New proof of Nagnibida’s theorem

Using the Duhamel product for holomorphic functions we give a new proof of Nagnibida’s theorem on unicellularity of integration operator Jα, (Jαf) (z) = z α f(t)dt , acting in the space Ho (Ω). Let α be a fixed complex number, Ω be a simple-connected domain in the plane containing the number α , and assume that Ω is star-shaped with respect to the α in the sense that z ∈ Ω implies λz+(1 − λ)α ∈ Ω for each λ ∈ (0, 1). Let Ho (Ω) be the vector space of all functions holomorphic in Ω with the topology of compact convergence, and let Jα denotes the integration operator acting in Ho (Ω) by formula (Jαf) (z) = ∫ z

Let α be a fixed complex number, Ω be a simple-connected domain in the plane containing the number α , and assume that Ω is star-shaped with respect to the α in the sense that z ∈ Ω implies λz +(1 − λ) α ∈ Ω for each λ ∈ (0, 1).Let Ho (Ω) be the vector space of all functions holomorphic in Ω with the topology of compact convergence, and let J α denotes the integration operator acting in Ho (Ω) by formula where integral is taken on the line segment with the ends α and z .
We denote by I p the set of functions with a zero at the point α of order greater than or equal p, i.e., It is clear that all the subspaces I p , p ≥ 1, are closed J α − invariant subspaces (i.e., I p ∈ Lat (J α ) , p ≥ 1) and if p > q, then I p ⊂ I q .The operator J α has been extensively studied by Nagnibida [?].
In this short communication we give a new proof of the following theorem of Nagnibida [3,Theorem 5] on unicellularity of the operator J α in the space Ho (Ω) .We recall that an operator A acting in the topological vector space X is called unicellular if its lattice of invariant subspaces Lat (A) is linearly ordered with respect to inclusion.

Proof.
In the proof of theorem we shall essentially use the similar arguments of the paper [?].For every f, g ∈ Ho (Ω) let us define the following product: (1) where integral is taken on the line segment with the ends α and z .(For α = 0 this is the so-called Duhamel product, see [?]).Clearly f α g also belongs to Ho (Ω), and with this multiplication Ho (Ω) becomes an algebra.One can use results of operational calculus [?], [?] and analytic continuation to show that Ho (Ω) is commutative and associative(it is actually clear from (1) that f (z) ≡ 1 is the unit for Ho (Ω) , and a simple change of variables shows commutativity).It is clear from (1) that Then by considering this and the fact is complete in Ho (Ω) (which is obtained from the well-known completeness criterion of Banach in the Frechet spaces and classical Runge's approximation theorem), it is not difficult to show that the problem of description of all closed J α − invariant subspaces is equivalent to the description of all closed ideals of the algebra Ho (Ω) with multiplication as α .We shall now describe the all closed ideals of the algebra Ho (Ω) , α by showing that any nontrivial closed ideals has the form I p for some p ≥ 1 .The proof will be done in several steps: Without loss of generality, we assume that f (α) = 1.
Then, obviously, f (z) = 1 − h(z), where h(z) = 1 − f (z) and h(0) = 0 .Let Ω be an arbitrary bounded domain, star-shaped with respect to the point α , and such that its closure is contained in Ω. Choose the number M > 0 such that h (z) ≤ M for z ∈ Ω .Then α denote the α − product of h with itself n times for n ≥ 0, where h α (z) ≡ 1 .We shall prove by induction that for z ∈ Ω .Indeed, for this purpose, assume that for z ∈ Ω .Then, by considering that h(α) = 0 , we obtain that α (z) are majorized respectively by the series where d is the diameter of Ω .Consequently, g is holomorphic in Ω and because of the arbitrariness of Ω , g is holomorphic in Ω. Finally Step 2. Let f, g ∈ Ho (Ω) .Suppose f has a zero at the point α of order p, g has a zero of order q , and p ≤ q .Then there exists h ∈ Ho (Ω) with a zero of order q − p such that (1) and integration by parts p times we obtain Step 3.
Ho (Ω) , In addition we observe that the sequence is a filtration of Ho (Ω) , α and that I p /I p+1 is isomorphic to the complex numbers.
Thus, by Step 3 the proof of theorem is established.
Remark.Note that the invertibility criterion with respect to the Duhamel product α can also be easily deduced from the corresponding result of Wigley [4,Theorem] for α = 0 .Indeed, there is a canonical algebra isomorphism Γ between the spaces Ho (Ω) and Ho (Ω 0 ) where Ω 0 = Ω − α , when endowed with their corresponding Duhamel products.
The map is obviously defined by Γf (z) = f (z + α) .Besides, Γ implements a similarity between the corresponding integration operators, so from the invariant subspace point of view, it is enough to assume that α = 0.

α
is a local ring and any nontrivial closed ideal in Ho (Ω) , α has the form I p for some p ≥ 1. Proof.If f ∈ Ho (Ω) , α and f (α) = 0 , then it follows from (1) that f α g (α) = 0 for g ∈ Ho (Ω) , α .Thus I 1 form an ideal.From the Step 1 this ideal must be maximal and is only maximal ideal of Ho (Ω) , α .Thus Ho (Ω) , α is local ring.Each I p is an ideal (closed) of Ho (Ω) , α as can be seen from (1).We now prove that other nontrivial ideals not exist.Indeed, if I is any nontrivial ideal in Ho (Ω) , α , then I has an element f of lowest order p.By Step 2, if g ∈ I p , then there exists h ∈ Ho (Ω) such that g = h α f , and hence I p ⊆ I .If I p = I , then I has an element of order less than or equal to p − 1 , a contradiction.Thus I = I p .Consequently, the only ideals of Ho (Ω) , α are the I p , p ≥ 1.