Proof. Step 1: a logarithmic potential with the required slope. For a holomorphic tuple, differentiation along a complex line gives \[\begin{align*} dd^c\tau(X,iX)&=|df(X)|^2,\\ dd^c\log\tau(X,iX) &=\frac{|f|^2|df(X)|^2 -\left|\sum_{j=1}^m\overline{f_j}\,df_j(X)\right|^2}{|f|^4} \quad\text{on }\Omega\setminus\{0\}. \tag{2}\end{align*}\] The second expression is nonnegative by Cauchy–Schwarz. Hence \(\tau\) and, away from its zero, \(\log\tau\) are plurisubharmonic.

Fix \(0<a<k\); we will embed the ball of capacity \(\pi a\). Choose \[\frac ak<b<1,\qquad a<\mu<bk.\] Choose numbers \(\log b<t_-<t_+<0\) and a smooth nondecreasing cutoff \(\beta:\mathbb R\to[0,1]\) which is \(0\) for \(t\le t_-\) and \(1\) for \(t\ge t_+\). Define \(\chi\) on \((-\infty,0)\) by \[\chi'(t)=b+\beta(t)(e^t-b),\] choosing its additive constant so that \(\chi(t)=e^t\) for \(t\ge t_+\). This is possible since \(\chi'=e^t\) there. Moreover, \(\chi'>0\) 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 \[ H(z)\le bk\log|z|^2+C_1 \quad\text{for all sufficiently small }z\ne0. \tag{3}\]

Step 2: smooth the singularity without changing the form near the boundary. The strict inequality \(\mu<bk\) lets a smooth radial logarithm dominate \(H\) close to zero while remaining below it on an outer sphere. Write \(B_r=\{z\in\mathbb C^n:|z|<r\}\), and choose \(r_2>0\) 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 \((B_r,dd^c(|z|^2+P_\delta))\) 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 (2), 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 \(P_\delta<H-2\) on \(|z|=r_2\) for every \(0<\delta\le1\). On the other hand, Equation (3) gives \[P_\delta(z)-H(z) \ge(\mu-bk)\log|z|^2-M-C_1.\] As \(z\to0\), this lower bound tends to \(+\infty\), uniformly in \(\delta\). We may therefore fix \(0<r_1<r_2\), independently of \(\delta\), such that \[ P_\delta>H+2\quad\text{on }0<|z|\le r_1 \quad\text{for every }0<\delta\le1. \tag{4}\] Fix from now on one positive \(\delta\) satisfying \[ 0<\delta<\min\left\{1,\ r_1^2\left(\frac\mu a-1\right)\right\}. \tag{5}\] 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 (Harvey et al. 2020, sec. 2). A related replacement of Kähler potentials appears in (Witt Nyström 2018, Proposition 3.3) 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 \(\int_{\mathbb R}\vartheta=1\), and set \[\rho(s)=\int_{\mathbb R}|s-t|\vartheta(t)\,dt, \qquad \operatorname{rmax}(h,p)=\frac{h+p+\rho(h-p)}2.\] 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 (4) 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 \[ C=\{z\in\Omega:\tau(z)\le\sigma\}. \tag{6}\] 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.

Step 3: transport the smoothed form. We now use Moser’s deformation argument (Moser 1965, sec. 4). The support estimate above supplies a complete flow even though \(V\) need not be compact. Consider the symplectic forms \[\omega_s=dd^c\bigl(|z|^2+(1-s)\tau+s\widetilde H\bigr), \qquad 0\le s\le1,\] 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 \[\iota_{Y_s}\omega_s=-\gamma, \qquad \gamma=d^c(\widetilde H-\tau).\] 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 \(\mathbb C^n\), 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 \(\mathbb C^n\setminus V\) 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 \[ \varphi_1^*\widetilde\omega=\omega, \qquad \varphi_1^{-1}:(V,\widetilde\omega)\longrightarrow(V,\omega) \text{ is symplectic}. \tag{7}\]

