Introduction

For holomorphic functions f and g in a given domain, we define the Volterra-type integral operator \(V_g\) by

$$\begin{aligned} V_gf(z)= \int _0^z f(w)g'(w) dw. \end{aligned}$$

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 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 \(1\le p<\infty \), the Fock spaces \(\mathcal {F}_p\) consist of all entire functions f on the complex plane \(\mathbb {C}\) for which

$$\begin{aligned} \Vert f\Vert _{p}^p= \frac{p}{2\pi }\int _{\mathbb {C}} |f(z)|e^{-\frac{p }{2}|z|^2} dA(z) <\infty , \end{aligned}$$

where A denotes the Lebesgue area measure on \(\mathbb {C}\). The bounded and compact \(V_g\) on Fock spaces had been identified in [7, 11]. In deed, for \(p\le q\), the operator \(V_g:\mathcal {F}_p \rightarrow \mathcal {F}_q \) is bounded if and only if \(g(z)= az^2+bz+c\) for some \(a, b, c\in \mathbb {C}\). In this case, compactness is described by the condition \(a=0\). On the other hand, for \( p>q\), \(V_g:\mathcal {F}_p \rightarrow \mathcal {F}_q \) is bounded if and only if it is compact, and this holds if and only if \(g(z)= az+b\) and \(q>2p/(p+2).\)

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 \(\mathcal {F}_2\) 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 \(1\le p, q < \infty \) and \(V_{g}:\mathcal {F}_p \rightarrow \mathcal {F}_q \) be bounded and hence \(g(z)= az^2+bz+c\) for some \(a, b, c\in \mathbb {C}\). Then \(V_g\) has a closed range if and only if \(a\ne 0\) and \(p=q\). The closed range is given by

$$\begin{aligned} \mathcal {R}(V_g )= \big \{f\in \mathcal {F}_p: f(0)=0\big \}. \end{aligned}$$
(1.1)

As will be explained latter, the result equivalently characterizes when \(V_g\) is bounded from below. That is, there exists a constant \(\epsilon >0\) such that \( \Vert V_g f\Vert _q\ge \epsilon \Vert f\Vert _p\) for all \(f \in \mathcal {F}_p\). 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 \(If(z)=\int _{0}^z f(w)dw \) and Hardy operator \(Hf(z)= \frac{1}{z}\int _{0}^z f(w)dw\) 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 \(\mathcal {R}(T)\) is closed and both \( \text {Ker}\ T\) and \( \text {Ker}\ T^*\) are finite dimensional. If T is Fredholm, its index is the number given by \(dim (Ker T)- dim(Ker T^*)\). It is known that every bounded operator with closed range has an inverse called the pseudo-inverse, or the Moore-Penrose inverse. Since \( \text {Ker}\ V_g^*\) is the orthogonal complement of the range of \(V_g\), (1.2) implies that \( \text {Ker}\ V_g^*= \mathbb {C}\).

Corollary 1.2

Let \(1\le p < \infty \) and \(V_{g}\) is bounded on \(\mathcal {F}_p \) and hence \(g(z)= az^2+bz+c\) for some \(a, b, c\in \mathbb {C}\). Then the following statements are equivalent.

  1. (i)

    \(a\ne 0\);

  2. (ii)

    \(V_g\) is Fredholm of index one and its Fredholm inverse is given by

    $$\begin{aligned} V_g^{-1} f(z)= {\left\{ \begin{array}{ll} \lim \limits _{\mathop {w\rightarrow \frac{-b}{2a}}}\frac{f'(w)}{2aw+b} \ \ \ \ \ z= -b/2a\\ \frac{f'(z)}{2az+b}, \ z\ne -b/2a. \end{array}\right. } \end{aligned}$$
    (1.2)

Note that the differential operator \(Df= f'\) is not bounded on Fock spaces [10]. Thus, the well definedness and the boundedness of \(V_g^{-1}\) on \(\mathcal {F}_p\) comes from the requirement \(a\ne 0\) 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

$$\begin{aligned} M_g(f)= f(0)g(0)+ V_g(f)+ J_f(g), \end{aligned}$$

where \(Mg(f)= gf\) is the multiplication operator and \(J_gf= V_f(g)\) is the Volterra companion integral operator. It is known that \(J_{g}:\mathcal {F}_p \rightarrow \mathcal {F}_q \) 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 \(\mathcal {F}_p\) and will not be a point of further discussion in the rest of our consideration.

