==== TABLE OF CONTENTS ====
Symplectic-Balls-in-Symmetric-Polar-Products//figures/lens-slices.tex
Symplectic-Balls-in-Symmetric-Polar-Products//main.tex
Symplectic-Balls-in-Symmetric-Polar-Products//sections/approximation.tex
Symplectic-Balls-in-Symmetric-Polar-Products//sections/ball.tex
Symplectic-Balls-in-Symmetric-Polar-Products//sections/history.tex
Symplectic-Balls-in-Symmetric-Polar-Products//sections/introduction.tex
Symplectic-Balls-in-Symmetric-Polar-Products//sections/planar.tex
Symplectic-Balls-in-Symmetric-Polar-Products//sections/realization.tex
Symplectic-Balls-in-Symmetric-Polar-Products//references.bib
===========================
==== BEGIN Symplectic-Balls-in-Symmetric-Polar-Products//figures/lens-slices.tex ====
\begin{figure}[htb]
\centering
\includegraphics[width=\linewidth]{figures/lens-slices.pdf}
\caption{A fixed horizontal slice and its preimage, illustrated for $t>0$.
Along the right half of the highlighted preimage curve in the left panel,
$r$ increases from
$r_0=T^{-1}(t)$ to $1$, while $v(r,t)$ increases from $0$ to $V(t)$.
The dashed circle and its image pass through the corresponding marked
points. Open markers indicate boundary limits.}
\label{fig:lens-slices}
\end{figure}
==== END Symplectic-Balls-in-Symmetric-Polar-Products//figures/lens-slices.tex ====
==== BEGIN Symplectic-Balls-in-Symmetric-Polar-Products//main.tex ====
\documentclass[11pt]{article}
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\title{Symplectic Balls in Symmetric Polar Products}
\author{OpenAI}
\date{September 22, 2026}
\hypersetup{pdftitle={Symplectic Balls in Symmetric Polar Products},pdfauthor={OpenAI},pdfsubject={Gromov width and the symmetric Mahler conjecture}}
\begin{document}
\maketitle
\begin{abstract}
For every integer $n\ge2$ and every origin-symmetric convex body
$K\subset\R^n$, we prove that the Gromov width of
$\interior K\times\interior K^\circ$ is $4$.
We construct smooth symplectic embeddings of standard balls of every
capacity $00$, there are finitely many vectors
$b_1,\ldots,b_m\in\R^n$ spanning $\R^n$ such that
\[
K'=\{x\in\R^n:|\ip{b_j}{x}|\le1,\quad 1\le j\le m\}
\]
satisfies
\begin{equation}\label{eq:polar-approximation}
K\subset K'\subset(1+\alpha)K,
\qquad (K')^\circ\subset K^\circ.
\end{equation}
\end{lemma}
\begin{proof}
First, $0\in\interior K$. Indeed, if the Euclidean ball $B(y,r)$
is contained in $K$, then so is $B(-y,r)$, and their midpoints contain
$B(0,r)$. Since $K$ is also bounded, there are $r,R>0$ with
$B(0,r)\subset K\subset B(0,R)$. The definition of the polar gives
\[
B(0,R^{-1})\subset K^\circ\subset\overline{B(0,r^{-1})}.
\]
Thus $K^\circ$ is a compact, origin-symmetric convex body with nonempty
interior.
We will use the bipolar identity $K^{\circ\circ}=K$. To recall its
justification here, if $x\notin K$, separation gives a nonzero vector
$v$ such that
\[
\ip{v}{x}>h_K(v),
\qquad h_K(v)=\max_{y\in K}\ip{v}{y}>0.
\]
The normalized vector $v/h_K(v)$ belongs to $K^\circ$ and pairs with
$x$ to a value greater than $1$, so $x\notin K^{\circ\circ}$.
The reverse inclusion follows directly from the definition. The same
argument applies to every convex body containing the origin in its
interior.
Put $\lambda=1+\alpha$ and
\[
\rho=\min_{|v|=1}h_{K^\circ}(v)>0.
\]
Choose $d>0$ with $d\le\alpha\rho/(1+\alpha)$, and choose a finite
$d$-net $\{b_1,\ldots,b_m\}$ in the compact set $K^\circ$, with all
net points in $K^\circ$. Let
\[
P=\operatorname{conv}\{\pm b_1,\ldots,\pm b_m\}.
