Exact value of the Bohr radius of the bidisc

A rational-inner witness gives a new exact upper bound

An exact rational-inner construction gives

$$K_2\le R=0.312739413109697160561285748489\ldots .$$

Here $R=P/10^{265}$, with the full integer $P$ recorded in the attached manuscript and Lean/Python certificate. This is an upper bound only; it does not determine the exact value of $K_2$.

With exact rational parameters $a,c,d$, define

$$Q(z,w)=1-az-cw+dzw,$$ $$G(z,w)=d-cz-aw+zw,$$

and

$$F(z,w)=\frac{G(z,w)}{Q(z,w)}.$$

The chosen parameters satisfy exact stability inequalities

$$0\le c\le d,\qquad ad\le c,\qquad (1-c)^2>(d-a)^2.$$

For fixed $|w|\le1$,

$$|1-cw|^2-|a-dw|^2 = ((1-c)^2-(d-a)^2) +(d^2-c^2)(1-|w|^2) +2(c-ad)(1-\operatorname{Re}w)>0.$$

Hence $Q$ has no zero on the closed bidisc. On the distinguished boundary $|z|=|w|=1$,

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

Therefore $F=G/Q$ is rational-inner and belongs to the bidisc Schur class.

Writing

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

the identity $QF=G$ gives the exact triangular recurrence

$$b_{jk}-ab_{j-1,k}-cb_{j,k-1}+db_{j-1,k-1}=g_{jk},$$

where $G=\sum g_{jk}z^jw^k$ and coefficients with negative indices are zero.

The attached exact certificate computes the true Taylor coefficients for

$$0\le j,k\le224$$

and proves

$$\sum_{0\le j,k\le224}|b_{jk}|R^{j+k}>1$$

using a strict common-denominator integer comparison. Since all omitted terms in the full absolute majorant are nonnegative, this finite violation implies

$$K_2\le R.$$

No floating-point sign decision and no Taylor-tail estimate is used.

Compared with the previous public endpoint $0.3174541$, the new endpoint is smaller by

$$0.004714686890302839\ldots ,$$

about $1.49\%$ of the previous upper endpoint. Relative to the lower bound $0.3006$, it reduces the unresolved interval width by about $27.97\%$.

Improved lower and upper bounds for the bidisc Bohr radius

Let $K_2$ denote the unrestricted bidisc Bohr radius: the supremum of the radii $r>0$ for which every holomorphic function

$$f(z,w)=\sum_{j,k\geq 0}a_{jk}z^jw^k, \qquad \sup_{(z,w)\in\mathbb D^2}|f(z,w)|\leq 1,$$

satisfies $\sum_{j,k\geq 0}|a_{jk}|r^{j+k}\leq 1$. The bounds proved here are

$$\boxed{\;0.30158758<K_2<0.302825279492.\;}$$

The exact value of $K_2$ remains open.

1. Lower bound

We first prove a uniform coefficient estimate for every holomorphic $H$ on $\mathbb D^2$ with $H(0,0)=1$ and $\operatorname{Re}H>0$. Write

$$H(z,w)=1+\sum_{n\ge1}H_n(z,w),\qquad H_n(z,w)=\sum_{j+k=n}h_{jk}z^jw^k,$$ $$S_n(H)=\sum_{j+k=n}|h_{jk}|, \qquad M_r(H)=\sum_{n\ge1}S_n(H)r^n,$$

and put

$$A_2=|h_{20}|+|h_{02}|,\quad B_2=|h_{11}|, \qquad A_3=|h_{30}|+|h_{03}|,\quad B_3=|h_{21}|+|h_{12}|.$$

Thus $S_2=A_2+B_2$ and $S_3=A_3+B_3$.

