# A Fischer type decomposition theorem from the apolar inner product

J. M. Aldaz* and H. Render  H. Render: School of Mathematical Sciences,
University College Dublin, Dublin 4, Ireland<EMAIL_ADDRESS>J.M.
Aldaz: Instituto de Ciencias Matemáticas (CSIC-UAM-UC3M-UCM) and Departamento
de Matemáticas, Universidad Autónoma de Madrid, Cantoblanco 28049, Madrid,
Spain<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We continue the study initiated by H. S. Shapiro on Fischer decompositions of
entire functions, showing that such decomposition exist in a weak sense (we do
not prove uniqueness) under hypotheses regarding the order of the entire
function $f$ to be expressed as $f=P\cdot q+r$, the polynomial $P$, and bounds
on the apolar norm of homogeneous polynomials of degree $m$.

These bounds, previously used by Khavinson and Shapiro, and by Ebenfelt and
Shapiro, can be interpreted as a quantitative, asymptotic strengthening of
Bombieri’s inequality. In the special case where both the dimension of the
space and the degree of $P$ are two, we characterize for which polynomials $P$
such bounds hold.

2020 Mathematics Subject Classification: _Primary: 31B05_ , _Secondary:
35A20,35A10_

Key words and phrases: _Fischer decomposition, Fischer pair, Bombieri norm,
entire function of finite order_

##  1\.  Introduction

Let us denote by $\mathcal{P}\left(\mathbb{C}^{d}\right)$ the set of all
polynomials with complex coefficients in the variable
$z=\left(z_{1},\dots,z_{d}\right)\in\mathbb{C}^{d}$, and by
$\mathcal{P}_{m}\left(\mathbb{C}^{d}\right)$ the subspace of all homogeneous
polynomials of degree $m.$ Recall that a polynomial $P\left(z\right)$ is
_homogeneous_ of degree $\alpha$ if
$P\left(tz\right)=t^{\alpha}P\left(z\right)$ for all $t>0$ and all
$z\in\mathbb{C}^{d}.$ In particular, the zero polynomial is homogeneous of
every degree. Given a polynomial $P\left(z\right)$, we denote by
$P^{\ast}\left(z\right)$ the polynomial obtained from $P\left(z\right)$ by
conjugating its coefficients, and by $P\left(D\right)$ the linear differential
operator obtained by replacing the variable $z_{j}$ with the differential
operator $\frac{\partial}{\partial z_{j}}$.

It is well known that a polynomial $P\left(z\right)$ of degree $k$ can be
written as a sum of homogeneous polynomials $P_{j}\left(z\right)$ of degree
$j$ for $j=0,\dots,k,$ so

$P\left(z\right)=P_{k}\left(z\right)+\cdots+P_{0}\left(z\right),$

and we call the homogeneous polynomial $P_{k}\left(z\right)$ the leading term.
Often the alternative notation $P\left(z\right)=P_{k}\left(z\right)-\cdots-
P_{0}\left(z\right)$ will be more convenient, in relation with formulas
regarding division of polynomials.

Fischer’s theorem states that if $P$ is a polynomial with leading term
$P_{k}$, then the following decomposition holds: for every polynomial
$f\in\mathcal{P}\left(\mathbb{C}^{d}\right)$ there exist _unique_ polynomials
$q\in\mathcal{P}\left(\mathbb{C}^{d}\right)$ and
$r\in\mathcal{P}\left(\mathbb{C}^{d}\right)$ such that

(1) $f=P\cdot q+r\text{ and }P_{k}^{\ast}\left(D\right)r=0.$

To indicate the dependency of $q$ on $f$ and $P$ we will often write
$q=T_{P}(f)$. H. S. Shapiro studied Fischer decompositions in a wider setting,
cf. [17], going beyond the case of polynomials to more general function
spaces, in particular, to the space $E\left(\mathbb{C}^{d}\right)$ of all
entire functions $f:\mathbb{C}^{d}\rightarrow\mathbb{C}$. It is useful to
adopt a notion introduced in [17, p. 522]: suppose that $E$ is a vector space
of infinitely differentiable functions $f:G\rightarrow\mathbb{C}$ (defined on
an open subset $G$ of $\mathbb{C}^{d}$) that is a module over
$\mathcal{P}\left(\mathbb{C}^{d}\right)$.

######  Definition 1.

We say that a polynomial $P$ and a differential operator $Q\left(D\right)$
form a _Fischer pair $(P,Q)$ for the space _$E$, if for each $f\in E$ there
exist _unique_ elements $q\in E$ and $r\in E$ such that

(2) $f=P\cdot q+r\text{ and }Q\left(D\right)r=0.$

We will speak of weak Fischer pairs when the decomposition $f=P\cdot q+r$ is
not assumed to be unique. But the expression “Fischer decomposition” will be
used even in the absence of uniqueness.

Shapiro proved in [17, Theorem 1] that the following version of Fischer’s
theorem is true when $\mathcal{P}\left(\mathbb{C}^{d}\right)$ is replaced by
$E\left(\mathbb{C}^{d}\right)$: for every homogeneous polynomial $P$ and every
entire function $f$ there exist _unique_ entire functions $q$ and $r$ such
that

$f=P\cdot q+r\text{ and }P^{\ast}\left(D\right)r=0.$

That is, $P$ and $P^{*}$ form a Fischer pair for the entire functions,
whenever $P$ is homogeneous. Shapiro conjectured that his theorem held even
for non-homogeneous $P$ and arbitrary entire functions, cf. [17, p. 517 (i)].
Below we obtain some partial results for an arbitrary polynomial $P$ with
leading term $P_{k}$, showing that $(P,P^{*}_{k})$ is a weak Fischer pair for
entire functions of sufficiently low order.

The following related conjecture, for the special case where the polynomial
under consideration is $P_{k}-1$, has been studied in the literature:

Conjecture: Let $P$ be a nonconstant homogeneous polynomial and define

$F_{P}\left(q\right):=P^{\ast}\left(D\right)\left[\left(P-1\right)q\right].$

Then $F_{P}$ is a bijection on the set of all entire functions; equivalently,
$P-1$ and $P^{\ast}\left(D\right)$ form a Fischer pair for
$E\left(\mathbb{C}^{d}\right)$.

We mention that the equivalence between the bijectivity of the Fischer
operator $F_{P}$ and the fact that $P-1$ and $P^{\ast}\left(D\right)$ form a
Fischer pair, is due to Meril and Struppa, cf. [11, Proposition 1].

In [9], D. Khavinson and H.S. Shapiro prove a partial case of the preceding
conjecture, restricting the class of entire functions under consideration and
assuming uniqueness from the outset, cf. [9, Theorem 3] (we mention that the
proof uses Fredholm theory).

The homogeneous polynomials $P$ in [9, Theorem 3] satisfy a certain property
called “amenability”, defined as follows in [9, p. 464]: a homogeneous
polynomial $P_{k}\in\mathbb{C}[z_{1},\dots,z_{d}]$ of degree $k>0$ is said to
be amenable if for each $j\in\\{1,\dots,d\\}$ there is a multi-index $\alpha$,
with $|\alpha|=k-1$, such that $D^{\alpha}P_{k}$ is a non-zero constant times
$z_{j}$ (note however that no polynomial of degree $k=1$ is amenable when
$d\geq 2$). Khavinson and Shapiro observe that amenability holds in the
important special case $P=\left|x\right|^{2}$, so
$P^{\ast}\left(D\right)=\Delta$. Furthermore, amenability entails the
following bounds, cf. [9, Lemma 11], which represent a weaker assumption: if
$P$ is an amenable homogeneous polynomial and $f$ is homogeneous of degree
$m$, that is, $f\in\mathcal{P}_{m}\left(\mathbb{C}^{d}\right)$, then there is
a constant $C>0$, independent of $m$, such that

(3) $\|Pf\|_{a}\geq Cm^{1/2}\|P\|_{a}\|f\|_{a},$

where $\|\cdot\|_{a}$ is the norm defined by the apolar inner product (see the
next section). Note that the term $\|P\|_{a}$ can be absorbed into the
constant $C$.

Satisfaction of the Khavinson-Shapiro bounds (3) appears to be a much more
generic property than amenability. We explore this question in the specific
instance where both the dimension and the degree are 2, so
$P\left(z_{1},z_{2}\right)=az_{1}^{2}+bz_{1}z_{2}+cz_{2}^{2}$ (with $a,b,c$
not all 0). While in this case the only amenable polynomials are
$az_{1}^{2}+cz_{2}^{2}$ when $a,c\neq 0$, and $bz_{1}z_{2}$ when $b\neq 0$, we
prove that the Khavinson-Shapiro bounds are satisfied precisely when $4ac\neq
b^{2}$, cf. Theorem 7 below.

Recalling Bombieri’s inequality

$\|Pf\|_{a}\geq\|P\|_{a}\|f\|_{a},$

we see that satisfaction of the Khavinson-Shapiro bounds entails a
quantitative strengthening of Bombieri’s bounds, as $m\to\infty$.

Let $P_{k}$ be a nonzero homogeneous polynomial of degree $k$. By Fischer’s
Theorem, for each homogeneous polynomial $f_{m}$ of degree $m$ there exist
unique polynomials $T_{P_{k}}$ and $r_{m}$, with
$P_{k}^{\ast}\left(D\right)r_{m}=0$ and

(4) $f_{m}=P_{k}\cdot T_{P_{k}}\left(f_{m}\right)+r_{m}.$

This decomposition is orthogonal with respect to the apolar inner product,
since by (11) below

$\langle P_{k}\cdot T_{P_{k}}\left(f_{m}\right),r_{m}\rangle_{a}=\langle
T_{P_{k}}\left(f_{m}\right),P_{k}^{\ast}\left(D\right)r_{m}\rangle_{a}=0.$

By the Pythagorean Theorem, and under the assumption that the Khavinson-
Shapiro bounds (3) hold, we obtain

$\left\|f_{m}\right\|_{a}^{2}=\left\|P_{k}\cdot
T_{P_{k}}\left(f_{m}\right)\right\|_{a}^{2}+\left\|r_{m}\right\|^{2}\geq\left\|P_{k}\cdot
T_{P_{k}}\left(f_{m}\right)\right\|_{a}^{2}\geq
C^{2}m\|T_{P_{k}}\left(f_{m}\right)\|_{a}^{2}.$

This shall be the type of assumption used in the theorem stated next. The main
differences with the Khavinson-Shapiro result lie in the fact that arbitrary
polynomials $P=P_{k}-P_{k-1}-\cdots-P_{0}$ are used for division, instead of
polynomials of the form $P=P_{k}-1$, and instead of bounds of the form
$O(m^{1/2})$ we use $O(m^{\tau/2})$, where $\tau\geq 0$.

Next we state the main result of the paper.

######  Theorem 2.

Let $P_{k}$ be a homogeneous polynomial of degree $k>0$ on $\mathbb{C}^{d}$,
and let us write $T:=T_{P_{k}}$, where $T_{P_{k}}$ is defined by (4). Assume
that there exist a $C=C(P_{k})>0$ ($C$ independent of $m$) and an
$\tau\in\\{0,\dots,k\\}$, such that for every $m>0$ and every homogeneous
polynomial $f_{m}$ of degree $m,$ the following inequality holds:

(5)
$\left\|Tf_{m}\right\|_{a}\leq\frac{C}{m^{\tau/2}}\left\|f_{m}\right\|_{a}.$

If for $0\leq j<k$ the polynomials $P_{j}\left(z\right)$ are homogeneous of
degree $j$, and for some $\beta<k$ and every $j$ with $\beta<j<k$ we have
$P_{j}=0$, then for every entire function $f:\mathbb{C}^{d}\to\mathbb{C}$ of
order $\rho$, where $\rho$ satisfies the inequality

$\rho(k-\tau)<2(k-\beta),$

there exist entire functions $q$ and $r$ of order $\leq\rho$ with

$f=\left(P_{k}-P_{\beta}-\cdots-P_{0}\right)q+r\text{ and
}P_{k}^{\ast}\left(D\right)r=0.$

Throughout this paper, given the homogeneous polynomial $P_{k}$ we will use
the abbreviation $T:=T_{P_{k}}$.