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 (Loi and Zuddas 2008, proof of Theorem 1.1, Equation (23)). In the pullback of \(\lambda_0\) from Equation (1), the terms differentiating the radial factor cancel. Hence \[R^*\lambda_0=\Psi'(|z|^2)\lambda_0=d^c\Psi(|z|^2), \qquad R^*\omega_0=\widetilde\omega\big|_{B_{r_1}}.\] 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 (5). 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 (7) and the radial pullback identity. Since \(a\) was arbitrary in \((0,k)\), the theorem follows. ◻

Proof. The map and its angular derivative. The series in Equation (8) is absolutely summable on the closed disk. Hence \(F\) is bounded, odd, and commutes with conjugation. Termwise differentiation in the open disk gives \[ wF'(w)=\frac{8}{\pi^2}\arctan w, \qquad \arctan w=\frac{1}{2i}\log\frac{1+iw}{1-iw}. \tag{11}\] Here the logarithm is the branch on the right half-plane that is real on the positive real axis. Indeed, \[\Re\frac{1+iw}{1-iw} =\frac{1-|w|^2}{|1-iw|^2}>0 \qquad (w\in\mathbb D).\] 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 \[ A(r,\theta)=\Re\bigl(wF'(w)\bigr) =\frac{4}{\pi^2} \arctan\frac{2r\cos\theta}{1-r^2}. \tag{12}\] To obtain the last identity, the imaginary part of the fraction in Equation (11), 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, \(0<A(r,\theta)<2/\pi\). The identities \[ \partial_r F(re^{i\theta})=\frac{wF'(w)}{r}, \qquad \partial_\theta F(re^{i\theta})=iwF'(w) \tag{13}\] therefore show that \(\Re F(w)>0\) 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 (11) on the imaginary axis gives Equation (10). In particular, \(T\) is strictly increasing, \(T(0)=0\), and \(T'(r)\) diverges as \(r\uparrow1\). Since \(F(r)\) is real, Equation (13) 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 \(T(r)\to1\).

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 \[ \frac{\partial v}{\partial r}(r,t) =\frac{|wF'(w)|^2}{rA(r,\theta(r,t))}>0. \tag{14}\] 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 \(0<V(t)<\infty\).

This parametrization proves injectivity on each positive horizontal slice. The reflection identity \(F(-\overline w)=-\overline{F(w)}\) gives the negative slices, and the strict monotonicity of \(T\) gives injectivity on the imaginary axis. The right half-disk, imaginary axis, and left half-disk have images with positive, zero, and negative real parts, respectively. Hence \(F\) is globally injective. Every point of its image has imaginary coordinate in \((-1,1)\), and every such coordinate occurs on the imaginary axis. This proves Equation (9). Since \(F'\ne0\), the local holomorphic inverses combine into a holomorphic inverse \(g:D\to\mathbb D\). ◻

Proof. Step 1: express the slice integral in the disk radius. For any fixed \(v+it\in D\), the horizontal segment from \(it\) to \(v+it\) is a compact subset of the open set \(D\); the same is true for all sufficiently nearby endpoints and heights. The formula \[J_k(v,t)=k^2v\int_0^1 |g(av+it)|^{2k-2}|g'(av+it)|^2\,da\] then proves smoothness by differentiation under the integral. The integrand is smooth even at the zero of \(g\), since \(|g|^{2k-2}=(|g|^2)^{k-1}\) and \(k\) is an integer. Equation (16) follows from the fundamental theorem of calculus. The symmetry \(g(-\overline\zeta)=-\overline{g(\zeta)}\) makes the density in Equation (15) even in its real coordinate. Thus \(J_k(-v,t)=-J_k(v,t)\), and it suffices to estimate \(v>0\).