We give a word on notation. The notion \(U(z)\lesssim V(z)\) (or equivalently \(V(z)\gtrsim U(z)\)) means that there is a constant C such that \(U(z)\le CV(z)\) holds for all z in the set of a question. We write \(U(z)\simeq V(z)\) if both \(U(z)\lesssim V(z)\) and \(V(z)\lesssim U(z)\).

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 \(\mathcal {F}_p\) such that \(V_{g}f_1= V_{g}f_2\). Taking derivative on both sides we notice \(f_1(z)= f_2(z)\) for all \(z\in \mathbb {C}\) 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

$$\begin{aligned} \Vert f\Vert _p^p \simeq |f(0)|^p +\int _{\mathbb {C}} |f'(z)|^p(1+|z|)^{-p}e^{-\frac{p\alpha }{2}|z|^2}dA(z) \end{aligned}$$
(1.3)

which holds for all entire functions f [7]. Suppose now that \(a\ne 0\) and \(p=q\). Then for every \(f\in \mathcal {F}_p\),

$$\begin{aligned} \Vert V_g f\Vert _p^p \simeq \int _{\mathbb {C}} \frac{|g'(z)|^p}{(1+|z|)^p}|f(z))|^p e^{-\frac{p}{2}|z|^2} dA(z) \simeq \int _{\mathbb {C}} |f(z)|^pe^{-\frac{p}{2}|z|^2} dA(z)\simeq \Vert f\Vert _p^p, \end{aligned}$$

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 \(k_n= K_n/\Vert K_n\Vert _2\) and estimate (1.3)

$$\begin{aligned} \Vert V_g k_n\Vert _q^q&\simeq \frac{1}{\Vert K_n\Vert _2^q} \int _{\mathbb {C}} \frac{|g'(z)|^q}{(1+|z|)^q}|K_n(z))|^q e^{-\frac{q}{2}|z|^2} dA(z) \nonumber \\&= \frac{n^{-q}|b|^q}{\Vert K_n\Vert _2^q} \int _{\mathbb {C}} \frac{ |nK_n(z))|^q}{(1+|z|)^q} e^{-\frac{q}{2}|z|^2} dA(z)\simeq \frac{|b|^q\Vert K_n\Vert _q^q}{n^q\Vert K_n\Vert _2^q}= \frac{|b|^q}{n^q} \rightarrow 0 \end{aligned}$$

as \(n\rightarrow \infty \). This contradicts boundedness from below. Therefore, \(a\ne 0\).

Next, we assume \(a\ne 0\) 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 \(a\ne 0\), note that for every \( f\in \mathcal {F}_q\) boundedness from below implies

$$\begin{aligned} \Vert V_g f\Vert _q^q \simeq \int _{\mathbb {C}} \frac{|g'(z)|^q}{(1+|z|)^q}f(z))|^q e^{-\frac{q}{2}|z|^2} dA(z)\simeq \Vert f\Vert _q^q\ge \epsilon \Vert f\Vert _p^q \end{aligned}$$
(1.4)

for some \(\epsilon >0\). The last inequality in (1.4) clearly indicates we need to compare the norm of functions in \(\mathcal {F}_p\) and \(\mathcal {F}_q\). We may consider the sequence \(f_n(z)= z^n, n= 1, 2,...\) in \(\mathcal {F}_p\). Using polar integration and Stirling’s approximation formula

$$\begin{aligned} \Vert f_n\Vert _p^p= p \int _{0}^\infty r^{np+1} e^{-pr^2/2} dr = \Big (\frac{1}{p}\Big )^{np/2} \Gamma \Big (\frac{np+2}{2}\Big ) \simeq \Big (\frac{n}{e}\Big )^{\frac{np}{2}} \sqrt{n}. \end{aligned}$$
(1.5)

See also [16, p.40]. It follows from this and (1.4), the estimate

$$\begin{aligned} \Vert V_g f_n\Vert _q \simeq \Vert f_n\Vert _q\ge \epsilon \Vert f_n\Vert _p \end{aligned}$$

holds only

$$\begin{aligned} \Vert f_n\Vert _q/\Vert f_n\Vert _p\simeq n^{\frac{1}{2q}-\frac{1}{2p}} \ge \epsilon \end{aligned}$$

for all \(n\in \mathbb {N}\). This gives a contradiction when \(n\rightarrow \infty \).