A quadratic estimate. For $|\zeta|=1$, the function $\lambda\mapsto H(\lambda\zeta,\lambda)$ has positive real part on $\mathbb D$ and constant coefficient $1$. The one-variable Carathéodory coefficient estimate therefore gives

$$|H_n(\zeta,1)|\le2\qquad(n\ge1).$$

For $n=2$, rotate $\zeta$ so that the two endpoint coefficients of the quadratic have the same argument. Its values at $1$ and $-1$ then have the form $a+b$ and $a-b$, where $|a|=A_2$ and $|b|=B_2$. Averaging their squared moduli proves

$$\boxed{A_2^2+B_2^2\le4.}$$

A cubic estimate. Let $p(\zeta)=\sum_{j=0}^3c_j\zeta^j$ satisfy $\max_{|\zeta|=1}|p(\zeta)|\le1$. Rotate its argument so that $c_1$ and $c_2$ have the same argument. Independently choose $|\xi|=1$ so that $|\overline\xi c_0+c_3|=|c_0|+|c_3|$. If any of these coefficients vanishes, make an arbitrary compatible choice. Define

$$u=(3,4,-4,-3)^{\mathsf T},\qquad v=(0,1,1,0)^{\mathsf T},\qquad w=(\xi,0,0,1)^{\mathsf T},$$ $$R=uu^*+28vv^*+35ww^*.$$

This is a positive semidefinite Toeplitz matrix: its first row is $(44,12,-12,-9+35\xi)$.

To see directly why $c^*Rc\le44$, realize $R$ as the Gram matrix of four vectors $e_0,e_1,e_2,e_3$. Toeplitz symmetry makes $e_j\mapsto e_{j+1}$, $0\le j\le2$, a well-defined isometry between the corresponding spans. Extend this isometry to a unitary $T$ on the finite-dimensional span of all four vectors. Then $e_j=T^je_0$, and the finite-dimensional spectral theorem gives

$$c^*Rc=\|p(T)e_0\|^2\le\|e_0\|^2=44.$$

Expanding $c^*Rc$ and discarding $|u^*c|^2$ yields

$$35(|c_0|+|c_3|)^2+28(|c_1|+|c_2|)^2\le44.$$

This Gram argument is a weighted variant of the identity in Haoxiang Yu's proof of the cubic Sidon inequality. Applying the displayed estimate to $p(\zeta)=H_3(\zeta,1)/2$ gives

$$\boxed{35A_3^2+28B_3^2\le176.}$$

A relation between the quadratic and cubic coefficients. The bidisc Schur realization theorem applies to the full Schur class, not just a proper subclass. The realization and the equality of the Schur and Schur–Agler classes on the bidisc are recalled in Knese, Sections 1–2. For completeness, its positive-real form is

$$H(z,w)=v^*(I+ZU)(I-ZU)^{-1}v, \qquad Z=zP+wQ,$$

where $P,Q$ are complementary orthogonal projections, $U$ is unitary, and $\|v\|=1$. The Hilbert space need not be finite dimensional.

Here is the passage from the Schur realization to this formula. Apply that theorem to $g=(H-1)/(H+1)$, which satisfies $g(0)=0$. It gives

$$g=B_0Z(I-D_0Z)^{-1}C_0, \qquad \begin{pmatrix}0&B_0\\ C_0&D_0\end{pmatrix} \text{ unitary}.$$

If the theorem is stated with an isometric colligation, take its unitary extension with the original space invariant, placing the added state space in one coordinate summand. The state operator and coordinate projections still preserve the original state space, so every transfer coefficient is unchanged.

Set $U=D_0+C_0B_0$ and $v=B_0^*$. The block unitary identities give

$$U^*U=UU^*=I,\qquad B_0B_0^*=1,\qquad D_0B_0^*=0.$$

Writing $T_0=I-ZD_0$, we have $T_0^{-1}v=v$ and $B_0T_0^{-1}ZC_0=g$. The rank-one resolvent identity therefore gives

$$B_0(I-ZU)^{-1}B_0^*=1+\frac{g}{1-g}=\frac1{1-g}.$$

Since $(I+ZU)(I-ZU)^{-1}=2(I-ZU)^{-1}-I$, this is the asserted positive-real formula.