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 (14) gives \[ \frac{J_k(v,t)}{k} =k\int_{r_0}^{R} \frac{r^{2k-1}}{A(r,\theta(r,t))}\,dr. \tag{18}\] 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 \(0<A<2/\pi\). The constant part in Equation (18) contributes \(\frac{\pi}{4}(R^{2k}-r_0^{2k})\). It therefore suffices to bound the remaining weighted integral uniformly over all heights and endpoints by a quantity tending to zero as \(k\to\infty\).

Step 2: estimate the loss near the two tips. We next control the possible small denominator near the ends of the right semicircle. Suppose \(1/2\le r<1\) and set \[s=\frac{\pi}{2}-|\theta|\in(0,\pi/2], \qquad X=\frac{2r\sin s}{1-r^2}.\] For every \(X>0\), \[ \frac{1}{\arctan X}-\frac{2}{\pi}\le\frac{2}{X}. \tag{19}\] 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 (19) with \(\sin s\ge2s/\pi\) and \((1+r)/(2r)\le3/2\), we obtain \[ \frac1{A(r,\theta)} \le\frac{\pi}{2}+\frac{3\pi^3}{8}\frac{1-r}{s}. \tag{20}\] On the other hand, the vertical gap to the semicircle endpoint is \[\begin{align*} \Delta(r,\theta) &:=T(r)-|\Im F(re^{i\theta})| =\int_0^s\frac{4}{\pi^2} \arctan\frac{2r\sin\xi}{1-r^2}\,d\xi \\ &\le\frac{4r}{\pi^2(1-r^2)}s^2 \le\frac{2}{\pi^2}\frac{s^2}{1-r}. \tag{21}\end{align*}\] 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 (20) implies \[ \frac1{A(r,\theta(r,t))} \le\frac{\pi}{2} +C_0\sqrt{\frac{1-r}{T(r)-|t|}} \qquad (r\ge1/2,\ r>r_0). \tag{22}\]

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 \[ C_*k r_*^{2k-2}, \qquad C_*:=\operatorname{diam}(D) \max_{|w|\le r_*}|F'(w)|^{-2}<\infty. \tag{23}\] 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 \[ T(r)-|t|\ge T(r)-T(r_a)\ge L_*(r-r_a). \tag{24}\] Equations (18) and (22) therefore bound the outer contribution by \[ \frac{\pi}{4}\bigl(R^{2k}-r_a^{2k}\bigr) +\frac{C_0}{\sqrt{L_*}}\, k\int_{r_a}^{1}r^{2k-1} \sqrt{\frac{1-r}{r-r_a}}\,dr. \tag{25}\] 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 \[ I(k,r_a)\le 1+\frac{\pi H}{2}e^{-H/2} \le 1+\frac{\pi}{e}. \tag{26}\] 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 (23), (25), and (26), and using symmetry when \(v<0\), we have, for every \(v+it\in D\), \[ \frac{|J_k(v,t)|}{k}-\frac{\pi}{4}|g(v+it)|^{2k} \le C_*k r_*^{2k-2} +\frac{C_0(1+\pi/e)}{\sqrt{L_*}}. \tag{27}\] 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 \(v=0\).

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 \[ \epsilon_k= \max\left\{0,\ \sup_{v+it\in D} \left(\frac{|J_k(v,t)|}{k} -\frac{\pi}{4}|g(v+it)|^{2k}\right)\right\}. \tag{28}\] Equation (27) makes this supremum finite for each \(k\). For fixed \(r_*\), its first error term tends to zero as \(k\to\infty\). Moreover, Equation (10) 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 (27) 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 (17). ◻

Proof. Step 1: encode the strips by holomorphic functions. The body \(K\) and its finite list of defining vectors are fixed throughout the proof. Extend each real functional \(b_j\cdot x\) complex-linearly to \(\ell_j(z)=b_j\cdot z\). Let \(F:\mathbb D\to D\) and \(g=F^{-1}\) be the maps of Lemma 3, and define \[ \Omega=\{z\in\mathbb C^n:\ell_j(z)\in D\text{ for every }j\}. \tag{29}\] This is an open neighborhood of the origin. The spanning condition makes the linear map \(z\mapsto(\ell_1(z),\ldots,\ell_m(z))\) injective, so there is a constant \(C_b\) such that \[|z|\le C_b\max_j|\ell_j(z)|.\] Since \(D\) is bounded, \(\Omega\) is bounded as well; neither set depends on the integer \(k\) chosen below. The same linear bound makes \(K\) compact, and a sufficiently small Euclidean ball lies in all its defining strips. In particular, \(0\in\operatorname{int}(K)\).