Note that no statement is made regarding the possible uniqueness of the entire
functions $q$ and $r$. However, we shall see that when $k=1$, the conjecture
is true and furthermore we do have uniqueness, even though the Khavinson-
Shapiro bounds do not hold, cf. Theorem 8 below. When the dimension is $d=1$,
it follows directly from polynomial interpolation that the result also holds,
cf. Theorem 9: for every entire function $f:\mathbb{C}\to\mathbb{C}$, there
exist unique entire functions $q$ and $r$ such that

$f=\left(P_{k}-P_{\beta}-\cdots-P_{0}\right)q+r\text{ and
}P_{k}^{\ast}\left(D\right)r=0.$

Thus, in the proof of Theorem 2 above, it can always be assumed that $k,d\geq
2$.

Finally let us mention that conditions of the type (3) had already been used
in the work of P. Ebenfelt and H.S. Shapiro about a generalized Cauchy-
Kowaleskaya theorem in [4], and the mixed Cauchy problem for differential
equations in [5], see also [6], [7]. Let $P$ be a polynomial of degree $k,$
written as a sum of homogeneous polynomials

$P=P_{k}+P_{k-1}+\cdots+P_{0}$

with leading coefficient $P_{k}$, and let the floor function $\lfloor
x\rfloor$ be the largest integer $\leq x$. Theorem 3.1.1 in [6] (see also
remark on p. 259 in [6]) implies that for each entire function $f$, there
exist entire functions $q$ and $r$ such that

(6) $f=P_{k}q+r\text{ and }P^{\ast}\left(D\right)r=0,$

provided, first, that there are constants $C>0$ and $\tau\geq 0$ such that

(7) $\left\|P_{k}f_{m}\right\|\geq Cm^{\tau/2}\left\|f_{m}\right\|_{a},$

and second, that $P_{j}=0$ for all $j$ with

$\left\lfloor\frac{k+\tau}{2}\right\rfloor<j<k.$

Note that (6) can be seen as the dual problem to the one formulated in (1).

##  2\. Basic properties of the apolar inner product

The variant of the Khavinson-Shapiro result obtained here also employs the
apolar inner product (introduced during the XIX century within the theory of
invariants) also known as Fischer’s inner product, and, with different
normalizations, as Bombieri’s inner product (more details can be found in
[14]). For easy reference we include some known information about it.

Denote the natural numbers by $\mathbb{N}_{0}$ (emphasizing the fact that they
include zero) and recall the standard notation for multi-indices: given
$\alpha=\left(\alpha_{1},\dots,\alpha_{d}\right)\in\mathbb{N}_{0}^{d}$, we
write $z^{\alpha}=z_{1}^{\alpha_{1}}\cdots z_{d}^{\alpha_{d}},$
$\alpha!=\alpha_{1}!\cdots\alpha_{d}!$, and
$\left|\alpha\right|=\alpha_{1}+\cdots+\alpha_{d}$.

Let $P$ and $Q$ be polynomials of degree $N$ and $M$, respectively given by

$P\left(z\right)=\sum_{\alpha\in\mathbb{N}_{0}^{d},\left|\alpha\right|\leq
N}c_{\alpha}z^{\alpha}\text{ and
}Q\left(z\right)=\sum_{\alpha\in\mathbb{N}_{0}^{d},\left|\alpha\right|\leq
M}d_{\alpha}z^{\alpha}.$

The _apolar inner product_ $\langle\cdot,\cdot\rangle_{a}$ on
$\mathcal{P}\left(\mathbb{C}^{d}\right)$ is defined by

(8) $\left\langle
P,Q\right\rangle_{a}:=\left[Q^{*}\left(D\right)P\right]\;(0)=\sum_{\alpha\in\mathbb{N}_{0}^{d}}\alpha!c_{\alpha}\overline{d_{\alpha}},$

and the associated apolar norm, by $\left\|f\right\|_{a}=\sqrt{\left\langle
f,f\right\rangle_{a}}.$ Note that for the constant polynomial $1$ one has

(9) $\left\langle P,1\right\rangle_{a}=P\left(0\right).$

A fundamental property of the apolar inner product, is that the adjoint
operator of the multiplication operator $M_{Q}\left(g\right)=Qg$ is the
differential operator associated to the polynomial $Q^{\ast}.$

For the reader’s convenience we include the short proof of the following known
result.

######  Proposition 3.

The following formulae hold for polynomials $P,Q$ and $f,g$:

(10) $\left\langle
P,Q\right\rangle_{a}=[Q^{\ast}\left(D\right)P]\left(0\right)=\left\langle
Q^{\ast}\left(D\right)P,1\right\rangle_{a}.$

and

(11) $\left\langle Q^{\ast}\left(D\right)f,g\right\rangle_{a}=\left\langle
f,Q\cdot g\right\rangle_{a}.$

###### Proof.

Regarding (10), the first equality is the definition and the second follows
from (9). From (10) we conclude that

$\displaystyle\left\langle Q^{\ast}\left(D\right)f,g\right\rangle_{a}$
$\displaystyle=\left\langle
g^{\ast}\left(D\right)Q^{\ast}\left(D\right)f,1\right\rangle_{a},\mbox{ \ and
}$ $\displaystyle\left\langle f,Q\cdot g\right\rangle_{a}$
$\displaystyle=\left\langle
Q^{\ast}\left(D\right)g^{\ast}\left(D\right)f,1\right\rangle_{a}.$

Since
$g^{\ast}\left(D\right)Q^{\ast}\left(D\right)=Q^{\ast}\left(D\right)g^{\ast}\left(D\right)$
equation (11) follows. ∎

It is shown in [14, Theorem 38] that if $(P_{k},Q)$ is a Fischer pair, where
$P_{k}$ is the principal part of $P$ and $Q$ is homogeneous, then $(P,Q)$ is a
Fischer pair. However, from the fact that $(P,Q)$ is a Fischer pair we cannot
conclude that $(P_{k},Q)$ is a Fischer pair, as the following example from
[11] shows (cf. also [14, Example 34]): take $n=2$,
$P(z_{1},z_{2})=z_{1}-z_{2}^{2}$ and $Q(z_{1},z_{2})=z_{1}$. Then it can be
checked that $P$ and $Q$ form a Fischer pair, while
$P_{k}(z_{1},z_{2})=-z^{2}$ and $Q$ do not, since they are homogeneous of
different degrees. This is impossible by [14, Theorem 36].

##  3\. Apolar norms of products of linear factors

In addition to $\partial/\partial z_{j}$, we also use $D_{j}$ and
$\partial_{j}$ to denote the $j$-th partial derivative. The results in this
section will be used later, when comparing amenability to the Khavinson-
Shapiro bounds. For $a,b\in\mathbb{C}^{d}$ we denote by $\left\langle
a,b\right\rangle$ the standard inner product

$\left\langle
a,b\right\rangle=a_{1}\overline{b_{1}}+\cdots+a_{d}\overline{b_{d}}=\overline{\langle
b,\overline{a}\rangle}.$

Given $b,z\in\mathbb{C}^{d}$, consider the lineal polynomial

$Q_{b}\left(z\right):=\left\langle
z,\overline{b}\right\rangle=b_{1}z_{1}+\cdots+b_{d}z_{d},$

and write

$L_{\overline{b}}=Q_{b}^{\ast}\left(D\right)=\overline{b_{1}}\partial_{1}+\cdots+\overline{b_{d}}\partial_{d}.$

It is easy to see that for any pair of differentiable functions
$f\left(z\right),g\left(z\right)$,

$L_{\overline{b}}\left(fg\right)={\displaystyle\sum_{j=1}^{n}}\overline{b_{j}}\partial_{j}\left(f\cdot
g\right)=(L_{\overline{b}}f)\cdot g+f\cdot L_{\overline{b}}g.$

In the next result we assume that the nonzero vector $c$ is orthogonal to the
vectors $a_{1},\dots,a_{M}$ (they may be linearly dependent, or even
repeating):

######  Proposition 4.

Assume that $a_{1},\dots,a_{M}\in\mathbb{C}^{d}$, where
$a_{k}=\left(a_{k,1},\dots,a_{k,d}\right)$, and set
$\sigma_{k}\left(z\right)=\left\langle z,\overline{a_{k}}\right\rangle$ for
$k=1,\dots,M.$ Suppose there exists a vector
$c\in\mathbb{C}^{d}\setminus\\{0\\}$ such that for $k=1,\dots,M$,

${\displaystyle\sum_{j=1}^{d}}\overline{a_{k,j}}c_{j}=0.$

Let $g$ be a univariate polynomial. Then

$\left\|\sigma_{1}(z)\cdots\sigma_{M}(z)\cdot g\left(\left\langle
z,\overline{c}\right\rangle\right)\right\|_{a}^{2}=\left\|\sigma_{1}(z)\cdots\sigma_{M}(z)\right\|_{a}^{2}\left\|g\left(\left\langle
z,\overline{c}\right\rangle\right)\right\|_{a}^{2}.$

Taking $g\left(t\right)=t^{m}$ we see that

$\left\|\sigma_{1}(z)\cdots\sigma_{M}(z)\cdot\left\langle
z,\overline{c}\right\rangle^{m}\right\|_{a}^{2}=\left\|\sigma_{1}(z)\cdots\sigma_{M}(z)\right\|_{a}^{2}\left\|\left\langle
z,\overline{c}\right\rangle^{m}\right\|_{a}^{2}.$

so the product $\sigma_{1}(z)\cdots\sigma_{M}(z)$ does not satisfy the
Khavinson-Shapiro bounds.

###### Proof.

If $a=\left(a_{1},\dots,a_{d}\right)$ and $c$ are orthogonal, so $\langle
a,c\rangle=0$, then

$L_{\overline{a}}\left(g\left(\left\langle
z,\overline{c}\right\rangle\right)\right)={\displaystyle\sum_{j=1}^{d}}\overline{a_{j}}\partial_{j}\left(g\left(\left\langle
z,\overline{c}\right\rangle\right)\right)=g^{\prime}\left(\left\langle
z,\overline{c}\right\rangle\right){\displaystyle\sum_{j=1}^{d}}\overline{a_{j}}c_{j}=0.$

It follows that

$L_{\overline{a}}\left(f\cdot g\left(\left\langle
z,\overline{c}\right\rangle\right)\right)=L_{\overline{a}}f\cdot
g\left(\left\langle z,\overline{c}\right\rangle\right)+f\cdot
L_{\overline{a}}\left(g\left(\left\langle
z,\overline{c}\right\rangle\right)\right)=L_{\overline{a}}f\cdot
g\left(\left\langle z,\overline{c}\right\rangle\right).$

Since $\sigma_{k}\left(c\right)=\left\langle
c,\overline{a_{k}}\right\rangle=0$ for $k=1,\dots,M$, we have

$L_{\overline{a_{1}}}\cdots
L_{\overline{a_{M}}}\left(\sigma_{1}(z)\cdots\sigma_{M}(z)\cdot
g\left(\left\langle
z,\overline{c}\right\rangle\right)\right)=L_{\overline{a_{1}}}\cdots
L_{\overline{a_{M}}}\left(\sigma_{1}(z)\cdots\sigma_{M}(z)\right)\cdot
g\left(\left\langle z,\overline{c}\right\rangle\right).$

Furthermore, we know that

$\displaystyle\left\|\sigma_{1}(z)\cdots\sigma_{M}(z)\cdot g\left(\left\langle
z,\overline{c}\right\rangle\right)\right\|_{a}^{2}$
$\displaystyle=\left\langle L_{\overline{a_{1}}}\cdots
L_{\overline{a_{M}}}\left(\sigma_{1}(z)\cdots\sigma_{M}(z)\cdot
g\left(\left\langle
z,\overline{c}\right\rangle\right)\right),g\left(\left\langle
z,\overline{c}\right\rangle\right)\right\rangle_{a}$
$\displaystyle=L_{\overline{a_{1}}}\cdots
L_{\overline{a_{M}}}\left(\sigma_{1}(z)\cdots\sigma_{M}(z)\right)\left\|g\left(\left\langle
z,\overline{c}\right\rangle\right)\right\|_{a}^{2}$
$\displaystyle=\left\|\sigma_{1}(z)\cdots\sigma_{M}(z)\right\|_{a}^{2}\left\|g\left(\left\langle
z,\overline{c}\right\rangle\right)\right\|_{a}^{2}.$

∎

##  4\. Observations on amenability and the bounds of Khavinson and Shapiro

Let $P\in\mathbb{C}[z_{1},\dots,z_{d}]$. We say that $P$ is independent of
$z_{j}$ if $\partial/\partial z_{j}P=0$. If for no $j$ this holds, then we say
that $P$ depends on all the variables. Obviously if $P_{k}$ is amenable then
it depends on all the variables. The converse is not true when $d>1$: for
$k=d=2$, just consider $P_{2}(z_{1},z_{2})=z_{1}^{2}+z_{1}z_{2}+z_{2}^{2}$;
the only possibilities for $\alpha$ are $(1,0)$ and $(0,1)$; clearly, neither
of them satisfy the amenability condition. This example also shows that a
symmetric polynomial (i.e., invariant under permutations of the variables) may
fail to be amenable.