Write the resulting unitary in blocks relative to $P\mathcal H\oplus Q\mathcal H$:

$$U=\begin{pmatrix}A&B\\ C&D\end{pmatrix},\qquad v=\binom ab,\qquad Uv=\binom cd.$$

Both pairs $(a,b)$ and $(c,d)$ have total squared norm one. Expanding the resolvent gives

$$\frac{h_{11}}2=a^*Bd+b^*Cc,\qquad \frac{h_{30}}2=a^*A^2c,\qquad \frac{h_{03}}2=b^*D^2d.$$

Put

$$X=\|A^*a\|^2+\|D^*b\|^2,\qquad Y=\|Ac\|^2+\|Dd\|^2.$$

Cauchy–Schwarz and $UU^*=I$ imply

$$\left|\frac{h_{11}}2\right|^2 \le\|B^*a\|^2+\|C^*b\|^2=1-X.$$

Applying Cauchy–Schwarz in the other order and using $U^*U=I$ gives

$$\left|\frac{h_{11}}2\right|^2 \le\|Bd\|^2+\|Cc\|^2=1-Y.$$

Finally,

$$|h_{30}|\le\|A^*a\|^2+\|Ac\|^2,\qquad |h_{03}|\le\|D^*b\|^2+\|Dd\|^2,$$

so $A_3\le X+Y\le2-B_2^2/2$. We have proved, for every such $H$,

$$\boxed{A_3+\tfrac12B_2^2\le2.}$$

Combining the estimates. Let $\alpha,\beta,\gamma>0$ and define

$$\begin{aligned} \mathcal B(r)={}&4\alpha+176\beta+2\gamma +\frac1{4\alpha}+\frac1{4(\alpha+\gamma/2)}\\ &+\frac{(r-\gamma)^2}{140\beta} +\frac{r^2}{112\beta}. \end{aligned}$$

The following identity is a completion of four squares:

$$\begin{aligned} \mathcal B(r)-[A_2+B_2+r(A_3+B_3)] ={}&\alpha(4-A_2^2-B_2^2)\\ &+\beta(176-35A_3^2-28B_3^2)\\ &+\gamma(2-A_3-B_2^2/2)\\ &+\alpha\left(A_2-\frac1{2\alpha}\right)^2\\ &+(\alpha+\gamma/2) \left(B_2-\frac1{2(\alpha+\gamma/2)}\right)^2\\ &+35\beta\left(A_3-\frac{r-\gamma}{70\beta}\right)^2\\ &+28\beta\left(B_3-\frac r{56\beta}\right)^2. \end{aligned}$$

Every term on the right is nonnegative by the three boxed coefficient estimates. Consequently

$$S_2(H)+rS_3(H)\le\mathcal B(r).$$

We also use the homogeneous estimate

$$S_n(H)\le2\sqrt n\qquad(n\ge1).$$

It follows from Knese's Lemma 2.1: apply that lemma to $f_\varepsilon=(1-\varepsilon H)/(1+\varepsilon H)$. For each fixed positive degree its coefficients, divided by $-2\varepsilon$, tend to those of $H$, while

$$1-|f_\varepsilon(0)|^2=\frac{4\varepsilon}{(1+\varepsilon)^2}.$$

Divide the lemma's bound by $2\varepsilon$ and let $\varepsilon\downarrow0$. Thus the entire majorant is bounded by the continuous function

$$M_r(H)\le E(r):=2r+r^2\mathcal B(r) +2\sum_{n=4}^{\infty}\sqrt n\,r^n, \qquad 0<r<1.$$

The exact numerical inequality. Choose the following rational numbers:

$$r_* =\frac{15079379}{50000000},\qquad \alpha=\frac{264631}{800000},\qquad \beta=\frac{249627}{100000000},\qquad \gamma=\frac{10197021}{100000000}.$$

Substitution gives exactly

$$\mathcal B(r_*)= \frac{44915682119816917151345293949} {11769153837063865963000000000}.$$

Here is an explicit rational bound for every term of the remaining series. For $4\le n\le20$, let $q_n$ be the integers in the table. Each satisfies $q_n^2\ge10^{16}n$, hence $\sqrt n\le q_n/10^8$.

