A topological splitting of the space of meromorphic germs in several variables and continuous evaluators

Abstract

We prove a topological decomposition of the space of meromorphic germs at zero in several variables with prescribed linear poles as a sum of spaces of holomorphic and polar germs. Evaluating the resulting holomorphic projection at zero gives rise to a continuous evaluator (at zero) on the space of meromorphic germs in several variables. Our constructions are carried out in the framework of Silva spaces and use an inner product on the underlying space of variables. They generalise to several variables, the topological direct decomposition of meromorphic germs at zero as sums of holomorphic and polar germs previously derived by the first and third author and provide a topological refinement of a known algebraic decomposition of such spaces previously derived by the second author and collaborators.

A. Proof details for Sect. 2

In this appendix we provide the proof for Lemma 2.13 and Lemma 2.22.

1.1 A.1. Proof of Lemma 2.13

Recall the statement of the Lemma: Fix $k\in {{\mathbb {N}}}$ and assume that $f_1,f_2,\ldots$ is a sequence of elements in $\mathcal {M}_{\mathcal {L}}({\mathbb {C}}^k)$ which converges to f in $\mathcal {M}_{\mathcal {L}}({\mathbb {C}}^k)$. Then for any $\ell$ in ${{\,\textrm{Dep}\,}}(f)$, there is a subsequence $(f_{m_j})_{j\in {{\mathbb {N}}}}$ and a sequence $\ell _{m_j}$ in ${{\,\textrm{Dep}\,}}(f_{m_j}), j\in {{\mathbb {N}}}$ which converges to $\ell$.

Proof of Lemma 2.13

For f and $\ell$ and a sequence as in the statement of the Lemma let us build a subsequence of $(f_m)_{m\in {\mathbb {N}}}$ together with the sequence $\ell _m\in {{\,\textrm{Dep}\,}}(f_m)$.

Step 1: The independence subspaces ${{\,\textrm{Indep}\,}}(f_m)$ and ${{\,\textrm{Indep}\,}}(f)$.

By Corollary 2.11 there is an open convex set $U\subseteq {\mathbb {C}}^k\setminus \{0\}$ such that each $f_m$ is bounded and holomorphic on U and the sequence $f_m|_U$ converges uniformly to $f|_U\in {\text {BHol}}(U)$. With a slight abuse of notation we denote by $f_m$ and f again $f_m|_U$ and $f|_U$, respectively. Note that the (in)dependence subspaces of $f_m$ and f do not depend on U.

Step 2: An orthonormal basis for ${{\,\textrm{Indep}\,}}(f)$.

For each $m\in {{\mathbb {N}}}$ the dimension $d_m{:}{=}\dim {{\,\textrm{Indep}\,}}(f_m)$ of the space ${{\,\textrm{Indep}\,}}(f_m)$ is a number in the finite set $\{0,\ldots ,k\}$. Hence, after chosing a subsequence, we may assume without loss of generality that $d{:}{=}\dim {{\,\textrm{Indep}\,}}(f_m)$ is constant. For each $m\in {{\mathbb {N}}}$ we may chose a Q-orthonormal basis

$$\begin{aligned} B_m{:}{=}\left( b_1^{(m)},\ldots , b_k^{(m)} \right) \end{aligned}$$

of ${\mathbb {C}}^k$ such that the first d vectors form a Q-orthonormal basis of ${{\,\textrm{Indep}\,}}(f_m)$. The sequence $(B_m)_m$ can be regarded as a bounded sequence in the finite-dimensional space $({\mathbb {C}}^k)^k$ and therefore has a converging subsequence which we will again denote by $(B_m)_m$ to keep the notation relatively simple.

The defining properties of a Q-orthonormal basis are stable under limits (due to the continuity of Q), whence

$$\begin{aligned}{} & {} B=\left( b_1,\ldots , b_k\right) {:}{=}\lim _{m\rightarrow \infty }\\{} & {} B_m = \left( \lim _{m\rightarrow \infty } b_1^{(m)},\ldots , \lim _{m\rightarrow \infty } b_k^{(m)} \right) \end{aligned}$$

is also a Q-orthonormal basis of ${\mathbb {C}}^k$. By the definition of the independent subspaces (see Eq. (4)), for every $m\in {{\mathbb {N}}}$ and $j\le d$ we have $D_{b_j^{(m)}}f_m=0$. Since differentiation is continuous with respect to the topology of uniform convergence on holomorphic functions, we infer $D_{b_j}f=0$ and $b_j\in {{\,\textrm{Indep}\,}}(f)$ for all $j\le d$. This shows that $\textrm{span}(b_1,\ldots ,b_d)\subseteq {{\,\textrm{Indep}\,}}(f)$, from which it follows that $(b_1,\ldots ,b_d)$ is an orthonormal basis for ${{\,\textrm{Indep}\,}}(f)$.

Step 3: Approximating $\ell$ in ${{\,\textrm{Dep}\,}}(f)$.