Next we explore some cases of homogeneous polynomials that satisfy, or fail to
satisfy, the Khavinson-Shapiro bounds (3). Formula (12) below, where the sum
is taken over all multi-indices with non-negative entries, appears in [17, p.
523] (an earlier, and more general statement applying to certain entire
functions that include the polynomials, can be found in [13, Theorem 3]; in
[18] formula (12) is called Reznick identity). Note that when $|\alpha|>k$ the
corresponding terms are zero, while summing only over the multi-indices with
$|\alpha|=k$ yields Bombieri’s inequality. Refinements of Bombieri’s
inequality can be obtained by estimating other terms in the sum

(12)
$\|P_{k}f_{m}\|_{a}^{2}=\sum_{\alpha}\|\partial^{\alpha}P_{k}^{*}(D)f_{m}\|_{a}^{2}/\alpha!,$

as we do next in some special cases. Suppose the analogous statement to the
Khavinson-Shapiro bounds holds for some positive integer $\tau$ not
necessarily equal to one, i.e., for every homogeneous polynomial $f_{m}$ of
degree $m$, there is a constant $C>0$, independent of $m$, such that

(13) $\|Pf\|_{a}\geq Cm^{\tau/2}\|f_{m}\|_{a}.$

Then we call $\tau$ a Khavinson-Shapiro exponent, and the largest such $\tau$
the optimal Khavinson-Shapiro exponent.

######  Theorem 5.

Let $P_{k}:\mathbb{C}^{d}\to\mathbb{C}$ be a homogeneous polynomial of degree
$k>0$. Then:

1) When the dimension $d=1$, then the optimal Khavinson-Shapiro exponent is
always equal to $k$.

2) When the degree $k=1$ and the dimension $d>1$, no homogeneous polynomial
$P_{1}$ satisfies the Khavinson-Shapiro bounds. For $k>1$, the optimal
Khavinson-Shapiro exponent is bounded by $k-1$.

3) When all the exponents of the variables are even (and in particular, so is
the total degree $k$), the homogeneous polynomial $P_{k}$ is amenable, and
hence it satisfies the Khavinson-Shapiro bounds.

###### Proof.

When $d=1$, writing $P_{k}(z)=az^{k}$ and $f_{m}(x)=bz^{m}$ with $a,b\neq 0$,
we have

(14) $\|P_{k}f_{m}\|_{a}^{2}=|a|^{2}\frac{(k+m)!}{m!}\|f_{m}\|_{a}^{2}.$

For 2), suppose that

$P_{1}\left(z\right)=\sum_{1\leq i\leq d}c_{i}z_{i},$

where some but not all the coefficients $c_{i}$ might be $0$, and let $m>1$.
Since $P_{1}(D)$ maps the homogeneous polynomials of degree $m$ into those of
degree $m-1$, and the latter space has a smaller dimension (because $d>1$) it
follows that there is a homogeneous polynomial $f_{m}\neq 0$ of degree $m$
such that $P_{1}(D)f_{m}=0$. By the Newman-Shapiro identity (12), a.k.a.
Reznick identity, with $|\alpha|=1$,

$\|P_{1}f_{m}\|_{a}^{2}=\|P_{1}\|_{a}^{2}\|f_{m}\|_{a}^{2}+\|P_{1}(D)f_{m}\|_{a}^{2}=\|f_{m}\|_{a}^{2}\|P_{1}\|_{a}^{2}.$

For $k>1$, the same argument shows that there is a homogeneous polynomial
$f_{m}\neq 0$ of degree $m$ such that $P_{k}(D)f_{m}=0$, so the term in (12)
corresponding to the multi-index $\alpha=0$ vanishes, and hence the optimal
Khavinson-Shapiro exponent is bounded by $k-1$.

Next we check the assertion in 3). For each $j\in\\{1,\dots,d\\}$ choose a
monomial $c_{\alpha}z^{\alpha}$ with maximal $\alpha_{j}$, and let
$\alpha^{\prime}$ be obtained from $\alpha$ by replacing $\alpha_{j}$ with
$\alpha_{j}-1$, so $|\alpha^{\prime}|=k-1$. Then for some constat $c\neq 0$ we
have $D^{\alpha^{\prime}}c_{\alpha}z^{\alpha}=cz_{j}$. If $c_{\beta}z^{\beta}$
is another monomial with $\beta_{j}<\alpha_{j}$, so
$\beta_{j}\leq\alpha_{j}-2$, then $D^{\alpha^{\prime}}c_{\beta}z^{\beta}=0$,
while if $\beta_{j}=\alpha_{j}$ then for some $i\neq j$,
$\beta_{i}<\alpha_{i}$, so again $D^{\alpha^{\prime}}c_{\beta}z^{\beta}=0$. It
follows that $D^{\alpha^{\prime}}P_{k}$ is a non-zero constant times $z_{j}$.
∎

Note that when the remainder $r=0$, we have $f_{m}=P_{k}Tf_{m}$, so by
Beauzamy’s inequality (Lemma 13 below) we always have $\tau\leq k$ in (5).

######  Example 6.

By Lemma 13 below, for every $f_{m}$ we have

$\|P_{k}f_{m}\|_{a}^{2}\leq Cm^{k}\|f_{m}\|_{a}^{2},$

so $\tau\leq k$, and by part 2) of the preceding theorem, if $d>1$ it is
always possible to find an $f_{m}$ for which $\tau\leq k-1$. The condition

(15)
$\left\|Tf_{m}\right\|_{a}\leq\frac{C}{m^{\frac{\tau}{2}}}\left\|f_{m}\right\|_{a}$

used in Theorem 2 can be much weaker when applied to specific functions. In
the extreme case $Tf_{m}=0$ obviously every positive $\tau$ will work. For a
less extreme example, take $m,\ell\gg k$, and let
$P_{k}\in\mathbb{C}[z_{1},z_{d}]$ be $z_{1}^{k}$. Choosing
$f_{m}(z_{1},z_{2}):=z_{1}^{k}z_{2}^{m-k}+m^{\ell}z_{2}^{m}$, it is clear that
we can take $\tau=\ell$. But as noted above, when the remainder $r=0$ we have
$f_{m}=P_{k}Tf_{m}$, so since $\tau$ is required to work for all homogeneous
polynomials, $\tau\leq k$ also in (15), and when $d>1$, $\tau\leq k-1$.

Next we show that if $k=d=2$, so
$P\left(z_{1},z_{2}\right)=az_{1}^{2}+bz_{1}z_{2}+cz_{2}^{2}$, then $P$ will
satisfy the Khavinson-Shapiro bounds under the condition $4ab\neq b^{2}$,
which is distinctly more general than amenability. It is easy to check that
for $k=d=2$, the only amenable polynomials are
$Q\left(z_{1},z_{2}\right)=az_{1}^{2}+cz_{2}^{2}$ when $a,c\neq 0$ and
$R\left(z_{1},z_{2}\right)=bz_{1}z_{2}$ when $b\neq 0$.

######  Theorem 7.

Let $P\left(z_{1},z_{2}\right)=az_{1}^{2}+bz_{1}z_{2}+cz_{2}^{2}$ be a
polynomial with complex coefficients $a,b,c$, not all of them 0. Then the
following are equivalent:

1) $P$ does not satisfy the Khavinson-Shapiro bounds.

2) $4ac=b^{2}$.

3) $P\left(z_{1},z_{2}\right)=\left(rz_{1}+sz_{2}\right)^{2}$, where the only
condition on the complex coefficients $r$ and $s$ is that they cannot
simultaneously be 0.

###### Proof.

To obtain 1) $\implies$ 2), we shall prove an equivalent formulation: if
$4ab\neq b^{2}$, then $P$ satisfies the Khavinson-Shapiro bounds.

Given $P\left(z_{1},z_{2}\right)=az_{1}^{2}+bz_{1}z_{2}+cz_{2}^{2}$, we have

$\partial_{1}P^{\ast}\left(z_{1},z_{2}\right)=2\overline{a}z_{1}+\overline{b}z_{2}\text{
and
}\partial_{2}P^{\ast}\left(z_{1},z_{2}\right)=\overline{b}z_{1}+2\overline{c}z_{2}.$

Let

$f_{m}(z_{1},z_{2})=\sum_{0\leq i\leq m}c_{i}z_{1}^{i}z_{2}^{m-i}.$

A computation (alternatively, use [17, Lemma 3]) shows that

$m\|f_{m}\|_{a}^{2}=\|\partial_{1}f_{m}\|_{a}^{2}+\|\partial_{2}f_{m}\|_{a}^{2}.$

Let $0\leq t\leq 1$ satisfy

(16) $tm\|f_{m}\|_{a}^{2}=\|\partial_{1}f_{m}\|_{a}^{2},\mbox{ \ so \
}(1-t)m\|f_{m}\|_{a}^{2}=\|\partial_{2}f_{m}\|_{a}^{2}.$

It follows from (12) that

$\left\|Pf_{m}\right\|_{a}^{2}\geq A_{m},\text{ where
}A_{m}:={\displaystyle\sum_{k=1}^{2}}\left\|\left(\partial_{k}P^{\ast}\right)\left(D\right)f_{m}\right\|_{a}^{2}.$

Now

$\displaystyle\left\|\left(2\overline{a}\partial_{1}+\overline{b}\partial_{2}\right)f_{m}\right\|_{a}^{2}$
$\displaystyle=4\left|a\right|^{2}\left\|\partial_{1}f_{m}\right\|_{a}^{2}+\left|b\right|^{2}\left\|\partial_{2}f_{m}\right\|_{a}^{2}$
$\displaystyle+2\overline{a}b\left\langle\partial_{1}f_{m},\partial_{2}f_{m}\right\rangle_{a}+2a\overline{b}\left\langle\partial_{2}f_{m},\partial_{1}f_{m}\right\rangle_{a},$

and the sum of the last two terms is

$4\operatorname{Re}\left(\overline{a}b\left\langle\partial_{1}f_{m},\partial_{2}f_{m}\right\rangle_{a}\right).$

The analogous formula holds for
$\left\|\left(\overline{b}\partial_{1}+2\overline{c}\partial_{2}\right)f_{m}\right\|_{a}^{2}.$
It follows that

$\displaystyle A_{m}$
$\displaystyle=\left\|\left(2\overline{a}\partial_{1}+\overline{b}\partial_{2}\right)f_{m}\right\|_{a}^{2}+\left\|\left(\overline{b}\partial_{1}+2\overline{c}\partial_{2}\right)f_{m}\right\|_{a}^{2}$
$\displaystyle=\left(4\left|a\right|^{2}+\left|b\right|^{2}\right)\left\|\partial_{1}f_{m}\right\|_{a}^{2}+\left(\left|b\right|^{2}+4\left|c\right|^{2}\right)\left\|\partial_{2}f_{m}\right\|_{a}^{2}+$
$\displaystyle+4\operatorname{Re}\left(\left(\overline{a}b+\overline{b}c\right)\left\langle\partial_{1}f_{m},\partial_{2}f_{m}\right\rangle_{a}\right)$

Using $\operatorname{Re}z\geq-\left|z\right|$ and the Cauchy-Schwarz
inequality, we conclude that

$A_{m}\geq\left(4\left|a\right|^{2}+\left|b\right|^{2}\right)\left\|\partial_{1}f_{m}\right\|_{a}^{2}$
$+\left(\left|b\right|^{2}+4\left|c\right|^{2}\right)\left\|\partial_{2}f_{m}\right\|_{a}^{2}-4\left|\overline{a}b+\overline{b}c\right|\left\|\partial_{1}f_{m}\right\|_{a}\left\|\partial_{2}f_{m}\right\|_{a}.$

It now follows from (16) that

$A_{m}\geq\left(4\left|a\right|^{2}+\left|b\right|^{2}\right)mt\left\|f_{m}\right\|_{a}^{2}$
$+\left(\left|b\right|^{2}+4\left|c\right|^{2}\right)m\left(1-t\right)\left\|f_{m}\right\|_{a}^{2}-4\left|\overline{a}b+\overline{b}c\right|m\sqrt{t}\sqrt{1-t}\left\|f_{m}\right\|_{a}^{2}.$

Thus $A_{m}\geq m\left\|f_{m}\right\|_{a}^{2}f\left(t\right)$, where

$f\left(t\right)=\left(4\left|a\right|^{2}+\left|b\right|^{2}\right)t+\left(\left|b\right|^{2}+4\left|c\right|^{2}\right)\left(1-t\right)-4\left|\overline{a}b+\overline{b}c\right|\sqrt{t}\sqrt{1-t}.$

Letting $C:=\min\\{f(t):t\in[0,1]\\}$, since $A_{m}\geq
m\left\|f_{m}\right\|_{a}^{2}f\left(t\right)$, for the Khavinson-Shapiro
bounds to be satisfied it is enough that $f$ be strictly positive on
$\left[0,1\right].$

Recall our assumption $4ab\neq b^{2}.$ If $b=0$, then $ac\neq 0,$ and

$f\left(t\right)=4\left|a\right|^{2}t+4\left|c\right|^{2}\left(1-t\right).$

Since $f$ is affine and $f\left(0\right)>0$, $f\left(1\right)>0$, we conclude
that $C>0$.

