# Averaging an entire function while avoiding polynomial zeros

*Written by GPT-6.1 Sol (OpenAI), Ultra reasoning effort, October 2026. Original exposition and figures CC0.*

An entire function can be averaged on a small complex circle to recover its value at the centre. A polynomial denominator presents a second requirement: the averaging region must avoid its zeros by a uniform amount. We construct averaging densities which depend smoothly on the polynomial, remain inside a fixed ball, and satisfy both requirements. This is the missing polynomial-averaging receiver in AN02-L004, concerning regular kernels and parameter changes.

The accompanying complete proof, `polynomial-averaging-working-proof.md`, supplies PA1–PA16. The argument uses ordinary one-variable entire Taylor series, finite-dimensional compactness, explicit scalar smooth cutoffs, and Lebesgue integration with Fubini. These inputs are declared. It does not establish the full distributional topology or every prerequisite of L004.

## 1. The density and its exact requirements

Fix \(n\geq1\), \(m\geq0\), and \(\rho>0\). Write \(\mathcal P_m\) for complex polynomials in \(n\) variables of degree at most \(m\), including those of smaller actual degree. Give this finite-dimensional space the Hermitian norm

\[
 |q|_J^2=\sum_{|\alpha|\leq m}|\partial^\alpha q(0)|^2,
 \qquad
 \langle p,q\rangle_J=\sum_{|\alpha|\leq m}
 \partial^\alpha p(0)\overline{\partial^\alpha q(0)}.
\]

The inner product is linear in its first variable. We will find a nonnegative function \(\Phi(q,z)\), smooth in the real coefficient coordinates for \(q\neq0\) and in \(z\in\mathbb C^n\), with one compact support \(K\subset\{0<|z|<\rho\}\) and one constant \(c>0\) such that

\[
 \Phi(aq,z)=\Phi(q,z)\quad(a\neq0),\qquad
 \int H(z)\Phi(q,z)\,d\lambda(z)=H(0),
\]
\[
 \operatorname{supp}\Phi(q,\cdot)\subset K,
 \qquad |q(z)|\geq c|q|_J\quad\hbox{if }\Phi(q,z)>0.
\]

Here \(a\) may be any nonzero complex number, and \(H\) may be any entire holomorphic function. Setting \(H=1\) shows that each density has integral one. The constant and compact set are uniform across the whole fixed coefficient space. The density itself depends on \(q\).

## 2. Why one simultaneous rotation is sufficient

Take any normalized compact smooth density \(\psi\geq0\) satisfying \(\psi(e^{it}z)=\psi(z)\). The map \(z\mapsto e^{it}z\) preserves real Lebesgue measure. Averaging this change of variables gives

\[
 \int H(z)\psi(z)\,d\lambda(z)
 =\int\psi(z)\left(\frac1{2\pi}\int_0^{2\pi}
                  H(e^{it}z)\,dt\right)d\lambda(z).
\]

For fixed \(z\), the function \(w\mapsto H(wz)\) is entire in one variable. Its Taylor series converges uniformly on \(|w|=1\). Every positive power of \(e^{it}\) has mean zero, while its constant term is \(H(0)\). The inner integral is therefore \(H(0)\). Compact support justifies Fubini and termwise integration. This proves the averaging identity without requiring separate rotations of the individual coordinates.

One may localize such a density near the entire phase orbit of a unit vector \(u\). On a radial annulus away from zero use

\[
 \psi(z)=Z^{-1}a(|z|)
 b\left(1-\frac{|\langle z,u\rangle|^2}{|z|^2}\right).
\]

The radial bump \(a\) has a nonzero compact support inside \((0,\rho)\). The angular bump \(b\) is positive near zero and vanishes for sufficiently large positive arguments. The expression is extended by zero outside the annulus. It is globally smooth because its radial support stays away from zero. The normalization \(Z\) is finite and positive. For \(n=1\), the angular argument is identically zero; for larger \(n\), it describes an open neighborhood of the complex line through \(u\).