$n$	$q_n$	$n$	$q_n$
4	200000000	13	360555128
5	223606798	14	374165739
6	244948975	15	387298335
7	264575132	16	400000000
8	282842713	17	412310563
9	300000000	18	424264069
10	316227767	19	435889895
11	331662480	20	447213596
12	346410162	—	—

For all later degrees, use $\sqrt n\le n$ and sum the differentiated geometric series:

$$2\sum_{n=21}^{\infty}\sqrt n\,r_*^n \le2\sum_{n=21}^{\infty}n r_*^n =\frac{2r_*^{21}(21-20r_*)}{(1-r_*)^2}.$$

All quantities in the next two inequalities are rational. Substitution of the displayed values and multiplication by positive denominators give

$$2r_*+r_*^2\mathcal B(r_*) <\frac{950295182089920172}{10^{18}},$$ $$\frac2{10^8}\sum_{n=4}^{20}q_n r_*^n +\frac{2r_*^{21}(21-20r_*)}{(1-r_*)^2} <\frac{49704796478844948}{10^{18}}.$$

These provide the strict, entirely rational comparison

$$E(r_*) <\frac{999999978568765120}{10^{18}} <1-\frac1{50000000}.$$

In particular, no part of the infinite tail has been omitted.

Passage to the usual Bohr radius. Since the constants $\alpha,\beta,\gamma$ are fixed and $E$ is continuous on $(0,1)$, the strict inequality gives some $r'>r_*$ with $E(r')<1$. Let $f$ be any holomorphic function on the bidisc with $|f|\le1$. Rotate its values so that $f(0)=t\in[0,1]$, without changing its coefficient moduli. If $t=1$, then $f$ is constant. Otherwise

$$H=\frac{1-f}{1-t}$$

has positive real part, constant coefficient $1$, and $|f_{jk}|=(1-t)|h_{jk}|$ for $j+k>0$. Therefore

$$\sum_{j,k\ge0}|f_{jk}|(r')^{j+k} =t+(1-t)M_{r'}(H) \le t+(1-t)E(r')\le1.$$

Thus $r'$ is admissible for every bounded holomorphic bidisc function, and $K_2\ge r'>r_*=0.30158758$.

2. Upper bound

An explicit upper bound. We prove

$$K_2<\frac{302825279492}{10^{12}}=0.302825279492.$$

For a holomorphic function $f(z,w)=\sum_{j,k\geq0}c_{jk}z^jw^k$ on the open bidisc, write

$$B_r(f)=\sum_{j,k\geq0}|c_{jk}|r^{j+k}.$$

It suffices to construct a function with $|f|\leq1$ on the bidisc whose finite coefficient sum exceeds one at a radius strictly smaller than the displayed number.

Set

$$L=2500000000,\qquad T=3067398171,\qquad S=10^{15}, \qquad r_*=\frac{302825279492}{10^{12}},$$

and define

$$\begin{aligned} U(z,w)&=(1+z)(1-w),&V(z,w)&=1+zw,\\ P(z,w)&=LU(z,w)+iTV(z,w),& Q(z,w)&=LV(z,w)+iTU(z,w),\\ f(z,w)&=\frac{SQ(z,w)-P(z,w)}{SQ(z,w)+P(z,w)}. \end{aligned}$$