Assume next that $b\neq 0$. Then

$f\left(t\right)=\left(\left|b\right|^{2}+4\left|c\right|^{2}\right)\left(\left(1-t\right)-\frac{4\left|\overline{a}b+\overline{b}c\right|}{\left|b\right|^{2}+4\left|c\right|^{2}}\sqrt{t}\sqrt{1-t}+\frac{4\left|a\right|^{2}+\left|b\right|^{2}}{\left|b\right|^{2}+4\left|c\right|^{2}}t\right)$
$=\left(\left|b\right|^{2}+4\left|c\right|^{2}\right)\times$
$\left(\left(\sqrt{1-t}-\frac{2\left|\overline{a}b+\overline{b}c\right|}{\left|b\right|^{2}+4\left|c\right|^{2}}\sqrt{t}\right)^{2}+\left(\frac{4\left|a\right|^{2}+\left|b\right|^{2}}{\left|b\right|^{2}+4\left|c\right|^{2}}-\frac{4\left|\overline{a}b+\overline{b}c\right|^{2}\
}{\left(\left|b\right|^{2}+4\left|c\right|^{2}\right)^{2}}\right)t\right).$

Note that $f\left(0\right)>0.$ Thus $f\left(t\right)>0$ for all
$t\in\left[0,1\right]$, provided the numerator obtained when summing the last
two fractions is strictly positive, that is,

$B=\left(4\left|a\right|^{2}+\left|b\right|^{2}\right)\left(\left|b\right|^{2}+4\left|c\right|^{2}\right)-4\left|\overline{a}b+\overline{b}c\right|^{2}>0.$

Now

$\left|\overline{a}b+\overline{b}c\right|^{2}=\left(\overline{a}b+\overline{b}c\right)\left(a\overline{b}+b\overline{c}\right)=\left|a\right|^{2}\left|b\right|^{2}+\overline{a}bb\overline{c}+\overline{b}ca\overline{b}+\left|b^{2}\right|\left|c^{2}\right|,$

so

$B=\
16\left|ac\right|^{2}+\left|b\right|^{4}-8\operatorname{Re}\left(\overline{a}bb\overline{c}\right).$

Write $z=b^{2}$ and $w=\overline{a}\overline{c}$, with $z=u+iv$ and $w=x+iy.$
Then

$B=16\left(z^{2}+y^{2}\right)+u^{2}+v^{2}-8\left(ux-
vy\right)=\left(4x-u\right)^{2}+\left(4y+v\right)^{2}.$

Hence $B=0$ entails that $y=-v/4$ and $x=u/4,$ so

$\overline{a}\overline{c}=w=x+iy=\frac{1}{4}\left(u-iv\right)=\frac{1}{4}\overline{z}=\frac{1}{4}\overline{b}^{2}.$

Thus $B=0$ if and only if $4ac=b^{2}.$ Since we assume that $b^{2}\neq 4ab$ we
see that $A_{m}\geq Cm\left\|f_{m}\right\|_{a}^{2}.$

For $b)\implies c)$, write
$P(z_{1},z_{2})=r^{2}z_{1}^{2}+bz_{1}z_{2}+s^{2}z_{2}^{2}$, and assume that
$b^{2}=4r^{2}s^{2}$. Then $P(z_{1},z_{2})=(rz_{1}+sz_{2})^{2}$. Finally, for
$c)\implies a)$, given the nontrivial polynomial
$P(z_{1},z_{2})=(rz_{1}+sz_{2})^{2}$, choose the nonzero vector $c=(s,-r)$ and
let $f_{m}(z_{1},z_{2}):=(sz_{1}-rz_{2})^{m}$. Then the equality
$\|Pf_{m}\|_{a}^{2}=\|P\|_{a}^{2}\|f_{m}\|_{a}^{2}$ is a special case of
Proposition 4. ∎

##  5\. Special cases: when $k=1$ or $d=1$ the conjecture is true

In view of the fact that the Khavinson-Shapiro bounds never hold when $k=1$,
this case requires separate treatment. It is noted next that the Khavinson-
Shapiro bounds are not needed for lineal polynomials.

######  Theorem 8.

Let $P_{1}(z_{1},\dots,z_{d})$ be a non-zero homogeneous polynomial of degree
$1$, and let $P=P_{1}-P_{0}$. Then the Fischer operator $F$ defined by

$F\left(q\right):=P_{1}^{\ast}\left(D\right)\left[\left(P_{1}-P_{0}\right)q\right]$

is a bijection on the set of all entire functions in $d$ variables, i.e.,
$P=P_{1}-P_{0}$ and $P_{1}^{\ast}\left(D\right)$ form a Fischer pair for this
space.

###### Proof.

We prove that $P_{1}-P_{0}$ and $P_{1}^{\ast}\left(D\right)$ form a Fischer
pair. By Shapiro’s theorem [17, Theorem 1], $P_{1}$ and
$P_{1}^{\ast}\left(D\right)$ form a Fischer pair for the entire functions.
Since $P_{1}\left(z\right)$ is homogeneous of degree $1$, for some $b\neq 0$
we have

$P_{1}\left(z\right)=\left\langle z,b\right\rangle.$

Since $b\neq 0$, the linear funtion
$\left\langle\cdot,b\right\rangle:\mathbb{C}^{d}\to\mathbb{C}$ is surjective,
so there is a $z_{0}$ with $\left\langle z_{0},b\right\rangle=P_{0}$, and
hence $P_{1}\left(z-z_{0}\right)=P_{1}\left(z\right)-P_{0}.$

Let $f$ be entire, and set $F\left(z\right)=f\left(z+z_{0}\right).$ By
Shapiro’s theorem there are unique entire functions $q$ and $h$ such that
$P_{1}^{\ast}\left(D\right)h=0$ and

$F\left(z\right)=P_{1}\left(z\right)q\left(z\right)+h\left(z\right).$

Replace $z$ with $z-z_{0}.$ Then

$\displaystyle f\left(z\right)$
$\displaystyle=F\left(z-z_{0}\right)=P_{1}\left(z-z_{0}\right)q\left(z-z_{0}\right)+h\left(z-z_{0}\right)$
$\displaystyle=\left(P_{1}\left(z\right)-P_{0}\right)q\left(z-z_{0}\right)+h\left(z-z_{0}\right).$

Clearly $\widetilde{h}\left(z\right):=h\left(z-z_{0}\right)$ satisfies
$P_{1}^{\ast}\left(D\right)\widetilde{h}=0$, since
$P_{1}^{\ast}\left(D\right)h=0.$ Uniqueness of $\widetilde{h}\left(z\right)$
and $\widetilde{q}\left(z\right):=q\left(z-z_{0}\right)$ follows from the
corresponding uniqueness statements for $h$ and $q$. ∎

For $d=1$, the conjecture can be proven directly:

######  Theorem 9.

Let $P:\mathbb{C}\to\mathbb{C}$ be a non-zero polynomial of degree $k$, with
homogeneous principal part $P_{k}$. Then $P$ and $P_{k}^{\ast}\left(D\right)$
form a Fischer pair for $E(\mathbb{C})$. Furthermore, if $f$ has order $\rho$,
writing $f=Pq+r$ we find that $q$ also has order $\rho$ and $r$ is a
polynomial of degree $k-1$.

###### Proof.

Let $P$ be a non-zero polynomial of degree $k$ with complex zeros
$\alpha_{1},\dots,\alpha_{k}$, listed according to their multiplicity. Given
an arbitrary entire function $f$, define $I\left(f\right)$ to be the unique
polynomial of degree $k-1$ interpolating $f$ at the zeros of $P$ (using the
Lagrange interpolation polynomial if all the zeros are different, or more
generally, using Hermite interpolation in the case of repeated roots, so not
only the values of $f$ and $I(f)$ coincide at the roots, but also an
appropriate number of derivatives do so). Then $f-I\left(f\right)$ vanishes at
the points $\alpha_{1},\dots,\alpha_{k}$ and we can write
$f-I\left(f\right)=Pq$ for some entire function $q.$ This yields the Fischer
decomposition, because trivially $P_{k}^{\ast}\left(D\right)(I(f))=0$.
Uniqueness follows from the uniqueness of the interpolating polynomial, since
given any decomposition $f=Pq^{\prime}+r^{\prime}$, in order for
$P_{k}^{\ast}\left(D\right)(r^{\prime})=0$ to hold, $r^{\prime}$ must be a
polynomial of degree strictly smaller than $k$.

When $f$ is entire of order $\rho$, then both $f-I\left(f\right)$ and $q$ also
have order $\rho$. ∎

We recall next some additional results regarding the conjecture. It also holds
when $P_{k}^{\ast}\left(z\right)$ is of the form $z_{1}^{k}$ for
$z=\left(z_{1},z^{\prime}\right)\in\mathbb{C}\times\mathbb{C}^{d-1}.$ In [11]
Meril and Struppa have proven the following result:

######  Theorem 10.

Let $P\left(z\right)$ be a polynomial of degree $k,$ let
$Q_{k}\left(z\right)=z_{1}^{k}$, where
$z=\left(z_{1},z^{\prime}\right)\in\mathbb{C}\times\mathbb{C}^{d-1}$, let
$C\neq 0$ be a complex number, and let $p_{0},\dots,p_{k-1}$ be polynomials in
the variable $z^{\prime}\in\mathbb{C}^{d-1}$. Then the polynomial $P$ and the
differential operator $Q_{k}\left(D\right)=\partial_{1}^{k}$ form a Fischer
pair if and only if $P$ is of the form

$P\left(z\right)=Cz_{1}^{k}+p_{k-1}\left(z^{\prime}\right)z_{1}^{k-1}+\cdots+p_{0}\left(z^{\prime}\right).$

Finally we mention that the following conjecture:

######  Conjecture 11 (II).

Let $P$ be a polynomial. Then the Fischer operator
$F_{P}:E\left(\mathbb{C}^{d}\right)\rightarrow E\left(\mathbb{C}^{d}\right)$
defined by

$F_{P}\left(q\right)=P^{\ast}\left(D\right)\left(Pq\right)$

is a bijection.

A. Meril and A. Yger have shown that $F_{p}$ is injective when $P$ is a
polynomial of degree $\leq 2$, cf. [12]. In dimension $d=2$, they have also
proven that the Fischer operator
$F_{P}:E\left(\mathbb{C}^{2}\right)\rightarrow E\left(\mathbb{C}^{2}\right)$
is bijective for any polynomial of degree $\leq 2$. In [8] it is shown that
conjecture II holds for the polynomial $P\left(z\right)=1+z^{\alpha}$, where
$\alpha\in\mathbb{N}_{0}^{d}$ has only positive entries. In general,
conjecture II is still open.

For more information about Fischer operators and their relationship to
problems in analysis we refer to [10], [15] and the classical paper [17].

##  6\. Proof of Theorem 2

Let us start with some preliminary bounds.

