sufficiently large

log M_f(t)

>(α−2)Ψ(t), r ≤t≤2r. (3.21) Again, because Ψ is slowly varying, we get Ψ(kr)≥(1−₂)Ψ(r), and hence

−Ψ(r)≥ −

1−₂Ψ(t)≥ −2Ψ(t), r≤t≤2r

Applying this, and by the mean value theorem for integrals, it follows from Theorem 3.25 that we can chooset, such thatr ≤t≤2r and

log m_f(t)

>log M_f(t)

−Ψ(r)≥log M_f(t)

−2Ψ(t), ifr is sufficiently large.

On combining this with (3.21) we obtain log m_f(t)

>(α−4)Ψ(t).

Since is arbitrarily small we obtain lim sup

t→∞

log m_f(t) Ψ(t) ≥α.

Since m_f(r)≤M_f(r) we have from (3.13) lim sup

t→∞

log m_f(t) Ψ(t) ≤α, which proves the corollary.

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3.4 A Formula for Type

The next theorem will yield a formula for the type of an entire function with respect to a proximate order ρ(r). Since by Proposition 3.12, r^ρ(r) is an increasing function for r > 0, if ρ >0 the equationt=r^ρ(r) admits a unique solution fort >0. We will denote this solution by r =φ(t). Therefore φ(t) is just the inverse function of r^ρ(r). Of course φ(t) depends on ρ(r), but we will not denote this dependence. The proof we present is based on the proofs of this theorem given in [5], [6].

Theorem 3.30. Let f(z) = P∞

n=0c_nzⁿ, z ∈ C, be the Taylor series expansion of the entire function f(z) of finite order ρ >0 and of proximate order ρ(r). Then the type σf with respect to the proximate order ρ(r) is given by

1

ρlog(σf) = lim sup

n→∞

1

nlog |c_n|

+ log φ(n)

− 1

ρ− log(ρ)

ρ , ρ >0, (3.22) where φ(t) is the inverse function of r^ρ(r).

Proof. We divide the proof in several steps.

Step 1: We show that lims→∞ φ(ks)

φ(s) =k¹^ρ, k ∈ (0,∞). Set t = r^ρ(r). Since, by definition, log(t(r)) =ρ(r) log(r), we get

d(log(t))

d(log(r)) =ρ⁰(r)rlog(r) +ρ(r),

which, by Definition 3.4, tends to ρ when r tends to infinity. Furthermore d(log(t(r)))

d(log(r)) =

d

dtlog(t(r))

d

dtlog(r) =

d

dtlog(t(r))

d

dtlog φ(t(r)). So combining these two equalities yields, forr sufficiently large,

1 ρ −

d

dtlog t(r)

< d

dtlog φ(t(r))

<

1 ρ +

d

dtlog t(r) . If we integrate from stoks, we obtain

1 ρ −

log(k)<log

φ(ks) φ(s)

<

1 ρ +

log(k), and hence taking exponentials yields

s→∞lim φ(ks)

φ(s) =k¹^ρ. (3.23)

Step 2: By the Cauchy Integral Formula we have the following estimate for the coefficients of the power series

|c_n| ≤ sup_|z|≤r|f(z)|

rⁿ .

Since f(z) is of finite order ρ, there existsr0 such that for r≥r0,k≥ρand ˜σ > σ_f^r^k sup

|z|≤r

|f(z)| ≤e^˜^σr^k

Master Thesis Canonical Systems

Master Thesis

Canonical Systems

Contents

1 Introduction

2 Exponential Type

3 Growth Functions and λ-type