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\title{A direct proof of the complete Crouzeix inequality}
\author{OpenAI}
\date{September 2026}
\begin{document}
\maketitle
\begin{abstract}
Every square matrix $A$ comes with a compact convex subset $W(A)$ of
the complex plane, called its numerical range. The complete Crouzeix
conjecture says: if you plug $A$ into a matrix-valued polynomial $P$,
the result is never larger than twice the biggest value of $P$ on
$W(A)$, and this is uniform in all matrix dimensions. We give a
self-contained direct proof. It uses an exterior conformal map of an
analytic convex domain containing $W(A)$, a positive kernel on the
unit circle, and a two-by-two positive block comparison. Degenerate
cases, where $W(A)$ is a single point or a line segment, are included.
\end{abstract}
\section{Introduction}
\subsection*{How to read this paper}
Sections~\ref{sec:coords}--\ref{sec:limit} prove the theorem. The one
technical input is the exterior conformal map of Appendix~\ref{sec:collar};
a reader who wants to accept it may take that appendix as a black box.
\subsection*{The numerical range}
Fix a positive integer $n$ and a matrix $A \in M_n(\C)$. A vector
$u \in \C^n$ with $u^*u = 1$ is a \emph{unit vector}. Its
\emph{Rayleigh quotient} is the complex number $u^* A u$. The
\emph{numerical range} of $A$ is the set
\begin{equation}\label{eq:wr}
W(A) \;=\; \{\,u^* A u \,:\, u \in \C^n,\ u^*u = 1\,\}.
\end{equation}
The numerical range of $A$ is always a compact convex subset of $\C$, and
it always contains the spectrum of $A$: a unit eigenvector $u$ of an
eigenvalue $\lambda$ satisfies $u^* A u = \lambda u^* u = \lambda$.
Two extremes: (a) if $A = \diag(\lambda_1, \dots, \lambda_n)$ with all
$\lambda_i \in \mathbb{R}$, then $W(A) = [\min \lambda_i, \max \lambda_i]$;
(b) if $A = \begin{pmatrix}0 & 2\\0 & 0\end{pmatrix}$, then
$|u^*Au| = 2|u_1 u_2| \le |u_1|^2 + |u_2|^2 = 1$, so $W(A)$ is the
closed unit disk $\overline\D$, even though the spectrum is just $\{0\}$.
For a normal matrix (one with $A A^* = A^* A$), the spectral theorem
gives a unitary $U$ with $A = U\Lambda U^*$, and then $u^*Au$ is a
convex combination of the eigenvalues. So $W(A)$ is exactly the convex
hull of the eigenvalues. For a general matrix, this statement fails
badly: the numerical range is genuinely larger than the spectrum,
and that extra size is what makes the spectral-set constant $2$
possible.
\subsection*{The main theorem}
We allow \emph{matrix-valued} polynomials. If
$P(z) = \sum_{k=0}^d B_k z^k$ with coefficient matrices $B_k \in M_m(\C)$, define
\begin{equation}\label{eq:PA}
P[A] \;=\; \sum_{k=0}^d A^k \otimes B_k \;\in\; M_n(\C) \otimes M_m(\C) \;\cong\; M_{nm}(\C).
\end{equation}
Here $\otimes$ is the Kronecker product: the factor $A^k$ acts on
$\C^n$ and the factor $B_k$ acts on $\C^m$, and the two actions commute.
Equivalently, $P[A]$ is obtained by replacing the scalar variable $z$
with the matrix $A$, keeping the coefficients $B_k$ in the exact order
written. The two tensor factors are the \emph{base} $\C^n$ and the
\emph{coefficient} $\C^m$.
All norms are the operator norms from the Euclidean inner products and
their tensor products (equivalently, the largest singular value). The
Hilbert--Schmidt norm of a matrix $C$ is denoted $\|C\|_2 = (\tr C^* C)^{1/2}$.
\begin{theorem}[Complete Crouzeix inequality]\label{thm:main}
Let $n,m \ge 1$, let $A\in M_n(\C)$, let $d \ge 0$, and let
$P(z) = \sum_{k=0}^d B_k z^k$ with $B_k \in M_m(\C)$. Then
\begin{equation}\label{eq:main}
\|P[A]\| \;\le\; 2 \, \max_{z \in W(A)} \, \|P(z)\|.
\end{equation}
The constant $2$ is optimal: it cannot be lowered uniformly in $n, m, d$.
\end{theorem}
The word \emph{complete} refers to the uniformity in the coefficient
dimension $m$. A scalar proof (the case $m = 1$) does not automatically
extend, because scalar coefficients commute but matrix coefficients
need not: $F(z)G(z)$ and $G(z)F(z)$ are different $M_m(\C)$-valued
functions in general, and $F[A]G[A]$ is not $G[A]F[A]$. The proof below
keeps these two products separate all the way through, which is why it
works uniformly in $m$.
\subsection*{Why the constant cannot be less than $2$}
Take $n = 2$, $m = 1$, $A = \begin{pmatrix}0 & 2\\0 & 0\end{pmatrix}$,
and $P(z) = z$. The eigenvalue $0$ of $A$ has a two-dimensional
generalised eigenvector space, and $\|A\| = 2$. Meanwhile the previous
subsection showed that $W(A)$ is the closed unit disk, so
$\max_{W(A)} |P(z)| = \max_{W(A)} |z| = 1$. The ratio
$\|P[A]\|/\max_{W(A)}|P| = 2$, so no universal constant below $2$
works. The rest of the paper proves that $2$ is enough.
\subsection*{Roadmap}
There are five steps. First, enclose $W(A)$ in a bounded convex
domain $\Omega$ with real-analytic Jordan boundary. Second, build an
exterior conformal map $h : \{|\lambda|>1\} \to \C \setminus \overline\Omega$
parametrising $\partial\Omega$ by the unit circle. Third, expand the
Cauchy kernel $q(\lambda)/(h(\lambda)-z)$, $q(\lambda) = \lambda h'(\lambda)$,
at infinity to obtain scalar \emph{Faber polynomials} $b_k(z)$ and a
positive kernel $K$ on $\T$; convexity enters here. Fourth, deduce from
$K$ the two-sided coefficient estimate
\begin{equation}\label{eq:road-coef}
\sum_k \|C_k\|_2^2 \;\le\; \int_\T \|G(h(\mu))\|_2^2\, d\tau(\mu) \;\le\; \|C_0\|_2^2 + 2 \sum_{k\ge1} \|C_k\|_2^2,
\end{equation}
whose factor of $2$ is the sole source of the sharp constant. Fifth,
pair with the exterior resolvent $R(\lambda) = q(\lambda)(h(\lambda)I-A)^{-1}$
and use positivity of $R+R^*$ to pass the coefficient estimate to
$F[A]$ and force $\|F[A]\| \le 2$. Finally, shrink $\Omega$ to $W(A)$.
\subsection*{Historical context}
Crouzeix posed the scalar problem in 2004 \cite{Crouzeix2004} and gave
a universal complete bound of $11.08$ in 2007 \cite{Crouzeix2007}.
Delyon and Delyon introduced positive boundary representations for the
numerical range \cite{DelyonDelyon1999}. Crouzeix and Palencia obtained
the complete bound $1+\sqrt2$ \cite{CrouzeixPalencia2017}, later
simplified by Ransford and Schwenninger \cite{RansfordSchwenninger2018}.
The complete constant $2$ was obtained for base dimension $2$ by
Badea, Crouzeix, Delyon \cite{BadeaCrouzeixDelyon2006} and for base
dimension $3$ by \AA hag, Czy{\.z}, Virtanen \cite{AhagCzyzVirtanen2026}.
The scalar case $m=1$ was settled by Jin \cite{Jin2026},
Lorist and Schwenninger \cite{LoristSchwenninger2026}, and Luo
\cite{Luo2026}. A structural proof of the complete case is in
\cite{OpenAI2026Complete}. The present paper gives a direct,
self-contained argument with all matrix dimensions independent.
\section{Exterior coordinates and Faber coefficients}\label{sec:coords}
Throughout this section, $\Omega \subset \C$ is a bounded open convex
set whose boundary is a \emph{regular real-analytic Jordan curve}:
a simple closed curve parametrised real-analytically with nowhere-zero
derivative. The unit circle is $\T = \{\lambda : |\lambda|=1\}$. The
exterior unit disk is $\C \setminus \overline{\D} = \{\lambda : |\lambda| > 1\}$.
