SYMPLECTIC BALLS IN SYMMETRIC POLAR PRODUCTS
============================================
A complete, self-contained proof that the Gromov width of
int(K) x int(K^o) equals 4, and that the symmetric Mahler
conjecture holds in every dimension.
Rewritten for a reader who has had multivariable calculus and
linear algebra but no symplectic geometry. Every step is
proved. Exactly one result is quoted rather than proved:
Gromov's nonsqueezing theorem, stated precisely in Section 6.2
and used exactly once. (Section 6.6 records one add-on that
also invokes an external packing theorem; nothing else in the
document depends on it.)
0. HOW TO READ THIS
-------------------
Three moving parts, worth knowing before the details arrive.
(A) A local machine (Section 3). A holomorphic map vanishing
to order k at the origin makes an explicit potential on an
explicit domain contain symplectic balls of every capacity
below pi*k.
(B) A one-dimensional estimate (Section 4). One conformal map
of the unit disk gives a slice coordinate whose primitive
grows at most (pi*k/4) times a power, uniformly on the
image, up to o(1).
(C) Assembly (Sections 5 and 6). Writing a symmetric convex
body as an intersection of strips, then approximating in the
polar, turns (A) and (B) into balls of every capacity below
4 inside int(K) x int(K^o).
Part (A) is several complex variables. Part (B) is
one-variable calculus. Part (C) is linear algebra and
bookkeeping. Only (B) is fussy, and it is fussy for an honest
reason: the estimate must survive when a parameter runs off to
the boundary of the disk.
Notation. R^n carries the standard dot product <.,.> and norm
|.|. C^n is identified with R^{2n} by z = u + ix with u and x
in R^n. Balls and domains are open unless stated otherwise.
"Smooth" means C-infinity. int(S) is the interior of S.
1. THE STATEMENT
----------------
1.1 Convex bodies and their polars.
A convex body in R^n is a compact convex set with nonempty
interior. It is origin-symmetric if K = -K. Its polar is
K^o = { p in R^n : <q,p> <= 1 for every q in K }.
If K contains a ball of radius r and lies in a ball of radius
R, then B(0,1/R) is contained in K^o and K^o is contained in
B(0,1/r). So K^o is again a convex body, and it is again
origin-symmetric. Also K^{oo} = K; the nontrivial half of
this is proved in Lemma 6.1.
The volume product vol_n(K) vol_n(K^o) is unchanged when K is
subjected to an invertible linear map T, since (TK)^o = T^{-T}
K^o and the two volume factors move in opposite directions.
Mahler asked for the minimum of this product. The symmetric
case is the following.
Conjecture (symmetric Mahler).
vol_n(K) vol_n(K^o) >= 4^n / n!
for every origin-symmetric convex body K in R^n.
Equality holds for the cube, whose polar is the crosspolytope.
1.2 Phase space, balls, width.
Put position in the first factor and momentum in the second:
R^{2n} = R^n_q x R^n_p,
omega_0 = sum_{j=1}^n dq_j ^ dp_j .
The form omega_0 is closed (constant coefficients) and
nondegenerate (its n-th exterior power is n! times the volume
form). A smooth map e between open subsets of R^{2n} is
symplectic if e^*(omega_0) = omega_0.
The standard ball of capacity c > 0 is
B^{2n}(c) = { (q,p) : pi (|q|^2 + |p|^2) < c }.
Its Euclidean radius is sqrt(c/pi), so
vol_{2n}( B^{2n}(c) ) = pi^n (c/pi)^n / n! = c^n / n!.
The Gromov width of an open set U in R^{2n} is
c_G(U) = sup { c : there is a symplectic embedding
B^{2n}(c) -> U }.
A symplectic map preserves the n-th exterior power of omega_0,
hence preserves volume. Therefore
c_G(U) <= ( n! vol_{2n}(U) )^{1/n}.
1.3 The theorem.
THEOREM 1.1. Let n >= 2 and let K be an origin-symmetric
convex body in R^n. Set U_K = int(K) x int(K^o). Then
c_G(U_K) = 4.
More precisely, for every c with 0 < c < 4 there is a
smooth symplectic embedding B^{2n}(c) -> U_K.
Two consequences follow at once. First, by volume preservation,
for every c < 4,
c^n/n! = vol_{2n}(B^{2n}(c)) <= vol_{2n}(U_K)
= vol_n(K) vol_n(K^o),
and letting c tend to 4 gives the symmetric Mahler inequality.
Second, for the cube the product is exactly 4^n/n!, so the
constant 4 cannot be improved. Section 6.4 does the measure
bookkeeping and Section 6.5 records the cube computation.
Why 4, and not something larger? A single dual pair of
coordinates already saturates the bound: after a linear
symplectic change of coordinates, U_K sits inside the product
of the open square (-1,1) x (-1,1) with R^{2n-2}, and that
square has area 4 (Section 6.2). Duality is what makes one
pair of coordinates enough: if K is long in some direction then
K^o is thin in the same direction, so the two factors can never
both be large. The content of Theorem 1.1 is that this
trade-off is exact, not merely an upper bound.
The rest of the paper proves the two halves separately. The
lower bound c_G(U_K) >= 4 is Sections 3 to 5; it is the hard
half and it produces actual embeddings, not just an inequality.
The upper bound c_G(U_K) <= 4 is Section 6.2; it is short.
