Improve C_59 upper bound

[sebastian-griego]

Summary

This updates the upper bound for the Bohr radius of the bidisc from

$$ K_2 < 0.3177 $$

to

$$ K_2 < 0.3174541. $$

The bound comes from a degree-$(250,250)$ polynomial obtained as a double Fejer mean of the rational-inner function

$$ F(z,w)=\frac{981359-660190z-641550w+1000000zw} {1000000-641550z-660190w+981359zw}. $$

At

$$ r=\frac{3174541}{10000000}, $$

the exact verifier below proves

$$ B_r(p)>1+\frac1{7307638490}>1. $$

Since $B_s(p)$ is continuous in $s$, the Bohr inequality fails at some $s&lt;r$, and hence $K_2&lt;r$.

Certificate

Let

$$ L=1000000, \quad A=641550, \quad C=660190, \quad D=981359, \quad N=250, \quad M=251. $$

Define

$$ Q(z,w)=L-Az-Cw+Dzw, \qquad G(z,w)=D-Cz-Aw+Lzw. $$

For $\lvert w\rvert=1$,

$$ |L-Cw|^2-|Dw-A|^2 =L^2+C^2-D^2-A^2-2(LC-AD)\operatorname{Re}w. $$

Here

$$ LC-AD=30599133550>0, $$

so the minimum on $\lvert w\rvert=1$ occurs at $w=1$. The exact integer check is

$$ (L-C)^2-(D-A)^2=679619>0. $$

Thus $\lvert Dw-A\rvert&lt;\lvert L-Cw\rvert$ on $\lvert w\rvert=1$. Since $(Dw-A)/(L-Cw)$ is analytic on $\lvert w\rvert\le 1$, the same strict inequality holds on $\lvert w\rvert\le 1$. Hence $Q$ has no zero on the closed bidisc. On the torus,

$$ G(z,w)=zw\overline{Q(z,w)}. $$

Therefore $F=G/Q$ is analytic on the closed bidisc and satisfies $\lvert F\rvert=1$ on the torus, hence $\lvert F\rvert\le 1$ on the bidisc.

Write

$$ F(z,w)=\sum_{j,k\ge0}a_{j,k}z^jw^k. $$

The coefficient integers $\nu_{j,k}$ are defined by

$$ a_{j,k}=\frac{\nu_{j,k}}{L^{j+k+1}} $$

and the recurrence

$$ \nu_{j,k}=A\nu_{j-1,k}+C\nu_{j,k-1}-DL\nu_{j-1,k-1}+e_{j,k}L^{j+k}, $$

with missing-index terms taken to be zero, where

$$ e_{0,0}=D, \quad e_{1,0}=-C, \quad e_{0,1}=-A, \quad e_{1,1}=L, $$

and all other $e_{j,k}$ are zero.

Define the polynomial

$$ p(z,w)=\sum_{0\le j,k\le N}c_{j,k}z^jw^k $$

by

$$ c_{j,k}=\frac{(M-j)(M-k)\nu_{j,k}}{M^2L^{j+k+1}}. $$

This is the double Fejer mean of $F$. Since the Fejer kernel is a nonnegative probability kernel, $p$ is a convex average of boundary values of $F$ on the torus. Hence $\lvert p\rvert\le 1$ on the torus, and by the maximum principle $\lvert p\rvert\le 1$ on the bidisc.

Verification

The following script uses integer arithmetic for every assertion. Decimal arithmetic is used only for human-readable output. It also has an optional --write-coefficients flag that writes the full rational coefficient list for $p$.

#!/usr/bin/env python3
"""
Exact certificate for the upper bound K_2 < 3174541/10000000.

All assertions use integer arithmetic only. Decimal arithmetic is used only for
human-readable output.
"""

from __future__ import annotations

import argparse
from decimal import Decimal, getcontext
from math import gcd

L = 10**6
A = 641550
C = 660190
D = 981359
R = 3174541
T = 10**7
N = 250
M = N + 1
MARGIN_DEN = 7307638490


def compute_nu() -> list[list[int]]:
    """Return nu[j][k] where a[j][k] = nu[j][k] / L**(j+k+1)."""
    nu = [[0] * (N + 1) for _ in range(N + 1)]
    for j in range(N + 1):
        for k in range(N + 1):
            v = 0
            if j > 0:
                v += A * nu[j - 1][k]
            if k > 0:
                v += C * nu[j][k - 1]
            if j > 0 and k > 0:
                v -= D * L * nu[j - 1][k - 1]
            if j == 0 and k == 0:
                v += D
            elif j == 1 and k == 0:
                v -= C * L
            elif j == 0 and k == 1:
                v -= A * L
            elif j == 1 and k == 1:
                v += L**3
            nu[j][k] = v
    return nu