######  Lemma 12.

Given a multi-index $\alpha\in\mathbb{N}_{0}^{d}$, the estimate

(17) $\left\|f_{m}\right\|_{a}\leq\left\|z^{\alpha}f_{m}\right\|_{a}\leq
C_{\alpha,m}\left\|f_{m}\right\|_{a}$

holds for all homogeneous polynomials $f_{m}$ of degree $m$, where

$C_{\alpha,m}=\sup_{\beta\in\mathbb{N}_{0}^{d},\left|\beta\right|=m}\sqrt{\frac{\left(\alpha+\beta\right)!}{\beta!}},$

and this is the smallest constant such that (17) holds for all homogeneous
polynomials of degree $m.$

###### Proof.

We consider $f_{m}(z)=z^{\beta}$ with $\left|\beta\right|=m.$ Then

$\left\|z^{\alpha}f_{m}(z)\right\|_{a}^{2}=\left\|z^{\alpha+\beta}\right\|_{a}^{2}=\left(\alpha+\beta\right)!=\frac{\left(\alpha+\beta\right)!}{\beta!}\left\|z^{\beta}\right\|_{a}^{2}=\frac{\left(\alpha+\beta\right)!}{\beta!}\left\|f_{m}\right\|_{a}^{2}.$

It follows that

$C_{\alpha,m}\geq\sup_{\beta\in\mathbb{N}_{0}^{d},\left|\beta\right|=m}\sqrt{\frac{\left(\alpha+\beta\right)!}{\beta!}}.$

Now we show that $C_{\alpha,m}$ is a suitable constant. Let us write
$f_{m}(z)={\displaystyle\sum_{\left|\beta\right|=m}}a_{\beta}z^{\beta}.$ Then
$z^{\alpha}f_{m}(z)={\displaystyle\sum_{\left|\beta\right|=m}}a_{\beta}z^{\alpha+\beta}$
and

$\displaystyle\left\|f_{m}\right\|_{a}^{2}\leq\left\|z^{\alpha}f_{m}(z)\right\|_{a}^{2}$
$\displaystyle={\displaystyle\sum_{\left|\beta\right|=m}}\left(\alpha+\beta\right)!\left|a_{\beta}\right|^{2}={\displaystyle\sum_{\left|\beta\right|=m}}\frac{\left(\alpha+\beta\right)!}{\beta!}\beta!\left|a_{\beta}\right|^{2}$
$\displaystyle\leq{\displaystyle\sum_{\left|\beta\right|=m}}C_{\alpha,m}^{2}\beta!\left|a_{\beta}\right|^{2}=C_{\alpha,m}^{2}\left\|f_{m}\right\|_{a}^{2}.$

∎

With different normalizations, the following result is essentially due to B.
Beauzamy, cf. [3, Formula (6)].

######  Lemma 13.

If
$P\left(z\right)={\displaystyle\sum_{\left|a\right|=k}}c_{\alpha}z^{\alpha}$
is a homogeneous polynomial of degree $k$, then

$\left\|Pf_{m}\right\|_{a}\leq\left\|f_{m}\right\|_{a}\left(1+m\right)^{\frac{k}{2}}{\displaystyle\sum_{\left|\alpha\right|=k}}\left|c_{\alpha}\right|\sqrt{\alpha!}.$

###### Proof.

By the preceding lemma,

$\left\|Pf_{m}\right\|_{a}\leq{\displaystyle\sum_{\left|a\right|=k}}\left|c_{\alpha}\right|\left\|z^{\alpha}f_{m}\right\|_{a}\leq\left\|f_{m}\right\|_{a}{\displaystyle\sum_{\left|a\right|=k}}\left|c_{\alpha}\right|\sup_{\beta\in\mathbb{N}_{0}^{d},\left|\beta\right|=m}\sqrt{\frac{\left(\alpha+\beta\right)!}{\beta!}}.$

Note next that

$\displaystyle\frac{\left(\alpha+\beta\right)!}{\alpha!\beta!}$
$\displaystyle=\frac{\left(\alpha_{1}+\beta_{1}\right)!\cdots\left(\alpha_{d}+\beta_{d}\right)!}{\alpha_{1}!\cdots\alpha_{d}!\beta_{1}!\cdots\beta_{d}!}$
$\displaystyle=\frac{\left(\beta_{1}+1\right)\left(\beta_{1}+2\right)\cdots\left(\beta_{1}+\alpha_{1}\right)}{\alpha_{1}!}\cdots\frac{\left(\beta_{d}+1\right)\cdots\left(\beta_{d}+\alpha_{d}\right)}{\alpha_{d}!}$
$\displaystyle\leq\left(1+\beta_{1}\right)^{\alpha_{1}}\cdots\left(1+\beta_{d}\right)^{\alpha_{d}}\leq\left(1+m\right)^{\left|\alpha\right|}.$

∎

We will use the following result, which appears in [14, Theorem 17]. More
details regarding its proof are presented in [16, Theorem 7]. While in the
latter reference the result is stated for even degrees $2k$, the argument
presented there works for general values of $k$.

######  Theorem 14.

Let $Q$ be a homogeneous polynomial of degree $k>0$, let $P$ be a polynomial
of degree $k$ of the form

(18) $P=P_{k}-P_{k-1}-\cdots-P_{0},\ $

and assume that $(P_{k},Q)$ is a Fischer pair for
$\mathcal{P}\left(\mathbb{C}^{d}\right)$. Setting $T:=T_{P_{k}}$, we have

(19)
$T_{P}\left(f_{m}\right)=\sum_{j=-1}^{m}\sum_{s_{0}=0}^{k-1}\sum_{s_{1}=0}^{k-1}\dots\sum_{s_{j}=0}^{k-1}T(P_{s_{j}}T(\cdots
P_{s_{1}}T(P_{s_{0}}T(f_{m}))\cdots))$

for all homogeneous polynomials $f_{m}$ of degree $m$, with the convention
that the summand for $j=-1$ is $Tf_{m}$.

We have noted above that when the remainder $r=0$ we have $f_{m}=P_{k}Tf_{m}$,
and thus assuming the Khavinson-Shapiro bounds is equivalent to the condition

(20)
$\left\|Tf_{m}\right\|_{a}^{2}\leq\frac{C}{m^{\tau}}\left\|f_{m}\right\|_{a}^{2}$

actually used in the proof of Theorem 2. However, it is conceivable that for a
particular pair $(f,P)$ and all the homogeneous polynomials appearing in (19),
bounds of type (20) might actually be strictly weaker than the Khavinson-
Shapiro conditions. But we will not pursue these elaborations here.

To simplify expressions such as (19) parentheses shall often be omitted. We
will also use the following result of H.S. Shapiro [17, p. 519]:

######  Lemma 15.

(H.S. Shapiro) Suppose that
$f_{k}(z)=\sum_{\left|\alpha\right|=k}c_{\alpha}z^{\alpha}$ is a homogeneous
polynomial of degree $k$. Then for every complex vector $z\in\mathbb{C}^{d}$
the following estimate holds:

$\left|f_{k}\left(z\right)\right|^{2}\leq\frac{1}{k!}\left|z\right|^{2k}\left\|f_{k}\right\|_{a}^{2}.$

Let us recall some well known definitions and facts about entire functions
(additional details and references can be found in [16], cf. also [1]). The
order $\rho_{\mathbb{C}^{d}}\left(f\right)$ of a continuous function
$f:\mathbb{C}^{d}\rightarrow\mathbb{C}$ is defined by setting

$M_{\mathbb{C}^{d}}\left(f,r\right):=\sup\left\\{\left|f\left(z\right)\right|:z\in\mathbb{C}^{d},\left|z\right|=r\right\\},$

and then

$\rho_{\mathbb{C}^{d}}\left(f\right):=\lim_{r\rightarrow\infty}\sup\frac{\log\log
M_{\mathbb{C}^{d}}\left(f,r\right)}{\log r}\in\left[0,\infty\right].$

Given an entire function $f$, we write $f=\sum_{m=0}^{\infty}f_{m}$, where the
homogeneous polynomials $f_{m}$ are those given by the Taylor expansion about
0, that is,

(21) $f_{m}\left(z\right)=\sum_{\left|\alpha\right|=m}\frac{1}{\alpha!}\
\partial^{\alpha}f\left(0\right)\;z^{\alpha}\text{ for }m\in\mathbb{N}_{0}.$

It is well known that for $\rho\geq 0$, we have
$\rho_{\mathbb{C}^{d}}\left(f\right)\leq\rho$ if and only if for every
$\varepsilon>0$, there exists an $m_{0}\geq 0$ such that for every $m\geq
m_{0}$ the following bounds hold:

(22)
$\max_{\theta\in\mathbb{S}^{d-1}}\left|f_{m}\left(\theta\right)\right|\leq\frac{1}{m^{m/\left(\rho+\varepsilon\right)}}.$

We shall also use an old result due to V. Bargmann, cf. [Bar].

######  Theorem 16.

Let $P$ and $Q$ be polynomials in $d$ complex variables. Then

(23) $\left\langle
P,Q\right\rangle_{a}=\frac{1}{\pi^{d}}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}P\left(x+iy\right)\overline{Q\left(x+iy\right)}e^{-\left|x\right|^{2}-\left|y\right|^{2}}dxdy<\infty,$

where $dxdy$ is Lebesgue measure on $\mathbb{R}^{2d}$.

######  Lemma 17.

Let $f_{m}$ be a homogeneous polynomial of degree $m$, and denote by
$\mathbb{S}^{2d-1}$ the unit sphere in $\mathbb{R}^{2n}$. There is a
dimensional constant $C_{d}>0$ such that, identifying $\mathbb{C}^{n}$ with
$\mathbb{R}^{2n}$ as a measure space, we have

$\left\|f_{m}\right\|_{a}\leq
C_{d}\sqrt{\left(m+d-1\right)!}\max_{\theta\in\mathbb{S}^{2d-1}}\left|f_{m}\left(\eta\right)\right|.$

###### Proof.

Let $f_{m}:\mathbb{C}^{d}\rightarrow\mathbb{C}$ be a homogeneous polynomial of
degree $m$, and recall that for $x>0$, the Gamma function is defined as
$\Gamma(x):=\int_{0}^{\infty}e^{-t}\ t^{x-1}dt$. Integrating in polar
coordinates and using the change of variables $t=r^{2}$ we get

$\left\|f_{m}\right\|_{a}^{2}=\frac{1}{\pi^{d}}\int_{\mathbb{C}^{d}}\left|f_{m}\left(z\right)\right|^{2}e^{-\left|z\right|^{2}}dz=\frac{1}{\pi^{d}}\int_{0}^{\infty}e^{-r^{2}}r^{2m+2d-1}dr\int_{\mathbb{S}^{2d-1}}\left|f_{m}\left(\eta\right)\right|^{2}d\eta$
$\leq C_{d}\
\Gamma(m+d)\max_{\theta\in\mathbb{S}^{2d-1}}\left|f_{m}\left(\eta\right)\right|.$

∎

Next we present the proof of our main result, Theorem 2.

###### Proof.

In view of Theorem 8 we may suppose that $k\geq 2$.

Set $P=P_{k}-P_{\beta}-\cdots-P_{0}$. Then $T_{P}\left(f_{m}\right)$ is either
the zero polynomial or a polynomial of degree $<m$ (not necessarily
homogeneous). Our strategy is to show that

(24) $g:={\displaystyle\sum_{m=0}^{\infty}}T_{P}\left(f_{m}\right)$

defines an entire function $g:\mathbb{C}^{d}\rightarrow\mathbb{C}$ of order
bounded by $\rho$, by writing
$g\left(z\right)={\displaystyle\sum_{M=0}^{\infty}}G_{M}\left(z\right),$ where
each $G_{M}$ is a homogeneous polynomial of degree $M$, and then applying the
criterion presented in (22). As a reminder, we mention that when the
representation of $g$ as $\sum_{M=0}^{\infty}G_{M}$ exists, it is unique.