2. BACKGROUND, BRIEFLY
----------------------
Mahler raised the volume-product question in 1939. The planar
case and the unconditional case are classical; the symmetric
three-dimensional case is Iriyeh-Shibata (2020). Bourgain and
Milman (1987) proved vol_n(K) vol_n(K^o) >= a^n/n! for some a > 0
independent of dimension, via the reverse Santalo inequality.
That shows the conjecture holds up to the value of the base; it
does not give a = 4.
Artstein-Avidan, Karasev and Ostrover (2014) proved that the
Hofer-Zehnder capacity of K x K^o equals 4, hence that
c_G(K x K^o) <= 4, hence the symmetric Mahler inequality with the
sharp constant. A capacity bound of that kind does not produce a
symplectic ball: producing actual embeddings of large balls is
strictly stronger, and that is what Theorem 1.1 does. Ball
embeddings into special polar products were known earlier
(Karasev for ell_p balls, Ramos-Sepe for the cube-crosspolytope
product, Vicente for functional-dual products); the general
polar-product question was open.
Two cautionary facts. Viterbo's volume-capacity conjecture for
general convex domains is false: Haim-Kislev and Ostrover (2026)
built a four-dimensional counterexample from a regular pentagon
and a rotated copy. That pentagon is not centrally symmetric and
their domain is not a polar product, so nothing here conflicts
with it. Second, the planar map F of Section 4 is Gross's
conformal Skorokhod map, rotated; the smoothing in Section 3 is
Richberg's; the transport is Moser's; the radial change of
coordinates is the flat case of the Loi-Zuddas formula. All are
reproved here, so no external reading is needed.
3. BALLS FROM A HOLOMORPHIC ZERO
--------------------------------
This section proves the local machine (A). The statement is
purely about a neighbourhood of the origin in C^n; convex bodies
do not appear until Section 5.
3.1 The normalization of d^c and dd^c.
Identify C^n with R^{2n} via z = u + ix. For a smooth real
function phi define
d^c phi (X) = -(1/4) dphi(iX),
dd^c phi = d( d^c phi ),
where iX is the tangent vector obtained by multiplying by i.
The constant 1/4 is a convention; it is fixed so that the
following two lines come out without stray factors. Doing the
calculation once, in one complex variable zeta, with
phi(z + zeta X) as a function of zeta, gives
dd^c phi (X, iX) = (1/4) Laplacian_zeta phi(z + zeta X)
evaluated at zeta = 0.
Consequences, used constantly below.
(i) dd^c |z|^2 = omega_0 = sum_j du_j ^ dx_j, with primitive
d^c|z|^2 = (1/2) sum_j ( u_j dx_j - x_j du_j ).
(ii) A smooth phi is plurisubharmonic (psh) exactly when
dd^c phi (X, iX) >= 0 for every real X and every point.
For such phi the form dd^c(|z|^2 + phi) is symplectic:
it is closed, and on the pair (X, iX) it is at least
|X|^2 + 0, hence positive.
We also need a closure property.
LEMMA 3.1. Let h_1,...,h_N be smooth psh functions on an
open set and let Gamma: R^N -> R be smooth, convex, and
nondecreasing in each variable. Then
Gamma(h_1,...,h_N) is psh.
Proof. Restrict to a complex line, parametrized by a real
variable t, and let Delta be the second derivative in t.
The chain rule gives
Delta Gamma(h) = sum_j Gamma_j Delta h_j
+ sum_{i,j} Gamma_{ij} <grad h_i, grad h_j>.
The first sum is nonnegative because Delta h_j >= 0 and
Gamma_j >= 0. The second sum is a quadratic form with a
positive semidefinite matrix applied to the gradients,
hence nonnegative. Both statements hold at every point and
on every complex line, which is what psh means.
Finally the fundamental computation. Let f: Omega -> C^m be
holomorphic on an open set Omega in C^n and put tau = |f|^2.
Along a complex line z + zeta X, expand
f(z + zeta X) = f(z) + zeta v + O(zeta^2), v = df(z)(X).
Then |f|^2 = |f|^2 + 2 Re( conj(f) zeta v ) + |zeta|^2 |v|^2,
whose Laplacian in zeta is 4|v|^2. Dividing by 4 as in (ii),
dd^c tau (X, iX) = |df(X)|^2 >= 0,
dd^c log tau (X, iX) = ( |f|^2 |v|^2 - |conj(f) . v|^2 ) / |f|^4,
the second formula valid away from the zero set of f. The
numerator is nonnegative by Cauchy-Schwarz, so log tau is psh
wherever it is defined. Notice that for m = 1 the numerator
vanishes identically: a single holomorphic function contributes
nothing. Everything below genuinely needs m >= 2, which is why
Section 5 writes a body using several functionals at once.
3.2 The theorem.
THEOREM 3.2. Let n, m >= 1. Let Omega be an open set in
C^n containing 0, and let f = (f_1,...,f_m): Omega -> C^m
be holomorphic. Suppose that for some integer k >= 1:
(H1) the common zero set f^{-1}(0) is exactly {0}, and
|f(z)| = O(|z|^k) as z -> 0;
(H2) for every s with 0 < s < 1, the sublevel set
{ z in Omega : |f(z)|^2 <= s } is compact in Omega.
Set
tau = |f|^2,
V = { z in Omega : tau(z) < 1 },
omega = dd^c( |z|^2 + tau ).
