Series expansions of analytic functions. II

Theorem 1. Suppose that (2.1)-(2.4) hold and let h be a complex sequence. Then the following are equivalent: - (i) $\sum_{k=0}^{\infty} \alpha_k^{(\lambda h)_k} converges$ ; - (ii) h belongs to the domain of B; - (iii) $\sum_{k=0}^{\infty} h_k p_k(z)$ converges for all z in some uniqueness set for $\varphi_1, \varphi_2, \dots, \varphi_d$ ; - (iv) $\sum_{k=0}^{\infty} h_k p_k(z)$ is uniformly convergent on compact subsets of $\Omega$ .

Theorem 2. Suppose that (2.1)-(2.4) hold and let h and y be complex sequences. Then y=Bh if and only if $\sum_{k=0}^{\infty} h_k p_k(z)$ is uniformly convergent on compact subsets of $\Omega$ to the function f whose power series at 0 is given by $$f(z) = \sum_{j=0}^{\infty} y_j z^j / j!.$$ The problem of explicitly determining those sequences y which belong to the space $\mathcal{Y}$ is extremely difficult, even for very simple sequences $\{p_k\}_0^{\infty}$ (cf. [2]). The coefficient space $\mathcal{H}$ is, however, much more accessible, and conditions (2.1)-(2.4) allow us to characterize it completely. For each integer $i, 1 \le i \le d$ , let $S^{(i)}$ denote the complex sequence with terms given by $$S_{kd+i-1}^{(i)} = (\alpha_k)_{ii}, \ 1 \le j \le d, \ 0 \le k < \infty.$$

Theorem 3. Suppose that (2.1)-(2.4) hold. A complex sequence h belongs to $\mathcal{H}$ if and only if each of the series (2.5) $$\sum_{k=0}^{\infty} S_k^{(i)} h_k, \ 1 \le i \le d,$$ is convergent. We note here for later use that if (2.5) converges, then (2.6) $$\sum_{k=0}^{\infty} S_k^{(i)} h_k = \left(\sum_{k=0}^{\infty} \alpha_k (\lambda h)_k\right)_i, \quad 1 \leq i \leq d.$$ The expansions discussed in [2] and [3] share two interesting and apparently unrelated properties. The first of these is uniqueness: if $\sum_{k=0}^{\infty}h_kp_k(z)=0$ for all $z\in\Omega$ , then $h_k=0, k=0,1,2,\cdots$ . The second is the existence of an underlying Banach space structure; that is, one can define a norm $\|\cdot\|$ on $\mathscr{F}$ , under which $\mathscr{F}$ is complete, such that convergence with respect to $\|\cdot\|$ implies uniform convergence on compact subsets of $\Omega$ , and such that $\{p_k\}_0^{\infty}$ is a strong basis for the Banach space $\mathscr{F}$ . In the present setting we show that the first of these properties implies the second, and also that $\mathscr{F}$ is isomorphic to $c_0$ .

Theorem 4. Suppose in addition to (2.1)-(2.4) that (2.7) $$\sum_{k=0}^{\infty} h_k p_k(z) = 0 \text{ for all } z \in \Omega \text{ implies}$$ that $h_k = 0, \ k = 0, 1, 2, \cdots$ For each $f \in \mathcal{F}$ with $$f(z) = \sum_{k=0}^{\infty} h_k p_k(z),$$ set $$||f|| = \sup_{0 \le n < \infty} \left\| \sum_{k=n}^{\infty} \alpha_k (\lambda h)_k \right\|_{\tau}.$$ Then $\mathscr{F}$ is a Banach space, $\{p_k\}_0^{\infty}$ is a basis for $\mathscr{F}$ , and the mapping $$\Lambda f = \lambda^{-1} \left\{ \sum_{k=n}^{\infty} \alpha_k (\lambda h)_k \right\}_{n=0}^{\infty}$$ defines an isomorphism between $\mathcal{F}$ and $c_0$ . Note that the uniqueness assumption (2.7) is, in view of Theorem 2, equivalent to the assertion that the operator B is one-to-one. 3. Proofs of theorems. The following lemma will be used in the proof of Theorem 1.