For each integer \(k\ge2\), put \[f_j(z)=g(\ell_j(z))^k,\qquad \tau(z)=\sum_{j=1}^m|f_j(z)|^2,\qquad V=\{z\in\Omega:\tau(z)<1\},\qquad \omega=dd^c\bigl(|z|^2+\tau\bigr).\] The tuple \(f=(f_1,\ldots,f_m)\) is holomorphic on \(\Omega\). Since \(g\) has its only zero at the origin, the common zero set of \(f\) is \[\{z\in\Omega:\ell_j(z)=0\text{ for every }j\}=\{0\}.\] Holomorphicity at the origin gives \(|f(z)|=O(|z|^k)\). To verify the compactness hypothesis of Theorem 2, fix \(0<s<1\). On \(\{\tau\le s\}\), for every \(j\) we have \[\ell_j(z)\in F\bigl(\{w\in\mathbb C:|w|\le s^{1/(2k)}\}\bigr),\] a compact subset of \(D\). The linear bound above makes this sublevel bounded. Moreover, the limit of any convergent sequence in it still has all its images under the functionals \(\ell_j\) in that compact subset of \(D\), hence belongs to \(\Omega\) and satisfies \(\tau\le s\) by continuity. Thus the sublevel is compact in \(\Omega\). Theorem 2 now supplies symplectic embeddings into \((V,\omega)\) of every standard ball of capacity less than \(\pi k\).

Step 2: realize the form by a globally injective map. Write \(z=u+ix\) and use the signed integrals \(J_k\) of Proposition 4 to define \[ \Phi_k:V\longrightarrow\mathbb R^n_q\times\mathbb R^n_p, \qquad q=x,\qquad p=-u-\sum_{j=1}^m J_k(b_j\cdot u,b_j\cdot x)b_j. \tag{30}\] The signed integrals \(J_k\) are smooth on \(D\) by Proposition 4, so \(\Phi_k\) is smooth.

Set \(v_j=b_j\cdot u\) and \(t_j=b_j\cdot x\). Pulling back the target form gives \[\begin{align*} \Phi_k^*\omega_0 &=\sum_{\ell=1}^n dx_\ell\wedge(-du_\ell) -\sum_{j=1}^m dt_j\wedge dJ_k(v_j,t_j)\\ &=\sum_{\ell=1}^n du_\ell\wedge dx_\ell +\sum_{j=1}^m \partial_vJ_k(v_j,t_j)\,dv_j\wedge dt_j\\ &=dd^c\bigl(|z|^2+\tau\bigr)=\omega. \tag{31}\end{align*}\] Indeed, the terms containing \(\partial_tJ_k\) vanish because \(dt_j\wedge dt_j=0\), while \[\partial_vJ_k(v,t)=k^2|g(v+it)|^{2k-2}|g'(v+it)|^2.\] This is precisely the coefficient of \(dd^c|g(v+it)^k|^2\) in front of \(dv\wedge dt\), with the convention for \(d^c\) fixed above.