Every circle integral uses $d\tau(e^{it}) = dt/(2\pi)$, so constant
functions of modulus one have integral one.
\subsection*{The exterior coordinate}
By Lemma~\ref{lem:exterior-collar} in the Appendix, there is a number
$0 < r_0 < 1$ and a map
\begin{equation}\label{eq:h}
h(\lambda) \;=\; c\lambda + c_0 + \sum_{j \ge 1} c_j \lambda^{-j}, \qquad c \ne 0,
\end{equation}
meromorphic on $\{\lambda : |\lambda| > r_0\} \cup \{\infty\}$, one-to-one
there as a map of the Riemann sphere, with a simple pole at infinity,
mapping $|\lambda|>1$ conformally onto $\C \setminus \overline\Omega$
and $|\lambda|=1$ onto $\partial\Omega$. This is the
\emph{exterior conformal map}, or \emph{exterior coordinate} of $\Omega$.
The Appendix gives a complete derivation from Riemann's mapping theorem.
The reader who wants to accept it can take \eqref{eq:h} and the
statements below as basic inputs.
\subsection*{The supporting line and the outward normal}
Define
\begin{equation}\label{eq:q}
q(\lambda) \;=\; \lambda h'(\lambda).
\end{equation}
The two partial derivatives of $h$ at a point $e^{i\theta} \in \T$ are
$\partial_\theta h = i q$ (tangent to the boundary) and
$\partial_r h = q$ where $r = |\lambda|$ (radial direction, pointing
outward from $\overline\Omega$). So $q(\lambda)$ is a nonzero outward
normal vector at $h(\lambda) \in \partial\Omega$, and $i q(\lambda)$ is
a positively oriented tangent vector.
Because $\Omega$ is convex, the entire set $\overline\Omega$ lies on the
inner side of the supporting line through $h(\lambda)$ with normal
direction $q(\lambda)$:
\begin{equation}\label{eq:support}
\Re\!\left( \overline{q(\lambda)} \bigl( h(\lambda) - z \bigr) \right) \;\ge\; 0
\qquad (\lambda \in \T,\ z \in \overline\Omega).
\end{equation}
This is the only geometric input from convexity, and it is what makes
the kernel below nonnegative. Figure~\ref{fig:normal} shows the picture.
\begin{figure}[ht]
\centering
\begin{tikzpicture}[scale=.83,>=Latex]
\draw[thick] (0,0) circle (1.05);
\node at (0,0) {$\D$};
\fill (1.05,0) circle (2pt);
\node[below right] at (1.05,0) {$\lambda$};
\draw[->] (1.05,0) -- (1.8,0);
\draw[->] (2.1,.5) to[bend left=15] node[above] {$h$} (4,.5);
\draw[thick] (5.35,0) ellipse (1.1 and .85);
\node at (5.25,-.15) {$\Omega$};
\draw[dashed] (6.45,-1.3) -- (6.45,1.3);
\fill (6.45,0) circle (2pt);
\draw[->,thick] (6.45,0) -- (7.6,0) node[right] {$q(\lambda)$};
\node[above right] at (6.45,.08) {$h(\lambda)$};
\fill (4.95,.35) circle (1.5pt);
\node[above] at (4.95,.35) {$z$};
\end{tikzpicture}
\caption{The map $h$ takes the exterior unit circle to the exterior of
$\overline\Omega$. The vector $q(\lambda) = \lambda h'(\lambda)$ is the
outward normal at $h(\lambda)$. Convexity of $\Omega$ places every
$z \in \overline\Omega$ on the inner side of the supporting line
$\Re(\overline{q(\lambda)}(w - h(\lambda))) = 0$.}
\label{fig:normal}
\end{figure}
\subsection*{Faber polynomials}
Define scalar polynomials $b_k(z)$, $k \ge 0$, by the expansion at infinity
\begin{equation}\label{eq:faber}
\frac{q(\lambda)}{h(\lambda) - z}
\;=\; \sum_{k\ge0} b_k(z)\, \lambda^{-k}, \qquad |\lambda| \gg 1.
\end{equation}
The left side is holomorphic at $\lambda = \infty$ (since
$h(\lambda) \sim c \lambda$ with $c \ne 0$), so the expansion makes sense.
These $b_k$ are the \emph{Faber polynomials} of the exterior coordinate $h$;
they are the natural generalisation of the monomials $z^k$ to the domain
$\Omega$. In the simplest case $\Omega = \D$ with $h(\lambda) = \lambda$,
one has $b_k(z) = z^k$; for a general $\Omega$, the $b_k$ have
corrections coming from the shape of $\partial\Omega$.
We can be more precise: after factoring $c \lambda$ from the denominator,
we find $b_0 = 1$ and, for $k \ge 1$, $b_k$ is a polynomial of degree $k$
with leading coefficient $c^{-k}$. So $(b_k)_{k \ge 0}$ really is a basis
of the space of polynomials, and polynomial degrees match indices.
This is useful because it lets us rewrite any matrix polynomial
$G(z) = \sum_{k=0}^N b_k(z) C_k$ uniquely in the Faber basis, with
$C_k \in M_m(\C)$ in the natural way.
\subsection*{Boundary values and the mixed coefficient}
The boundary restrictions $b_k \circ h$ of the Faber polynomials have a
remarkably clean Fourier form. There are scalars $s_{kj}$, indexed by
$k,j \ge 1$, such that on the exterior collar one has
\begin{equation}\label{eq:boundary-basis}
b_k(h(\mu)) \;=\; \mu^k \;+\; \sum_{j\ge1} s_{kj}\, \mu^{-j}, \qquad k \ge 1.
\end{equation}
Equivalently, the \emph{mixed correction kernel}
\begin{equation}\label{eq:s}
s(\lambda,\mu) \;=\; \frac{q(\lambda)}{h(\lambda) - h(\mu)} \;-\; \frac{\lambda}{\lambda - \mu}
\;=\; \sum_{k,j \ge 1} s_{kj}\, \lambda^{-k} \mu^{-j}
\end{equation}
collects exactly the difference between the true boundary kernel and
the boundary kernel of the unit disk.
The function $s$ has a nice structure. It extends holomorphically
across the diagonal $\lambda = \mu$ and across both infinities, so it
is regular on a full product collar of the exterior. Its mixed Taylor
coefficients decay geometrically:
\begin{equation}\label{eq:s-decay}
|s_{kj}| \;\le\; C_\rho\, \rho^{k+j} \qquad (r_0 < \rho < 1,\ k,j \ge 1).
\end{equation}
Consequently the double series converges normally on the exterior
product collar, hence absolutely and uniformly on $\T^2$. So every
interchange of sum and integral below is legal.
The kernel we will need is
\begin{equation}\label{eq:K}
K(\lambda,\mu) \;=\; 1 + s(\lambda,\mu) + \overline{s(\lambda,\mu)}.
\end{equation}
Since $s$ has only strictly mixed Fourier modes
($\lambda^{-k}\mu^{-j}$ with $k,j \ge 1$), neither $s$ nor its conjugate
contributes to the integral in $\lambda$ alone or in $\mu$ alone. So
\[
\int_\T K(\lambda,\mu)\, d\tau(\lambda) = 1 \quad (\text{for each fixed } \mu),
\qquad
\int_\T K(\lambda,\mu)\, d\tau(\mu) = 1 \quad (\text{for each fixed } \lambda).
\]
These are the \emph{marginal identities}. They say: $K$ is like the
kernel of the identity operator, but with a real, symmetric, nonnegative
correction.
\subsection*{The positive kernel theorem}
\begin{theorem}[Exterior Faber coefficients and the positive kernel]\label{lem:faber}
Let $\Omega$ be a bounded convex domain with regular real-analytic Jordan
boundary. Let $h$ be as in \eqref{eq:h}, and let $q$, $(b_k)_{k\ge0}$,
and $s_{kj}$ be defined by \eqref{eq:q}, \eqref{eq:faber}, and
\eqref{eq:s}. Then:
\begin{enumerate}
\item $b_0 = 1$, and for each $k \ge 1$, $b_k$ is a polynomial of
degree $k$ with leading coefficient $c^{-k}$ in \eqref{eq:h}.
\item The mixed coefficients $s_{kj}$ satisfy the geometric decay
estimate \eqref{eq:s-decay}.