\]
Symmetry of $K^\circ$ gives $P\subset K^\circ$. For every unit vector
$v$, a maximizer of $\ip{v}{p}$ over $p\in K^\circ$ lies within
distance $d$ of a net point. Consequently
\[
h_P(v)\ge h_{K^\circ}(v)-d
\ge\lambda^{-1}h_{K^\circ}(v).
\]
The characterization of inclusion by support functions, which follows
from convex separation, now gives
\[
\lambda^{-1}K^\circ\subset P\subset K^\circ.
\]
In particular, $P$ has interior, so its generating vectors span
$\R^n$. Taking polars and using the bipolar identity yields
\[
K\subset P^\circ\subset\lambda K,
\qquad (P^\circ)^\circ=P\subset K^\circ.
\]
Finally, the definition of $P$ shows that $P^\circ$ is exactly the
finite strip body $K'$ in the statement.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:main}]
Fix $00$ such that $\lambda c<4$, where
$\lambda=1+\alpha$, and choose one body $K'$ from
Lemma~\ref{lem:polar-approximation}. Proposition~\ref{prop:finite-strips}
gives a smooth symplectic embedding
\[
e:B^{2n}(\lambda c)\longrightarrow
\interior K'\times\interior\bigl((K')^\circ\bigr).
\]
If one body is contained in another, every interior ball in the first
is contained in the second. Thus Equation~\eqref{eq:polar-approximation}
implies
\[
\interior K'\subset\lambda\interior K,
\qquad \interior\bigl((K')^\circ\bigr)\subset\interior K^\circ,
\]
even when their boundaries meet. Regard $e$ as an embedding into
$\lambda\interior K\times\interior K^\circ$. Define
\[
D_{\sqrt\lambda}(z)=\sqrt\lambda\,z,
\qquad Q_\lambda(q,p)=(q/\lambda,p).
\]
The composition
\begin{equation}\label{eq:general-rescaling}
Q_\lambda\circ e\circ D_{\sqrt\lambda}:
B^{2n}(c)\longrightarrow U_K
\end{equation}
is a smooth embedding. Since
\[
D_{\sqrt\lambda}^*\omega_0=\lambda\omega_0,
\qquad Q_\lambda^*\omega_0=\lambda^{-1}\omega_0,
\]
its pullback of $\omega_0$ equals $\omega_0$. This proves
$c_G(U_K)\ge4$.
For the upper bound, we use the supporting-slab construction of
Artstein-Avidan, Karasev, and Ostrover
\cite[Remark~4.2]{AAKO2014}. Choose $q_0\in K$ and $p_0\in K^\circ$
with $\ip{q_0}{p_0}=1$. Such a pair is obtained by maximizing any
nonzero linear functional on $K$ and normalizing its maximum to $1$.
Complete $q_0$ to a basis of $\R^n$ by vectors in $\ker p_0$, and let
$A$ be the matrix of this basis. The linear change
\[
Q=A^{-1}q,\qquad P=A^Tp
\]
is symplectic and has first coordinates
$Q_1=\ip{p_0}{q}$ and $P_1=\ip{q_0}{p}$.
Both functionals are nonzero, so symmetry, polarity, and the interior
condition give $|Q_1|<1$ and $|P_1|<1$ on $U_K$.
Thus this change embeds $U_K$ into
$(-1,1)^2\times\R^{2n-2}$, with the square in its first conjugate
coordinate pair.
For completeness, this open square is area-preservingly diffeomorphic
to the disk of area $4$. Put $R=2/\sqrt\pi$,
$h(X)=\sqrt{R^2-X^2}$ for $-R0$,
$H$ is a smooth diffeomorphism from $(-R,R)$ onto $(-1,1)$.
The map
\[
(u,v)\longmapsto
\bigl(X=H^{-1}(u),\ Y=v\,h(H^{-1}(u))\bigr)
\]
has image $B^2(4)$ and Jacobian $X'(u)h(X)=1$.
Applying it to the first conjugate pair gives a symplectic embedding
$U_K\hookrightarrow B^2(4)\times\R^{2n-2}$.
Gromov's nonsqueezing theorem \cite{Gromov1985} therefore bounds the
capacity of every ball in $U_K$ by $4$. Combined with the lower bound,
this proves $c_G(U_K)=4$.
\end{proof}
\begin{remark}[Order of choices]
\label{rem:parameter-order}
For each prescribed $00$ so that
$\lambda c<4$, where $\lambda=1+\alpha$. Then fix the finite body
$K'$ and its $m$ defining constraints. Choose $\eta>0$ so that
$(1+\eta)\lambda c<4$, then choose $k$ large enough that
\[
\frac{C_{K'}}k+m\epsilon_k<\frac{\eta\pi}{4}.
