Closed range Volterra-type integral operators and dynamical sampling

We solve the closed range problem for Volterra-type integral operator on Fock spaces. Several applications of the result related to the operators invertibility, Fredholm, and dynamical sampling structures from frame perspectives are provided. We further prove a bounded Volterra-type integral operator preserves no frame property. On the contrary, the adjoint operator preserves frame if and only if it is noncompact but fails to preserve both tight frames and Riesz basis.


Introduction
For holomorphic functions f and g in a given domain, we define the Volterra-type integral operator V g by Various aspects of V g have been widely studied since 1997 mainly on spaces of holomorphic functions: see for example on Hardy spaces [4,5,15], Bergman spaces [6,8,14], and Fock spaces [7,11,12] and the respective reference therein. The purpose of this note is to take the study further and solve the closed range problem for V g on the Communicated by Adrian Constantin.

B Tesfa Mengestie
Tesfa.Mengestie@hvl.no 1 Mathematics Section, Western Norway University of Applied Sciences, Klingenbergvegen 8,5414 Stord, Norway Fock spaces. We further provide several application of the result concerning invertibility and dynamical sampling structures of the operator from frame perspectives. Note that the closed range problem is one of the basic problems in operator theory which finds lots of connections in various parts of mathematics.
For 1 ≤ p < ∞, the Fock spaces F p consist of all entire functions f on the complex plane C for which where A denotes the Lebesgue area measure on C. The bounded and compact V g on Fock spaces had been identified in [7,11]. In deed, for p ≤ q, the operator In this case, compactness is described by the condition a = 0. On the other hand, for p > q, V g : F p → F q is bounded if and only if it is compact, and this holds if and only if g(z) = az + b and q > 2 p/( p + 2). The rest of this note is organized into two parts. In the first part we study the closed range problem for V g on the Fock spaces. Theorem 1.1 provides a complete answer to this problem. We apply this result and identify conditions under which the operator becomes Fredholm and draw the conclusion that it fails to be surjective and hence not invertible. In the second part, we further apply the result to study more applications related to the dynamical sampling behaviours of the operator from frame perspectives. It is proved that there exists no function f in F 2 for which its orbits under V g or its adjoint represents a frame family for the space. In addition, we show that the operator fails to preserves frames structures. On the contrary, the adjoint operator preserves frame if and only if it is noncompact but fails to preserve both tight frames and Riesz basis.
Note that if g is a constant, then the operator V g reduces to the zero operator, and we excluded this case in the rest of the manuscript. We may now state the first main result. Theorem 1.1 Let 1 ≤ p, q < ∞ and V g : F p → F q be bounded and hence g(z) = az 2 + bz + c for some a, b, c ∈ C. Then V g has a closed range if and only if a = 0 and p = q. The closed range is given by (1.1) As will be explained latter, the result equivalently characterizes when V g is bounded from below. That is, there exists a constant > 0 such that V g f q ≥ f p for all f ∈ F p . In contrast to conditions often given in terms of sampling sets or reverse Carelson measures, our condition here is quite simple to apply.
As first immediate consequence of the result, we observe that the classical integral operator I f (z) = z 0 f (w)dw and Hardy operator H f (z) = 1 z z 0 f (w)dw have no closed ranges on Fock spaces. As another consequence of Theorem 1.1, we record the next corollary about Fredholm Volterra-type integral operators. Recall that a bounded operator T in a Banach space is said to be Fredholm if its range R(T ) is closed and both Ker T and Ker T * are finite dimensional. If T is Fredholm, its index is the number given by dim(K er T ) − dim(K er T * ). It is known that every bounded operator with closed range has an inverse called the pseudo-inverse, or the Moore-Penrose inverse. Since Ker V * g is the orthogonal complement of the range of V g , (1.2) implies that Ker V * g = C.
Corollary 1.2 Let 1 ≤ p < ∞ and V g is bounded on F p and hence g(z) = az 2 +bz+c for some a, b, c ∈ C. Then the following statements are equivalent.
(i) a = 0; (ii) V g is Fredholm of index one and its Fredholm inverse is given by Note that the differential operator D f = f is not bounded on Fock spaces [10]. Thus, the well definedness and the boundedness of V −1 g on F p comes from the requirement a = 0 and the estimate in (1.3) below.
We remark that applying integration by part in the definition of the operator V g above gives the relation where Mg( f ) = g f is the multiplication operator and J g f = V f (g) is the Volterra companion integral operator. It is known that J g : F p → F q is bounded if and only if M g is bounded, and this holds only when g is a constant where the constant being zero for p > q. Thus, these operators have obviously closed ranges on F p and will not be a point of further discussion in the rest of our consideration.
We give a word on notation. The notion U (z)