\item The kernel $K$ in \eqref{eq:K} is continuous on $\T^2$,
nonnegative everywhere, and has integral one in each variable
with the other fixed.
\end{enumerate}
\end{theorem}
\begin{proof}
\emph{Part 1 (degree and leading coefficient).}
Write $h(\lambda) - z = c\lambda \cdot (1 + (c_0 - z)/(c\lambda) + O(\lambda^{-2}))$.
For large $|\lambda|$,
\[
\frac{q(\lambda)}{h(\lambda) - z}
= \frac{c\lambda + O(\lambda^{-1})}{c\lambda} \cdot
\left(1 + \frac{c_0 - z}{c\lambda} + \cdots\right)^{-1}
= 1 + O(\lambda^{-1}).
\]
Expanding the inverse as a geometric series in $(c_0-z)/(c\lambda)$
produces, for the coefficient of $\lambda^{-k}$, a polynomial in $z$
of degree $k$. Inspecting the largest power of $z$ shows it is $c^{-k} z^k$.
So $b_0 = 1$ and $b_k(z) = c^{-k} z^k + (\text{lower degree})$ for $k \ge 1$.
In particular the $b_k$ form a basis of the scalar polynomials.
\emph{Part 2 (geometric decay and joint extension).}
Consider
\begin{equation}\label{eq:H}
H(\lambda,\mu) \;=\; \frac{h(\lambda) - h(\mu)}{\lambda - \mu}.
\end{equation}
Write $u = \lambda^{-1}$, $v = \mu^{-1}$, and $h(\lambda) = c\lambda + g(u)$
with $g$ holomorphic on $|u| < r_0^{-1}$. Then
\[
H(\lambda,\mu) = c - uv \, \frac{g(u) - g(v)}{u - v}.
\]
The divided difference $(g(u)-g(v))/(u-v)$ is jointly holomorphic in
$(u,v)$, with diagonal value $g'(u)$, so this formula covers the diagonal
and infinities. Direct differentiation gives
\[
s(\lambda,\mu) \;=\; \frac{\lambda \partial_\lambda H}{H} \;=\; -\frac{u \partial_u H}{H}.
\]
Since $H$ is nonzero on the reciprocal bidisk $|u|,|v| < r_0^{-1}$
(injectivity of $h$, and $H \to c$ at infinity), $s$ is holomorphic
there and vanishes on both axes $\{u=0\}$ and $\{v=0\}$. Its Taylor
expansion therefore contains exactly the strictly mixed powers
$\lambda^{-k}\mu^{-j}$ with $k,j \ge 1$.
For any $r_0 < \rho < 1$, apply Cauchy estimates on the closed disk
$|u|,|v| \le \rho^{-1}$: there is a constant $C_\rho$ with
$|s_{kj}| \le C_\rho \rho^{k+j}$. Summing the resulting geometric series
shows the Laurent expansion \eqref{eq:s} converges normally (absolutely
and uniformly on compacta, including those meeting infinity) throughout
the product collar $|\lambda|,|\mu| > r_0$. Restricting to $|\lambda| = |\mu| = 1$
gives absolute uniform convergence on $\T^2$.
Comparing \eqref{eq:faber} and \eqref{eq:s} coefficient-by-coefficient
in $\lambda^{-k}$, with $\mu$ fixed in the exterior collar, yields
\eqref{eq:boundary-basis}. Indeed, $\lambda/(\lambda-\mu) = \sum_{k \ge 0} \mu^k \lambda^{-k}$,
so the difference of the two sides matches $b_k(h(\mu)) - \mu^k$.
\emph{Part 3 (positivity of $K$ and marginal identities).}
For distinct $\lambda,\mu \in \T$, split $K$ as
\[
K(\lambda,\mu) \;=\; 2\,\Re \frac{q(\lambda)}{h(\lambda) - h(\mu)} \;+\; \left(1 - \frac{\lambda}{\lambda-\mu} - \frac{\bar\lambda}{\bar\lambda-\bar\mu}\right).
\]
On $|\lambda| = |\mu| = 1$, we have $\bar\lambda = 1/\lambda$, $\bar\mu = 1/\mu$, and
one verifies directly that $\lambda/(\lambda-\mu) + \bar\lambda/(\bar\lambda-\bar\mu) = 1$,
so the bracketed term vanishes. Hence
\[
K(\lambda,\mu) \;=\; 2\,\Re \frac{q(\lambda)}{h(\lambda) - h(\mu)}.
\]
Now $\Re(q/(h(\lambda) - h(\mu))) = |q|^2 / \Re(\overline{q}(h(\lambda) - h(\mu)))$
where $|q|^2 > 0$, and the denominator is nonnegative by \eqref{eq:support}
with $z = h(\mu)$. So $K(\lambda,\mu) \ge 0$ for distinct $\lambda,\mu$,
and continuity gives the same on the diagonal.
For the marginal identities: every term in $s$ has Fourier mode
$\lambda^{-k}\mu^{-j}$ with $k,j \ge 1$, so integrating against
$d\tau(\lambda)$ picks out $k=0$, which never occurs; and similarly for
the conjugate. Hence
$\int_\T s(\lambda,\mu) d\tau(\lambda) = \int_\T \overline{s(\lambda,\mu)} d\tau(\lambda) = 0$,
so $\int_\T K(\lambda,\mu) d\tau(\lambda) = 1$. Symmetric argument in $\mu$.
\end{proof}
\subsection*{Identification with Faber polynomials}
Since \eqref{eq:boundary-basis} gives $b_k(z) - h^{-1}(z)^k = O(z^{-1})$
as $z \to \infty$, the $b_k$ are the classical \emph{Faber polynomials}
of the exterior coordinate $h$ \cite{Beckermann2005,BeckermannCrouzeix2014}.
\section{The coefficient norm estimate}\label{sec:coef}
The positive kernel $K$ from Theorem~\ref{lem:faber} is the entire source
of the constant $2$ in the main theorem. This section extracts one clean
inequality from it. Fix any dimension $m$ and any finite-family of
coefficient matrices $C_0, C_1, \dots, C_N \in M_m(\C)$, and put
\[
G(z) \;=\; \sum_{k=0}^N b_k(z)\, C_k.
\]
That is, $G$ is an $M_m(\C)$-valued polynomial written in the Faber basis
of $\Omega$. We want to bound the Hilbert--Schmidt $L^2$-norm of
$G \circ h$ on the unit circle from above and below by the coefficient
Hilbert--Schmidt norms $\|C_k\|_2$.
\subsection*{The operator $T$ defined by $K$}
Consider the bounded operator
\begin{equation}\label{eq:T}
(Tu)(\mu) \;=\; \int_\T K(\lambda,\mu)\, u(\lambda)\, d\tau(\lambda),
\qquad u \in L^2(\T;\mathcal H),
\end{equation}
where $\mathcal H$ is any separable Hilbert space (in practice
$\mathcal H = M_m(\C)$ with the Hilbert--Schmidt inner product).
\begin{lemma}\label{lem:T}
The operator $T$ extends to a contraction of $L^2(\T;\mathcal H)$.
\end{lemma}
\begin{proof}
Nonnegativity $K \ge 0$ and the marginal identity
$\int_\T K(\lambda,\mu)\, d\tau(\lambda) = 1$ together give, for each
fixed $\mu$, by Cauchy--Schwarz,
\[
\|(Tu)(\mu)\|^2
\;\le\; \left(\int_\T K(\lambda,\mu)\, d\tau(\lambda)\right)
\left(\int_\T K(\lambda,\mu)\, \|u(\lambda)\|^2\, d\tau(\lambda)\right)
\;=\; \int_\T K(\lambda,\mu)\, \|u(\lambda)\|^2\, d\tau(\lambda).
\]
Integrate over $\mu$ and use the other marginal identity:
\[
\|Tu\|_{L^2}^2
\;\le\; \int_\T \left(\int_\T K(\lambda,\mu)\, d\tau(\mu)\right)\, \|u(\lambda)\|^2\, d\tau(\lambda)
\;=\; \|u\|_{L^2}^2.
\]
\end{proof}
\subsection*{What $T$ does on monomials}
For $u(\lambda) = \sum_{k=1}^N C_k \lambda^k$ (each $C_k \in M_m(\C)$),
only the $s$ summand of $K$ contributes to $(Tu)(\mu)$: the term $1$ picks
up $\int_\T \lambda^k\, d\tau = 0$, and the $\overline{s}$ term has only
positive powers of $\lambda$. Hence
\begin{equation}\label{eq:Tu-monom}
(Tu)(\mu) \;=\; \sum_{j\ge1} \Bigl(\sum_{k=1}^N s_{kj}\, C_k\Bigr)\, \mu^{-j}.