Holomorphy and boundedness. For all complex $z,w$,

$$2\operatorname{Re}(U\overline V) =(1-|w|^2)|1+z|^2+(1-|z|^2)|1-w|^2.$$

The terms involving $iLT$ in $P\overline Q$ are purely imaginary, so

$$\operatorname{Re}(P\overline Q) =(L^2+T^2)\operatorname{Re}(U\overline V).$$

It follows that

$$\begin{aligned} |SQ+P|^2-|SQ-P|^2 &=4S\operatorname{Re}(P\overline Q)\\ &=2S(L^2+T^2)\bigl[ (1-|w|^2)|1+z|^2 +(1-|z|^2)|1-w|^2\bigr]. \end{aligned}$$

The right side is strictly positive when $|z|,|w|<1$. In particular, $SQ+P$ cannot vanish there, since a zero would make the left side nonpositive. Thus $f$ is holomorphic on the entire open bidisc and $|f(z,w)|<1$ there.

The Taylor coefficients as Gaussian integers. Put $\kappa=L+iT$, $\widetilde p=SQ-P$, and $\widetilde q=SQ+P$. Their nonzero coefficients are

$$\begin{array}{c|cc} \text{monomial}&\widetilde p&\widetilde q\\ \hline 1&(S-1)\kappa&(S+1)\kappa\\ z&-L+iST&L+iST\\ w&L-iST&-L-iST\\ zw&(S+1)\overline\kappa&(S-1)\overline\kappa. \end{array}$$

Let $q_0=(S+1)\kappa$, and multiply both polynomials by $\overline{q_0}$. Thus

$$p=\overline{q_0}\widetilde p,\qquad q=\overline{q_0}\widetilde q =D+q_{10}z+q_{01}w+q_{11}zw,$$

where

$$\begin{aligned} D&=(S+1)^2(L^2+T^2)\\ &=15658931539454176558863078908306140931539454145241>0. \end{aligned}$$

All coefficients of $p,q$ belong to $\mathbb Z[i]$. Write $p_{jk}=0$ outside $\{0,1\}^2$.

Define $v_{jk}\in\mathbb Z[i]$ for $j,k\geq0$, taking every negative-index term to be zero, by

$$v_{jk} =p_{jk}D^{j+k} -q_{10}v_{j-1,k}-q_{01}v_{j,k-1} -Dq_{11}v_{j-1,k-1}.$$

Then the actual Taylor coefficients of $f$ satisfy

$$c_{jk}=\frac{v_{jk}}{D^{j+k+1}}.$$

Indeed, comparison of coefficients in $qf=p$ gives

$$Dc_{jk}+q_{10}c_{j-1,k}+q_{01}c_{j,k-1} +q_{11}c_{j-1,k-1}=p_{jk}.$$

Multiplication by $D^{j+k}$ and induction on $j+k$ prove the formula, including $c_{00}=(S-1)/(S+1)$.

These integers also have an explicit finite expression. With the same zero convention, put

$$\begin{aligned} W_{jk}={}&\sum_{m=0}^{\min(j,k)} (-1)^{j+k-m} \binom{j+k-m}{m}\binom{j+k-2m}{j-m}\\ &\hspace{15mm}\cdot q_{10}^{j-m}q_{01}^{k-m}q_{11}^{m}D^m. \end{aligned}$$

Expanding $1/q$ as a geometric series near the origin shows that $[z^jw^k](1/q)=W_{jk}/D^{j+k+1}$. Consequently,

$$v_{jk}=p_{00}W_{jk}+Dp_{10}W_{j-1,k} +Dp_{01}W_{j,k-1}+D^2p_{11}W_{j-1,k-1}.$$

A finite integer comparison. Set

$$N=28,\qquad R=302825279492,\qquad E=10^{12}.$$

For each of the $(N+1)^2=841$ pairs $0\leq j,k\leq N$, let $n_{jk}$ be the unique nonnegative integer satisfying

$$n_{jk}^2\leq (\operatorname{Re}v_{jk})^2+(\operatorname{Im}v_{jk})^2 <(n_{jk}+1)^2.$$