It remains to verify (1.1). From the proof made above, we already have \(p=q\) and hence \(\mathcal {R}(V_g )\subseteq \mathcal {F}_p\). On the other hand, for each \(h\in \mathcal {F}_p\), we consider the function \(f_h\) defined by

$$\begin{aligned} f_h(z)= {\left\{ \begin{array}{ll} \lim \limits _{\mathop {w\rightarrow \frac{-b}{2a}}} \frac{h'(w)}{2aw+b}\ \ \ \ \ z= -b/2a\\ \frac{h'(z)}{2az+b}, \ z\ne -b/2a. \end{array}\right. } \end{aligned}$$

Observe that \(f_h\) is entire and by (1.3), it belongs to \( \mathcal {F}_p\) and \(V_g f_h= h\). Therefore, the other inclusion \(\mathcal {F}_p\subseteq \mathcal {R}(V_g )\) holds and completes the proof.

Dynamical sampling with \(V_g\) and \(V_g^*\)

We now turn our attentions to some applications of Theorem 1.1 on dynamical sampling from frame perspectives. Dynamical sampling deals with representations of frames \( \{ f_n\}_{n=0}^\infty \) in the form \( \{ T^n f\}_{n=0}^\infty \) for some linear operator T defined on a given Hilbert space \(\mathcal {H}\) where

$$\begin{aligned} \{ T^n f\}_{n=0}^\infty = \big \{ f,\ Tf,\ T^2f,\ T^3 f, \ ...\big \} \end{aligned}$$

is the orbit of \(f\in \mathcal {H}\) under the operator T. Recall that a family \((f_j), j\in I\) of vectors in a Hilbert space \(\mathcal {H}\) is a frame if there exist positive constants A and B such that for any \(g\in \mathcal {H}\)

$$\begin{aligned} A \Vert g\Vert _{\mathcal {H}}^2 \le \sum _{j\in I} |\langle g, f_j\rangle _{\mathcal {H}}|^2\le B \Vert g\Vert _{\mathcal {H}}^2. \end{aligned}$$
(2.1)

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 \((f_j), j\in I\) of vectors in \(\mathcal {H}\) is a Riesz basis if it is complete and there exist constants \(0<A\le B<\infty \) such that for any \(c_j\in \ell ^2(I)\)

$$\begin{aligned} A \sum _{j\in I} |c_j|^2 \le \bigg \Vert \sum _{j\in I} c_j f_j \bigg \Vert _{\mathcal {H}}^2\le B\sum _{j\in I} |c_j|^2. \end{aligned}$$
(2.2)

We may start with the following important lemma which connects the closed range problem with dynamical sampling in frame theory.

Lemma 2.1

Let \(\mathcal {H}\) be a Hilbert space and T be a bounded linear operator on \(\mathcal {H}\). If \(\{T^n f\}_{n=0}^\infty \) is a frame for some \(f\in \mathcal {H}\), then

  1. (i)

    T is surjective.

  2. (ii)

    \(\Vert (T^*)^n g\Vert _{\mathcal {H}} \rightarrow 0\) as \(n\rightarrow \infty \) for all \(g\in \mathcal {H}\).

While the proof of part (ii) is available in [3], part (i) follows easily since \(\{T^n f\}_{n=0}^\infty \) is a frame for each \(h\in \mathcal {H}\), there exists sequence \((c_n)\) such that

$$\begin{aligned} h= \sum _{n=1}c_n T^n f= T\bigg (\sum _{n=1}c_n T^{n-1} f\bigg ) \end{aligned}$$

from which the claim follows. Note here that the frame property implies \(R(T)= \mathcal {H}\) is closed and this in particular connects us with our result in Theorem 1.1.

Theorem 2.2

Let \(V_g\) is bounded on \(\mathcal {F}_2\) and hence \(g(z)= az^2+bz+c\) for some \(a, b, c\in \mathbb {C}\). Then neither \( \{V_g^n f\}_{n=0}^\infty \) nor \( \{(V_g^*)^n f\}_{n=0}^\infty \) is a frame for any choice of f in \(\mathcal {F}_2\).