\end{equation}
(The normal convergence of Theorem~\ref{lem:faber} justifies every
interchange.) Since the $\mu^{-j}$ are orthonormal in $L^2(\T)$,
\[
\|Tu\|_{L^2}^2 \;=\; \sum_{j\ge1} \Bigl\|\sum_{k=1}^N s_{kj}\, C_k\Bigr\|_2^2,
\]
where $\|\cdot\|_2$ denotes the Hilbert--Schmidt norm on $M_m(\C)$.
Combined with Lemma~\ref{lem:T},
\begin{equation}\label{eq:coef-contraction}
\sum_{j\ge1} \Bigl\|\sum_{k=1}^N s_{kj}\, C_k\Bigr\|_2^2 \;\le\; \sum_{k=1}^N \|C_k\|_2^2.
\end{equation}
\subsection*{The two-sided coefficient inequality}
By \eqref{eq:boundary-basis},
\[
G(h(\mu)) \;=\; C_0 \;+\; \sum_{k=1}^N C_k \mu^k \;+\; \sum_{j\ge1} \Bigl(\sum_{k=1}^N s_{kj} C_k\Bigr) \mu^{-j}
\;=\; C_0 + u(\mu) + (Tu)(\mu).
\]
The three summands have disjoint Fourier supports (constant, positive,
negative), so Parseval and \eqref{eq:coef-contraction} give
\[
\|C_0\|_2^2 + \sum_{k=1}^N \|C_k\|_2^2
\;\le\; \int_\T \|G(h(\mu))\|_2^2\, d\tau(\mu)
\;\le\; \|C_0\|_2^2 + 2 \sum_{k=1}^N \|C_k\|_2^2.
\]
Passing from coefficients to boundary values never shrinks the coefficient
Hilbert--Schmidt norm, and can enlarge it by at most a factor of $2$.
\begin{proposition}[The coefficient norm estimate]\label{prop:coefficients}
For every positive integer $m$, every integer $N \ge 0$, and every
$M_m(\C)$-valued polynomial $G(z) = \sum_{k=0}^N b_k(z)\, C_k$ in the
Faber basis of $\Omega$, one has
\begin{equation}\label{eq:coefficient-estimate}
\sum_{k=0}^N \|C_k\|_2^2
\;\le\; \int_\T \|G(h(\mu))\|_2^2\, d\tau(\mu)
\;\le\; \|C_0\|_2^2 + 2 \sum_{k=1}^N \|C_k\|_2^2.
\end{equation}
\end{proposition}
\begin{proof}
This is exactly the combined identity above. The lower bound uses only
$\|Tu\|^2 \ge 0$. The upper bound uses Lemma~\ref{lem:T} to shoehorn
$\|Tu\|_{L^2} \le \sum_k \|C_k\|_2$ into the Parseval identity.
\end{proof}
\subsection*{A note on the literature}
In the notation of \cite[Proposition~3.1]{OpenAI2026Complete}, our
$h,\lambda,\mu$ correspond to the companion's $G,t,w$, and
$s(\lambda,\mu) = B(\mu,\lambda)$ (i.e., $s_{kj} = \beta_{jk}$).
Equation~\eqref{eq:Tu-monom} uses this orientation and does not
require symmetry of the kernel.
\section{The resolvent and the double-layer positivity}\label{sec:resolvent}
From now on the matrix $A \in M_n(\C)$ is fixed, with $W(A) \subset \Omega$.
We will also fix the exterior coordinate $h$ of $\Omega$ from
section~\ref{sec:coords}. The purpose of this section is to define the
\emph{exterior resolvent} and show that it has a positive Hermitian part.
\subsection*{The resolvent function}
Define, on the exterior circle $\T$,
\begin{equation}\label{eq:R-def}
R(\lambda) \;=\; q(\lambda) \left( h(\lambda) I - A \right)^{-1}
\;\in\; M_n(\C).
\end{equation}
Here $h(\lambda) I - A$ is invertible whenever $h(\lambda) \notin \sigma(A)$.
Since $\sigma(A) \subset W(A) \subset \Omega$ (compact) and the circle
$\T$ parametrises $\partial\Omega$ (compact, disjoint from $\sigma(A)$),
the resolvent exists and is continuous on $\T$, and extends
holomorphically to a collar $\{|\lambda| > r_A\}$ for some $r_0 < r_A < 1$
with $r_A$ closer to one than $r_0$.
\begin{theorem}[Resolvent expansion and double-layer positivity]\label{lem:resolvent}
Let $\Omega, h, q, (b_k)$ be as above. For every $n \ge 1$ and every
$A \in M_n(\C)$ with $W(A) \subset \Omega$, the following hold.
\begin{enumerate}
\item \emph{Expansion.} $R$ has the Laurent series
\begin{equation}\label{eq:R-expansion}
R(\lambda) \;=\; \sum_{k\ge0} b_k(A)\, \lambda^{-k}, \qquad |\lambda| \ge 1,
\end{equation}
converging absolutely and uniformly on $\T$. The coefficient
matrices $b_k(A)$ decay geometrically: there are constants
$C < \infty$ and $0 < \theta < 1$ (depending on $A$ and $h$) with
$\|b_k(A)\| \le C \theta^k$ for all $k \ge 0$.
\item \emph{Means.} $\int_\T R(\lambda)\, d\tau(\lambda) = I$ and
$\int_\T (R(\lambda) + R(\lambda)^*)\, d\tau(\lambda) = 2I$.
\item \emph{Positivity.} On the exterior circle,
$R(\lambda) + R(\lambda)^* \succeq 0$.
\end{enumerate}
\end{theorem}
\begin{proof}
\emph{Part 1 (expansion and decay).}
Every eigenvalue of $A$ belongs to $W(A) \subset \Omega$, by testing
the corresponding unit eigenvector. Hence the resolvent
$(h(\lambda) I - A)^{-1}$ is holomorphic on and outside $\T$, including
a small exterior collar. Expanding at $\infty$ in $\lambda$ gives a
Laurent series in $\lambda^{-1}$, and comparing with \eqref{eq:faber}
with $z$ replaced by $A$ shows that the coefficient of $\lambda^{-k}$
is exactly $b_k(A)$. The series converges absolutely and uniformly on
$\T$ by compactness of the exterior collar. To get the geometric
bound, pick $r_A < \theta < 1$ with $R$ holomorphic on
$|\lambda| > r_A$, apply Cauchy estimates on the circle
$|\lambda| = \theta^{-1}$.
\emph{Part 2 (means).}
The constant coefficient $b_0(A) = b_0(z)\big|_{z=A} = I$
(Theorem~\ref{lem:faber}, part 1). So by \eqref{eq:R-expansion} and
orthogonality of the modes $\lambda^{-k}$,
$\int_\T R\, d\tau = b_0(A) = I$. Taking adjoints and adding gives
$\int_\T (R+R^*)\, d\tau = 2I$.
\emph{Part 3 (positivity).}
Let $B(\lambda) = h(\lambda) I - A$ so that $R = q B^{-1}$. Then
$R^* = \overline{q}\, (B^*)^{-1}$ and
\[
R + R^* \;=\; q\, B^{-1} + \overline{q}\, (B^*)^{-1}
\;=\; (B^*)^{-1} \left( \overline{q}\, B + q\, B^* \right) B^{-1}.
\]
Since $(B^*)^{-1}$ and $B^{-1}$ are adjoints of each other, they give a
congruence that preserves positivity. So it suffices to show that
$\overline{q(\lambda)}\, B(\lambda) + q(\lambda)\, B(\lambda)^* \succeq 0$.
For any unit vector $v \in \C^n$, the quadratic form on the left at $v$
is
\[
v^* \left( \overline{q}\, B + q\, B^* \right) v
\;=\; \overline{q}\, v^* B v + q\, v^* B^* v
\;=\; 2 \Re\!\left( \overline{q(\lambda)}\, v^* B(\lambda) v \right).
\]
Substituting $B = h(\lambda) I - A$:
\[
v^* B v \;=\; h(\lambda) - v^* A v.
\]
So the quadratic form is
$\Re\!\left( \overline{q(\lambda)} (h(\lambda) - v^* A v) \right)$.