Lemma 1. Suppose that $\{u_k\}_0^{\infty}$ is a sequence of $d \times d$ invertible matrices such that (3.1) $$\sum_{k=0}^{\infty} \|u_{k+1} - u_k\|_G < \infty,$$ and that $\{u_k\}_0^{\infty}$ converges elementwise to an invertible matrix. Then if $\{v_k\}_0^{\infty}$ is a sequence of column vectors, $\sum_{k=0}^{\infty} v_k$ converges if and only if $\sum_{k=0}^{\infty} u_k v_k$ converges. Proof. Suppose $\sum_{0}^{\infty} v_{k}$ converges. Writing $s_{n} = v_{o} + \cdots + v_{n}$ , we have $\sum_{k=0}^{n} u_{k} v_{k} = u_{n} s_{n} + \sum_{k=0}^{n-1} (u_{k} - u_{k+1}) s_{k}$ . Since $\sum_{k=0}^{\infty} ||(u_{k} - u_{k+1}) s_{k}||_{\tau} \le d \sum_{k=0}^{\infty} ||u_{k} - u_{k+1}||_{G} ||s_{k}||_{\tau} < \infty$ , the convergence of $\sum_{0}^{\infty} u_{k} v_{k}$ follows. Now suppose that $\sum_{k=0}^{\infty} u_k v_k$ converges, set $\sigma_n = u_0 v_0 + \cdots + u_n v_n$ , and note that $\sum_{k=0}^{n} v_k = \sum_{k=0}^{n} u_k^{-1} (u_k v_k) = u_n^{-1} \sigma_n + \sum_{k=0}^{n-1} u_k^{-1} (u_{k+1} - u_k) u_{k+1}^{-1} \sigma_k$ . Since $\{u_k\}_0^{\infty}$ converges to an invertible matrix, say u, then $\{u_k^{-1}\}_0^{\infty}$ must converge to $u^{-1}$ . Consequently, the sequence $\{u_n^{-1} \sigma_n\}_0^{\infty}$ converges and the sequence $\{\|u_k^{-1}\|_{G}\}_0^{\infty}$ is bounded. In view of $$\|u_{k}^{-1}(u_{k+1}-u_{k})u_{k+1}^{-1}\sigma_{k}\|_{\tau} \leq d^{3}\|u_{k+1}-u_{k}\|_{G}\|u_{k+1}^{-1}\|_{G}\|u_{k+1}^{-1}\|_{G}\|\sigma_{k}\|_{\tau},$$ (3.1) shows that $\sum_{k=0}^{\infty} v_k$ converges, and this completes the proof. We will require the following lemma on linearly independent sets of functions. The proof is omitted.

Lemma 2. Let E be a uniqueness set for $f_1, f_2, \dots, f_m$ . Then there are points $z_1, z_2, \dots, z_m$ in E such that the matrix $(f_j(z_k)), 1 \leq j, k \leq d$ , is invertible. If $f_1, \dots, f_m$ are analytic in $\Omega$ , then there exist nonnegative integers $n_1, n_2, \dots, n_m$ such that the matrix $(f_j^{(n_k)}(0)), 1 \leq j, k \leq d$ , is invertible.

Lemma 3. Suppose that (2.1)-(2.3) hold and let h be a complex sequence such that $\sum_{k=0}^{\infty} \alpha_k(\lambda h)_k$ converges. Then $$(3.2) \sum_{k=0}^{\infty} h_k p_k(z)$$ converges if and only if $$(3.3) \sum_{k=0}^{\infty} P_k(z) (\lambda h)_k$$ converges. Moreover, (3.2) converges uniformly on compact subsets of $\Omega$ if and only if the same is true of (3.3). Proof. Since $$\sum_{k=0}^{\infty} P_k(z) (\lambda h)_k = \sum_{k=0}^{\infty} \sum_{t=0}^{d-1} h_{kd+t} p_{kd+t} (z),$$ then clearly convergence of (3.2) implies convergence of (3.3). In the other direction, it is sufficient to prove that, for fixed t, (3.4) $$\lim_{n \to \infty} h_{nd+t} p_{nd+t}(z) = 0$$ uniformly on compact subsets of $\Omega$ . Let $\{x_k\}_0^{\infty}$ be the sequence of column vectors defined by $$x_k = \sum_{j=k}^{\infty} \alpha_j (\lambda h)_j,$$ $k = 0, 1, 2, \cdots,$ and define (3.5) $$Q_k(z) = P_k(z)\alpha_k^{-1} - P_{k-1}(z)\alpha_{k-1}^{-1}, \qquad k = 0, 1, 2, \cdots.$$ Then $(\lambda h)_k = \alpha_k^{-1}(x_k - x_{k+1})$ and $P_k(z) = \sum_{j=0}^k Q_j(z)\alpha_k$ , $0 \le k < \infty$ . Therefore $$\begin{aligned} |h_{nd+t}p_{nd+t}(z)| &= \left| ((\lambda h)_n)_t (P_n(z))_t \right| \\ &= \left| (\alpha_n^{-1}(x_n - x_{n+1}))_t \right| \left| \left( \sum_{j=0}^n Q_j(z) \alpha_n \right)_t \right| \\ &\leq \|\alpha_n^{-1}(x_n - x_{n+1})\|_{\tau} \left\| \sum_{j=0}^n Q_j(z) \alpha_n \right\|_{\rho} \\ &\leq d^2 \|\alpha_n^{-1}\|_{G} \|x_n - x_{n+1}\|_{\tau} \|\alpha_n\|_{G} \left\| \sum_{j=0}^n Q_j(z) \right\|. \end{aligned}$$ Since $\{(x_n - x_{n+1})\}_0^{\infty}$ converges to the zero vector, the result follows from (2.3). The uniform convergence is a consequence of (2.2). In view of Lemma 3, Theorem 1 is equivalent to the following result.