\]
This is possible because the constraint list is fixed before $k$
varies. Proposition~\ref{prop:finite-strips}, applied at capacity
$\lambda c$, and the final rescaling then give the required ball.
Each prescribed capacity is obtained by one finite construction; no
limit of embeddings is required.
\end{remark}
\begin{corollary}[Symmetric Mahler inequality]
\label{cor:mahler}
For every integer $n\ge1$ and every origin-symmetric convex body
$K\subset\R^n$,
\[
\vol_n(K)\,\vol_n(K^\circ)\ge\frac{4^n}{n!}.
\]
\end{corollary}
\begin{proof}
For $n=1$, write $K=[-a,a]$ with $a>0$. Then
$K^\circ=[-a^{-1},a^{-1}]$, and the product of their lengths is $4$.
Assume henceforth that $n\ge2$.
We first justify that taking interiors does not change the volumes in
the statement. If a convex body $C\subset\R^n$ contains $B(0,r)$,
then convexity gives, for $00$ satisfy
\[
\sum_{i=1}^k R_i^n<4^n,
\qquad R_i+R_j<4\quad(1\le i0$ and
\[
\chi''(t)=\beta'(t)(e^t-b)+\beta(t)e^t\ge0,
\]
because the transition occurs where $e^t>b$. For $t\le t_-$, the
function $\chi$ is affine with slope $b$.
It follows that
\[
H=\chi(\log\tau)
\]
is smooth and plurisubharmonic on $V\setminus\{0\}$ and satisfies
$H=\tau$ wherever $\tau\ge s_0:=e^{t_+}$. The vanishing assumption gives
$\tau(z)\le C_0|z|^{2k}$ near zero. Since $\chi$ is eventually affine,
there is a constant $C_1$ such that
\begin{equation}
H(z)\le bk\log|z|^2+C_1
\quad\text{for all sufficiently small }z\ne0.
\label{eq:ball-singularity}
\end{equation}
\medskip
\noindent\textit{Step 2: smooth the singularity without changing the form near the boundary.}
The strict inequality $\mu0$ with
$\overline{B}_{r_2}\subset V$. For $0<\delta\le1$, consider
\[
P_\delta(z)=\mu\log(|z|^2+\delta)-M.
\]
For each fixed $r>0$, the radial calculation in Step~4 identifies
$(B_r,\ddc(|z|^2+P_\delta))$ with a standard ball whose squared radius
is $r^2+\mu r^2/(r^2+\delta)$. Since $\mu>a$, this exceeds $a$ for
all sufficiently small $\delta$. We will therefore arrange that the
modified potential agrees with $P_\delta$ on a ball whose radius is
fixed before $\delta$ is chosen.
The functions $P_\delta$ are plurisubharmonic by
Equation~\eqref{eq:ball-log-levi}, applied to the nowhere-zero
holomorphic tuple $(\sqrt\delta,z_1,\ldots,z_n)$. Choose $M$ so that
\[
M>\mu\log(r_2^2+1)-\min_{|z|=r_2}H(z)+2.
\]
The minimum is finite because $f$ has no zero on this sphere. Then
$P_\deltaH+2\quad\text{on }0<|z|\le r_1
\quad\text{for every }0<\delta\le1.
\label{eq:ball-inner-domination}
\end{equation}
Fix from now on one positive $\delta$ satisfying
\begin{equation}
0<\delta<\min\left\{1,\ r_1^2\left(\frac\mu a-1\right)\right\}.
\label{eq:ball-delta-choice}
\end{equation}
The upper bound is positive because $\mu>a$.
We next replace the singularity of $H$ by $P_\delta$ using a
regularized maximum, the standard gluing device in Richberg's smoothing
method; see \cite[Section~2]{HarveyLawsonPlis2020}. A related replacement
of K\"ahler potentials appears in
\cite[Proposition~3.3]{WittNystrom2018} in a projective setting.
We give the elementary two-function construction explicitly. Let
$\vartheta\in C_c^\infty((-1,1))$ be nonnegative and even, with
$\int_{\R}\vartheta=1$, and set
\[
\rho(s)=\int_{\R}|s-t|\vartheta(t)\,dt,
\qquad
\operatorname{rmax}(h,p)=\frac{h+p+\rho(h-p)}2.