By the support inequality \eqref{eq:support} with $z = v^* A v \in W(A) \subset \overline\Omega$,
this is $\ge 0$. The congruence argument finishes the proof.
\end{proof}
\section{The main estimate on an analytic convex domain}\label{sec:matrix}
Now the payoff. In this section $\Omega, h, q, (b_k)$ are as in
Section~\ref{sec:coords}, $A \in M_n(\C)$ satisfies $W(A) \subset \Omega$,
and $R(\lambda) = q(\lambda)(h(\lambda) I - A)^{-1}$ is the exterior
resolvent from Theorem~\ref{lem:resolvent}.
\subsection*{Statement and strategy}
\begin{theorem}[Bound on an analytic convex domain]\label{thm:domain}
Let $n, m \ge 1$. Let $\Omega \subset \C$ be bounded, open, convex,
with regular real-analytic Jordan boundary, and let $A \in M_n(\C)$ with
$W(A) \subset \Omega$. Let $F$ be an $M_m(\C)$-valued polynomial with
$\|F(z)\| \le 1$ on $\partial\Omega$ (equivalently, on $\overline\Omega$,
since $F$ is a polynomial and the boundary controls the closure). Then
\[
\|F[A]\| \;\le\; 2.
\]
\end{theorem}
Three moves: top singular vectors $x, y$ turn the norm into a scalar
$y^* F[A] x = \|F[A]\|$; the two products $FG$ and $GF$ turn the diagonal
functionals $x^* G[A] x$ and $y^* G[A] y$ into an off-diagonal one; and
$R+R^* \succeq 0$ compares the resulting $P+P^*$, $Q+Q^*$, $M+L^*$ fields.
\subsection*{Setup: singular vectors and block notation}
Put $\gamma = \|F[A]\|$. If $\gamma > 0$, choose unit vectors $x, y$
in $\C^n \otimes \C^m$ forming a top singular pair:
\begin{equation}\label{eq:singular}
F[A] x \;=\; \gamma y, \qquad F[A]^* y \;=\; \gamma x.
\end{equation}
If $\gamma = 0$, there is nothing to prove.
Write $x = \sum_{i=1}^m X e_i \otimes e_i$ and
$y = \sum_{i=1}^m Y e_i \otimes e_i$ for $n \times m$ matrices $X, Y$,
where $e_1,\dots,e_m$ is the standard basis of $\C^m$. Then
$\|X\|_2^2 = \|Y\|_2^2 = 1$ (with $\|X\|_2^2 = \tr(X^* X)$).
Define the $m \times m$ matrix functions
\begin{equation}\label{eq:PQML}
P = X^* R X, \quad Q = Y^* R Y, \quad M = Y^* R X, \quad L = X^* R Y.
\end{equation}
Their Laurent coefficients in the variable $\lambda^{-1}$ will be
denoted with subscripts: $M_k = Y^* b_k(A) X$, $P_k = X^* b_k(A) X$,
$Q_k = Y^* b_k(A) Y$, $L_k = X^* b_k(A) Y$. Since $\|b_k(A)\| \le C \theta^k$
(Theorem~\ref{lem:resolvent}), all four series converge absolutely and
uniformly on $\T$, and all four are square-summable in the Hilbert-Schmidt
norm.
Important bookkeeping at $k = 0$. Since $b_0(A) = I$,
\begin{equation}\label{eq:PQML-0}
P_0 = X^* X, \quad Q_0 = Y^* Y, \quad L_0^* = M_0.
\end{equation}
In particular $P_0$ and $Q_0$ are Hermitian and positive semidefinite.
Finally, set
\begin{equation}\label{eq:a}
a^2 \;=\; \sum_{k\ge0} \|M_k\|_2^2, \qquad\text{finite and } a > 0.
\end{equation}
The finiteness is from the geometric decay of $b_k(A)$. The positivity
$a > 0$ follows from applying the cross-functional bound below to $G = F$:
the value $y^* F[A] x = \gamma > 0$ cannot be achieved by a Cauchy-Schwarz
bound with a zero right-hand side.
\subsection*{Step A. The off-diagonal functional}
For any $M_m(\C)$-valued polynomial $G(z) = \sum_{k=0}^N b_k(z) C_k$, direct
expansion in the tensor product gives
\begin{equation}\label{eq:pairing}
y^* G[A] x \;=\; \sum_{k=0}^N \sum_{i,j=1}^m (M_k)_{ij}\, (C_k)_{ij}.
\end{equation}
Indeed, $G[A] x = \sum_{k,j} (b_k(A) X e_j) \otimes (C_k e_j)$, so
\[
y^* G[A] x \;=\; \sum_{k,i,j} (Y e_i)^* b_k(A) (X e_j) \cdot e_i^* C_k e_j,
\]
which is \eqref{eq:pairing}.
Cauchy-Schwarz on \eqref{eq:pairing} gives
\begin{equation}\label{eq:crossfunctional}
|y^* G[A] x|
\;\le\; \left( \sum_{k\ge0} \|M_k\|_2^2 \right)^{1/2}
\left( \sum_{k=0}^N \|C_k\|_2^2 \right)^{1/2}
\;=\; a \left( \sum_{k=0}^N \|C_k\|_2^2 \right)^{1/2}.
\end{equation}
Plugging in $G = F$ and using $y^* F[A] x = \gamma$ from \eqref{eq:singular}
shows that $a > 0$.
\subsection*{Step B. Two ordered tests}
Polynomial evaluation with matrix coefficients respects multiplication
in the given order. If $F(z) = \sum_i D_i z^i$ and $G(z) = \sum_j E_j z^j$,
then
\[
(FG)[A] \;=\; \sum_{i,j} A^{i+j} \otimes D_i E_j \;=\; F[A] G[A].
\]
Similarly $(GF)[A] = G[A] F[A]$ when factors are in the reversed order.
Multiplying \eqref{eq:singular} on the left by suitable pieces,
\begin{equation}\label{eq:ordered-products}
y^* (FG)[A] x \;=\; y^* F[A] G[A] x \;=\; \gamma\, x^* G[A] x,
\qquad
y^* (GF)[A] x \;=\; \gamma\, y^* G[A] y.
\end{equation}
So the diagonal functionals $x^* G[A] x$ and $y^* G[A] y$ can be turned
into the off-diagonal functional by ordering the two factors differently.
Write the two products in the Faber basis:
\[
FG \;=\; \sum_{k\ge0} b_k U_k, \qquad GF \;=\; \sum_{k\ge0} b_k V_k,
\]
with finitely many nonzero coefficient matrices $U_k, V_k \in M_m(\C)$.
Define the boundary norm
\[
\|H\|_\partial^2 \;=\; \int_\T \|H(h(\mu))\|_2^2\, d\tau(\mu),
\qquad H \in M_m(\C)[z].
\]
Because $\|F(z)\| \le 1$ on the boundary, submultiplicativity of the
pointwise operator norm gives
\[
\|FG\|_\partial \;\le\; \|F\|_\partial\, \|G\|_\partial \;\le\; \|G\|_\partial,
\qquad
\|GF\|_\partial \;\le\; \|G\|_\partial.
\]
(Using $\|F(z)G(z)\|_2 \le \|F(z)\| \|G(z)\|_2$ and its reverse-order
analog.)
Apply \eqref{eq:crossfunctional} to $FG$ and $GF$, and use
\eqref{eq:ordered-products}, the lower coefficient estimate from
Proposition~\ref{prop:coefficients}, and the boundary multiplier bounds:
\begin{align*}
\gamma\, |x^* G[A] x|
&= |y^* (FG)[A] x|
\le a \left( \sum_k \|U_k\|_2^2 \right)^{1/2}
\le a \|FG\|_\partial
\le a \|G\|_\partial, \\
\gamma\, |y^* G[A] y|
&= |y^* (GF)[A] x|
\le a \left( \sum_k \|V_k\|_2^2 \right)^{1/2}
\le a \|GF\|_\partial
\le a \|G\|_\partial.
\end{align*}
The upper coefficient estimate for $G$ in Proposition~\ref{prop:coefficients}
turns $\|G\|_\partial$ into a norm of the coefficient array:
$\|G\|_\partial \le (\|C_0\|_2^2 + 2 \sum_{k\ge1} \|C_k\|_2^2)^{1/2}$.