By Riesz’ theorem, there is a vector $v\in {\mathbb {C}}^k$ such that $\ell =Q(\cdot ,v)$. Since ${{\,\textrm{Dep}\,}}(f)$ is the annulator of ${{\,\textrm{Indep}\,}}(f)$, we obtain:

$$\begin{aligned} v\in {{\,\textrm{Indep}\,}}(f)^\perp \subseteq \textrm{span}(b_1,\ldots ,b_d)^\perp = \textrm{span}(b_{d+1},\ldots , b_k). \end{aligned}$$

Thus there exist scalars $\alpha _{d+1},\ldots ,\alpha _k\in {\mathbb {C}}$ such that $v = \sum _{j=d+1}^k \alpha _j b_j.$ For each $m\in {{\mathbb {N}}}$ we define $v_m{:}{=}\sum _{j=d+1}^k \alpha _j b_j^{(m)}\in {{\,\textrm{Indep}\,}}(f_m)^\perp$ and obtain a sequence converging to v. Since each $v_m$ is Q-orthogonal to ${{\,\textrm{Indep}\,}}(f_m)$, it corresponds to an element $\ell _m$ in ${{\,\textrm{Dep}\,}}(f_m)$ using Riesz’ representation theorem. We have therefore built a sequence $(\ell _m)_{m\in {{\mathbb {N}}}}$ in the dual space $({\mathbb {C}}^k)^*$ which converges to $\ell$ as desired. $\square$

1.2 A.2 Proof of Lemma 2.22

Recall the statement of the Lemma: Given $f\in {\text {BHol}}^{\mathbb {C}}(B_{1/n}({\mathbb {C}}^k))$, $L \in \mathcal {L}$ and $n \in {{\mathbb {N}}}$ there is an integer $m > n$ such that the maps $h:= f\circ {{\,\textrm{pr}\,}}_L|_{B_{1/m}({\mathbb {C}}^k)}$ and $g=\left. \left( f- f\circ {{\,\textrm{pr}\,}}_L\right) / L\right| _{B_{1/m}({\mathbb {C}}^k)}$ are bounded holomorphic functions on $B_{1/m}({\mathbb {C}}^k)$ and the associated mapping

$$\begin{aligned} \begin{aligned} \theta _{n,m,Q}^L :{\text{ BHol }}^{\mathbb {C}}(B_{1/n}({\mathbb {C}}^k))&\rightarrow {\text{ BHol }}^{\mathbb {C}}(B_{1/m}({\mathbb {C}}^k))^2,\\ \quad f = L\cdot g + h&\mapsto (g,h), \end{aligned} \end{aligned}$$

is continuous linear.

Proof of Lemma 2.22

To distinguish supremum norms we write

$$\begin{aligned} \Vert f \Vert _{R, \infty } {:}{=}\sup _{z\in B_{R}({\mathbb {C}}^k)} |f(z)| \text { for } R>0. \end{aligned}$$

Pick now f in ${\text {BHol}}^{\mathbb {C}}(B_{1/n}({\mathbb {C}}^k))$ and define $h = f \circ {{\,\textrm{pr}\,}}_L|_{B_{1/n}({\mathbb {C}}^k)}$. Clearly h is holomorphic and bounded by

$$\begin{aligned} \Vert h\Vert _{1/n,\infty } \le \Vert f \Vert _{1/n,\infty }. \end{aligned}$$

(16)

By definition $f-f\circ {{\,\textrm{pr}\,}}_L$ takes values in the closed linear subspace

$$\begin{aligned} V_L {:}{=}\{g \in {\text {BHol}}^{\mathbb {C}}(B_{1/n}({\mathbb {C}}^k))\mid g|_{\ker L} \equiv 0 \}. \end{aligned}$$

Now the subtraction map $s_L :{\text {BHol}}^{\mathbb {C}}(B_{1/n}({\mathbb {C}}^k)) \rightarrow V_L, s_L(f) = f - f\circ {{\,\textrm{pr}\,}}_L \in V_L$, is continuous linear since $\Vert s_L(f)\Vert _{1/n,\infty } \le 2\Vert f\Vert _{1/n,\infty }$ by (16).

Step 1: For $\kappa \in V_L$, $z \mapsto \kappa (z)/L(z)$ is holomorphic on $B_{1/(n+1)}({\mathbb {C}}^k)$.

The map L is a linear form on ${\mathbb {C}}^k$, whence the Riesz representation theorem yields $v \in {\mathbb {C}}^k$ with $L(x) = Q(x,v)$. Since L is continuous we can pick $0< s < 1/n$ such that for $z \in B_{s}({\mathbb {C}}^k)$ the Q-orthogonal decomposition $z = {{\,\textrm{pr}\,}}_L (z) + (L(z)/\Vert v\Vert ^2)v$ satisfies $\Vert {{\,\textrm{pr}\,}}_L(z)\Vert + |L(z)|/\Vert v\Vert < 1/n$. Hence we may apply for every z in $B_s({\mathbb {C}}^k)$ and $\kappa$ in $V_L$ the mean value theorem:

$$\begin{aligned} \kappa (z)&= \kappa ({{\,\textrm{pr}\,}}_L (z) + (L(z)/\Vert v\Vert ^2)v) -\kappa ({{\,\textrm{pr}\,}}_L (z)) \\ {}&= \int _0^1 d\kappa ({{\,\textrm{pr}\,}}_L (z)+\lambda (L(z)/\Vert v\Vert ^2)v;(L(z)/\Vert v\Vert ^2)v) \textrm{d}\lambda \\ {}&= L(z) \int _0^1 d\kappa ({{\,\textrm{pr}\,}}_L (z)+\lambda (L(z)/\Vert v\Vert ^2)v;v/\Vert v\Vert ^2)\textrm{d}\lambda \end{aligned}$$

hence $\kappa (z)/L(z)$ makes sense as a holomorphic mapping on $B_{s}({\mathbb {C}}^k)$ and is clearly bounded on $B_{R}({\mathbb {C}}^k)$ for every $R<s$. Thus it makes sense to define for $R <s$ the map

$$\begin{aligned} \delta _R :V_L \rightarrow {\text {BHol}}^{\mathbb {C}}(B_{R}({\mathbb {C}}^k)), \kappa \mapsto \kappa /L|_{B_{R}({\mathbb {C}}^k)}. \end{aligned}$$

(17)

Step 2: The map $\delta _R(\kappa ){:}{=}\kappa / L|_{B_{R}({\mathbb {C}}^k)}$ is continuous for a suitable R. let $f \in V_L$ and write $f = L\cdot g$, i.e. $g|_{B_{R}({\mathbb {C}}^k)} = \delta _R (f)$. Let now $z \in {\mathbb {C}}^k$ with $\Vert z\Vert =r <s$ for the s as in Step 1. As the projection has operator norm 1, we have by construction that $\Vert {{\,\textrm{pr}\,}}_L (z)\Vert \le r < 1/n$. Setting $\varepsilon _r = \Vert v\Vert (1/n - r)$ a quick computation yields $\Vert {{\,\textrm{pr}\,}}_L (z)+(\lambda /\Vert v\Vert ^2) v\Vert \le 1/n$ for $|\lambda |< \varepsilon _r$. Hence it makes sense to define the following holomorphic function of one variable

$$\begin{aligned} \phi _z :\{\lambda \in {\mathbb {C}}\mid | \lambda | < \varepsilon _r\} \rightarrow {\mathbb {C}}, \quad \lambda \mapsto g({{\,\textrm{pr}\,}}_L (z) + (\lambda /\Vert v\Vert ^2) v). \end{aligned}$$

From (7) in Lemma 2.20 we infer that $\sup _{|\lambda | < \varepsilon _r} |\lambda \phi _z(\lambda )| = \varepsilon _r\Vert \phi _z\Vert _{\varepsilon _r,\infty }.$ Let us assume that $\Vert z\Vert \le r < s$. Then from $\Vert z-{{\,\textrm{pr}\,}}_L (z)\Vert =|L(z)|/\Vert v\Vert$ we deduce that $|L(z)| \le 2r\Vert v\Vert$. As the constant $\varepsilon _r$ is growing for smaller r, we can choose $m \in {{\mathbb {N}}}$ such that $R:= 1/m < \min \{s,\varepsilon _R, \frac{\varepsilon _r}{2\Vert v\Vert }\}$. Summing up this yields the estimate

$$\begin{aligned} | g(z)|&= |\phi _z(L(z))| \le \frac{\varepsilon _r}{\varepsilon _r} \Vert \phi _z\Vert _{\varepsilon _r,\infty } = \frac{1}{\varepsilon _R} \sup _{|\lambda | \le \varepsilon _r}| \lambda \phi _z(\lambda )| \\&\quad \le \frac{\Vert L \cdot g\Vert _{1/n,\infty }}{1/n-s}. \end{aligned}$$

As z was arbitrary with $\Vert z \Vert < 1/m$, we infer that

$$\begin{aligned} \Vert g\Vert _{1/m,\infty } \le \frac{\Vert L \cdot g\Vert _{1/n,\infty }}{1/n-s} = \frac{\Vert f\Vert _{1/n,\infty }}{1/n-s}. \end{aligned}$$

Hence the map $\delta _R$ defined in (17) is continuous for every $R \le 1/m$.

Step 3: $\theta _{n,Q}^L$ is continuous linear. First recall that the restriction map $r^n_m :{\text {BHol}}^{\mathbb {C}}(B_{1/n}({\mathbb {C}}^k)) \rightarrow {\text {BHol}}^{\mathbb {C}}(B_{1/m}({\mathbb {C}}^k))$ is continuous linear. We can then write $\theta _{n,Q}^L = (\delta _{1/m} \circ s_L, r^n_m -L|_{B_{1/m}({\mathbb {C}}^k)} \cdot \delta _{1/m} \circ s_L)$. This mapping makes sense and the first component is continuous by Step 1 and 2. Exploiting that ${\text {BHol}}^{\mathbb {C}}(B_{1/m}({\mathbb {C}}^k))$ is a Banach algebra with multiplication given by the pointwise multiplication of functions, Step 1-2 show that $\theta ^L_{n,Q}$ is indeed continuous. $\square$