\]
The function $\rho$ is smooth and convex, with $|\rho'|\le1$, and
equals $|s|$ for $|s|\ge1$. Thus $\operatorname{rmax}$ is smooth,
convex, nondecreasing in both arguments, and equals $\max(h,p)$ when
$|h-p|\ge1$. Define $\widetilde H$ to be
$\operatorname{rmax}(H,P_\delta)$ on $B_{r_2}\setminus\{0\}$, to be
$H$ outside $B_{r_2}$, and to take the value $P_\delta(0)$ at zero.
The outer comparison gives equality with $H$ in a neighborhood of
$\partial B_{r_2}$, and Equation~\eqref{eq:ball-inner-domination} gives
equality with $P_\delta$ on $B_{r_1}$. Consequently $\widetilde H$ is
smooth and plurisubharmonic on all of $V$.
The difference $\widetilde H-\tau$ has compact support in $V$.
Indeed, set $s_2=\max_{\overline B_{r_2}}\tau<1$, choose
$\max(s_0,s_2)<\sigma<1$, and put
\begin{equation}
C=\{z\in\Omega:\tau(z)\le\sigma\}.
\label{eq:ball-support}
\end{equation}
This is a compact subset of $V$ by hypothesis. Outside $C$ a point
lies outside $\overline B_{r_2}$ and has $H=\tau$, so that
$\widetilde H-\tau$ vanishes there.
\medskip
\noindent\textit{Step 3: transport the smoothed form.}
We now use Moser's deformation argument \cite[Section~4]{Moser1965}.
The support estimate above supplies a complete flow even though $V$
need not be compact. Consider the symplectic forms
\[
\omega_s=\ddc\bigl(|z|^2+(1-s)\tau+s\widetilde H\bigr),
\qquad 0\le s\le1,
\]
and write $\widetilde\omega=\omega_1$. Each form is symplectic because
the two non-Euclidean potentials are plurisubharmonic. Define the
smooth time-dependent vector field $Y_s$ by
\[
\iota_{Y_s}\omega_s=-\gamma,
\qquad \gamma=\dc(\widetilde H-\tau).
\]
Both $\gamma$ and every $Y_s$ are supported in the fixed compact set
$C$. Extending $Y_s$ by zero outside $V$ gives a smooth vector field
on $\C^n$, with its coefficients and first derivatives bounded
uniformly for $s\in[0,1]$. The ordinary differential equation for
this field has a flow $\varphi_s$ throughout this time interval.
The ambient flow fixes $\C^n\setminus V$ pointwise and hence restricts
to diffeomorphisms of $V$. This argument applies even if $V$ is
disconnected and places no regularity assumption on its boundary.
Since $d\omega_s=0$, differentiation of pullbacks gives
\[
\frac{d}{ds}\varphi_s^*\omega_s
=\varphi_s^*\bigl(d\gamma+d\iota_{Y_s}\omega_s\bigr)=0.
\]
Therefore
\begin{equation}
\varphi_1^*\widetilde\omega=\omega,
\qquad
\varphi_1^{-1}:(V,\widetilde\omega)\longrightarrow(V,\omega)
\text{ is symplectic}.
\label{eq:ball-moser}
\end{equation}
\medskip
\noindent\textit{Step 4: read off the ball in the radial model.}
The deformation has reduced the problem to an explicit radial form.
On $B_{r_1}$ the new potential, up to the additive constant $-M$, is
$\Psi(|z|^2)$, where
\[
\Psi(t)=t+\mu\log(t+\delta).
\]
Define the radial map
\[
R(z)=\sqrt{\Psi'(|z|^2)}\,z
=\sqrt{1+\frac\mu{|z|^2+\delta}}\,z.
\]
This is the radial case of the symplectic-coordinate construction in
\cite[proof of Theorem~1.1, Equation~(23)]{LoiZuddas2008}.
In the pullback of $\lambda_0$ from
Equation~\eqref{eq:ball-normalization}, the terms differentiating the
radial factor cancel. Hence
\[
R^*\lambda_0=\Psi'(|z|^2)\lambda_0=\dc\Psi(|z|^2),
\qquad R^*\omega_0=\widetilde\omega\big|_{B_{r_1}}.
\]
The squared radial coordinate is strictly increasing, because
\[
\frac{d}{dt}\bigl(t\Psi'(t)\bigr)
=\frac{d}{dt}\left(t+\frac{\mu t}{t+\delta}\right)
=1+\frac{\mu\delta}{(t+\delta)^2}>0.