Therefore
\begin{equation}\label{eq:diagonal-functional}
\max\bigl\{ |x^* G[A] x|, |y^* G[A] y| \bigr\}
\;\le\; \frac{a}{\gamma}\,
\left( \|C_0\|_2^2 \;+\; 2 \sum_{k\ge1} \|C_k\|_2^2 \right)^{1/2}.
\end{equation}
This is the crucial estimate that produces the weight of two.
\subsection*{Step C. Weighted dual norms}
The functional $x^* G[A] x$ pairs $C_k$ against $P_k$
(repeat the calculation of \eqref{eq:pairing}, but with $y$ replaced by
$x$; the coefficient array is $P_k = X^* b_k(A) X$). Similarly $y^* G[A] y$
pairs $C_k$ against $Q_k$. So \eqref{eq:diagonal-functional} constrains
the size of the sequences $(P_k)$ and $(Q_k)$, with weights that reverse
the ones in \eqref{eq:diagonal-functional}.
Concretely, for each $N \ge 0$ set
\[
d_N^{(P)} \;=\; \|P_0\|_2^2 \;+\; \tfrac12 \sum_{k=1}^N \|P_k\|_2^2.
\]
Test \eqref{eq:diagonal-functional} with the coefficient sequence
\[
C_0 = \overline{P_0}, \qquad C_k = \overline{P_k}/2 \;(1 \le k \le N), \qquad C_k = 0 \;(k > N),
\]
where the bar denotes entrywise complex conjugation. The scalar paired
value is exactly $d_N^{(P)}$. The weighted norm on the right is
$\|C_0\|_2^2 + 2 \sum_{k=1}^N \|C_k\|_2^2 = \|P_0\|_2^2 + \tfrac12 \sum_{k=1}^N \|P_k\|_2^2 = d_N^{(P)}$.
So
\[
d_N^{(P)} \;\le\; \frac{a}{\gamma} \,(d_N^{(P)})^{1/2}.
\]
If $d_N^{(P)} = 0$ this is trivial. Otherwise divide and square:
$d_N^{(P)} \le a^2/\gamma^2$. Letting $N \to \infty$:
\begin{equation}\label{eq:weighted-dual}
\|P_0\|_2^2 + \tfrac12 \sum_{k\ge1} \|P_k\|_2^2 \;\le\; \frac{a^2}{\gamma^2},
\qquad
\|Q_0\|_2^2 + \tfrac12 \sum_{k\ge1} \|Q_k\|_2^2 \;\le\; \frac{a^2}{\gamma^2}.
\end{equation}
The same finite test with $Q_k$ in place of $P_k$ gives the second bound.
\subsection*{Step D. Boundary $L^2$ norms of the blocks}
Now compute the square-integrated Hilbert-Schmidt norms of the three
extreme matrix fields $P+P^*$, $Q+Q^*$, and $M+L^*$ on the circle.
\emph{Diagonal blocks.} The function $P+P^*$ has Laurent expansion
\[
(P+P^*)(\lambda) \;=\; 2 P_0 \;+\; \sum_{k\ge1} P_k\, \lambda^{-k} \;+\; \sum_{k\ge1} P_k^*\, \lambda^{k}
\]
where we used $P_0^* = P_0$ and $\overline{\lambda}^k = \lambda^{-k}$ on
$|\lambda| = 1$. The three groups (constant, negative powers, positive
powers) have disjoint Fourier support, so Parseval gives
\[
\int_\T \|P+P^*\|_2^2\, d\tau(\lambda)
\;=\; \|2 P_0\|_2^2 \;+\; \sum_{k\ge1} \|P_k\|_2^2 \;+\; \sum_{k\ge1} \|P_k^*\|_2^2
\;=\; 4 \|P_0\|_2^2 \;+\; 2 \sum_{k\ge1} \|P_k\|_2^2
\;=\; 4 d_\infty^{(P)},
\]
where $d_\infty^{(P)} = \|P_0\|_2^2 + \tfrac12 \sum_{k\ge1} \|P_k\|_2^2$
is the limit of the sequence $d_N^{(P)}$ from Step C.
Combined with \eqref{eq:weighted-dual},
\begin{equation}\label{eq:diag-L2}
\int_\T \|P+P^*\|_2^2\, d\tau \;\le\; 4 a^2 / \gamma^2,
\qquad
\int_\T \|Q+Q^*\|_2^2\, d\tau \;\le\; 4 a^2 / \gamma^2.
\end{equation}
The second inequality follows from the second half of \eqref{eq:weighted-dual}
by the same calculation with $Q$ in place of $P$.
\emph{Cross block.} The function $M+L^*$ has Laurent expansion
\[
(M+L^*)(\lambda) \;=\; \sum_{k\ge0} M_k\, \lambda^{-k} \;+\; \sum_{k\ge0} L_k^*\, \lambda^{k}.
\]
The constant coefficient is $M_0 + L_0^* = 2 M_0$, using $L_0^* = M_0$
from \eqref{eq:PQML-0}. Grouping by Fourier mode,
\[
\int_\T \|M+L^*\|_2^2\, d\tau
\;=\; \|2 M_0\|_2^2 \;+\; \sum_{k\ge1} \|M_k\|_2^2 \;+\; \sum_{k\ge1} \|L_k^*\|_2^2.
\]
Since $\|L_k^*\|_2 = \|L_k\|_2$ (the Hilbert-Schmidt norm is invariant
under adjoint) and every term is nonnegative,
\begin{equation}\label{eq:cross-L2}
\int_\T \|M+L^*\|_2^2\, d\tau
\;\ge\; 4 \|M_0\|_2^2 \;+\; \sum_{k\ge1} \|M_k\|_2^2
\;\ge\; \sum_{k\ge0} \|M_k\|_2^2
\;=\; a^2.
\end{equation}
The final inequality holds because the coefficient of $4\|M_0\|_2^2$
exceeds the coefficient of $\|M_0\|_2^2$; dropping the $\|L_k\|_2^2$ terms
only makes the left smaller.
\subsection*{Step E. Positive block comparison}
The proof now closes with linear algebra.
From Theorem~\ref{lem:resolvent},
$R(\lambda) + R(\lambda)^* \succeq 0$ for every $\lambda \in \T$.
Compress by the $n \times 2m$ matrix $[X\ Y]$. The $2m \times 2m$ matrix
of blocks
\[
\Delta(\lambda) \;=\; \begin{pmatrix} P+P^* & L+M^* \\ M+L^* & Q+Q^* \end{pmatrix}(\lambda)
\;=\; \begin{pmatrix} X \\ Y \end{pmatrix}^* (R+R^*)(\lambda)\, \begin{pmatrix} X & Y \end{pmatrix}
\;\succeq\; 0
\]
is positive semidefinite for every $\lambda$. A compression of a positive
semidefinite matrix is positive semidefinite, so this is immediate.
Let $J = \diag(I_m, -I_m)$. If $\Delta \succeq 0$, then $J \Delta J \succeq 0$
too (the conjugation by $J$ keeps positivity). If two matrices $S$
and $T$ are positive semidefinite, then $\tr(S T) \ge 0$: writing
$S = A A^*$ and $T = B B^*$, $\tr(S T) = \tr(A A^* B B^*) = \tr((A^* B)^* (A^* B)) = \|A^* B\|_2^2 \ge 0$.
So
\[
0 \;\le\; \tr(\Delta (J \Delta J)) \;=\; \tr((\Delta J)^2).
\]
Compute $(\Delta J)^2$ directly. $\Delta J = \begin{pmatrix} P+P^* & -(L+M^*) \\ M+L^* & -(Q+Q^*) \end{pmatrix}$.
Then
\[
\tr\bigl((\Delta J)^2\bigr)
\;=\; \tr\bigl((P+P^*)^2\bigr) \;-\; \tr\bigl((L+M^*)(M+L^*)\bigr) \;-\; \tr\bigl((M+L^*)(L+M^*)\bigr) \;+\; \tr\bigl((Q+Q^*)^2\bigr).
\]
Both $(L+M^*)(M+L^*) = AA^*$ and $(M+L^*)(L+M^*) = A^*A$
(with $A = L+M^*$) have the same trace $\|A\|_2^2 = \|M+L^*\|_2^2 \ge 0$,
so the two middle terms sum to
\[
-\tr\bigl((L+M^*)(M+L^*) + (M+L^*)(L+M^*)\bigr) \;=\; -2 \tr(L+M^*)^*(L+M^*) \;=\; -2 \|M+L^*\|_2^2.