Theorem 5. Suppose that (2.1)-(2.4) hold and let h be a complex sequence. Then the following are equivalent: - (i) $\sum_{k=0}^{\infty} \alpha_k (\lambda h)_k$ converges; - (ii) $\sum_{k=0}^{\infty} P_k^{(j)}(0)(\lambda h)_k$ converges for $j=0,1,2,\cdots$ , - (iii) $\sum_{k=0}^{\infty} P_k(z)(\lambda h)_k$ converges for all z in some uniqueness set for $\varphi_1, \varphi_2, \dots, \varphi_d$ ; - (iv) $\sum_{k=0}^{\infty} P_k(z)(\lambda h)_k$ converges uniformly on compact subsets of $\Omega$ . Proof. We first show that (i) and (iii) are equivalent. Using Lemma 2 and (2.4), choose points $z_1, z_2, \dots, z_d$ in $\Omega$ such that the matrix $(\varphi_j(z_k))$ , $1 \le j, k \le d$ , is invertible. In proving that (i) implies (iii) we can take $\{z_1, \dots, z_d\}$ as our uniqueness set; for the reverse implication we will suppose that $z_1, \dots, z_d$ are chosen from some arbitrary uniqueness set. Define $$u_{k} = \begin{bmatrix} P_{k}(z_{1}) \\ \vdots \\ P_{k}(z_{d}) \end{bmatrix} \alpha_{k}^{-1}, \qquad k = 0, 1, 2, \cdots,$$ $$u = \begin{bmatrix} \Phi(z_{1}) \\ \vdots \\ \Phi(z_{d}) \end{bmatrix},$$ and let $v_k = \alpha_k(\lambda h)_k$ . Then u is invertible, $\{u_k\}_0^\infty$ converges to u, and (iii) holds if and only if $\Sigma_0^\infty u_k v_k$ converges. Note also that $u_k$ is invertible for k sufficiently large, and we suppose without loss of generality that this is true for all k. By Lemma 1, then, we can prove the equivalence of (i) and (iii) by showing that $$\sum_{k=0}^{\infty} \|u_{k+1} - u_k\|_G < \infty.$$ For $0 \le k < \infty$ , we have $$\|u_{k+1} - u_k\|_G = \sum_{i=1}^d \sum_{t=1}^d |(P_{k+1}(z_t)\alpha_{k+1}^{-1} - P_k(z_t)\alpha_k^{-1})_i|$$ $$\leq d \sum_{t=1}^d \|P_{k+1}(z_t)\alpha_{k+1}^{-1} - P_k(z_t)\alpha_k^{-1}\|_{\rho},$$ and thus $$\sum_{k=0}^{\infty} \|u_{k+1} - u_k\|_G \leq d \sum_{t=1}^{d} \sum_{k=1}^{\infty} \|P_k(z_t)\alpha_k^{-1} - P_{k-1}(z_t)\alpha_{k-1}^{-1}\|_{\rho} < \infty.$$ We now prove that (i) $\Rightarrow$ (iv) $\Rightarrow$ (ii) $\Rightarrow$ (i). Thus suppose that $\sum_{k=0}^{\infty} \alpha_{k}(\lambda h)_{k}$ converges and define $Q_{k}(z)$ , $0 \le k < \infty$ , as in (3.5). Then (3.6) $$\sum_{k=0}^{n} P_{k}(z)(\lambda h)_{k} = \sum_{k=0}^{n} \left\{ \sum_{j=0}^{k} Q_{j}(z) \alpha_{k} \right\} (\lambda h)_{k}$$ $$= \sum_{j=0}^{n} \sum_{k=j}^{n} Q_{j}(z) \alpha_{k} (\lambda h)_{k} = \sum_{j=0}^{n} Q_{j}(z) \left\{ \sum_{k=j}^{\infty} \alpha_{k} (\lambda h)_{k} - \sum_{k=n+1}^{\infty} \alpha_{k} (\lambda h)_{k} \right\}$$ $$= \sum_{j=0}^{n} Q_{j}(z) \sum_{k=j}^{\infty} \alpha_{k} (\lambda h)_{k} - \sum_{j=0}^{n} Q_{j}(z) \sum_{k=n+1}^{\infty} \alpha_{k} (\lambda h)_{k}$$ $$= \sum_{j=0}^{n} Q_{j}(z) \sum_{k=j}^{\infty} \alpha_{k} (\lambda h)_{k} - (P_{n}(z)\alpha_{n}^{-1}) \sum_{k=n+1}^{\infty} \alpha_{k} (\lambda h)_{k}.