Proof

Suppose, for the sake of contradiction, that there exists an \(f\in \mathcal {F}_2\) such that \( \{V_g^n f\}_{n=0}^\infty \) 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 \(h\in \mathcal {F}_2\) such that \( \{(V_g^*)^n h\}_{n=0}^\infty \) 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 \(a\ne 0\). On the other hand, set \(f(z)=z\) and consider the iterations

$$\begin{aligned} V_g f(z)&= \int _{0}^z (2aw+b)w dw= \frac{2a}{3} z^3 + \frac{b}{2}z^2,\\ V_g^2 f(z)&= \int _{0}^z (2aw+b)\Big (\frac{2a}{3} w^3 + \frac{b}{2}w^2\Big ) dw= \frac{2^2a^2}{3\cdot 5} z^5 +\frac{10ab}{2\cdot 3\cdot 4}z^4+ \frac{b^2}{2\cdot 3}z^3,\\ V_g^3 f(z)&= \int _{0}^z (2aw+b)\bigg (\frac{2^2a^2}{3\cdot 5} w^5 +\frac{10ab}{2\cdot 3\cdot 4}w^4+ \frac{b^2}{2\cdot 3}w^3 \bigg ) dw\\&= \frac{2^3a^3}{3\cdot 5\cdot 7} z^7 +\frac{132a^2b}{6!}z^6+ \frac{18ab^2}{5!}z^5+ \frac{b^3}{4!}z^4, \end{aligned}$$

and

$$\begin{aligned} V_g^4 f(z)&= \int _{0}^z (2aw+b)\bigg ( \frac{2^3a^3}{3\cdot 5\cdot 7} w^7 +\frac{132a^2b}{6!}w^6+ \frac{18ab^2}{5!}w^5+ \frac{b^3}{4!}w^4 \bigg ) dw \\&= \frac{2^4a^4}{3\cdot 5\cdot 7\cdot 9} z^9 +\frac{a^3(384+1148b)}{8!}z^8+\frac{324a^2b^2}{7!}z^7+ \frac{28ab^3}{6!}z^6+ \frac{b^4}{ 5!}z^5. \end{aligned}$$

Continuing the iteration,

$$\begin{aligned} V_g^n f(z)&= \int _{0}^z (2aw+b)V_g^{n-1} f(w) dw = c_{2n+1}a^n z^{2n+1} + c_{2n}a^{n-1}z^{2n}\\&\quad + \cdot \cdot \cdot + c_{n+2}ab^{n-1}z^{n+2}+ c_{n+1} b^{n}z^{n+1}, \end{aligned}$$

where the sequence \(c_k\) is of the form \(p_k/q_k\), \( q_k\le k!\) 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,

$$\begin{aligned} \Vert V_g^n f\Vert _2&= \Vert c_{2n+1}a^n z^{2n+1} + c_{2n}a^{n-1}z^{2n}+ \cdot \cdot \cdot + c_{n+2}az^{n+2}+ c_{n+1}z^{n+1}\Vert _2\nonumber \\&\ge \frac{1}{(2n+1)!}\Big | |2^na^n|\Vert _2 z^{2n+1}\Vert _2 -|p_{2n}a^{n-1}|\Vert z^{2n}\Vert _2\nonumber \\&\qquad - \cdot \cdot \cdot -| p_{n+2}a|\Vert z^{n+2}\Vert _2- |p_{n+1}|\Vert z^{n+1}\Vert _2\Big |\nonumber \\&= \frac{\Vert z^{2n+1}\Vert _2}{(2n+1)!}\bigg | |2^na^n| -\frac{|p_{2n}a^{n-1}|\Vert z^{2n}\Vert _2}{\Vert z^{2n+1}\Vert _2}- \cdot \cdot \cdot -\frac{| p_{n+2}a|\Vert z^{n+2}\Vert _2}{\Vert z^{2n+1}\Vert _2}- \frac{ |b^{n}|\Vert z^{n+1}\Vert _2}{\Vert z^{2n+1}\Vert _2}\bigg |.\nonumber \\ \end{aligned}$$
(2.3)

On the other hand, by (1.5)

$$\begin{aligned} \Vert z^n\Vert _2\simeq \Big (\frac{n}{e}\Big )^{n} \sqrt{n} \end{aligned}$$

which obviously grows much faster than exponential and factorial sequences. Setting this in (2.3), we observe that

$$\begin{aligned} \Vert V_g^n f\Vert _2 \rightarrow 0,\ \ n\rightarrow \infty \end{aligned}$$

as required by part (ii) of Lemma 2.1 only when \(a=b=0\) which is a contradiction, and the claim is proved. \(\square \)

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 \( V_gf_n\) 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 Hilbert space \(\mathcal {H}\). Then T preserves

  1. (i)

    frames on \(\mathcal {H}\) if and only if \(T^*\) is bounded below on \(\mathcal {H}\), and the latter happens if and only if T is surjective on \(\mathcal {H}\).

  2. (ii)

    tight frames if and only if there exists a positive constant \(\lambda \) such that \(\Vert T^* f\Vert _{\mathcal {H}}=\lambda \Vert f\Vert _{\mathcal {H}}\) for all \(f\in \mathcal {H}\).