Then for every c with 0 < c < pi*k there is a smooth
symplectic embedding (B^{2n}(c), omega_0) -> (V, omega).
Remark on normalization. The capacity of the target ball is
pi*k and the source is written in capacity units; the number pi
enters only through the definition of B^{2n}(c) in Section 1.2.
Readers who prefer radius conventions should divide every
capacity by pi.
3.3 Strategy.
There are four steps and each has one job.
Step 1. Build a psh function H that equals tau away from the
origin and has a log singularity of controlled slope bk there,
with bk strictly smaller than k. The point of using bk < k is
that it leaves room for a second, explicit singularity whose
slope mu lies strictly between the target capacity a and bk.
Step 2. Replace the singularity of H by an explicit radial
model P_delta. The model is bigger than H near the origin and
smaller than H on an outer sphere, so a smooth maximum glues
them into a global psh Htilde that agrees with tau outside a
compact set.
Step 3. Moser's deformation: because Htilde - tau is compactly
supported, the form dd^c(|z|^2 + tau) can be deformed into the
form dd^c(|z|^2 + Htilde) by a flow, without changing anything
near the boundary of V.
Step 4. On a small ball the new potential is exactly
Psi(|z|^2) with Psi(t) = t + mu*log(t + delta). A radial change
of variables turns that into a standard ball, and reading off
its radius gives capacity pi*a with a as close to k as we like.
Two things are worth stressing now. First, H2 (compact
sublevels) is used only to make the flow complete and to keep
the deformation away from the boundary; no regularity of the
level sets {tau = const} is assumed anywhere. Second, the
capacity obtained in Step 4 is pi*a for an arbitrary a < k, but
it depends on a only through a single inequality
(delta < r_1^2 (mu/a - 1)). There is no limit of embeddings.
3.4 Proof of Theorem 3.2, Step 1: a psh logarithm with slope bk.
Fix a with 0 < a < k; we will construct an embedding of
B^{2n}(pi*a) and then let a increase. Choose numbers
a/k < b < 1, a < mu < b*k, log b < t_- < t_+ < 0.
All four choices are possible because a/k < 1, because
a < bk, and because log b < 0.
LEMMA 3.3. There is a smooth function chi on (-infinity, 0)
with the following properties:
(C1) chi(t) = e^t whenever t >= t_+;
(C2) chi(t) = b*t + c_0 whenever t <= t_-, for a constant c_0;
(C3) chi is strictly increasing and convex.
Proof. Let beta: R -> [0,1] be smooth and nondecreasing,
with beta = 0 on (-infinity, t_-] and beta = 1 on
[t_+, infinity). Define chi by prescribing its derivative,
chi'(t) = b + beta(t) ( e^t - b ), t < 0,
and choose the additive constant so that chi(t) = e^t for
t >= t_+. This is consistent: on [t_+, 0) the derivative
equals e^t, so chi minus e^t is constant there, and the
constant is fixed to be zero.
Positivity of chi': on t <= t_- we have chi' = b > 0; on
t >= t_+ we have chi' = e^t > 0; in between, both b > 0 and
e^t - b > 0 (because t > t_- > log b), and beta lies in
[0,1], so chi' >= b > 0.
Convexity: differentiating once more,
chi''(t) = beta'(t) ( e^t - b ) + beta(t) e^t.
Both terms are nonnegative: beta' >= 0 and e^t > b on the
range where beta' can be nonzero, while beta >= 0 and
e^t > 0 everywhere. QED
The reason for the condition t_- > log b is exactly the first
term above. It is the only place where it is used.
Now put
H = chi(log tau), defined on V minus the origin.
Then H is smooth and psh there: log tau is psh by Section 3.1,
and tau = |f|^2 > 0 on that domain, while chi is smooth, convex
and increasing, so Lemma 3.1 applies. Two estimates.
(E1) H = tau wherever tau >= s_0 := e^{t_+}. Indeed, there
log tau >= t_+, so chi(log tau) = e^{log tau} = tau.
(E2) There is a constant C_1 with H(z) <= b*k*log|z|^2 + C_1
for all sufficiently small z other than 0. Indeed, by
(H1) we have tau(z) <= C_0 |z|^{2k} near 0, so
log tau <= log C_0 + 2k log|z|, which is below t_- once
|z| is small. There chi is affine of slope b, giving
H = b log tau + c_0 <= b*k*log|z|^2 + C_1.
The exponent b*k > a is the whole reason for introducing b. A
singularity of slope strictly less than k but strictly greater
than the target capacity a is what lets Step 2 glue.
3.5 Step 2: replace the singularity by a radial model.
Here is the model. For delta > 0 put
P_delta(z) = mu * log( |z|^2 + delta ) - M,
with mu as chosen in Step 1 and M a constant to be fixed. It
is psh for every delta > 0, because it equals mu times the
logarithm of the squared modulus of the nowhere-vanishing
holomorphic tuple (sqrt(delta), z_1, ..., z_n), and Section 3.1
applied to that tuple gives nonnegativity. Its point is that
dd^c(|z|^2 + P_delta) is the pullback of a standard ball form
under an explicit radial map (Step 4).
Write B_r = { z in C^n : |z| < r }. Choose r_2 > 0 so small
that the closed ball of radius r_2 lies in V; this is possible
because tau(0) = 0 < 1 and V is open. On the sphere |z| = r_2
the function tau is strictly positive, since the only common
zero of f is the origin, so H has a finite minimum there.