$$ Let $\epsilon > 0$ and choose an integer N > 0 such that $j \ge N$ implies $\|\sum_{k=j}^{\infty} \alpha_k(\lambda h)_k\|_{\tau} < \epsilon$ . Then $n \ge N$ implies (3.7) $$\sum_{j=0}^{n} \left| Q_{j}(z) \left( \sum_{k=j}^{\infty} \alpha_{k} (\lambda h)_{k} \right) \right|$$ $$\leq \sum_{j=0}^{N-1} \left| Q_{j}(z) \left( \sum_{k=j}^{\infty} \alpha_{k} (\lambda h)_{k} \right) \right| + d\epsilon \sum_{j=N}^{n} \| Q_{j}(z) \|_{\rho}.$$ Since $P_n(z)\alpha_n^{-1}$ converges to $\Phi(z)$ , the last term in (3.6) converges to 0. Combining this with (3.7) we see that $\sum_{k=0}^{\infty} P_k(z)(\lambda h)_k$ converges uniformly on compact subsets of $\Omega$ . This proves that (i) $\Rightarrow$ (iv). The implication (iv) $\Rightarrow$ (ii) is obvious, so there remains only to show that (ii) $\Rightarrow$ (i). Choose, via Lemma 2, integers $j_1, j_2, \dots, j_d$ such that the matrix $$u = \begin{bmatrix} \Phi^{(i_l)}(0) \\ \vdots \\ \Phi^{(i_d)}(0) \end{bmatrix}$$ is invertible. Let $$u_{k} = \begin{bmatrix} P_{k}^{(j_{1})}(0) \\ \vdots \\ P_{k}^{(j_{d})}(0) \end{bmatrix} \alpha_{k}^{-1}, \ v_{k} = \alpha_{k}(\lambda h)_{k}, \qquad k = 0, 1, 2, \cdots,$$ and note that (ii) implies that $\sum_{k=0}^{\infty} u_k v_k$ converges. Moreover, $\{u_k\}_0^{\infty}$ converges to u, $u_k$ is invertible for k sufficiently large, and we may again assume that $u_k$ is invertible for all k. To show that $\sum_{k=0}^{\infty} v_k$ converges, we use Lemma 1. First, the convergence of $$\sum_{k=0}^{\infty} \|P_k^{(j)}(0)\alpha_k^{-1} - P_{k-1}^{(j)}(0)\alpha_{k-1}^{-1}\|_{\rho}, \qquad 0 \leq j < \infty,$$ follows easily from (2.2). Now observe that $$\|u_{k+1} - u_k\|_G = \sum_{i=1}^d \sum_{t=1}^d |(P_{k+1}^{(j_t)}(0) \alpha_{k+1}^{-1} - P_k^{(j_t)}(0) \alpha_k^{-1})_t|$$ $$\leq d \sum_{i=1}^d \|P_{k+1}^{(j_t)}(0) \alpha_{k+1}^{-1} - P_k^{(j_t)}(0) \alpha_k^{-1}\|_{\rho},$$ so that $$\sum_{k=0}^{\infty} \|u_{k+1} - u_k\|_G \leq d \sum_{i=1}^{d} \sum_{k=0}^{\infty} \|P_{k+1}^{(i,i)}(0)\alpha_{k+1}^{-1} - P_k^{(i,i)}(0)\alpha_k^{-1}\|_{\rho} < \infty,$$ and this completes the proof of the theorem. Proof of Theorem 2. Suppose y = Bh for complex sequences y and h. Since h belongs to the domain of B, Theorem 1 implies that $\sum_{k=0}^{\infty} h_k p_k(z)$ converges uniformly on compact subsets of $\Omega$ . Let $f(z) = \sum_{k=0}^{\infty} h_k p_k(z)$ . Then $$f^{(j)}(0) = \sum_{k=0}^{\infty} h_k p_k^{(j)}(0) = \sum_{k=0}^{\infty} B_{jk} h_k = (Bh)_j = y_j,$$ $0 \leq j < \infty$ . In the other direction, if $$\sum_{j=0}^{\infty} y_{j} \frac{z^{j}}{j!