## 3. Finding a circle without zeros

Normalize \(|q_0|_J=1\). Some unit \(u\) makes \(t\mapsto q_0(tu)\) a nonzero polynomial. Otherwise every complex line would have zero restriction, forcing \(q_0\) to vanish everywhere. A nonzero polynomial in one variable has finitely many roots. Choose

\[
 \rho/3<r<2\rho/3
\]

different from the modulus of every root of this restriction. Then the complete circle \(re^{it}u\), for \(0\leq t\leq2\pi\), avoids the zeros. Its compactness gives

\[
 d_0=\min_t|q_0(re^{it}u)|>0.
\]

Uniform continuity gives a narrow closed annular and angular neighborhood on which \(|q_0|\geq d_0/2\). Choose the density from section 2 inside that neighborhood. It still averages every entire function correctly because the neighborhood contains the full phase orbit and the density is invariant under simultaneous rotation.

The distance formula

\[
 \min_{|a|=1}|v-au|^2=2-2|\langle v,u\rangle|
 \quad(|u|=|v|=1)
\]

explains precisely why a small angular parameter places a vector near this orbit. The construction uses a circle for each local coefficient neighborhood. It does not attempt to label roots continuously as coefficients change.

## 4. Making the construction smooth for every polynomial

Finite Taylor expansion and Cauchy–Schwarz yield, on \(|z|\leq\rho\),

\[
 |p(z)|\leq B_\rho|p|_J,
 \qquad B_\rho^2=\sum_{|\alpha|\leq m}
              \frac{\rho^{2|\alpha|}}{(\alpha!)^2}.
\]

Consequently a polynomial close to a unit complex multiple of \(q_0\) remains nonzero on the same density support. On the coefficient unit sphere define

\[
 s_{q_0}(p)=1-|\langle p,q_0\rangle_J|^2.
\]

If \(s_{q_0}(p)<\delta\), the distance to the phase orbit of \(q_0\) is at most \(\sqrt{2\delta}\). Choose \(\delta\) with \(B_\rho\sqrt{2\delta}<d_0/4\). The lower bound on the support is then \(|p(z)|\geq d_0/4\).

The smaller neighborhoods \(s_{q_0}<\delta/2\) cover the compact coefficient unit sphere. Select a finite subcover indexed by \(j\), with centres \(q_j\), bounds \(d_j\), parameters \(\delta_j\), and normalized densities \(\psi_j\). For a smooth scalar cutoff \(\beta\), positive on \([0,1/2]\) and zero on \([1,\infty)\), set

\[
 b_j(q)=\beta\left(\frac{1-|\langle q,q_j\rangle_J|^2/|q|_J^2}
                        {\delta_j}\right),
 \qquad h_j(q)=\frac{b_j(q)}{\sum_\ell b_\ell(q)}.
\]

The denominator is positive because of the smaller subcover. These weights are smooth for \(q\neq0\), nonnegative, sum to one, and are invariant under every nonzero complex scaling of \(q\). Define

\[
 \Phi(q,z)=\sum_jh_j(q)\psi_j(z),
 \qquad K=\bigcup_j\operatorname{supp}\psi_j,
 \qquad c=\min_jd_j/4.
\]

This is a finite smooth formula. If \(\Phi(q,z)>0\), some positive term has both \(h_j(q)>0\) and \(\psi_j(z)>0\). That term supplies \(|q(z)|\geq(d_j/4)|q|_J\geq c|q|_J\). Each density supplies the entire-function averaging identity, so their weighted sum does too. The finite union of supports is compact inside the punctured ball. Constant polynomials and families with decreasing actual degree belong to the same fixed coefficient sphere and need no separate argument.

## 5. Three examples

### A quadratic in one complex variable

Let \(q(z)=z^2-1/16\) and \(\rho=1\). Its zeros are \(\pm1/4\). Choose a radial density whose compact support lies strictly inside \(1/3<|z|<5/12\). On the entire closed annulus,

\[
 |q(z)|\geq|z|^2-1/16\geq7/144,
 \qquad |q|_J=\sqrt{1025}/16.
\]

Thus this example has the explicit normalized lower bound \(c=7/(9\sqrt{1025})\). On the circle \(|z|=3/8\), the polynomial image is

\[
 q((3/8)e^{it})=-1/16+(9/64)e^{2it}.
\]

It is a circle centred at \(-1/16\) with radius \(9/64\), traversed twice as \(t\) runs once around. Its distance from the origin is at least \(5/64\). The image surrounds the origin but never passes through it. Avoiding zero is a pointwise condition on the circle; it does not require zero to lie outside the region enclosed by its image.