We may now state our main result for the section.

Theorem 2.4

Let \(V_g\) be bounded on \(\mathcal {F}_2\) and hence \(g(z)= az^2+bz+c\) for some \(a, b, c\in \mathbb {C}\). Then

  1. (i)

    \(V_g\) fails to preserve frame in \(\mathcal {F}_2\).

  2. (ii)

    \(V_g^*\) preserves frame in \(\mathcal {F}_2\) if and only if \(a\ne 0\).

  3. (iii)

    \(V_g^*\) preserves neither tight frame nor Riesz basis in \(\mathcal {F}_2\).

Said differently, the result asserts that the adjoint of a bounded Volterra-type integral operator preserves frame structure on \(\mathcal {F}_2\) if and only if it has no compactness property.

Proof

Suppose \(V_g\) preserves frame on \(\mathcal {F}_2\). Then an application of Lemma 2.3 and Theorem 1.1 leads to a contradiction. Statement (ii) follows again from a simple application of Lemma 2.3, Theorem 1.1, and the Closed Range Theorem. Thus we proceed to verify (iii) and suppose \(V_g^*\) preserves tight frame. By Theorem 2.4, it follows that \(a\ne 0\). On the other hand, by Lemma 2.3, there exists a \(\lambda >0\) such that \(\Vert V_g f\Vert _2= \lambda \Vert f\Vert _2\) all \(f\in \mathcal {F}_2\). Using the function \(K_0\), we obtain

$$\begin{aligned} \lambda = \frac{\Vert V_g K_0\Vert _2}{\Vert K_0\Vert _2}= \Vert az^2+bz\Vert _2. \end{aligned}$$

Furthermore, considering the sequence of the monomials

$$\begin{aligned} V_g z^n= \frac{2a}{n+2} z^{n+2} + \frac{b}{n+1} z^{n+1} \end{aligned}$$

for all \(n\in \mathbb {N}\). Consequently,

$$\begin{aligned} \Vert az^2+bz\Vert _2=\lambda = \frac{\Vert V_g z^n\Vert _2}{\Vert z^n\Vert _2}= \frac{\Vert \frac{2a}{n+2} z^{n+2}+ \frac{b}{n+1} z^{n+1}\Vert _2}{\Vert z^n\Vert _2} \end{aligned}$$
(2.4)

Using orthogonality of the monomials, we simplify further to deduce that (2.4) holds if and only if

$$\begin{aligned}&\frac{(n^2+4n)|a|^2}{(n+2)^2}\Big ( \Vert z^2\Vert _2^2 \Vert z^n\Vert _2^2- \Vert z^{n+2}\Vert _2^2\Big )\\&\quad + \frac{(n^2+2n)|b|^2}{(n+1)^2}\Big ( \Vert z\Vert _2^2 \Vert z^n\Vert _2^2- \Vert z^{n+1}\Vert _2^2\Big )=0. \end{aligned}$$

Applying the norm of the monomials in (1.5), the above holds if and only if \(a= b=0\) and hence a contradiction.

Next, we show that \(V_g^*\) does not preserves Riesz basis either. Suppose on the contrary it does. Recall that a \((f_j), j\in I\) is a Riesz bases if and only it is a frame and \(\omega -\) independent. That is if

$$\begin{aligned} \sum _{j\in I} c_j f_j=0 \end{aligned}$$

for some sequence of scalars \((c_j)\), then \(c_j= 0\) for all \(j\in I\). In view of this, suppose \((f_j), j\in I\) is a Riesz basis and

$$\begin{aligned} \sum _{j\in I} c_j V_g^* f_j= 0. \end{aligned}$$

for some scalars \(c_j\). Using linearity, for each \(h\in \mathcal {F}_2\)

$$\begin{aligned} \Big \langle \sum _{j\in I} c_j V_g^* f_j, \ h\Big \rangle = \Big \langle \sum _{j\in I} V_g^* (c_j f_j), \ h\Big \rangle = \Big \langle \sum _{j\in I} c_j f_j, \ V_g h\Big \rangle =0. \end{aligned}$$

This shows \(\sum _{j\in I} c_j f_j\) belongs to the orthogonal complement of the range of \(V_g\). Then by (1.1), \(\sum _{j\in I} c_j f_j \in \mathbb {C}= Ker V_g^*\). On the other hand, \(f_j\) is a Riesz basis. Hence, \(\sum _{j\in I} c_j f_j\) is not necessarily zero.