It has been noted above that

(25)
$T_{P}\left(f_{m}\right)=\sum_{j=-1}^{m}\sum_{s_{0}=0}^{k-1}\sum_{s_{1}=0}^{k-1}\cdots\sum_{s_{j}=0}^{k-1}TP_{s_{j}}\cdots
TP_{s_{0}}Tf_{m},$

and clearly, $TP_{s_{j}}\cdots TP_{s_{0}}Tf_{m}$ is a homogeneous polynomial
of degree

$[s_{0}+\cdots+s_{j}+m-k\left(j+2\right)]_{+}.$

If $s_{0},\dots,s_{n}\in\left\\{0,\dots,\beta\right\\}$ are given, where by
definition $\beta\leq k-1$, and if $n>m,$ then $TP_{s_{n}}\cdots
TP_{s_{0}}Tf_{m}$ is zero by inspection of its degree:

$\displaystyle m+s_{0}+\cdots+s_{n}-k\left(n+2\right)$ $\displaystyle\leq
m+\left(n+1\right)\left(k-1\right)-k\left(n+2\right)$
$\displaystyle=m-k-n-1<0,$

so the $m$ in the first summatory of (25) can be replaced by $\infty.$ By
hypothesis, $P_{s}=0$ for all $s\in\left\\{\beta+1,\dots,k-1\right\\},$ with
the convention that this set is presented in increasing order, so it is empty
when $\beta=k-1$. Hence, we can write

$T_{P}\left(f_{m}\right)=\sum_{j=-1}^{\infty}\sum_{s_{0}=0}^{\beta}\sum_{s_{1}=0}^{\beta}\cdots\sum_{s_{j}=0}^{\beta}TP_{s_{j}}\cdots
TP_{s_{0}}Tf_{m}.$

In order to show that the sum in (24) defines an entire function it suffices
to prove that

$G=\sum_{j=-1}^{\infty}\sum_{s_{0}=0}^{\beta}\sum_{s_{1}=0}^{\beta}\cdots\sum_{s_{j}=0}^{\beta}{\displaystyle\sum_{m=0}^{\infty}}TP_{s_{j}}\cdots
TP_{s_{0}}Tf_{m}$

does so. Then the sum can be reordered and shown to be equal to $g.$ Next we
collect all summands having degree $M\geq 0.$ The requirement

$\deg TP_{s_{j}}\cdots
TP_{s_{0}}Tf_{m}=s_{0}+\cdots+s_{j}+m-k\left(j+2\right)=M$

means that $m=M+k\left(j+2\right)-\left(s_{0}+\cdots+s_{j}\right),$ and
therefore we consider the sum

(26)
$G_{M}(z):=\sum_{j=-1}^{\infty}\sum_{s_{0}=0}^{\beta}\sum_{s_{1}=0}^{\beta}\cdots\sum_{s_{j}=0}^{\beta}TP_{s_{j}}\cdots
TP_{s_{0}}Tf_{M+k\left(j+2\right)-\left(s_{0}+\cdots+s_{j}\right)}(z).$

Next we show that $G_{M}$ converges absolutely everywhere. Note that while
$G_{M}$ will turn out to be a homogeneous polynomial of degree $M$,
contributions to it may come from infinitely many values of $j$, so a proof of
convergence is required.

By Lemma 15, for any complex vector $z\in\mathbb{C}^{d}$ and any homogeneous
polynomial $h_{M}$ of degree $M$ the following estimate holds:

$\left|h_{M}\left(z\right)\right|\leq\frac{1}{\sqrt{M!}}\left|z\right|^{M}\left\|h_{M}\right\|_{a}.$

Thus we have

(27) $\left|TP_{s_{j}}\cdots
TP_{s_{0}}Tf_{M+k\left(j+2\right)-\left(s_{0}+\cdots+s_{j}\right)}\left(z\right)\right|$
(28) $\leq\frac{\left|z\right|^{M}}{\sqrt{M!}}\left\|TP_{s_{j}}\cdots
TP_{s_{0}}Tf_{M+k\left(j+2\right)-\left(s_{0}+\cdots+s_{j}\right)}\right\|_{a}.$

Since

$\left\|Tf_{m}\right\|_{a}\leq\frac{C}{m^{\tau/2}}\left\|f_{m}\right\|_{a}$

and $P_{s_{j}}\cdots
TP_{s_{0}}Tf_{M+k\left(j+2\right)-\left(s_{0}+\cdots+s_{j}\right)}$ has degree
$M+k$, it follows that

(29) $S_{j,M}:=\left\|TP_{s_{j}}\cdots
TP_{s_{0}}Tf_{M+k\left(j+2\right)-\left(s_{0}+\cdots+s_{j}\right)}\right\|_{a}$
(30) $\leq\frac{C}{\left(M+k\right)^{\tau/2}}\left\|P_{s_{j}}\cdots
TP_{s_{0}}Tf_{M+k\left(j+2\right)-\left(s_{0}+\cdots+s_{j}\right)}\right\|_{a}$

(note that $S_{j,M}$ depends also on $\left(s_{0}+\cdots+s_{j}\right)$, but we
omit this fact from the notation). Now

$TP_{s_{j-1}}\cdots
TP_{s_{0}}Tf_{M+k\left(j+2\right)-\left(s_{0}+\cdots+s_{j}\right)}$

has degree $M+k-s_{j},$ so using the abbreviation

$D_{P}:={\displaystyle\sum_{\left|\alpha\right|=k}}\left|c_{\alpha}\right|\sqrt{\alpha!},$

and, to reduce the number of subindices, writing $D_{s_{j}}:=D_{P_{s_{j}}}$,
from Lemma 13 we get

$\left\|P_{s_{j}}f_{m}\right\|_{a}\leq
D_{s_{j}}\left\|f_{m}\right\|_{a}\left(m+1\right)^{s_{j}/2}.$

Hence we see that

$S_{j,M}\leq
CD_{s_{j}}\frac{\left(M+k-s_{j}+1\right)^{s_{j}/2}}{\left(M+k\right)^{\tau/2}}\left\|TP_{s_{j-1}}\cdots
TP_{s_{0}}Tf_{M+k\left(j+2\right)-\left(s_{0}+\cdots+s_{j}\right)}\right\|_{a}.$

Proceed inductively to obtain positive numbers $A_{j},\dots,A_{0}$ and
$B_{j},\dots,B_{-1}$ such that

$S_{j,M}\leq\frac{C^{j+1}D_{s_{j}}\cdots
D_{s_{0}}A_{j}^{s_{j}/2}A_{j-1}^{s_{j-1}/2}\cdots
A_{0}^{s_{0}/2}\left\|f_{M+k\left(j+2\right)-\left(s_{0}+\cdots+s_{j}\right)}\right\|_{a}}{B_{j}^{\tau/2}B_{j-1}^{\tau/2}\cdots
B_{-1}^{\tau/2}},$

where $B_{j}=M+k$, $B_{n-1}=B_{n}+k-s_{n}$ for $n=j,j-1,\dots,0$, and

$A_{n}=B_{n}-s_{n}+1=B_{n-1}-k+1\text{ for }n=j,j-1,\dots,0.$

Thus $B_{j-1}=B_{j}+k-s_{j}=M+2k-s_{j}$, and in general we have

$B_{j-n}=M+\left(n+1\right)k-\left(s_{j}+s_{j-1}+\cdots+s_{j-n+1}\right);$

note that the largest term is the last one:

$B_{-1}=M+\left(j+2\right)k-\left(s_{j}+s_{j-1}+\cdots+s_{0}\right)=m.$

Now $A_{n}<B_{n-1}$, so $A_{n}^{s_{n}}\leq B_{n-1}^{s_{n}}$ for all $s_{n}$,
where $0\leq s_{n}\leq\beta$. It follows that

$S_{j,M}\leq C^{j+1}D_{s_{j}}\cdots D_{s_{0}}B_{j-1}^{(s_{j}-\tau)/2}\cdots
B_{-1}^{(s_{0}-\tau)/2}\left\|f_{M+k\left(j+2\right)-\sum_{n=0}^{j}s_{n}}\right\|_{a}$
$\leq C^{j+1}D_{s_{j}}\cdots
D_{s_{0}}B_{-1}^{2^{-1}\sum_{n=0}^{j}(s_{n}-\tau)}\left\|f_{M+k\left(j+2\right)-\sum_{n=0}^{j}s_{n}}\right\|_{a}.$

By Lemma 17, there is a dimensional constant $C_{d}$ such that

$\left\|f_{m}\right\|_{a}\leq C_{d}\
\sqrt{\left(m+d-1\right)!}\max_{\eta\in\mathbb{S}^{2d-1}}\left|f_{m}\left(\eta\right)\right|,$

so we have

$\left\|f_{M+k\left(j+2\right)-\sum_{n=0}^{j}s_{n}}\right\|_{a}\leq C_{d}\
\sqrt{\left(B_{-1}+d-1\right)!}\max_{\eta\in\mathbb{S}^{2d-1}}\left|f_{B_{-1}}\left(\eta\right)\right|.$

Note that for every $M$ sufficiently large and all $j$, or for every $j$
sufficiently large and all $M$,

$\frac{\left(B_{-1}+d-1\right)!}{M!}\leq\left(B_{-1}+d-1\right)^{B_{-1}+d-1-M}$
$\leq\left(B_{-1}\right)^{B_{-1}+d-1-M}\left(1+\frac{d-1}{B_{-1}}\right)^{B_{-1}+d-1-M}$
$\leq e^{d^{2}}\left(B_{-1}\right)^{B_{-1}+d-1-M},$

so

$\frac{B_{-1}^{2^{-1}\sum_{n=0}^{j}(s_{n}-\tau)}\sqrt{\left(B_{-1}+d-1\right)!}}{\sqrt{M!}}$
$\leq
B_{-1}^{2^{-1}\sum_{n=0}^{j}(s_{n}-\tau)}e^{d^{2}}B_{-1}^{2^{-1}\left(M+\left(j+2\right)k-\sum_{n=0}^{j}s_{n}+d-1-M\right)}$
$=e^{d^{2}}B_{-1}^{2^{-1}\left(k+d-1+\left(j+1\right)(k-\tau)\right)},$

from whence it follows that

$\frac{S_{j,M}}{\sqrt{M!}}\leq e^{d^{2}}C_{d}\ C^{j+1}D_{s_{j}}\cdots
D_{s_{0}}B_{-1}^{2^{-1}(k+d-1+(j+1)(k-\tau))}\max_{\eta\in\mathbb{S}^{2d-1}}\left|f_{B_{-1}}\left(\eta\right)\right|.$

Since $f$ has order $\rho$, the bound (22) entails that for every
$\varepsilon>0$ sufficiently small there exists a constant $A_{\varepsilon}$
such that

(31)
$\max_{\eta\in\mathbb{S}^{2d-1}}\left|f_{m}\left(\eta\right)\right|\leq\frac{A_{\varepsilon}}{m^{\frac{m}{\rho+\varepsilon}}}$

for all natural numbers $m.$ Using (31) and then replacing
$B_{-1}^{-\frac{B_{-1}}{\rho+\varepsilon}}$ with the larger quantity
$B_{-1}^{-\frac{M+k+(j+1)(k-\beta))}{\rho+\varepsilon}}$, we get

(32) $\frac{S_{j,M}}{\sqrt{M!}}\leq e^{d^{2}}C_{d}A_{\varepsilon}\
C^{j+1}D_{s_{j}}\cdots
D_{s_{0}}B_{-1}^{\frac{(k+d-1+(j+1)(k-\tau))}{2}}B_{-1}^{-\frac{B_{-1}}{\rho+\varepsilon}}$
(33) $\leq e^{d^{2}}C_{d}A_{\varepsilon}\ C^{j+1}D_{s_{j}}\cdots
D_{s_{0}}B_{-1}^{\frac{(j+1)(k-\tau)}{2}}B_{-1}^{-\frac{(j+1)(k-\beta)}{\rho+\varepsilon}}B_{-1}^{\frac{(k+d-1)}{2}}B_{-1}^{-\frac{M+k}{\rho+\varepsilon}}.$

From the hypothesis $\rho<\frac{2(k-\beta)}{k-\tau}$, by selecting
$\varepsilon$ small enough we conclude that

$\frac{k-\tau}{2}<\frac{k-\beta}{\rho+\varepsilon}.$

Thus, replacing $B_{-1}$ in
$B_{-1}^{\frac{k-\tau}{2}-\frac{k-\beta}{\rho+\varepsilon}}$ by something
smaller, namely

$M+k+(j+1)(k-\beta),$

we get a larger quantity in (32), to wit,

(34) $\frac{S_{j,M}}{\sqrt{M!}}\leq e^{d^{2}}C_{d}A_{\varepsilon}\
C^{j+1}D_{s_{j}}\cdots
D_{s_{0}}\left(\left(M+k+(j+1)(k-\beta)\right)^{\frac{k-\tau}{2}-\frac{k-\beta}{\rho+\varepsilon}}\right)^{j+1}$
(35) $\times
B_{-1}^{\frac{(k+d-1)}{2}}B_{-1}^{-\frac{M+k}{\rho+\varepsilon}}.$

We can make the same substitution in
$B_{-1}^{\frac{(k+d-1)}{2}}B_{-1}^{-\frac{M+k}{\rho+\varepsilon}}$ when
$\frac{(k+d-1)}{2}\leq\frac{M+k}{\rho+\varepsilon},$ and use $B_{-1}\leq
M+(j+2)k$ when $\frac{(k+d-1)}{2}>\frac{M+k}{\rho+\varepsilon}.$ In either
case, the $j+1$-root of the corresponding quantity approaches 1 as
$j\to\infty$, so for $M\in\mathbb{N}_{0}$ fixed and $j$ sufficiently large we
obtain

(36) $B_{-1}^{\frac{(k+d-1)}{2}}B_{-1}^{-\frac{M+k}{\rho+\varepsilon}}\leq
2^{j+1}$

for all choices of $s_{0},\dots,s_{j}$ between $0$ and $\beta$.