Fix M so large that
M > mu * log(r_2^2 + 1) - min_{|z| = r_2} H(z) + 2.
Since |z|^2 + delta <= r_2^2 + 1 on the sphere for 0 < delta <= 1,
we get, for every such delta,
P_delta <= mu log(r_2^2 + 1) - M < min_{|z|=r_2} H - 2 < H - 2
on the sphere |z| = r_2. That is the outer comparison: the
model sits strictly below H on the outer boundary.
For the inner comparison use (E2) and mu < b*k:
P_delta(z) - H(z)
>= mu log|z|^2 - M - ( b*k*log|z|^2 + C_1 )
= (mu - b*k) log|z|^2 - M - C_1 .
Because mu - b*k < 0 and log|z|^2 tends to minus infinity, this
lower bound tends to plus infinity as z tends to 0, and it does
so uniformly in delta. We may therefore fix a radius r_1 with
0 < r_1 < r_2, independent of delta, such that
(3.1) P_delta > H + 2 on 0 < |z| <= r_1,
simultaneously for every 0 < delta <= 1.
That is the inner comparison: the model sits strictly above H
on a small punctured ball.
Now fix one value of delta. The inequalities (3.1) and the
outer comparison hold for all small delta, and the ball produced
in Step 4 has capacity increasing with delta decreasing, so we
take delta as small as we like subject to the Step 4 condition:
(3.2) 0 < delta < min { 1, r_1^2 (mu/a - 1) }.
The upper bound is positive because mu > a. This is the only
place where delta is constrained, and it is where the target
capacity a enters.
Next, glue. Let vartheta be a smooth even nonnegative function
supported in (-1,1) with integral 1, and set
rho(s) = integral_R |s - t| vartheta(t) dt,
rmax(h,p) = ( h + p + rho(h - p) ) / 2 .
The function rho is smooth and convex (it is a smoothed absolute
value), satisfies |rho'| <= 1, and equals |s| for |s| >= 1. So
rmax is smooth, convex, nondecreasing in each argument, and
equals max(h,p) whenever |h - p| >= 1.
Define Htilde on V by
Htilde = rmax(H, P_delta) on B_{r_2} minus {0},
Htilde = H on V minus B_{r_2},
Htilde(0) = P_delta(0) = mu log delta - M.
This is well defined and smooth. Why: near the outer sphere
|z| = r_2 we have P_delta < H - 2 < H - 1, so rmax(H,P_delta)
equals max(H,P_delta) = H there, and the formula for |z| < r_2
therefore joins the formula for |z| > r_2 smoothly. Near the
inner sphere |z| = r_1 we have P_delta > H + 2 > H + 1, so rmax
equals P_delta there; hence Htilde equals P_delta on the whole
ball B_{r_1}. In particular Htilde agrees with the smooth
function P_delta on a neighbourhood of the origin, and its value
there is the finite number mu log(delta) - M.
Moreover Htilde is psh on all of V. Being psh is a local
condition, so it suffices to check it on an open cover. On
B_{r_2} minus {0} it is a composition of the psh pair
(H, P_delta) with the smooth convex increasing rmax, so Lemma
3.1 applies. On B_{r_1} it is P_delta, which is psh. On the
remainder it is either H or rmax(H, P_delta), and both of those
are psh wherever they are defined. Hence every point of V has a
neighbourhood on which Htilde equals a known psh function.
Finally, the difference Htilde - tau has compact support in V.
To see this, put
s_2 = max over the closed ball of radius r_2 of tau,
which is less than 1 by the choice of r_2. Choose sigma with
max(s_0, s_2) < sigma < 1 and let
C = { z in Omega : tau(z) <= sigma }.
By (H2) this set is compact in Omega. It is also contained in
V, since sigma < 1. If z is not in C then tau(z) > sigma > s_0,
so H(z) = tau(z) by (E1); and tau(z) > sigma > s_2 forces
|z| > r_2, so Htilde(z) = H(z) = tau(z). Therefore
Htilde - tau vanishes off C, as claimed. This compactness is
the single point at which hypothesis (H2) is used.
3.6 Step 3: transport the smoothed form by a flow.
For 0 <= s <= 1 consider the family of two-forms
omega_s = dd^c( |z|^2 + (1-s) tau + s Htilde ),
on V, and write omegatilde = omega_1 = dd^c(|z|^2 + Htilde).
Each omega_s is symplectic: the potential is |z|^2 plus a convex
combination of two psh functions, which is psh because dd^c is
linear, so criterion (ii) of Section 3.1 applies.
Let
gamma = d^c( Htilde - tau ).
This is a smooth 1-form on V with compact support, since
Htilde - tau has compact support (end of Section 3.5). Because
omega_s is nondegenerate, for each s and each point there is a
unique vector Y_s with
iota_{Y_s} omega_s = -gamma.
The dependence of Y_s on (s,z) is smooth (solve a linear system
whose coefficients depend smoothly on s and z), and Y_s is
supported in the same compact set C. Extend Y_s by zero outside
V; the result is a smooth time-dependent vector field on R^{2n}
with compact support, uniformly in s. Consequently its flow
phi_s exists for all s in [0,1]. Since Y_s vanishes outside V,
the flow fixes every point of the complement of V and restricts
to a diffeomorphism of V.