The map \(\Phi_k\) is globally injective. To see this, equality of its \(q\) coordinates first fixes \(x\). For two possible real coordinates \(u_1,u_2\) at that \(x\), write \[\Delta u=u_2-u_1,\qquad v_{ji}=b_j\cdot u_i,\qquad t_j=b_j\cdot x,\] and denote the corresponding momenta by \(p_1,p_2\). Both \(v_{j1}\) and \(v_{j2}\) belong to the same horizontal interval of \(D\) at height \(t_j\). Since \(\partial_vJ_k\ge0\) on this entire interval, \[\begin{align*} (p_2-p_1)\cdot\Delta u &=-|\Delta u|^2 -\sum_{j=1}^m \bigl(J_k(v_{j2},t_j)-J_k(v_{j1},t_j)\bigr) (v_{j2}-v_{j1})\\ &\le -|\Delta u|^2. \end{align*}\] Thus equal momenta force \(u_1=u_2\). This comparison applies to any two points of the fiber, even if that fiber of \(V\) is disconnected. Equation (31) and nondegeneracy of \(\omega\) show that \(\Phi_k\) is a local diffeomorphism. It is therefore open, and its injectivity gives a smooth inverse onto its image. Consequently \(\Phi_k\) is a symplectic embedding.

Step 3: control the momentum by the polar body. We next bound its image. All imaginary parts of points of \(D\) lie in \((-1,1)\), so \(|b_j\cdot x|<1\) for every \(j\). The finite system of strict inequalities implies \(q=x\in\operatorname{int}(K)\). Introduce the support function \[h_K(p)=\max_{y\in K}\langle p,y\rangle.\] It satisfies \(h_K(\pm b_j)\le1\), and \(\operatorname{int}(K^\circ)=\{p:h_K(p)<1\}\). Because \(\Omega\) is bounded, there is a finite constant \(C_K\), independent of \(k\), such that \(h_K(-u)\le C_K\) on \(\Omega\). Subadditivity and positive homogeneity of \(h_K\), followed by Proposition 4, give \[\begin{align*} \frac{h_K(p)}{k} &\le \frac{C_K}{k} +\sum_{j=1}^m\frac{|J_k(v_j,t_j)|}{k}\\ &\le \frac{C_K}{k} +\frac{\pi}{4}\sum_{j=1}^m|g(\ell_j(z))|^{2k} +m\epsilon_k =\frac{C_K}{k}+\frac{\pi}{4}\tau(z)+m\epsilon_k. \tag{32}\end{align*}\] Here \(\epsilon_k\ge0\) tends to zero, uniformly in the planar argument. The number \(m\) and the constant \(C_K\) are fixed before \(k\) varies. It follows that, for any \(\eta>0\) and all sufficiently large \(k\), \[\frac{C_K}{k}+m\epsilon_k<\frac{\eta\pi}{4}.\] Since \(\tau<1\) on \(V\), Equation (32) then gives the strict inclusion \[ \Phi_k(V)\subset \operatorname{int}(K)\times S\operatorname{int}(K^\circ), \qquad S=(1+\eta)\frac{\pi k}{4}. \tag{33}\]

Step 4: rescale the ball and the momentum together. Finally, fix \(0<c<4\). Choose \(\eta>0\) with \((1+\eta)c<4\), and then fix an integer \(k\) large enough for Equation (33). 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 \[ L_S\circ\Phi_k\circ E\circ D_{\sqrt S}: B^{2n}(c)\longrightarrow \operatorname{int}(K)\times\operatorname{int}(K^\circ) \tag{34}\] 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. ◻

Proof. First, \(0\in\operatorname{int}K\). 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 \[\langle v,x\rangle>h_K(v), \qquad h_K(v)=\max_{y\in K}\langle v,y\rangle>0.\] 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 \(\langle v,p\rangle\) 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 \(\mathbb R^n\). 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. ◻