After these preliminary bounds, we prove that the sum in (26) is well defined
and converges absolutely for every $z\in\mathbb{C}^{d}\setminus\\{0\\}$ and
every natural number $M$. So fix $M$ and set $z=r\eta$, where $r>0$ and
$\eta\in\mathbb{S}^{2d-1}$. Let us write

$G_{M}^{\left(j\right)}\left(r\eta\right):=r^{M}\sum_{s_{0}=0}^{\beta}\sum_{s_{1}=0}^{\beta}\cdots\sum_{s_{j}=0}^{\beta}\left|TP_{s_{j}}....TP_{s_{0}}Tf_{M+k\left(j+2\right)-\left(s_{0}+\cdots+s_{j}\right)}\left(\eta\right)\right|,$

and $\widetilde{D}:=D_{s_{0}}+\cdots+D_{s_{\beta}}.$ Note that we have

$\sum_{s_{0}=0}^{\beta}\sum_{s_{1}=0}^{\beta}\cdots\sum_{s_{j}=0}^{\beta}D_{s_{j}}\cdots
D_{s_{0}}=\left(D_{s_{0}}+\cdots+D_{s_{\beta}}\right)^{j+1}=\widetilde{D}^{j+1}.$

Putting together (34)-(35) with the bound from (36), and using (27)-(28), we
obtain for sufficiently large values of $j$,

$G_{M}^{\left(j\right)}\left(r\eta\right)\leq
r^{M}e^{d^{2}}C_{d}A_{\varepsilon}\cdot\left((M+k+(j+1)(k-\beta))^{\frac{k-\tau}{2}-\frac{k-\beta}{\rho+\varepsilon}}2C\widetilde{D}\right)^{j+1}$
$\leq\left(((M+k+(j+1)(k-\beta))^{\frac{k-\tau}{2}-\frac{k-\beta}{\rho+\varepsilon}}3C\widetilde{D}\right)^{j+1}.$

Hence, we can find a natural number $j_{0}$ such that for all $j\geq j_{0}$
the last inequality is satisfied and furthermore

$(M+k+(j+1)(k-\beta))^{\frac{k-\tau}{2}-\frac{k-\beta}{\rho+\varepsilon}}3C\widetilde{D}\leq\frac{1}{2}.$

It follows that ${\displaystyle\sum_{j=0}^{\infty}}G_{M}^{\left(j\right)}$
converges on $\mathbb{C}^{d}$, so $G_{M}$ is a well-defined homogeneous
polynomial for each $M\in\mathbb{N}_{0}.$

Next we show that for sufficiently large values of $M$, the polynomial $G_{M}$
on the unit sphere satisfies the bounds given in (22). Recall that

$|G_{M}(\eta)|\leq{\displaystyle\sum_{j=0}^{\infty}}G_{M}^{\left(j\right)}(\eta).$

We want to estimate $G_{M}^{\left(j\right)}(\eta)$ for every
$j\in\mathbb{N}_{0}$, under the assumption that $M$ is “large”. It follows
from our previous discussion that whenever
$\frac{(k+d-1)}{2}<\frac{M+k}{\rho+\varepsilon},$ and $M$ is sufficiently
large, we have

$G_{M}^{\left(j\right)}(\eta)\leq\left(\frac{\left(e^{d^{2}}C_{d}A_{\varepsilon}\right)^{1/(j+1)}C\widetilde{D}}{\left(M+k+\left(k-\beta\right)\left(j+1\right)\right)^{\frac{k-\beta}{\rho+\varepsilon}-\frac{k-\tau}{2}}}\right)^{j+1}$
$\times\left(M+k+\left(k-\beta\right)\left(j+1\right)\right)^{\frac{(k+d-1)}{2}-\frac{M+k}{\rho+\varepsilon}}.$

We may assume that $e^{d^{2}}C_{d}A_{\varepsilon}\geq 1$ (otherwise we remove
the term from the inequality). Then

$G_{M}^{\left(j\right)}(\eta)\leq\left(\frac{\left(e^{d^{2}}C_{d}A_{\varepsilon}\right)C\widetilde{D}}{\left(M+k\right)^{\frac{k-\beta}{\rho+\varepsilon}-\frac{k-\tau}{2}}}\right)^{j+1}\left(M+k\right)^{\frac{(k+d-1)}{2}-\frac{M+k}{\rho+\varepsilon}}.$

Thus we can select $M_{0}$ so large that for all $M\geq M_{0}$,

$\frac{\left(e^{d^{2}}C_{d}A_{\varepsilon}\right)C\widetilde{D}}{\left(M+k\right)^{\frac{k-\beta}{\rho+\varepsilon}-\frac{k-\tau}{2}}}\leq\frac{1}{2}$

and additionally, so that for all $M\geq M_{0}$ the last inequality below is
satisfied:

$|G_{M}(\eta)|\leq{\displaystyle\sum_{j=0}^{\infty}}G_{M}^{\left(j\right)}(\eta)\leq
2\left(M+k\right)^{\frac{(k+d-1)}{2}-\frac{M+k}{\rho+\varepsilon}}\leq
M^{-\frac{M}{\rho+2\varepsilon}}.$

∎

##  7\. Fischer decompositions on certain Banach spaces of entire functions

As in [9], we use $\Lambda$ to denote the set of all decreasing sequences
$\lambda=\left(\lambda_{m}\right)_{m\in\mathbb{N}_{0}}$ of positive numbers
$\lambda_{m}\leq 1$ which converge to $0.$ Then $B_{\lambda}$ is defined as
the space of all entire functions $f$ on $\mathbb{C}^{n}$ such that the
homogeneous expansion $f=\sum_{m=0}^{\infty}f_{m}$ satisfies

(37)
$\frac{\left\|f_{m}\right\|_{a}}{m^{m/2}\left(\lambda_{m}\right)^{m}}\rightarrow
0.$

One can show that $B_{\lambda}$ is a Banach space with respect to the norm

(38)
$\left\|f\right\|_{\lambda}:=\sup_{m\in\mathbb{N}_{0}}\frac{\left\|f_{m}\right\|_{a}}{m^{m/2}\lambda_{m}^{m}}.$

Actually, the assumption $\lambda_{m}\leq 1$ is not made in [9], but it will
be convenient for us later on, so we include it into the definition; clearly
replacing a decreasing sequence
$\lambda=\left(\lambda_{m}\right)_{m\in\mathbb{N}_{0}}$ with the decreasing
sequence $\lambda\wedge 1:=\left(\lambda_{m}\wedge
1\right)_{m\in\mathbb{N}_{0}}$ leads to the same space $B_{\lambda}$, with a
comparable norm.

Motivated by an anonymous referee’s comments, we show that certain
modifications of the preceding arguments allow us to partially deal with a
question formulated in [9, Remark 5.2]: can the _uniqueness_ assumption in [9,
Theorem 3] be omitted? The answer is yes. We will assume that the sequence
$\lambda$ converges to 0 sufficiently fast, and more precisely, that condition
(39) below is satisfied. This condition adapts hypothesis (4.10) of [9, Lemma
12] to the more general setting considered here, where $\tau$ can be larger
than 1 and $\beta$ larger than 0.

######  Theorem 18.

Let $P_{k}$ be a homogeneous polynomial of degree $k>0$ on $\mathbb{C}^{d}$,
and let us write $T:=T_{P_{k}}$. Suppose that there exist a $C>0$ and a
$\tau\in\\{0,\dots,k\\}$ such that for every $m>0$ and every homogeneous
polynomial $f_{m}$ of degree $m,$ the following inequality holds:

$\left\|Tf_{m}\right\|_{a}\leq\frac{C}{m^{\tau/2}}\left\|f_{m}\right\|_{a}.$

Assume that for $0\leq j<k$ the polynomials $P_{j}\left(z\right)$ are
homogeneous of degree $j$, and for some $\beta<k$ and every $j$ with
$\beta<j<k$ we have $P_{j}=0$. Let $\lambda\in\Lambda$ be such that

(39)
$\lim_{m\rightarrow\infty}m^{\frac{(k-\tau)}{2}}\lambda_{m}^{(k-\beta)}=0.$

Then for every $f\in B_{\lambda}$, there exist $q\in B_{\lambda}$ and $r$
entire such that

$f=\left(P_{k}-P_{\beta}-\cdots-P_{0}\right)q+r\text{ and
}P_{k}^{\ast}\left(D\right)r=0.$

It is not clear to us whether $\left(P_{k}-P_{\beta}-\cdots-P_{0}\right)q$,
and hence $r$, must be in $B_{\lambda}$. The question whether $r\in
B_{\lambda}$ seems to be a delicate one: it is not neccesarily true that the
product of a function in $B_{\lambda}$ with a polynomial must again be in
$B_{\lambda}$.

###### Proof.

We proceed as in the proof of Theorem 2: our strategy is to show that

(40) $g:={\displaystyle\sum_{m=0}^{\infty}}T_{P}\left(f_{m}\right)$

defines an entire function $g:\mathbb{C}^{d}\rightarrow\mathbb{C}$ which
belongs to $B_{\lambda},$ by writing
$g\left(z\right)={\displaystyle\sum_{M=0}^{\infty}}G_{M}\left(z\right)$, where
each $G_{M}$ is a homogeneous polynomial of degree $M$, and then proving that
$\|g\|_{\lambda}<\infty$ by showing that

$\frac{\left\|G_{M}\right\|_{a}}{M^{M/2}\left(\lambda_{M}\right)^{M}}\rightarrow
0.$

It has been noted in the proof of Theorem 2 that

$T_{P}\left(f_{m}\right)=\sum_{j=-1}^{\infty}\sum_{s_{0}=0}^{\beta}\sum_{s_{1}=0}^{\beta}\cdots\sum_{s_{j}=0}^{\beta}TP_{s_{j}}\cdots
TP_{s_{0}}Tf_{m}.$

In order to show that the sum in (40) defines an entire function it suffices
to prove that

$G=\sum_{j=-1}^{\infty}\sum_{s_{0}=0}^{\beta}\sum_{s_{1}=0}^{\beta}\cdots\sum_{s_{j}=0}^{\beta}{\displaystyle\sum_{m=0}^{\infty}}TP_{s_{j}}\cdots
TP_{s_{0}}Tf_{m}$

does so. Then the sum can be reordered and shown to be equal to $g.$ As before
(cf. 26) we collect all summands having degree $M\geq 0$ and we consider the
sum

(41)
$G_{M}(z):=\sum_{j=-1}^{\infty}\sum_{s_{0}=0}^{\beta}\sum_{s_{1}=0}^{\beta}\cdots\sum_{s_{j}=0}^{\beta}TP_{s_{j}}\cdots
TP_{s_{0}}Tf_{M+k\left(j+2\right)-\left(s_{0}+\cdots+s_{j}\right)}(z).$

Next we show that $G_{M}$ converges absolutely everywhere. As in the proof of
Theorem 2 and with the same notation (cf. 29) setting