Now differentiate. Because d(omega_s) = 0 for every s,
d/ds ( phi_s^* omega_s )
= phi_s^* ( L_{Y_s} omega_s + d omega_s/ds )
= phi_s^* ( d iota_{Y_s} omega_s + iota_{Y_s} d omega_s
+ d omega_s/ds )
= phi_s^* ( -d gamma + 0 + d gamma )
= 0.
Here L is the Lie derivative, and d omega_s/ds = dd^c(Htilde - tau)
= d d^c(Htilde - tau) = d gamma. So the pullback is constant in s;
at s = 0 it is phi_0^*omega_0 = omega, since phi_0 is the identity.
Hence
phi_1^* omegatilde = omega,
that is, phi_1: (V, omegatilde) -> (V, omega) is symplectic via
its inverse phi_1^{-1}.
Two remarks. First, this is the only place where the completeness
of the flow matters, and it is guaranteed by compact support, not
by any property of the boundary of V. Second, nothing here
assumes that V is connected or that its boundary is smooth; the
flow never sees the boundary, because it is the identity on a
neighbourhood of it.
3.7 Step 4: read the ball off the radial model.
Restrict everything to the ball B_{r_1}. There Htilde equals
P_delta, so
omegatilde = dd^c( |z|^2 + mu log(|z|^2 + delta) - M )
= dd^c( Psi(|z|^2) )
with the additive constant dropped, where
Psi(t) = t + mu log(t + delta), t >= 0.
Since Psi'(t) = 1 + mu/(t + delta) > 0, the squared-radius
function t -> t Psi'(t) is strictly increasing:
d/dt ( t Psi'(t) ) = 1 + mu delta / (t + delta)^2 > 0.
LEMMA 3.4. Let Psi be smooth on [0, r^2) with Psi' > 0 and
with t -> t Psi'(t) strictly increasing. Define
R(z) = sqrt( Psi'(|z|^2) ) z.
Then R is a diffeomorphism of B_r onto the open ball of
squared radius r^2 Psi'(r^2), and
R^* omega_0 = dd^c( Psi(|z|^2) ).
Proof. In one complex dimension write w = v + i t. Then
d^c|w|^2 = (1/2)( v dt - t dv ), so
omega_0 = dd^c|w|^2 = dv ^ dt,
d^c|w|^2 = (1/2) Im( conj(w) dw ).
Write lambda(z) = sqrt(Psi'(|z|^2)), so R(z) = lambda(z) z and
lambda is real-valued. Then
conj(R) dR = lambda conj(z) ( lambda dz + z dlambda )
= lambda^2 conj(z) dz + lambda |z|^2 dlambda,
and dlambda is real, so the second term is real and has zero
imaginary part. Therefore
Im( conj(R) dR ) = lambda^2 Im( conj(z) dz ),
which after the factor 1/2 says R^*(d^c|z|^2) = lambda^2
d^c|z|^2 = Psi'(|z|^2) d^c|z|^2. Apply d to both sides:
the left side becomes R^*(dd^c|z|^2) = R^*omega_0, and the
right side becomes
d( Psi'(|z|^2) d^c|z|^2 ), which equals dd^c(Psi(|z|^2))
because d^c(Psi(|z|^2)) = Psi'(|z|^2) d^c|z|^2. That is the
claimed identity.
For bijectivity: |R(z)| = lambda(|z|^2)|z| is a strictly
increasing function of |z|, because |R|^2 = |z|^2 Psi'(|z|^2),
which is strictly increasing by hypothesis. So R maps the
sphere of radius t onto the sphere of radius
sqrt(t^2 Psi'(t^2)), it maps concentric spheres to concentric
spheres injectively, and it is a bijection of B_r onto the
ball of squared radius r^2 Psi'(r^2). It is smooth with
smooth inverse since the radius map is a strictly increasing
diffeomorphism. QED
Apply the lemma with r = r_1. The image is the ball of squared
radius
Q_delta = r_1^2 Psi'(r_1^2) = r_1^2 + mu r_1^2 / ( r_1^2 + delta ).
By the choice (3.2) of delta this is strictly bigger than a:
Q_delta > a
<=> mu r_1^2 / (r_1^2 + delta) > a
<=> delta < r_1^2 (mu/a - 1).
Hence the standard ball B^{2n}(pi a), which is by definition the
set of points of Euclidean squared radius less than a, is
contained in the open ball R(B_{r_1}). Therefore
R^{-1} is defined on B^{2n}(pi a), takes values in B_{r_1},
and (R^{-1})^* omegatilde = omega_0.
Now compose. Let iota: B_{r_1} -> V be the inclusion. Then
E = phi_1^{-1} o iota o R^{-1} | B^{2n}(pi a) :
B^{2n}(pi a) -> V
is a smooth map, and
E^* omega = (R^{-1})^* ( iota^* ( phi_1^{-1} )^* omega )
= (R^{-1})^* omegatilde
= omega_0,
using phi_1^{-1}(V, omega) -> (V, omegatilde) symplectic. It is
an immersion because E^*omega = omega_0 is nondegenerate, and an
injection because each of the three maps is injective. A smooth
injective immersion is an embedding, so E is a smooth symplectic
embedding of the standard ball of capacity pi*a into (V, omega).
Finally, a was any number with 0 < a < k, and the capacity pi*a
then covers every c with 0 < c < pi*k. This proves Theorem 3.2.