Define the positive integers

$$A=\sum_{j,k=0}^{28} n_{jk}R^{j+k}(ED)^{56-j-k}, \qquad B=D(ED)^{56}.$$

Substitution of the stated integers into these finite expressions gives the exact inequalities

$$10668060774835\,B <10^{39}(A-B) <10668060774836\,B.$$

In particular,

$$10^{26}(A-B)>B.$$

Every exponent in the definitions is nonnegative. The square-root inequalities imply $|v_{jk}|\geq n_{jk}$, and clearing denominators therefore gives

$$\sum_{j,k=0}^{28}|c_{jk}|\left(\frac RE\right)^{j+k} \geq\frac AB>1+10^{-26}.$$

This comparison includes the constant coefficient; all omitted Taylor terms have nonnegative contributions to $B_{r_*}(f)$.

Finally, the polynomial

$$M(s)=\sum_{j,k=0}^{28}|c_{jk}|s^{j+k}$$

is continuous and satisfies $M(r_*)>1$. Hence $M(s)>1$ for some $0<s<r_*$. Its coefficients are nonnegative, so every radius at least $s$ fails the Bohr inequality for this same bounded function. Thus $K_2\leq s<r_*$, as claimed.

3. A coefficient comparison for two-dimensional realizations

Let $U$ be a unitary $2\times2$ matrix, let $v\in\mathbb C^2$ be a unit vector, and put

$$Z=\operatorname{diag}(z,w),\qquad H_{U,v}(z,w)=v^*(I+ZU)(I-ZU)^{-1}v.$$

These functions are holomorphic on the open bidisc because $\|ZU\|<1$. If $x=(I-ZU)^{-1}v$, then

$$\operatorname{Re}H_{U,v} =x^*(I-U^*Z^*ZU)x =(Ux)^*(I-Z^*Z)(Ux)>0.$$

Also $H_{U,v}(0,0)=1$.

Independent rotations of the variables multiply the rows of $U$ by unit scalars and preserve every Taylor coefficient modulus. Diagonal unitary conjugation commutes with $Z$ and preserves the function when applied also to $v$. These operations put $U$ into the normal form

$$U=\begin{pmatrix}a&-b\\ b&a\end{pmatrix}, \qquad 0\leq a\leq1,\quad b=\sqrt{1-a^2}.$$

Indeed, the two diagonal entries of a unitary $2\times2$ matrix have the same modulus. For $0<a<1$, row rotations make them both equal to $a$; orthogonality then makes the off-diagonal entries negative conjugates, and diagonal conjugation removes their phase. The cases $a=0,1$ follow directly.

Matrix inversion gives

$$H_{a,q}(z,w) =\frac{1+qz-\overline q\,w-zw}{Q_a(z,w)}, \qquad Q_a(z,w)=1-a(z+w)+zw,$$

where

$$q=a(|v_1|^2-|v_2|^2)-2b\overline v_1v_2.$$

For $b>0$, the attainable values are exactly the ellipse

$$(\operatorname{Re}q)^2+ \frac{(\operatorname{Im}q)^2}{b^2}\leq1.$$

To see this, put

$$d=|v_1|^2-|v_2|^2,\qquad u=2\operatorname{Re}(\overline v_1v_2),\qquad s=2\operatorname{Im}(\overline v_1v_2).$$

Then $d^2+u^2+s^2=1$, $\operatorname{Re}q=ad-bu$, and $\operatorname{Im}q=-bs$. Projection of this unit sphere onto the orthogonal coordinates $ad-bu,s$ fills the unit disk. At $b=0$, the attainable set is the real interval $[-1,1]$.