\]
Putting it all together:
\begin{equation}\label{eq:block-trace}
0 \;\le\; \|P+P^*\|_2^2 \;+\; \|Q+Q^*\|_2^2 \;-\; 2 \|M+L^*\|_2^2.
\end{equation}
This holds for every $\lambda \in \T$.
Integrate \eqref{eq:block-trace} over $\mu \in \T$, use \eqref{eq:diag-L2}
for the two diagonal terms and \eqref{eq:cross-L2} for the cross term:
\[
0 \;\le\; 4 a^2/\gamma^2 + 4 a^2/\gamma^2 \;-\; 2 a^2
\;=\; 8 a^2/\gamma^2 \;-\; 2 a^2.
\]
So
\[
2 a^2 \;\le\; 8 a^2 / \gamma^2.
\]
Since $a > 0$, divide by $a^2$: $\gamma^2 \le 4$, hence $\gamma \le 2$.
This finishes the proof that $\|F[A]\| \le 2$.
$\square$
\subsection*{Remarks on the argument}
The factor of two in \eqref{eq:coefficient-estimate} is sharp
(Section~\ref{sec:limit}). Convexity enters only through
\eqref{eq:support}. For scalar $m=1$ one could use $FG=GF$; here the
two products are genuinely different, and using them separately is the
point of Step B.
\section{Enclosing domains and the general case}\label{sec:limit}
We have proved $\|F[A]\| \le 2$ under the hypothesis
$W(A) \subset \Omega$ with $\Omega$ an analytic convex domain. We now
remove that hypothesis: for an arbitrary compact convex $W(A)$ we
construct analytic convex neighborhoods $\Omega_j$ shrinking to $W(A)$,
apply the previous section, and pass to the limit.
\subsection*{Analytic convex outer approximation}
\begin{lemma}[Analytic convex outer approximation]\label{lem:approx}
Let $K \subset \C$ be nonempty, compact, and convex. There exist bounded
open convex domains $\Omega_j \subset \C$ with regular real-analytic
Jordan boundaries such that $K \subset \Omega_j$ for every $j$ and
\[
\sup_{z \in \overline\Omega_j} \dist(z, K) \;\xrightarrow[j\to\infty]{}\; 0.
\]
\end{lemma}
\begin{proof}
The construction is a smooth convex interpolation between $K$ and a
small thickening. Fix $0 < \delta < 1$. Choose a finite $\delta$-net
$D_\delta$ of unit directions that includes $1, -1, i, -i$. Let
$s_K(\nu) = \max_{z \in K} \Re(\overline\nu z)$ be the support function
of $K$. Choose $t_\delta > 0$ small enough that $|D_\delta| e^{-t_\delta \delta} < 1$.
Define the smooth, strictly convex function
\[
\Phi_\delta(z) \;=\; \sum_{\nu \in D_\delta} \exp\bigl(t_\delta\, (\Re(\overline\nu z) - s_K(\nu) - \delta)\bigr)
\]
and set $\Omega_\delta = \{z \in \C : \Phi_\delta(z) < 1\}$.
\emph{(i) $K \subset \Omega_\delta$.} For $z \in K$ and any $\nu$,
$\Re(\overline\nu z) \le s_K(\nu)$, so each exponential is at most
$e^{-t_\delta \delta}$, and the sum is at most $|D_\delta| e^{-t_\delta \delta} < 1$.
\emph{(ii) $\overline\Omega_\delta$ lies in a fixed rectangle.} If
$\Phi_\delta(z) \le 1$, then every summand is $\le 1$ (since each is
nonnegative), hence $\Re(\overline\nu z) \le s_K(\nu) + \delta$ for every
$\nu \in D_\delta$. In particular, for $\nu = 1$ and $\nu = -1$, we get
$\Re z$ in $[\min_{w \in K}\Re w - \delta, \max_{w \in K}\Re w + \delta]$;
and similarly the imaginary part. So all $\overline\Omega_\delta$ for
$0 < \delta < 1$ sit inside one fixed bounded rectangle $R \subset \C$.
\emph{(iii) $\Phi_\delta$ is strictly convex with positive definite Hessian.}
The Hessian is a sum of positive multiples of $\nu \nu^{\mathsf T}$ (outer
products), where $\nu$ ranges over $D_\delta$. Each $\nu \nu^{\mathsf T}$ is
positive semidefinite; the four directions $1, -1, i, -i$ together span
$\mathbb R^2$, so the sum is strictly positive definite. Hence $\Phi_\delta$
is strictly convex and real analytic.
\emph{(iv) Level set is regular and a Jordan curve.} If $z_0$ were a
critical point of $\Phi_\delta$ with $\Phi_\delta(z_0) = 1$, then by
convexity $\Phi_\delta$ would attain a global minimum at $z_0$, contradicting
$\Phi_\delta(\text{any } z \in K) < 1$. So $\nabla\Phi_\delta \ne 0$ on
$\Phi_\delta^{-1}(1)$. The implicit function theorem makes that level a
regular real-analytic curve. Pick an interior point $w_0 \in \Omega_\delta$
(any point of $K$ will do). Each ray from $w_0$ crosses $\partial\Omega_\delta$
exactly once, because $\Phi_\delta$ is strictly convex and bounded. Radial
projection gives a continuous bijection $\partial\Omega_\delta \to \mathbf S^1$,
so $\partial\Omega_\delta$ is a Jordan curve and $\overline\Omega_\delta = \{\Phi_\delta \le 1\}$.
\emph{(v) Distance to $K$.} For $z \in \overline\Omega_\delta$ let
$w \in K$ be the nearest point and $d = |z - w|$. If $d > 0$, set
$\nu = (z - w)/d$, pick $\mu \in D_\delta$ with $|\mu - \nu| \le \delta$,
and let $v \in K$ attain $s_K(\mu) = \Re(\overline\mu v)$.
By minimality of $w$, $\Re(\overline\nu(w' - w)) \le 0$ for every
$w' \in K$; in particular $\Re(\overline\nu(v - w)) \le 0$.
By the support inequality at $\mu$, $\Re(\overline\mu(v - w)) \ge 0$.
Then
\begin{itemize}
\item $\Re(\overline\mu(z - w)) = d \Re(\overline\mu\nu) \ge (1 - \delta)\,d$
(since $|\mu - \nu| \le \delta$).
\item $\Re(\overline\mu(z - v)) = \Re(\overline\mu z - s_K(\mu)) \le \delta$
(from $\Phi_\delta(z) \le 1$ and $s_K(\mu) = \Re(\overline\mu v)$).
\item $\Re(\overline\mu(v - w)) \le \delta\, |v - w|$
(from $\Re(\overline\nu(v - w)) \le 0$ and $|\mu - \nu| \le \delta$).
\end{itemize}
So
\[
(1 - \delta)\, d
\;\le\; \Re(\overline\mu(z - w))
\;\le\; \delta + \delta\, |v - w|
\;\le\; \delta(1 + \operatorname{diam}K),
\]
i.e., $\dist(z, K) = d \le \delta(1 + \operatorname{diam}K)/(1 - \delta)$.
Points of $K$ have distance zero, so the same bound holds throughout.
Taking $\delta_j = 1/(j+1)$ gives the $\Omega_j$.
\end{proof}
\subsection*{Proof of the main theorem}
\begin{proof}[Proof of Theorem~\ref{thm:main}]
First recall the Toeplitz--Hausdorff convexity theorem
\cite{Toeplitz1918,Hausdorff1919}: $W(A)$ is convex.
\emph{Compactly.} $W(A) = \{u^* A u : u^*u = 1\}$ is the image of the unit
sphere (compact) under a continuous map, hence is compact.
\emph{Convexly.} Take any $u, v \in \C^n$ with $|u| = |v| = 1$. We want
to show that the segment $[u^*Au, v^*Av]$ lies in $W(A)$. Let
$V = \mathrm{span}\{u, v\}$, of complex dimension 1 or 2.
If $\dim V = 1$, then $v = e^{i\theta} u$ and $v^*Av = u^*Au$, so the segment is trivial.