\]
At zero the map is smooth and has derivative
$DR(0)=\sqrt{1+\mu/\delta}\,I$, so its inverse is smooth there as
well. Thus $R$ is a diffeomorphism from $B_{r_1}$ onto the standard
ball $B^{2n}(\pi Q_\delta)$, where
\[
Q_\delta=r_1^2+\frac{\mu r_1^2}{r_1^2+\delta}>a
\]
by Equation~\eqref{eq:ball-delta-choice}. If
$\iota:B_{r_1}\hookrightarrow V$ denotes inclusion, the map
\[
\varphi_1^{-1}\circ\iota\circ
R^{-1}\big|_{B^{2n}(\pi a)}:
B^{2n}(\pi a)\longrightarrow V
\]
is a smooth embedding, and its pullback of $\omega$ is $\omega_0$ by
Equation~\eqref{eq:ball-moser} and the radial pullback identity.
Since $a$ was arbitrary in $(0,k)$, the theorem follows.
\end{proof}
==== END Symplectic-Balls-in-Symmetric-Polar-Products//sections/ball.tex ====
==== BEGIN Symplectic-Balls-in-Symmetric-Polar-Products//sections/history.tex ====
\subsection*{Background and prior constructions}
The symmetric volume-product problem goes back to Mahler's work on
transference in the geometry of numbers \cite{Mahler1939}.
Earlier sharp volume inequalities cover the plane
\cite[Theorem~13]{BoroczkyMakaiMeyerReisner2013}, unconditional bodies
(those invariant under coordinate sign changes)
\cite[Sections~1--2]{Reisner1987}, and dimension three
\cite{IriyehShibata}. Bourgain and Milman's asymptotic reverse
Santal\'o inequality gives a lower bound $a^n/n!$ for some $a>0$
independent of dimension \cite{BourgainMilman}; the sharp target is
$4^n/n!$. The companion paper \cite{SymmetricCompanion2026} discusses
the wider convex-geometric history and the equality problem. Our focus
is the stronger geometric assertion that the polar product contains
large symplectic balls.
Gromov's nonsqueezing theorem established that ball embeddings detect
symplectic restrictions beyond volume preservation \cite{Gromov1985}.
Viterbo later proposed a sharp volume--capacity inequality for convex
domains \cite{Viterbo2000}. Artstein-Avidan, Karasev, and Ostrover
connected this proposed inequality to the symmetric Mahler conjecture
and proved
\[
c_{\mathrm{HZ}}(K\times K^\circ)=4
\]
for every origin-symmetric convex body
\cite[Theorems~1.6--1.7 and Section~4]{AAKO2014}.
Here $c_{\mathrm{HZ}}$ is the Hofer--Zehnder capacity, normalized to
have value $\pi R^2$ on a ball of radius $R$
\cite{HoferZehnder1994}. This computation implies the upper bound for
Gromov width. We use the supporting-cylinder argument of
\cite[Remark~4.2]{AAKO2014} and nonsqueezing to prove that bound directly.
The matching lower bound requires actual ball embeddings; it does not
follow from the Hofer--Zehnder capacity computation.
Earlier constructions establish this lower bound for particular
families. Karasev proves it for $\ell_p$ unit balls, $10$, write
\[
B^{2n}(c)=\{(q,p):\pi(|q|^2+|p|^2)0\) such that
\begin{equation}\label{eq:horizontal-slices}
\{v\in\R:v+it\in D\}=(-V(t),V(t)).
\end{equation}
Write \(F(ir)=iT(r)\) for \(0\le r<1\). The function \(T\) increases
strictly from \(0\) to \(1\), with
\begin{equation}\label{eq:planar-T}
T'(r)=\frac{8}{\pi^2}\frac{\operatorname{arctanh}r}{r}
\quad(00
\qquad (w\in\D).
\]
This also shows that \(\arctan w=0\) only at \(w=0\).
Thus \(F'\) has no zeros away from the origin, and the series gives
\(F'(0)=8/\pi^2\ne0\).
For \(w=re^{i\theta}\), define
\begin{equation}\label{eq:planar-A}
A(r,\theta)=\Re\bigl(wF'(w)\bigr)
=\frac{4}{\pi^2}
\arctan\frac{2r\cos\theta}{1-r^2}.
\end{equation}
To obtain the last identity, the imaginary part of the fraction in
Equation~\eqref{eq:planar-arctan}, divided by its real part, is
\(2r\cos\theta/(1-r^2)\), and its argument lies in
\((-\pi/2,\pi/2)\). On the right semicircle,
\(00\) in the right half-disk: its radial
derivative is positive, and its value tends to zero at the origin.
They also show that \(\Im F(re^{i\theta})\) increases strictly with
\(\theta\) on each right semicircle.