Proof of Theorem 1.1
Assuming that g is not a constant, we first show V g is an injective map. Let f 1 and f 2 in Taking derivative on both sides we notice f 1 (z) = f 2 (z) for all z ∈ C except possibly at points where g vanishes. But since f 1 and f 2 are entire, it follows that f 1 = f 2 . Consequently, as known from an application of Open Mapping Theorem, an injective bounded operator has closed range if and only if it is bounded from below (see for example [1,Theorem 2.5]). Thus, we proceed to use this equivalent reformulation as a tool to prove the claim. Another important tool in our work is the estimate which holds for all entire functions f [7]. Suppose now that a = 0 and p = q. Then for every f ∈ F p , which readily shows V g is bounded from below. Hence, the conditions in the theorem are sufficient. Conversely, suppose for the sake of contradiction V g is bounded from below and a = 0. Applying the operator to the normalized kernel function k n = K n / K n 2 and estimate (1.3) as n → ∞. This contradicts boundedness from below. Therefore, a = 0. Next, we assume a = 0 and show that p = q whenever range of V g is closed. If p > q, then as already indicated above, boundedness of the operator is achieved only when a = 0. Thus, it remains to check for p < q. Since a = 0, note that for every f ∈ F q boundedness from below implies for some > 0. The last inequality in (1.4) clearly indicates we need to compare the norm of functions in F p and F q . We may consider the sequence f n (z) = z n , n = 1, 2, ... in F p . Using polar integration and Stirling's approximation formula  1). From the proof made above, we already have p = q and hence R(V g ) ⊆ F p . On the other hand, for each h ∈ F p , we consider the function f h defined by Observe that f h is entire and by (1.3), it belongs to F p and V g f h = h. Therefore, the other inclusion F p ⊆ R(V g ) holds and completes the proof.
is the orbit of f ∈ H under the operator T . Recall that a family ( f j ), j ∈ I of vectors in a Hilbert space H is a frame if there exist positive constants A and B such that for any g ∈ H The constants A and B are called the lower and upper bounds of the frame respectively. It is called a tight frame when A = B. Frames are generalizations of bases and their main applications comes from the fact that a frame can be designed to be redundant while still providing a reconstruction formula for each vector in the space. Thus, identifying methods that generate new frames has been an interesting problem in frame theory. A special type of frame is Riesz basis. A family ( f j ), j ∈ I of vectors in H is a Riesz basis if it is complete and there exist constants 0 < A ≤ B < ∞ such that for any c j ∈ 2 (I ) We may start with the following important lemma which connects the closed range problem with dynamical sampling in frame theory.

Lemma 2.1 Let H be a Hilbert space and T be a bounded linear operator on H.
If {T n f } ∞ n=0 is a frame for some f ∈ H, then (i) T is surjective.
While the proof of part (ii) is available in [3], part (i) follows easily since {T n f } ∞ n=0 is a frame for each h ∈ H, there exists sequence (c n ) such that h = n=1 c n T n f = T n=1 c n T n−1 f from which the claim follows. Note here that the frame property implies R(T ) = H is closed and this in particular connects us with our result in Theorem 1.1.

Theorem 2.2
Let V g is bounded on F 2 and hence g(z) = az 2 + bz + c for some a, b, c ∈ C. Then neither Proof Suppose, for the sake of contradiction, that there exists an f ∈ F 2 such that {V n g f } ∞ n=0 is a frame. Then by Lemma 2.1, V g is surjective which contradicts Theorem 1.1.
Next we consider the case with the adjoint operator. Suppose that there exists an h ∈ F 2 such that {(V * g ) n h} ∞ n=0 is a frame again. Then the range of V * g is closed. By Theorem 1.1 and Closed Range Theorem, this holds if and only if a = 0. On the other hand, set f (z) = z and consider the iterations where the sequence c k is of the form p k /q k , q k ≤ k! and p k is a sequence of numbers some of them involve multiples of b with p 2n+1 = 2 n and p n+1 = b n . Now, V n g f 2 = c 2n+1 a n z 2n+1 + c 2n a n−1 z 2n + · · · + c n+2 az n+2 + c n+1 z n+1 |2 n a n | 2 z 2n+1 2 − |p 2n a n−1 | z 2n 2 − · · · − |p n+2 a| z n+2 On the other hand, by (1.5) z n 2 n e n √ n which obviously grows much faster than exponential and factorial sequences. Setting this in (2.3), we observe that as required by part (ii) of Lemma 2.1 only when a = b = 0 which is a contradiction, and the claim is proved.

Frame preserving V g and V * g
Another interesting operator related question on frame property is as to when V g preserves frame;in the sense that V g f n is a frame whenever f n is. This is known to be one of the approaches used to construct new frames using tools in operator theory. A useful result connecting the closed range and the frame preserving problems is the following [2,9,13].

Lemma 2.3 Let T be a bounded linear operator on a
This shows j∈I c j f j belongs to the orthogonal complement of the range of V g . Then by (1.1), j∈I c j f j ∈ C = K er V * g . On the other hand, f j is a Riesz basis. Hence, j∈I c j f j is not necessarily zero.
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