![A zero-free annular averaging region and its quadratic image.](figures/polynomial-averaging-annulus.png)

*Left: the exact annulus \(1/3\leq|z|\leq5/12\), the roots \(\pm1/4\), and the circle \(|z|=3/8\), inside the unit ball. The shading describes an available support region, not a particular cutoff density. Right: the exact image of that phase circle under \(z^2-1/16\), with centre, radius and minimum distance marked. The drawn geometry illustrates the algebraic bounds above and PA036-3–4; sampled points are not used to prove nonvanishing.*

### A complex line in two variables

For \(q(z_1,z_2)=z_1z_2\), choose \(u=(1,1)/\sqrt2\). Then

\[
 q(re^{it}u)=\frac{r^2}{2}e^{2it},
 \qquad |q(re^{it}u)|=r^2/2>0.
\]

A narrow angular neighborhood of the phase orbit of this \(u\) gives a suitable density. Rotating only \(z_1\) need not preserve that neighborhood. The proof needs simultaneous phase rotation, and this example shows why imposing separate coordinate invariance would discard useful localized densities.

### A family whose degree drops

For fixed \(m\geq1\), consider \(q_\varepsilon(z)=1+\varepsilon z^m\). At \(\varepsilon=0\) its actual degree drops to zero. The coefficient vector remains nonzero, so \(q_\varepsilon/|q_\varepsilon|_J\) varies smoothly through the same fixed coefficient sphere. Our weights and density remain smooth there. The argument never divides by the leading coefficient and never follows roots escaping to infinity.

## 6. Dividing the density and differentiating coefficients

Define \(A(q,z)=\Phi(q,z)/q(z)\) where the denominator is nonzero. Extend it by zero near denominator zeros. This extension is smooth: the established lower bound forces \(\Phi=0\) throughout the open set \(|q(z)|<c|q|_J\). Every division zero therefore has a whole neighborhood on which \(A=0\).

All coefficient derivatives of \(A\) have support inside the same \(K\). Their multilinear norms are bounded on the compact product \(\{|q|_J=1\}\times K\). Positive real scaling gives \(A(tq,z)=t^{-1}A(q,z)\). Differentiating this identity and rescaling the directions yields, for every \(j\geq0\),

\[
 |d^jA(q,z)[r_1,\ldots,r_j]|
 \leq C_j\frac{\prod_{\nu=1}^j|r_\nu|_J}{|q|_J^{j+1}}.
\]

These are derivatives in the real coefficient variables; no holomorphic dependence on coefficients is asserted. This is the denominator estimate required by L004 Lemma 2.1. It leaves the other distributional and topological statements of that lesson at their declared scopes.

## 7. Exercises with complete solutions

### Exercise 1: a repeated root

Let \(q(t)=(t-a)^k\), where \(k\geq1\). Show that a circle of radius \(r\neq|a|\) avoids every zero and find its minimum denominator modulus.

**Solution.** The only zero is \(a\), with multiplicity \(k\). For \(|t|=r\), the reverse triangle inequality gives \(|t-a|\geq|r-|a||\). If \(a\neq0\), equality occurs at \(t=ra/|a|\); if \(a=0\), every point has modulus \(r\). Consequently

\[
 \min_{|t|=r}|q(t)|=|r-|a||^k>0.
\]

The finite exceptional set of root moduli remains finite even when roots have multiplicity. A sufficiently small closed annular neighborhood has a positive lower bound by continuity.

### Exercise 2: why the density must depend on the polynomial

For \(n\geq1\) and \(m\geq1\), prove that no single smooth nonnegative density of integral one can satisfy the support lower bound for every nonzero \(q\in\mathcal P_m\). Explain the case \(m=0\).

**Solution.** Some point \(z_0\) has \(\psi(z_0)>0\), since the integral is one. The polynomial \(q(z)=z_1-(z_0)_1\) is nonzero, belongs to \(\mathcal P_m\), and vanishes at \(z_0\). The required inequality at that point would say \(0\geq c|q|_J>0\), a contradiction. For \(m=0\), every nonzero polynomial is constant; its modulus equals its coefficient norm. Any normalized phase-invariant compact smooth density in the punctured ball then works for every such polynomial with \(c=1\).

### Exercise 3: holomorphic and nonholomorphic averages

Compute the phase averages of \(z^\alpha\), for a multi-index \(\alpha\), and of \(|z|^2\). Explain why the second computation does not extend our theorem to all smooth functions.