Proof of Theorem 1. Fix \(0<c<4\). Choose \(\alpha>0\) such that \(\lambda c<4\), where \(\lambda=1+\alpha\), and choose one body \(K'\) from Lemma 6. Proposition 5 gives a smooth symplectic embedding \[e:B^{2n}(\lambda c)\longrightarrow \operatorname{int}K'\times\operatorname{int}\bigl((K')^\circ\bigr).\] If one body is contained in another, every interior ball in the first is contained in the second. Thus Equation (35) implies \[\operatorname{int}K'\subset\lambda\operatorname{int}K, \qquad \operatorname{int}\bigl((K')^\circ\bigr)\subset\operatorname{int}K^\circ,\] even when their boundaries meet. Regard \(e\) as an embedding into \(\lambda\operatorname{int}K\times\operatorname{int}K^\circ\). Define \[D_{\sqrt\lambda}(z)=\sqrt\lambda\,z, \qquad Q_\lambda(q,p)=(q/\lambda,p).\] The composition \[ Q_\lambda\circ e\circ D_{\sqrt\lambda}: B^{2n}(c)\longrightarrow U_K \tag{36}\] 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 (Artstein-Avidan et al. 2014, Remark 4.2). Choose \(q_0\in K\) and \(p_0\in K^\circ\) with \(\langle q_0,p_0\rangle=1\). 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 \(\mathbb R^n\) 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 \(Q_1=\langle p_0,q\rangle\) and \(P_1=\langle q_0,p\rangle\). 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 \((-1,1)^2\times\mathbb R^{2n-2}\), 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 \(-R<X<R\), and \[H(X)=-1+\int_{-R}^{X}h(s)\,ds.\] Since \(\int_{-R}^{R}h(s)\,ds=2\) and \(H'=h>0\), \(H\) 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 \(U_K\hookrightarrow B^2(4)\times\mathbb R^{2n-2}\). Gromov’s nonsqueezing theorem (Gromov 1985) therefore bounds the capacity of every ball in \(U_K\) by \(4\). Combined with the lower bound, this proves \(c_G(U_K)=4\). ◻

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 \(C\subset\mathbb R^n\) contains \(B(0,r)\), then convexity gives, for \(0<t<1\), \[tC+B(0,(1-t)r)\subset C.\] Hence \(tC\subset\operatorname{int}C\), and \[t^n\operatorname{vol}_n(C)\le\operatorname{vol}_n(\operatorname{int}C)\le\operatorname{vol}_n(C).\] Letting \(t\uparrow1\) proves \(\operatorname{vol}_n(\operatorname{int}C)=\operatorname{vol}_n(C)\). Apply this to \(K\) and \(K^\circ\), both of which contain a neighborhood of the origin. The product formula for Lebesgue measure then gives \[ \operatorname{vol}_{2n}(U_K)=\operatorname{vol}_n(K)\,\operatorname{vol}_n(K^\circ). \tag{37}\]

For every \(0<c<4\), Theorem 1 supplies a smooth symplectic embedding \(f:B^{2n}(c)\to U_K\). Taking the \(n\)th exterior power of \(f^*\omega_0=\omega_0\) shows that \(\det Df=1\), so the change of variables formula gives \[\frac{c^n}{n!} =\operatorname{vol}_{2n}\bigl(B^{2n}(c)\bigr) =\operatorname{vol}_{2n}\bigl(f(B^{2n}(c))\bigr) \le\operatorname{vol}_{2n}(U_K).\] Here the first equality follows from the Euclidean ball volume \(\pi^nR^{2n}/n!\) at radius \(R=\sqrt{c/\pi}\). Combining this with Equation (37) and letting \(c\uparrow4\) proves the claimed inequality. ◻

Proof. Strictness and finiteness permit a choice of \(0<c<4\) such that \(\sum_i R_i^n<c^n\) and \(R_i+R_j<c\) for every \(i<j\). The ball-packing theorem (OpenAI 2026a, Theorem 1.1) embeds these closed balls into \(B^{2n}(c)\), with embeddings defined on neighborhoods. Their compact images are pairwise disjoint and lie in the open target, so the neighborhoods can be shrunk to have pairwise disjoint images contained in \(B^{2n}(c)\). Composing with the embedding in Theorem 1 proves the assertion. ◻