Oddness and conjugation symmetry give \(F(ir)=iT(r)\) with real \(T\).
Differentiating Equation~\eqref{eq:planar-arctan} on the imaginary axis
gives Equation~\eqref{eq:planar-T}. In particular, \(T\) is strictly
increasing, \(T(0)=0\), and \(T'(r)\) diverges as \(r\uparrow1\).
Since \(F(r)\) is real, Equation~\eqref{eq:planar-polar-derivatives}
also gives
\[
T(r)=\int_0^{\pi/2}A(r,\theta)\,d\theta.
\]
The integrand is bounded by \(2/\pi\) and tends to \(2/\pi\) at every
\(\theta\in[0,\pi/2)\) as \(r\uparrow1\). Dominated convergence
therefore gives \(T(r)\to1\).
\medskip
\noindent\textit{Horizontal slices and global invertibility.}
Fix \(|t|<1\) and let \(T(r_0)=|t|\). For every \(r\in(r_0,1)\),
the strictly increasing imaginary coordinate on the right semicircle
ranges from \(-T(r)\) to \(T(r)\). There is consequently exactly one
angle \(\theta(r,t)\) there with imaginary coordinate \(t\).
It depends smoothly on \(r\), since its angular derivative is \(A>0\).
Writing \(wF'(w)=A+iB\), implicit differentiation at fixed \(t\) gives
\[
\partial_r\theta=-\frac{B}{rA},
\qquad
\partial_r v=\frac{A}{r}-B\,\partial_r\theta
=\frac{A^2+B^2}{rA}.
\]
Thus
\begin{equation}\label{eq:horizontal-radius-derivative}
\frac{\partial v}{\partial r}(r,t)
=\frac{|wF'(w)|^2}{rA(r,\theta(r,t))}>0.
\end{equation}
As \(r\downarrow r_0>0\), the angle tends to the appropriate imaginary-axis
endpoint of the semicircle, so \(v(r,t)\to0\). If \(r_0=0\), the same
limit follows from \(F(0)=0\). Boundedness of \(F\) now shows that
\(v(r,t)\) maps \((r_0,1)\) continuously and strictly increasingly onto
an interval \((0,V(t))\), where \(00\).
Fix such a point and put
\[
R=|g(v+it)|,\qquad r_0=T^{-1}(|t|).
\]
Along its positive horizontal slice, the radius increases from \(r_0\)
to \(R\). Since \(g'(F(w))=1/F'(w)\),
Equation~\eqref{eq:horizontal-radius-derivative} gives
\begin{equation}\label{eq:slice-radial-integral}
\frac{J_k(v,t)}{k}
=k\int_{r_0}^{R}
\frac{r^{2k-1}}{A(r,\theta(r,t))}\,dr.
\end{equation}
Indeed, \(|g'|^2\,dv=(r/A)\,dr\).
At the lower endpoint this is understood as an improper integral:
perform the change of variables first on a segment beginning at a
positive real coordinate, and then let that coordinate decrease to zero.
The original integrand is smooth on the full compact segment, so its
integral is finite; the nonnegative transformed integrals converge to
the same value. This argument also covers \(t=0\), when \(r_0=0\).
The leading constant comes from splitting \(1/A\) into \(\pi/2\) and
the nonnegative excess \(1/A-\pi/2\), since \(00\),
\begin{equation}\label{eq:reciprocal-arctan}
\frac{1}{\arctan X}-\frac{2}{\pi}\le\frac{2}{X}.
\end{equation}
For \(X\le1\), this follows from \(\arctan X\ge X/2\); for \(X\ge1\),
use \(\pi/2-\arctan X=\arctan(1/X)\le1/X\) and
\(\arctan X\ge\pi/4\). Combining
Equation~\eqref{eq:reciprocal-arctan} with
\(\sin s\ge2s/\pi\) and \((1+r)/(2r)\le3/2\), we obtain
\begin{equation}\label{eq:A-angular-bound}
\frac1{A(r,\theta)}
\le\frac{\pi}{2}+\frac{3\pi^3}{8}\frac{1-r}{s}.
\end{equation}
On the other hand, the vertical gap to the semicircle endpoint is
\begin{align}
\Delta(r,\theta)
&:=T(r)-|\Im F(re^{i\theta})|
=\int_0^s\frac{4}{\pi^2}
\arctan\frac{2r\sin\xi}{1-r^2}\,d\xi \notag\\
&\le\frac{4r}{\pi^2(1-r^2)}s^2
\le\frac{2}{\pi^2}\frac{s^2}{1-r}.