3.8 What was used, and what was not.
For a reader checking dependencies: (H1) is used three times, to
make the common zero set a single point (so tau > 0 on spheres),
to give the vanishing estimate in (E2), and to make the local
behaviour of f at the origin flat of order k. Hypothesis (H2)
is used exactly once, in the compact-support argument. Lemma 3.1
is used twice, for H and for Htilde. Moser's argument is used
once. No regularity of the level sets of tau, and no pseudoconvexity
or Stein property of V, is ever used.
4. THE PLANAR COORDINATE
------------------------
This section proves the one-dimensional estimate (B). Everything
happens in the unit disk and its image under one explicit
conformal map. The estimate has to be uniform over the whole
image domain, including a neighbourhood of the two boundary
points where the map blows up; that is the only hard part.
4.1 The map.
Set D = { w in C : |w| < 1 } and define
F(w) = (8/pi^2) * sum_{l >= 0} (-1)^l w^{2l+1} / (2l+1)^2,
|w| < 1.
LEMMA 4.1 (the lens map). The following hold.
(a) The series converges absolutely on the closed unit disk.
Its sum F is holomorphic on D, continuous on the closed
disk, odd, satisfies F(0) = 0 and F(conj w) = conj F(w),
and has F'(0) = 8/pi^2.
(b) For every w in D,
w F'(w) = (8/pi^2) * arctan w,
where arctan w = (1/2i) log( (1+iw)/(1-iw) ) with the
logarithm taken on the right half-plane, real on the
positive real axis. Consequently F' has no zeros in D.
(c) Write w = r e^{i theta} with 0 <= r < 1. On the closed
right half-disk, that is |theta| <= pi/2, define
A(r,theta) = Re( w F'(w) ).
Then
A(r,theta) = (4/pi^2) arctan( 2 r cos(theta) / (1 - r^2) )
with 0 <= A(r,theta) < 2/pi, and A(r,theta) > 0 exactly
when |theta| < pi/2.
(d) F maps the right half-disk into the right half-plane,
and on each right semicircle the imaginary part of F is
strictly increasing in theta.
(e) F(ir) = i T(r) with T real-valued, smooth, strictly
increasing, T(0) = 0, T(1) = 1, and
T'(r) = (8/pi^2) * arctanh(r) / r, 0 < r < 1,
which tends to +infinity as r tends to 1.
(f) F is injective on D. Its image D = F(D) is a bounded
domain with the following property: the imaginary parts
of its points fill (-1,1), and for each t in (-1,1)
there is a finite V(t) > 0 with
{ v in R : v + it in D } = ( -V(t), V(t) ).
(g) Fix t with |t| < 1 and let r_0 = T^{-1}(|t|). For
r_0 < r < 1 there is a unique theta = theta(r,t) in
(-pi/2, pi/2) with Im F( r e^{i theta} ) = t, and the
curve r -> v(r,t) := Re F( r e^{i theta(r,t)} ) is
strictly increasing from 0 to V(t). Moreover
dv/dr = |w F'(w)|^2 / ( r A(r,theta) ) > 0,
w = r e^{i theta(r,t)}.
4.2 Proof of Lemma 4.1.
Absolute convergence on the closed disk is clear from
sum_l 1/(2l+1)^2 < infinity. Oddness and the conjugation
identity are immediate from the series, as is F(0) = 0, and
termwise differentiation at 0 gives F'(0) = 8/pi^2.
For (b), differentiate termwise in the open disk:
w F'(w) = (8/pi^2) sum_{l >= 0} (-1)^l w^{2l+1} / (2l+1).
The right side is the Taylor series of (8/pi^2) arctan w about 0,
so the identity holds where both sides are defined, that is on D.
For the stated branch: if w is in D then
Re( (1+iw)/(1-iw) ) = (1 - |w|^2) / |1 - iw|^2 > 0,
so the principal logarithm is defined and the formula for
arctan w is unambiguous, agreeing with the power series at w = 0.
The series shows wF'(w) vanishes only at w = 0, and F'(0) is
nonzero, so F' has no zeros at all in D.
For (c). Put w = r e^{i theta}. Multiply numerator and
denominator of (1+iw)/(1-iw) by the conjugate of the denominator:
(1+iw) / (1-iw) = ( 1 - r^2 + 2 i r cos(theta) ) / |1-iw|^2.
The denominator is real and positive, so the argument of the
quotient is arctan( 2 r cos(theta)/(1-r^2) ), and it lies in
(-pi/2, pi/2). Now arctan w = (1/2i) log(quotient), so
Re( arctan w ) = (1/2) arg( quotient )
= (1/2) arctan( 2 r cos(theta)/(1-r^2) ).
Multiplying by 8/pi^2 gives the displayed formula for A. The
bounds 0 < A < 2/pi follow from 0 < arctan(X) < pi/2 for X > 0.
For (d). Differentiate with respect to r along a ray:
d/dr Re F( r e^{i theta} ) = Re( (wF'(w))/r ) = A/r > 0.
So on each right ray starting at 0 the real part of F strictly
increases, and since Re F(0) = 0 it stays positive. Differentiate
with respect to theta along a semicircle:
d/dtheta Im F( r e^{i theta} ) = Im( i w F'(w) ) = Re( w F'(w) ) = A > 0.
So on each right semicircle the imaginary part strictly increases.