Theorem. Fix $0\leq a\leq1$ and $b=\sqrt{1-a^2}$. For every admissible $q$, write $h_{jk}(q)=[z^jw^k]H_{a,q}$. Then

$$|h_{jk}(q)|+|h_{kj}(q)| \leq |h_{jk}(ib)|+|h_{kj}(ib)| \qquad(j,k\geq0).$$

Consequently, every finite coefficient majorant with nonnegative weights invariant under $(j,k)\mapsto(k,j)$ is maximized over $q$ at $q=ib$. This includes all square and total-degree Bohr truncations at every nonnegative radius.

Proof. Let $d_{jk}$ be the real coefficient of $z^jw^k$ in $1/Q_a$, taking negative-index coefficients to be zero. Geometric expansion gives

$$d_{jk}= \sum_{\ell=0}^{\min(j,k)} (-1)^\ell(1-a^2)^\ell a^{j+k-2\ell} \binom j\ell\binom k\ell,$$

and coefficient comparison in $Q_a/Q_a=1$ gives

$$d_{jk}=a(d_{j-1,k}+d_{j,k-1})-d_{j-1,k-1} \qquad(j+k>0).$$

In particular, $d_{jk}=d_{kj}$.

We first establish the adjacent-minor identity

$$\begin{aligned} &d_{j-1,j+m}d_{j,j+m-1} -d_{j-1,j+m-1}d_{j,j+m}\\ &\quad= \frac{(1-a^2)a^{2m}}{j(j+m)} \sum_{\ell=0}^{j-1}(2\ell+m+1) \left[P_\ell^{(0,m)}(2a^2-1)\right]^2 \geq0 \end{aligned}$$

for $j\geq1$ and $m\geq0$, where $P_\ell^{(\alpha,\beta)}$ denotes the Jacobi polynomial with its standard normalization.

For $0<a<1$, set

$$x=2a^2-1,\quad n=j,\quad d=2n+m,\quad p=P_n^{(0,m)}(x),\quad s=P_{n-1}^{(0,m)}(x).$$

The finite binomial formula gives $d_{j,j+m}=a^mP_j^{(0,m)}(x)$. Thus the left side of the adjacent-minor identity is

$$a^{2m}\left[ P_{n-1}^{(0,m+1)}(x)P_n^{(0,m-1)}(x)-sp \right].$$

For $m=0$, this follows using the polynomial identity

$$P_n^{(0,-1)}(x) =\frac{1+x}{2}P_{n-1}^{(0,1)}(x);$$

no orthogonality assertion at parameter $-1$ is needed.

Specializing the classical contiguous and derivative relations for Jacobi polynomials gives the following identities; see NIST DLMF, §18.9:

$$\begin{aligned} (1+x)P_{n-1}^{(0,m+1)}(x) &=\frac{2np+2(n+m)s}{d},\\ P_n^{(0,m-1)}(x) &=\frac{(n+m)p+ns}{d},\\ d(1-x^2)p' &=-n(m+dx)p+2n(n+m)s,\\ d(1-x^2)s' &=(n+m)(dx-m)s-2n(n+m)p. \end{aligned}$$

Multiplication and subtraction give

$$P_{n-1}^{(0,m+1)}(x)P_n^{(0,m-1)}(x)-sp =\frac{1-x}{d}(p's-s'p).$$

The Jacobi squared norms and leading coefficients are

$$h_\ell=\frac{2^{m+1}}{2\ell+m+1},\qquad \kappa_\ell=2^{-\ell}\binom{2\ell+m}{\ell}.$$

The confluent Christoffel–Darboux identity therefore becomes

$$\sum_{\ell=0}^{n-1}(2\ell+m+1) \left[P_\ell^{(0,m)}(x)\right]^2 =\frac{2n(n+m)}{2n+m}(p's-s'p).$$

This is the specialization of NIST DLMF, equation 18.2.13. Combining the last two identities with $1-x=2(1-a^2)$ proves the adjacent-minor formula. The endpoints $a=0,1$ follow by polynomial continuity. Symmetry supplies the case in which the second index is smaller than the first; if either index is zero, the corresponding adjacent minor is zero by convention.