\label{eq:vertical-gap}
\end{align}
Here we used \(\arctan y\le y\) and \(\sin\xi\le\xi\).
The last inequality supplies the lower bound
\(s\ge(\pi/\sqrt2)\sqrt{\Delta(r,\theta)(1-r)}\).
Consequently, with the absolute constant
\(C_0=3\sqrt2\pi^2/8\), Equation~\eqref{eq:A-angular-bound} implies
\begin{equation}\label{eq:A-slice-bound}
\frac1{A(r,\theta(r,t))}
\le\frac{\pi}{2}
+C_0\sqrt{\frac{1-r}{T(r)-|t|}}
\qquad (r\ge1/2,\ r>r_0).
\end{equation}
\medskip
\noindent\textit{Step 3: separate a compact part from the tip region.}
Fix a radius \(r_*\in(1/2,1)\). The part of the original horizontal
integral for which \(|g|\le r_*\) is uniformly bounded, after division
by \(k\), by
\begin{equation}\label{eq:slice-inner-bound}
C_*k r_*^{2k-2},
\qquad
C_*:=\operatorname{diam}(D)
\max_{|w|\le r_*}|F'(w)|^{-2}<\infty.
\end{equation}
To see this, the radius is strictly increasing along the positive
slice, so the relevant portion is exactly its initial segment ending
at radius \(\min(R,r_*)\), when \(r_0\le r_*\), and is empty otherwise.
Every point of this segment lies in \(F(\{|w|\le r_*\})\), and its
horizontal length is at most \(\operatorname{diam}(D)\).
This includes the case \(R\le r_*\), when it is the whole integral.
If an outer portion remains, put
\[
r_a=\max(r_0,r_*),\qquad
L_*:=\inf_{r_*\le r<1}T'(r)>0.
\]
Since \(T(r_a)\ge T(r_0)=|t|\), for \(r>r_a\) we have
\begin{equation}\label{eq:gap-tail-bound}
T(r)-|t|\ge T(r)-T(r_a)\ge L_*(r-r_a).
\end{equation}
Equations~\eqref{eq:slice-radial-integral} and
\eqref{eq:A-slice-bound} therefore bound the outer contribution by
\begin{equation}\label{eq:slice-outer-bound}
\frac{\pi}{4}\bigl(R^{2k}-r_a^{2k}\bigr)
+\frac{C_0}{\sqrt{L_*}}\,
k\int_{r_a}^{1}r^{2k-1}
\sqrt{\frac{1-r}{r-r_a}}\,dr.
\end{equation}
The extension of the error integral from \(R\) to \(1\) uses its
nonnegativity and removes any dependence on the distance \(R-r_a\).
The last integral has an absolute bound, including when its lower
endpoint depends on \(k\). Write
\[
I(k,r_a)=k\int_{r_a}^{1}r^{2k-1}
\sqrt{\frac{1-r}{r-r_a}}\,dr,
\qquad H=k(1-r_a).
\]
The substitution \(y=k(1-r)\) gives
\begin{align*}
I(k,r_a)
&=\int_0^H(1-y/k)^{2k-1}\sqrt{\frac{y}{H-y}}\,dy\\
&\le\int_0^H e^{-y}\sqrt{\frac{y}{H-y}}\,dy,
\end{align*}
since \(\log r\le-(1-r)\) and \(2k-1\ge k\).
On \([0,H/2]\) the square-root factor is at most one. On
\([H/2,H]\), bound the exponential by \(e^{-H/2}\) and use
\[
\int_0^H\sqrt{\frac{y}{H-y}}\,dy=\frac{\pi H}{2}.
\]
It follows that
\begin{equation}\label{eq:uniform-endpoint-integral}
I(k,r_a)\le 1+\frac{\pi H}{2}e^{-H/2}
\le 1+\frac{\pi}{e}.
\end{equation}
The same integral without its exponential factor also gives
\(I(k,r_a)\le\pi H/2\). These estimates hold for every \(H>0\),
covering both \(H\downarrow0\) and \(H\to\infty\).
Combining Equations~\eqref{eq:slice-inner-bound},
\eqref{eq:slice-outer-bound}, and
\eqref{eq:uniform-endpoint-integral}, and using symmetry when \(v<0\),
we have, for every \(v+it\in D\),
\begin{equation}\label{eq:slice-uniform-error-bound}
\frac{|J_k(v,t)|}{k}-\frac{\pi}{4}|g(v+it)|^{2k}
\le C_*k r_*^{2k-2}
+\frac{C_0(1+\pi/e)}{\sqrt{L_*}}.