$S_{j,M}:=\left\|TP_{s_{j}}\cdots
TP_{s_{0}}Tf_{M+k\left(j+2\right)-\left(s_{0}+\cdots+s_{j}\right)}\right\|_{a}$

we see that

(42)
$\left|G_{M}\left(z\right)\right|\leq\frac{\left|z\right|^{M}}{\sqrt{M!}}\sum_{j=-1}^{\infty}\sum_{s_{0}=0}^{\beta}\sum_{s_{1}=0}^{\beta}\cdots\sum_{s_{j}=0}^{\beta}S_{j,M}.$

Thus (41) converges absolutely to a homogeneous polynomial of degree $M$ if

(43)
$\sum_{j=-1}^{\infty}\sum_{s_{0}=0}^{\beta}\sum_{s_{1}=0}^{\beta}\cdots\sum_{s_{j}=0}^{\beta}S_{j,M}.$

converges. As in the proof of Theorem 2 (and with the same notation) we have

$S_{j,M}\leq C^{j+1}D_{s_{j}}\cdots D_{s_{0}}\
B_{-1}^{2^{-1}\sum_{n=0}^{j}(s_{n}-\tau)}\left\|f_{M+k\left(j+2\right)-\sum_{n=0}^{j}s_{n}}\right\|_{a}.$

Recall that

$B_{-1}=M+k\left(j+2\right)-\sum_{n=0}^{j}s_{n}\geq
M+k+\left(k-\beta\right)\left(j+1\right).$

Let us fix $M\geq 0$ and write
$\widetilde{D}:=D_{s_{0}}+\cdots+D_{s_{\beta}}$. We want to prove the
convergence of (43). Now

$S_{j,M}\leq C^{j+1}D_{s_{j}}\cdots D_{s_{0}}\
B_{-1}^{2^{-1}\sum_{n=0}^{j}(s_{n}-\tau)}\
\frac{\left\|f_{B_{-1}}\right\|_{a}}{B_{-1}^{B_{-1}/2}\lambda_{B_{-1}}^{B_{-1}}}\
B_{-1}^{B_{-1}/2}\lambda_{B_{-1}}^{B_{-1}}.$ $=C^{j+1}D_{s_{j}}\cdots
D_{s_{0}}\
\frac{\left\|f_{B_{-1}}\right\|_{a}}{B_{-1}^{B_{-1}/2}\lambda_{B_{-1}}^{B_{-1}}}\
B_{-1}^{2^{-1}(M+k+\left(k-\tau\right)\left(j+1\right))}\lambda_{B_{-1}}^{B_{-1}}.$

Since $\lambda_{m}\leq 1$ for every $m$, and $f\in B_{\lambda}$, it follows
that

$\sum_{s_{0}=0}^{\beta}\sum_{s_{1}=0}^{\beta}\cdots\sum_{s_{j}=0}^{\beta}S_{j,M}\leq
C^{j+1}\widetilde{D}^{j+1}\ \left\|f\right\|_{\lambda}\
B_{-1}^{2^{-1}(M+k+\left(k-\tau\right)\left(j+1\right))}\lambda_{B_{-1}}^{M+k+\left(k-\beta\right)\left(j+1\right)}.$

Recalling the hypothesis
$\lim_{m\rightarrow\infty}m^{\frac{(k-\tau)}{2}}\lambda_{m}^{(k-\beta)}=0$,
cf. (39), we see that

$\lim_{j\rightarrow\infty}B_{-1}^{2^{-1}\left(\frac{M+k}{j+1}+\left(k-\tau\right)\right)}\lambda_{B_{-1}}^{\frac{M+k}{j+1}+\left(k-\beta\right)}=0,$

since $\lim_{j\rightarrow\infty}B_{-1}^{\frac{1}{j+1}}=1$ and
$\sup_{j}\lambda_{B_{-1}}^{\frac{1}{j+1}}\leq 1$. Choosing $J\gg 1$ so that
for every $j\geq J$ we have

$C\widetilde{D}\
B_{-1}^{2^{-1}\left(\frac{M+k}{j+1}+\left(k-\tau\right)\right)}\lambda_{B_{-1}}^{\frac{M+k}{j+1}+\left(k-\beta\right)}<1/2,$

it becomes clear that the series (43) converges, and hence, that each $G_{M}$
is a homogeneous polynomial of degree $M$. Once we know $G_{M}$ is a
polynomial, we can estimate its apolar norm using (41) and the triangle
inequality:

(44)
$\left\|G_{M}\right\|_{a}\leq\sum_{j=-1}^{\infty}\sum_{s_{0}=0}^{\beta}\sum_{s_{1}=0}^{\beta}\cdots\sum_{s_{j}=0}^{\beta}S_{j,M}.$

The next step consists in showing that

$\frac{\left\|G_{M}\right\|_{a}}{M^{M/2}\left(\lambda_{M}\right)^{M}}\rightarrow
0.$

Arguing as in the proof of Theorem 2 we obtain

$S_{j,M}\leq C^{j+1}D_{s_{j}}\cdots
D_{s_{0}}\frac{B_{-1}^{2^{-1}\sum_{n=0}^{j}(s_{n}-\tau)}}{B_{j}^{\tau/2}}\left\|f_{M+k\left(j+2\right)-\sum_{n=0}^{j}s_{n}}\right\|_{a},$

the only difference being the additional factor $1/B_{j}^{\tau/2}$, which
previously was estimated by $1$. This factor becomes important here, since
$M\to\infty$ in this part of the argument, while previously $M$ was fixed.

It follows that

$\frac{S_{j,M}}{M^{M/2}\lambda_{M}^{M}}\leq C^{j+1}D_{s_{j}}\cdots
D_{s_{0}}\frac{B_{-1}^{2^{-1}\sum_{n=0}^{j}(s_{n}-\tau)}}{B_{j}^{\tau/2}}\frac{\left\|f_{B_{-1}}\right\|_{a}}{M^{M/2}\lambda_{M}^{M}}\frac{B_{-1}^{B_{-1}/2}\lambda_{B_{-1}}^{B_{-1}}}{B_{-1}^{B_{-1}/2}\lambda_{B_{-1}}^{B_{-1}}}.$

As before,

$B_{-1}^{2^{-1}\sum_{n=0}^{j}(s_{n}-\tau)}B_{-1}^{B_{-1}/2}=B_{-1}^{M/2+k/2+\left(k-\tau\right)\left(j+1\right)/2},$

so using $B_{j}=M+k$ and

$\frac{\left\|f_{B_{-1}}\right\|_{a}}{B_{-1}^{B_{-1}/2}\lambda_{B_{-1}}^{B_{-1}}}\leq\left\|f\right\|_{\lambda},$

we get

$\frac{S_{j,M}}{M^{M/2}\lambda_{M}^{M}}\leq C^{j+1}D_{s_{j}}\cdots
D_{s_{0}}\frac{B_{-1}^{M/2+k/2+\left(k-\tau\right)\left(j+1\right)/2}}{\left(M+k\right)^{\tau/2}M^{M/2}}\frac{\lambda_{B_{-1}}^{B_{-1}}}{\lambda_{M}^{M}}\left\|f\right\|_{\lambda}.$

For every $M\geq 0$, since $B_{-1}\geq M$ and all terms in $\lambda$ are
bounded by 1, we have $\lambda_{B_{-1}}\leq\lambda_{M}$ and

$\frac{\lambda_{B_{-1}}^{B_{-1}}}{\lambda_{M}^{M}}=\left(\frac{\lambda_{B_{-1}}}{\lambda_{M}}\right)^{M}\left(\lambda_{B_{-1}}\right)^{k\left(j+2\right)-\sum_{n=0}^{j}s_{n}}\leq\left(\lambda_{B_{-1}}\right)^{k+\left(k-\beta\right)\left(j+1\right)}.$

Furthermore, if $M\geq 1$ then

$\left(\frac{B_{-1}}{M}\right)^{M/2}\leq\left(1+\frac{k\left(j+2\right)}{M}\right)^{M/2}\leq
e^{\frac{1}{2}k\left(j+2\right)},$

so

$\frac{S_{j,M}}{M^{M/2}\lambda_{M}^{M}}\leq C^{j+1}D_{s_{j}}\cdots
D_{s_{0}}\frac{B_{-1}^{k/2+\left(k-\tau\right)\left(j+1\right)/2}}{\left(M+k\right)^{\tau/2}}\
e^{\frac{1}{2}k\left(j+2\right)}\left(\lambda_{B_{-1}}\right)^{k+\left(k-\beta\right)\left(j+1\right)}\left\|f\right\|_{\lambda}.$

Next we observe that

$\displaystyle\frac{B_{-1}^{\tau/2}}{\left(M+k\right)^{\tau/2}}$
$\displaystyle\leq$
$\displaystyle\left(\frac{M+k\left(j+2\right)}{M+k}\right)^{\tau/2}=\left(1+\frac{k\left(j+1\right)}{M+k}\right)^{\tau/2}$
$\displaystyle\leq$
$\displaystyle\left(1+\left(j+1\right)\right)^{\tau/2}\leq\left(1+2^{j}\right)^{\tau/2}\leq
2^{\left(j+1\right)\tau/2}.$

Thus

$\displaystyle\frac{S_{j,M}}{M^{M/2}\lambda_{M}^{M}}$ $\displaystyle\leq$
$\displaystyle C^{j+1}D_{s_{j}}\cdots
D_{s_{0}}e^{\frac{1}{2}k\left(j+2\right)}\ 2^{\left(j+1\right)\tau/2}\
B_{-1}^{\left(k-\tau\right)/2+\left(k-\tau\right)\left(j+1\right)/2}\left(\lambda_{B_{-1}}\right)^{k+\left(k-\beta\right)\left(j+1\right)}\left\|f\right\|_{\lambda}$
$\displaystyle=$ $\displaystyle D_{s_{j}}\cdots
D_{s_{0}}e^{\frac{1}{2}k}\left(Ce^{\frac{1}{2}k}2^{\tau/2}B_{-1}^{\left(k-\tau\right)/2}\left(\lambda_{B_{-1}}\right)^{\left(k-\beta\right)}\right)^{j+1}B_{-1}^{\left(k-\tau\right)/2}\left(\lambda_{B_{-1}}\right)^{k}\left\|f\right\|_{\lambda}.$

Writing again $\widetilde{D}:=D_{s_{0}}+\cdots+D_{s_{\beta}}$, by (39), for
every $M\gg 1$ sufficiently large we have

$Ce^{\frac{1}{2}k}2^{\tau/2}\cdot
B_{-1}^{\frac{(k-\tau)}{2}}\lambda_{B_{-1}}^{(k-\beta)}\leq\frac{1}{2\widetilde{D}},$

from whence it follows that

$\frac{S_{j,M}}{M^{M/2}\lambda_{M}^{M}}\leq D_{s_{j}}\cdots
D_{s_{0}}e^{\frac{1}{2}k}\left(\frac{1}{2\widetilde{D}}\right)^{j+1}B_{-1}^{\left(k-\tau\right)/2}\left(\lambda_{B_{-1}}\right)^{k}\left\|f\right\|_{\lambda}.$

Therefore

$\frac{\left\|G_{M}\right\|_{a}}{M^{M/2}\lambda_{M}^{M}}\leq\sum_{j=-1}^{\infty}\sum_{s_{0}=0}^{\beta}\sum_{s_{1}=0}^{\beta}\cdots\sum_{s_{j}=0}^{\beta}\frac{S_{j,M}}{M^{M/2}\lambda_{M}^{M}}\leq\left\|f\right\|_{\lambda}B_{-1}^{\left(k-\tau\right)/2}\left(\lambda_{B_{-1}}\right)^{k}e^{\frac{1}{2}k}\sum_{j=-1}^{\infty}\frac{1}{2^{j+1}}.$

Recalling (39), we see that

$\lim_{M\rightarrow\infty}B_{-1}^{2^{-1}\left(k-\tau\right)}\lambda_{B_{-1}}^{k}=0,$

and thus

$\frac{\left\|G_{M}\right\|_{a}}{M^{M/2}\lambda_{M}^{M}}\rightarrow 0.$

∎

Data availability Data sharing is not applicable to this article since no data
sets were generated or analyzed.

Declarations

Conflict of interest The authors declare that they have no competing
interests.

Acknowlegement The first named author was partially supported by Grant
PID2019-106870GB-I00 of the MICINN of Spain, by ICMAT Severo Ochoa project
CEX2019-000904-S (MICINN), and by the Madrid Government (Comunidad de Madrid -
Spain) V PRICIT (Regional Programme of Research and Technological Innovation),
2022-2024.

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