The constant coefficient of $H_{a,q}$ is one. For $j+k>0$, put

$$A=d_{jk}-d_{j-1,k-1},\qquad B=d_{j-1,k}-d_{j,k-1},\qquad C=d_{j-1,k}+d_{j,k-1}.$$

Writing $q=u+iy$, the paired coefficients are

$$h_{jk}(q)=A+uB+iyC,\qquad h_{kj}(q)=A-uB+iyC.$$

With $M=A^2+b^2C^2$, the recurrence and the adjacent-minor identity imply

$$M-B^2 =4\bigl(d_{j-1,k}d_{j,k-1} -d_{j-1,k-1}d_{jk}\bigr)\geq0.$$

Suppose first that $b>0$. For fixed $u$, each modulus increases with $y^2$, so the ellipse condition allows us to replace $y^2$ by $b^2(1-u^2)$. Set

$$t=u^2\in[0,1],\qquad E=B^2-b^2C^2.$$

The squared sum of the two resulting moduli is

$$2(M+tE)+2\sqrt{(M+tE)^2-4A^2B^2t}.$$

Since $B^2\leq M$, we have $E\leq A^2$ and $M-tE\geq0$. Moreover,

$$\begin{aligned} (M-tE)^2- \bigl[(M+tE)^2-4A^2B^2t\bigr] &=4tb^2C^2(M-B^2)\\ &\geq0. \end{aligned}$$

The squared sum is therefore at most $4M$. Taking its nonnegative square root bounds the sum by $2\sqrt M$, exactly its value at $q=ib$. The case $b=0$ follows by continuity, or by the same algebra with $y=0$. Grouping a symmetric finite index set into pairs and diagonal terms proves the majorant assertion.

The resulting one-parameter family is

$$H_a(z,w)= \frac{1+i\sqrt{1-a^2}(z+w)-zw}{1-a(z+w)+zw}, \qquad0\leq a\leq1.$$

Thus a symmetric coefficient majorant on the two-dimensional unitary realization class can be maximized within this family. The statement concerns precisely this realization class; it does not identify the extremals among all normalized positive-real functions on the bidisc.

For $\eta=a+i\sqrt{1-a^2}$, direct simplification gives

$$g_a(z,w):=\frac{H_a(z,w)-1}{H_a(z,w)+1} =\frac{\eta(z+w)-2zw}{2-\overline\eta(z+w)}.$$

Thus $\overline\eta\,g_a=F_{\overline\eta}$, where

$$F_\omega(z,w)=\frac{z+w-2\omega zw}{2-\omega(z+w)}, \qquad |\omega|=1.$$

These functions satisfy $F_\omega(z,z)=z$. They are the negatives of the Agler–Young magic functions composed with the symmetrization $(z,w)\mapsto(z+w,zw)$: in their notation,

$$F_\omega(z,w)=-\Phi(\omega,z+w,zw).$$

See J. Agler and N. J. Young, The magic functions and automorphisms of a domain, equation (1.1).

4. Earlier bounds and references

Knese's Theorem 1.1 gives $K_2\geq r_0$, where

$$\sum_{n=1}^{\infty}\sqrt n\,r_0^n=\frac12, \qquad r_0\approx 0.3006282829859.$$

Corollary 1.2 states the rounded lower bound $K_2\geq 0.3006$. See Greg Knese, Three radii associated to Schur functions on the polydisk, Proceedings of the American Mathematical Society, Series B 12 (2025), 48–63: journal article, author's text.

Baran, Pikul, Woerdeman, and Wojtylak prove $K_2<0.3177$ in Theorem 6.4 of Contractive realization theory for the annulus and other intersections of disks on the Riemann sphere, Journal of Functional Analysis 290 (2026), no. 8, article 111346: journal article, author's text.

Sebastian Griego's publicly available May 2026 contribution gives the further upper bound $K_2<0.3174541$: mathematical statement and certificate. This reference is a public contribution rather than a journal article.