} = \sum_{k=0}^{\infty} h_{k} p_{k}(z),$$ then differentiating this equation m times and setting z = 0 yields $$y_m = \sum_{k=0}^{\infty} h_k p_k^{(m)}(0) = (Bh)_m,$$ so that y = Bh. Proof of Theorem 3. If $h \in \mathcal{H}$ , then h belongs to the domain of B. By Theorem 1, $\sum_{k=0}^{\infty} \alpha_k (\lambda h)_k$ converges. By definition of the sequences $\{S_k^{(i)}\}_{k=0}^{\infty}$ , $1 \le i \le d$ , it follows that each of the series $$\sum_{k=0}^{\infty} \sum_{j=1}^{d} S_{kd+j-1}^{(i)} h_{kd+d-1}, \qquad 1 \leq i \leq d,$$ is convergent. To show that the series (2.5) all converge, it suffices to show that, for fixed $j, 1 \le j \le d$ , and fixed i, the terms $S_{kd+j-1}^{(i)} h_{kd+j-1} \to 0$ as $k \to \infty$ . As in the proof of Lemma 3, let $x_k = \sum_{j=k}^{\infty} \alpha_j (\lambda h)_j$ , $k = 0, 1, 2, \cdots$ . Then $(\lambda h)_k = \alpha_k^{-1} (x_k - x_{k+1})$ , $0 \le k < \infty$ , and $$\begin{aligned} |S_{kd+j-1}^{(i)} h_{kd+j-1}| &= |(\alpha_k)_{ij} ((\lambda h)_k)_j| \\ &\leq |(\alpha_k)_{ij}| \|(\lambda h)_k\|_{\tau} = |(\alpha_k)_{ij}| \|\alpha_k^{-1} (x_k - x_{k+1})\|_{\tau} \\ &\leq d \|\alpha_k\|_G \|\alpha_k^{-1}\|_G \|x_k - x_{k+1}\|_{\tau}. \end{aligned}$$ Since $\{x_k - x_{k+1}\}_0^{\infty}$ converges to the zero vector, the desired result follows. If each of the series (2.5) converges, then each of the "grouped series" $$\sum_{k=0}^{\infty} \sum_{i=1}^{d} S_{kd+j-1}^{(i)} h_{kd+j-1}$$ converges. Therefore $\sum_{k=0}^{\infty} \alpha_k (\lambda h)_k$ converges and Theorem 1 implies $h \in \mathcal{H}$ . Proof of Theorem 4. Let n be a nonnegative integer and write n = md + i, $0 \le i < d$ . Then $$(\Lambda f)_n = \sum_{k=md}^{\infty} S_k^{(i+1)} h_k.$$ If $\|\cdot\|_{\infty}$ denotes the supremum norm on $c_0$ , then $$\|\Lambda f\|_{\infty} = \sup_{0 \le m < \infty} \left\{ \max_{0 \le i < d} \left| \sum_{k=md}^{\infty} S_k^{(i+1)} h_k \right| \right\}$$ $$= \sup_{0 \le m < \infty} \left\| \sum_{k=m}^{\infty} \alpha_k (\lambda h)_k \right\|_{\tau}$$ $$= \|f\|.$$ Thus $\Lambda$ is an isometry. Since $\Lambda$ maps $\mathcal{F}$ onto all of $c_0$ , it follows that $\mathcal{F}$ is a Banach space. Let $f \in \mathcal{F}$ , with (3.7) $$f(z) = \sum_{k=0}^{\infty} h_k p_k(z).$$ For each positive integer n, write n = jd + i, $j \ge 0$ , $0 \le i < d$ . Then $$\left\| f(z) - \sum_{k=0}^{n} h_k p_k(z) \right\| = \left\| \sum_{k=n+1}^{\infty} h_k p_k(z) \right\|$$ $$= \sup_{m \ge j} \left\| \sum_{k=m}^{\infty} \alpha_k (\lambda h)_k \right\|_{\tau},$$ and this last expression tends to 0 as $n \to \infty$ . Thus (3.7) converges in the norm $\|\cdot\|$ . From the uniqueness condition (2.7), it follows that $\{p_k\}_0^\infty$ is a basis for $\mathscr{F}$ .