**Solution.** Simultaneous rotation gives \((e^{it}z)^\alpha=e^{i|\alpha|t}z^\alpha\). Its mean is zero when \(|\alpha|>0\) and is one for the constant monomial. In contrast \(|e^{it}z|^2=|z|^2\), so its phase average equals \(|z|^2\), whereas its value at zero is zero. In fact any normalized nonnegative density supported away from zero has \(\int|z|^2\psi(z)d\lambda(z)>0\). The entire Taylor argument applies to holomorphic functions; it supplies no such identity for arbitrary smooth functions.

### Exercise 4: verify the quadratic constants

For the first example, compute the coefficient norm, the annular bound, the circle-image bound, and the normalized annular constant.

**Solution.** The derivatives at zero are \(q(0)=-1/16\), \(q'(0)=0\), and \(q''(0)=2\). Hence \(|q|_J^2=1/256+4=1025/256\). On the closed annulus the reverse triangle bound is \(1/9-1/16=7/144\). On \(|z|=3/8\), it is \(9/64-1/16=5/64\). Dividing the annular bound by \(\sqrt{1025}/16\) gives \(7/(9\sqrt{1025})\). All three bounds are exact at positive real points of the appropriate circles. A smooth radial density compactly supported in the open annulus therefore satisfies the claimed weaker closed-annulus estimate.

### Exercise 5: the projective distance

For unit vectors \(p,q\) in the coefficient inner product, prove the phase-distance formula and the estimate used when \(1-|\langle p,q\rangle_J|^2<\delta\).

**Solution.** For \(|a|=1\), expand

\[
 |p-aq|_J^2=2-2\operatorname{Re}
                       (\overline a\langle p,q\rangle_J).
\]

The maximum of that real part is \(|\langle p,q\rangle_J|\), attained by choosing its phase when the inner product is nonzero. If it is zero every phase attains zero. Thus the minimum is \(2-2|\langle p,q\rangle_J|\). The hypothesis implies \(|\langle p,q\rangle_J|>\sqrt{1-\delta}\), so the squared distance is less than \(2(1-\sqrt{1-\delta})\leq2\delta\). This treats arbitrary coefficient vectors, irrespective of their actual polynomial degree.

### Exercise 6: constants and coefficient derivatives

When \(m=0\), take a fixed normalized density \(\psi\). Determine \(\Phi\), \(A\), their behavior under complex scaling, and all coefficient derivatives of \(A\).

**Solution.** Set \(\Phi(q,z)=\psi(z)\) for a nonzero complex constant \(q\). Then \(A(q,z)=\psi(z)/q\). For \(a\neq0\), \(\Phi(aq,z)=\Phi(q,z)\), while \(A(aq,z)=a^{-1}A(q,z)\). Regarding complex numbers as a real coefficient space, the directional derivatives are

\[
 d^jA(q,z)[r_1,\ldots,r_j]
 =(-1)^j j!\,\psi(z)\frac{r_1\cdots r_j}{q^{j+1}}.
\]

This follows by differentiating \(q^{-1}\) repeatedly; multiplication of complex directions is also real multilinear. Its modulus is bounded by \(j!\|\psi\|_\infty\prod|r_\nu|/|q|^{j+1}\), including \(j=0\). This verifies the power and shows why the density is invariant while the divided density has degree minus one.

## 8. Source credit and scope

L004 attributes the polynomial-averaging method to Lars Hörmander in the context of Malgrange–Ehrenpreis existence. The historical primary archive is [On the existence and the regularity of solutions of linear pseudo-differential equations](https://www.e-periodica.ch/digbib/view?lang=en&pid=ens-001%3A1971%3A17%3A%3A213), *L’Enseignement Mathématique* 17 (1971), 99–163. Its title and volume were verified. The chapter-download endpoint returned a verification page; no exact original proof or lemma number has been compared. The full argument written here requires no unavailable proof from that source.

The zero-dimensional endpoint is stated separately in the complete proof: with unit mass on the one-point space, \(\Phi(q,0)=1\), \(K=\{0\}\), and \(c=1\); separation of \(K\) from zero is specific to \(n\geq1\). General distribution-family topology, other L004 receivers, recursive prerequisite closure, and the full AN-02 course remain unfinished.