If $\dim V = 2$, work in an orthonormal basis $\{e_1, e_2\}$ of $V$, and
let $B$ be the $2\times 2$ matrix of $P_V A|_{V}$ in this basis. For a
unit vector $x = x_1 e_1 + x_2 e_2$, writing
$X = 2\Re(\overline{x_1} x_2)$, $Y = 2\Im(\overline{x_1} x_2)$, $Z = |x_1|^2 - |x_2|^2$,
one has $X^2 + Y^2 + Z^2 = 1$ (because $4 |x_1 x_2|^2 + (|x_1|^2 - |x_2|^2)^2 = (|x_1|^2 + |x_2|^2)^2 = 1$),
so $(X, Y, Z)$ ranges over the unit sphere $\mathbf S^2 \subset \mathbb R^3$,
and $x^* B x$ is a real-linear function $f(X, Y, Z)$ into $\mathbf C \cong \mathbb R^2$.
The linear part of $f$ has a one-dimensional kernel. Every point of the
unit ball can be pushed along the kernel direction onto the sphere without
changing $f$, so $f(\mathbf S^2) = f(\overline{\mathbb B^3})$. The image of
a ball under a linear map is convex, so $f(\mathbf S^2)$ is convex. In
particular, the segment $[u^*Au, v^*Av] \subset f(\mathbf S^2) \subset W(A)$.
So $W(A)$ is compact and convex.
\emph{Enclosing.} Apply Lemma~\ref{lem:approx} with $K = W(A)$. Get
analytic convex domains $\Omega_j$ with $W(A) \subset \Omega_j$ and
$\sup_{\overline\Omega_j}\dist(\cdot, W(A)) \to 0$. In particular every
$\overline\Omega_j$ lies inside the fixed bounded rectangle from Lemma~\ref{lem:approx}$(ii)$.
\emph{Uniform bound.} Fix one $\Omega_j$. Let
$M_j = \sup_{z \in \overline\Omega_j} \|P(z)\|$. Two cases.
If $M_j = 0$, then $P$ vanishes on $\Omega_j$ (open), so $P$ is the zero
polynomial and $P[A] = 0$.
If $M_j > 0$, multiply $P$ by $1/M_j$. The normalized polynomial
satisfies $\|P/M_j\| \le 1$ on $\partial\Omega_j$, so Theorem~\ref{thm:domain}
applies and gives $\|(P/M_j)[A]\| \le 2$, i.e., $\|P[A]\| \le 2 M_j$.
In either case, $\|P[A]\| \le 2 M_j$.
\emph{Shrinking.} Since $\overline\Omega_j$ converges to $W(A)$ in
Hausdorff distance and all $\overline\Omega_j$ lie in a fixed compact set,
uniform continuity of the continuous function $z \mapsto \|P(z)\|$ implies
$M_j \to \sup_{z \in W(A)} \|P(z)\| = \max_{W(A)} \|P\|$. Passing to
the limit in $\|P[A]\| \le 2 M_j$,
\[
\|P[A]\| \;\le\; 2 \max_{z \in W(A)} \|P(z)\|.
\]
This is exactly \eqref{eq:main}.
\end{proof}
\subsection*{Sharpness of the constant $2$}
\begin{proof}[Sharpness in Theorem~\ref{thm:main}]
Take $n = 2$, $m = 1$, $P(z) = z$, and
$A = \begin{pmatrix}0 & 2\\0 & 0\end{pmatrix}$. Then $W(A)$ is the
closed unit disk and $\|A\| = 2$, so $\|P[A]\| = 2$ while
$\max_{W(A)} |P| = 1$: the ratio is exactly $2$.
\end{proof}
\appendix
\section{The exterior conformal collar}\label{sec:collar}
This appendix supplies the classical conformal input used in
Section~\ref{sec:coords}. Everything here is covered by
Ahlfors~\cite{Ahlfors1979}; we spell out the two ingredients that are
rarely stated explicitly: real-analytic boundary continuation and global
injectivity on a small exterior collar.
\begin{lemma}[Exterior coordinate and a univalent collar]\label{lem:exterior-collar}
Let $\Omega \subset \C$ be a bounded convex domain with regular
real-analytic Jordan boundary. Then there are $0 < r_0 < 1$ and a map
$h$ which is meromorphic and one-to-one on
$\{\lambda : |\lambda| > r_0\} \cup \{\infty\}$, has a simple pole at
$\infty$, and maps $|\lambda| > 1$ conformally onto
$\C \setminus \overline\Omega$, and $|\lambda| = 1$ onto
$\partial\Omega$. The expansion at infinity is
\[
h(\lambda) \;=\; c\lambda + c_0 + \sum_{j \ge 1} c_j \lambda^{-j}, \qquad c \ne 0.
\]
For $\lambda$ on the unit circle, $q(\lambda) = \lambda h'(\lambda)$ is a
nonzero outward normal vector at $h(\lambda)$, and the support inequality
\eqref{eq:support} holds.
\end{lemma}
\begin{proof}
\emph{Step 1: A bounded simply connected model.}
Choose a point $z_* \in \Omega$ and apply $z \mapsto 1/(z - z_*)$ (with
$\infty$ mapped to $0$). The spherical exterior
$(\C \setminus \overline\Omega) \cup \{\infty\}$ maps onto a bounded
simply connected domain $U$ containing $0$. Because $\Omega$ is convex,
each ray from $z_*$ meets $\partial\Omega$ in a unique point; after
inversion, the exterior portions of those rays become a domain that is
star-shaped about~$0$. Its boundary is a regular real-analytic Jordan
curve: the inversion of a regular real-analytic curve is again one.
\emph{Step 2: Riemann mapping.}
By the Riemann mapping theorem
\cite[Ch.~6, \S1.1, Thm.~1]{Ahlfors1979}, there is a conformal bijection
$g : \D \to U$ with $g(0) = 0$. By Carath\'eodory's Jordan-domain theorem
\cite[Ch.~6, \S1.2]{Ahlfors1979}, $g$ extends to a homeomorphism
$\overline\D \to \overline U$.
\emph{Step 3: Schwarz reflection.}
At each boundary point, Schwarz reflection
\cite[Ch.~4, \S6.5; Ch.~6, \S\S1.3--1.4]{Ahlfors1979} extends $g$
holomorphically across $\partial\D$; local continuations patch by
the identity theorem to a single holomorphic function on a
neighborhood of $\overline\D$.
\emph{Step 4: Injectivity on a neighborhood.}
Otherwise, distinct $w_n, w_n'$ with $\tilde g(w_n) = \tilde g(w_n')$
would converge (after subsequences) to a common $w^*$ in $\overline\D$.
Since $g$ is a homeomorphism on $\overline\D$, both sequences eventually
lie in a small neighborhood of $w^*$ on which $\tilde g$ is injective,
contradiction. So $\tilde g$ is injective on an open neighborhood of
$\overline\D$, which contains $\{w : |w| < R_0\}$ for some $R_0 > 1$.
\emph{Step 5: Assembly.}
The function $g$ is univalent on $\{w : |w| < R_0\} \supset \overline\D$.
Its only zero there is $0$, and that zero is simple ($g'(0) \ne 0$ by
conformality). Set
\[
h(\lambda) \;=\; z_* + \frac{1}{g(1/\lambda)}, \qquad r_0 \;=\; R_0^{-1}.
\]
Then $h$ is meromorphic on $\{|\lambda| > r_0\} \cup \{\infty\}$,
has a simple pole at $\infty$ (from the simple zero of $g$ at $0$), maps
$|\lambda| > 1$ conformally onto $\C \setminus \overline\Omega$ (from
$g$ mapping $\D$ to $U$ and $z \mapsto z_* + 1/z$ undoing the inversion),
and maps $|\lambda| = 1$ onto $\partial\Omega$. The expansion
$h(\lambda) = c\lambda + c_0 + O(1/\lambda)$ has $c = 1/g'(0) \ne 0$.
\emph{Step 6: The support inequality.}
At $\lambda = e^{i\theta} \in \T$, the radial and angular derivatives of
$h$ are $q(\lambda) = \lambda h'(\lambda)$ and $i q(\lambda)$, respectively.
They are perpendicular and nonzero. Increasing the radial parameter
$|\lambda|$ traces out the exterior of $\overline\Omega$, so $q(\lambda)$
points outward. Since $\Omega$ is convex with a smooth boundary, the
supporting-line property holds at $h(\lambda)$: for every $z \in \overline\Omega$,
\[
\Re\!\left( \overline{q(\lambda)} \bigl( h(\lambda) - z \bigr) \right) \;\ge\; 0.
\]
This is exactly \eqref{eq:support}.
\end{proof}
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