For (e). Oddness and F(conj w) = conj F(w) together with F(i r)
give F(i r) = i T(r) with T real, and reality of T on the
imaginary axis follows from F(-i r) = -F(i r) = -i T(r). The
derivative formula comes from (b) evaluated at w = i r: there
1 + i w = 1 - r and 1 - i w = 1 + r, so the quotient is real and
positive, whence arctan(i r) = (1/2i) log((1-r)/(1+r)) =
(i/2) log((1+r)/(1-r)) = i arctanh(r). Substituting into
wF'(w) = (8/pi^2) arctan w with w = i r and dividing by i gives
the claimed expression. Positivity of T' on (0,1) makes T
strictly increasing; T(0) = 0; and the classical identity
sum_l 1/(2l+1)^2 = pi^2/8 gives T(1) = 1. Finally
arctanh(r)/r tends to +infinity as r tends to 1, and
8/pi^2 > 0, so T' does too.
For (f). By (d), Re F > 0 on the right half-disk, Re F < 0 on
the left half-disk, and Re F = 0 on the imaginary axis. On the
imaginary axis F(i r) = i T(r) with T strictly increasing, so F
maps (-i, i) bijectively onto (-i, i). Fix t with |t| < 1 and
let r_0 = T^{-1}(|t|). On the closed right semicircle of radius
r > r_0 the imaginary part of F ranges from -T(r) to T(r),
strictly increasing in theta, so it takes the value t exactly
once; this is (g) and also proves injectivity on each positive
horizontal slice of F(D). Boundedness of F gives a finite
horizontal extent, hence V(t) < infinity. The identity
F(-conj w) = -conj F(w) shows that v + it lies in F(D) exactly
when -v + it does, so each horizontal slice is symmetric about
the imaginary axis and has the form (-V(t), V(t)) with V(t) > 0.
The same reflection handles negative t. Combining the three
cases Re F > 0, = 0, < 0 shows F is globally injective on D. By
(b) it has a nowhere-vanishing derivative, so its local inverses
patch into a holomorphic inverse g: D -> D.
For (g), the derivative formula. Write wF'(w) = A + i B on the
curve. Differentiating Im F( r e^{i theta(r,t)} ) = t in r gives
(B/r) + A * dtheta/dr = 0, so dtheta/dr = - B/(rA).
Then
dv/dr = d/dr Re F = A/r + (d Re F/d theta)( dtheta/dr )
= A/r + (-B) ( -B/(rA) )
= (A^2 + B^2)/(rA) = |w F'(w)|^2 / (r A) > 0.
Strictness uses A > 0. As r decreases to r_0 > 0 the angle tends
to +-pi/2, that is to the imaginary axis, where Re F = 0; if
r_0 = 0 the same conclusion follows from F(0) = 0. So v runs
strictly increasingly from 0 to a finite limit, which by the
definition of V(t) is V(t). QED
4.3 The slice primitive.
Let D = F(D) as in Lemma 4.1 and let g: D -> D be the inverse
of F. For an integer k >= 2 define, on the set of points
v + it in D,
J_k(v,t) = k^2 * integral_0^v
|g(h + it)|^{2k-2} |g'(h + it)|^2 dh.
The integral is along a horizontal segment ending at v + it.
Lemma 4.1(f) says the horizontal slice of D through height t is
the interval (-V(t), V(t)), and D is convex, being the image of
a convex set under a biholomorphism; so the whole segment from
it to v + it lies in D and the integral is well defined.
PROPOSITION 4.2 (the uniform slice estimate). Each J_k is
smooth on D, is odd in v, and satisfies
partial_v J_k(v,t) = k^2 |g(v+it)|^{2k-2} |g'(v+it)|^2 >= 0.
There is a sequence eps_k >= 0, depending only on F, with
eps_k -> 0, such that simultaneously for every v + it in D,
|J_k(v,t)| / k <= (pi/4) |g(v+it)|^{2k} + eps_k.
4.4 Why the constant is pi/4.
Before the proof, here is the idea in one paragraph, so that the
four steps below are not a fog. Change variables along the
horizontal slice, writing it in terms of the disk radius r.
Since |g'|^2 dv = (r/A) dr and A <= 2/pi, the integrand is
bounded by (pi/2) times r^{2k-1}, and integrating gives exactly
(pi/4) times the leading power |g|^{2k}. The bound A <= 2/pi is
strict, and the deficit 1/A - pi/2 is what we have to control.
It vanishes everywhere except near the two tips theta = +-pi/2,
where A can be small. Near a tip, A is small only if r is close
to 1, and there the weight r^{2k-1} is exponentially small in k.
The exponential decay beats the blow-up, uniformly, even when the
slice height and the endpoint both move with k. That is the
whole content of the estimate.
4.5 Proof of Proposition 4.2, Step 1: pass to the disk radius.
Fix v + it in D. The horizontal segment from it to v + it is a
compact subset of the open set D, and so is every nearby segment;
consequently
J_k(v,t) = k^2 * v * integral_0^1
|g(a v + it)|^{2k-2} |g'(a v + it)|^2 da,
and differentiating under the integral sign shows J_k is smooth
in (v,t). The integrand is smooth even where g vanishes, because
|g|^{2k-2} = (|g|^2)^{k-1} with k - 1 >= 1 and |g|^2 is smooth.
The formula for partial_v J_k is the fundamental theorem of
calculus.