def coefficient_fraction(nu: list[list[int]], j: int, k: int) -> tuple[int, int]:
    """Return the exact rational coefficient c[j][k] as numerator, denominator."""
    numerator = (M - j) * (M - k) * nu[j][k]
    denominator = M * M * L ** (j + k + 1)
    g = gcd(abs(numerator), denominator)
    return numerator // g, denominator // g


def compute_br_fraction(nu: list[list[int]]) -> tuple[int, int]:
    """Return the exact fraction B_{R/T}(p)."""
    Lpow = [1] * (2 * N + 2)
    Tpow = [1] * (2 * N + 1)
    Rpow = [1] * (2 * N + 1)
    for i in range(1, 2 * N + 2):
        Lpow[i] = Lpow[i - 1] * L
    for i in range(1, 2 * N + 1):
        Tpow[i] = Tpow[i - 1] * T
        Rpow[i] = Rpow[i - 1] * R

    numerator = 0
    for j in range(N + 1):
        for k in range(N + 1):
            n = j + k
            numerator += (
                abs(nu[j][k])
                * Rpow[n]
                * (M - j)
                * (M - k)
                * Lpow[2 * N - n]
                * Tpow[2 * N - n]
            )

    denominator = Lpow[2 * N + 1] * Tpow[2 * N] * M * M
    return numerator, denominator


def write_coefficients(nu: list[list[int]], filename: str) -> None:
    with open(filename, "w", encoding="utf-8") as out:
        out.write("j\tk\tnumerator\tdenominator\n")
        for j in range(N + 1):
            for k in range(N + 1):
                num, den = coefficient_fraction(nu, j, k)
                out.write(f"{j}\t{k}\t{num}\t{den}\n")


def main() -> None:
    parser = argparse.ArgumentParser()
    parser.add_argument(
        "--write-coefficients",
        metavar="PATH",
        help="write the complete coefficient list for p as a TSV file",
    )
    args = parser.parse_args()

    # Radius check: R/T < 3177/10000.
    assert R > 0
    assert R * 10000 < 3177 * T

    # Stability of Q(z,w)=L-Az-Cw+Dzw on the closed bidisc.
    # For |w|=1, |L-Cw|^2-|Dw-A|^2 is minimized at w=1 because LC-AD>0.
    assert C < L
    assert L * C - A * D > 0
    assert (L - C) ** 2 - (D - A) ** 2 > 0

    nu = compute_nu()
    numerator, denominator = compute_br_fraction(nu)

    assert numerator > denominator
    assert MARGIN_DEN * numerator > (MARGIN_DEN + 1) * denominator

    if args.write_coefficients:
        write_coefficients(nu, args.write_coefficients)

    getcontext().prec = 80
    value = Decimal(numerator) / Decimal(denominator)
    excess = Decimal(numerator - denominator) / Decimal(denominator)

    print(f"r = {R}/{T}")
    print(f"bidegree = ({N}, {N})")
    print("proved: R/T < 3177/10000")
    print("proved: denominator L-Az-Cw+Dzw is stable on the closed bidisc")
    print("proved: |p| <= 1 on the bidisc by rational-inner Fejer averaging")
    print(f"proved: B_r(p) > 1 + 1/{MARGIN_DEN}")
    print(f"B_r(p) = {value}")
    print(f"B_r(p)-1 = {excess}")


if __name__ == "__main__":
    main()

Expected output includes:

r = 3174541/10000000
bidegree = (250, 250)
proved: R/T < 3177/10000
proved: denominator L-Az-Cw+Dzw is stable on the closed bidisc
proved: |p| <= 1 on the bidisc by rational-inner Fejer averaging
proved: B_r(p) > 1 + 1/7307638490
B_r(p) = 1.0000000001368431130617046818753061666655003380903583500779450859192096580842119
B_r(p)-1 = 1.3684311306170468187530616666550033809035835007794508591920965808421187976372851E-10

Therefore

$$ B_{3174541/10000000}(p)>1, $$

with $\lvert p\rvert\le 1$ on the bidisc, so

$$ K_2<0.3174541. $$

AI assistance disclosure

The construction, patch, PR text, and verification script were prepared with assistance from ChatGPT 5.5 Pro. The submitter rechecked the construction and arithmetic before submission.