\end{equation}
When there is no outer portion, the first term alone bounds the excess,
so the same displayed inequality remains valid. It is also valid at
\(v=0\).
\medskip
\noindent\textit{Step 4: make the error uniform.}
The cutoff radius will be chosen before the power $k$, so the estimates
above allow the slice height and both radial endpoints to move with $k$.
Define explicitly
\begin{equation}\label{eq:slice-error-sequence}
\epsilon_k=
\max\left\{0,\ \sup_{v+it\in D}
\left(\frac{|J_k(v,t)|}{k}
-\frac{\pi}{4}|g(v+it)|^{2k}\right)\right\}.
\end{equation}
Equation~\eqref{eq:slice-uniform-error-bound} makes this supremum finite
for each \(k\). For fixed \(r_*\), its first error term tends to zero
as \(k\to\infty\). Moreover, Equation~\eqref{eq:planar-T} gives
\[
L_*\ge\frac{8}{\pi^2}\operatorname{arctanh}r_*
\longrightarrow\infty
\qquad\text{as }r_*\uparrow1.
\]
Given \(\varepsilon>0\), first choose \(r_*\) so that the second term
in Equation~\eqref{eq:slice-uniform-error-bound} is less than
\(\varepsilon/2\). For this fixed \(r_*\), choose \(N\) such that
\(C_*k r_*^{2k-2}<\varepsilon/2\) for every \(k\ge N\).
The resulting bound \(\epsilon_k<\varepsilon\) holds simultaneously
on all of \(D\). In particular it allows the height \(t\), the initial
radius \(r_0\), and the final radius \(R\) all to vary with \(k\), with
no separation from either endpoint. This proves
\(\epsilon_k\to0\) and Equation~\eqref{eq:uniform-slice}.
\end{proof}
==== END Symplectic-Balls-in-Symmetric-Polar-Products//sections/planar.tex ====
==== BEGIN Symplectic-Balls-in-Symmetric-Polar-Products//sections/realization.tex ====
\section{Realization in a polar product}
\label{sec:realization}
We now combine the ball construction with the uniform slice estimate.
Writing \(z=u+ix\), the imaginary coordinates \(x\) will give the position
in a convex body. The signed slice integrals will correct the standard
momentum \(-u\) to account for the holomorphic part of the symplectic form.
\begin{proposition}\label{prop:finite-strips}
Let $n\ge2$ be an integer, let $b_1,\ldots,b_m\in\R^n$ span $\R^n$,
and set
\[
K=\{x\in\R^n: |b_j\cdot x|\le 1,
\quad 1\le j\le m\}.
\]
For every $00$ and all sufficiently large $k$,
\[
\frac{C_K}{k}+m\epsilon_k<\frac{\eta\pi}{4}.
\]
Since $\tau<1$ on $V$, Equation~\eqref{eq:strip-support} then gives the
strict inclusion
\begin{equation}\label{eq:strip-scaled-image}
\Phi_k(V)\subset
\interior(K)\times S\interior(K^\circ),
\qquad S=(1+\eta)\frac{\pi k}{4}.
\end{equation}
\medskip
\noindent\textit{Step 4: rescale the ball and the momentum together.}
Finally, fix $00$ with $(1+\eta)c<4$, and then fix
an integer $k$ large enough for Equation~\eqref{eq:strip-scaled-image}.
The resulting constant $S$ satisfies $Sc<\pi k$, so there is a
symplectic embedding
\[
E:B^{2n}(Sc)\longrightarrow(V,\omega).
\]
Let $D_{\sqrt S}(z)=\sqrt S\,z$ and
$L_S(q,p)=(q,p/S)$. The composition
\begin{equation}\label{eq:strip-rescaling}
L_S\circ\Phi_k\circ E\circ D_{\sqrt S}:
B^{2n}(c)\longrightarrow
\interior(K)\times\interior(K^\circ)
\end{equation}
is a smooth embedding. Since $D_{\sqrt S}$ maps $B^{2n}(c)$ onto
$B^{2n}(Sc)$ and
\[
D_{\sqrt S}^*\omega_0=S\omega_0,
\qquad L_S^*\omega_0=S^{-1}\omega_0,
\]
its pullback of $\omega_0$ is exactly $\omega_0$. This proves the
proposition.
\end{proof}
==== END Symplectic-Balls-in-Symmetric-Polar-Products//sections/realization.tex ====
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