Oddness. Differentiating g(-conj z) = -conj g(z) gives
g'(-conj z) = conj g'(z). Applying this at z = v + it, whose
conjugate with a sign is -v + it, gives
|g(-v+it)| = |g(v+it)|, |g'(-v+it)| = |g'(v+it)|.
So the density in the definition of J_k is an even function of
the real coordinate, and integrating from 0 to -v gives
J_k(-v,t) = -J_k(v,t). It therefore suffices to estimate v > 0,
which we assume from now on.
Fix such a point and put
R = |g(v+it)|, r_0 = T^{-1}(|t|).
Along the positive horizontal slice through v + it, the radius
increases strictly from r_0 to R: this is Lemma 4.1(g), read
backwards through g, since the radius on the disk side is
strictly increasing along the slice. Change variables from h to
r. By Lemma 4.1(g),
|g'|^2 dv = (r / A) dr, A = A(r, theta(r,t)),
so
J_k(v,t)/k = k * integral_{r_0}^{R} r^{2k-1} / A(r,theta(r,t)) dr.
At the lower endpoint this is an improper integral, and it is
defined as follows: perform the change of variables on a segment
starting at a positive real coordinate, then let that coordinate
decrease to 0. The original integrand is smooth and bounded on
the compact segment, so its integral is finite, and the
nonnegative transformed integrals increase to a limit, which
must be that same value. This covers t = 0, when r_0 = 0, as
well.
Now split 1/A into its best constant and the rest:
1/A = pi/2 + ( 1/A - pi/2 ),
with the second term nonnegative because A < 2/pi. The constant
part contributes
(pi/2) * k * integral_{r_0}^{R} r^{2k-1} dr
= (pi/4) ( R^{2k} - r_0^{2k} ) <= (pi/4) R^{2k}
= (pi/4) |g(v+it)|^{2k}.
So the leading term is accounted for, and it remains to bound
E_k(v,t) = k * integral_{r_0}^{R}
( 1/A(r,theta(r,t)) - pi/2 ) r^{2k-1} dr
by a quantity that tends to 0 as k -> infinity, uniformly over
all v + it in D. That is Steps 2 to 4. Notice that we have
deliberately extended nothing yet: E_k is an integral over the
actual slice, whose endpoints r_0 and R both depend on (v,t) and
hence may depend on k.
4.6 Step 2: the loss near the two tips.
The only way 1/A can exceed pi/2 is that A is small, and by
Lemma 4.1(c) that means the argument 2r cos(theta)/(1-r^2) is
small. Since r < 1 and cos(theta) > 0 on the right half, this
can only happen when theta is close to +-pi/2, that is, close to
one of the two tips of the right half-disk. The next two
estimates quantify this.
Throughout this step we work on the right half-disk and put
s = pi/2 - |theta|, 0 < s <= pi/2,
so that cos(theta) = sin(s). The number s is the angular
distance to the nearest tip.
Sub-lemma 4.3. Suppose 1/2 <= r < 1. Then
(2.1) 1/arctan(X) - 2/pi <= 2/X
for every X > 0, where X = 2 r sin(s)/(1-r^2);
(2.2) 1/A(r,theta) <= pi/2 + (3 pi^3 / 8) (1-r)/s.
Proof of (2.1). Two cases.
If 0 < X <= 1: from the Taylor series with alternating
remainder, arctan(X) >= X - X^3/3 >= (2/3)X for X <= 1. Hence
1/arctan(X) <= 3/(2X) <= 2/X, and 1/arctan(X) - 2/pi <= 2/X
since 2/pi > 0.
If X >= 1: use pi/2 - arctan(X) = arctan(1/X) <= 1/X, and
arctan(X) >= pi/4 for X >= 1. Then
1/arctan(X) - 2/pi
= ( pi/2 - arctan(X) ) / ( (pi/2) arctan(X) )
<= (1/X) / ( (pi/2)(pi/4) ) = 8/(pi^2 X) <= 2/X,
using 8/pi^2 <= 2. This proves (2.1).
Proof of (2.2). Since A = (4/pi^2) arctan(X),
1/A = (pi^2/4) ( 1/arctan(X) )
<= (pi^2/4) ( 2/pi + 2/X ) by (2.1)
= pi/2 + (pi^2/2)/X.
Next bound 1/X. Because sin(s) >= 2s/pi on [0, pi/2],
1/X = (1-r^2) / (2 r sin(s))
<= (1-r^2) pi / (4 r s)
= (pi/4) (1-r) (1+r)/(r s)
<= (3 pi/4) (1-r)/s,
where the last step uses (1+r)/r <= 3, valid for r >= 1/2.
Substituting gives (2.2). QED
Sub-lemma 4.4 (the vertical gap). For |theta| <= pi/2 define
Delta(r,theta) = T(r) - | Im F( r e^{i theta} ) |.
This is nonnegative, and it is the vertical distance from the
point F(r e^{i theta}) to the nearer of the two tips
+- i T(r) of the right semicircle. Moreover
(2.3) Delta(r,theta) <= (2/pi^2) s^2 / (1-r),
(2.4) s >= (pi/sqrt(2)) sqrt( Delta(r,theta) (1-r) ).
Proof. The reflection identity F(-conj w) = -conj F(w) gives
Im F( r e^{-i theta} ) = - Im F( r e^{i theta} ), so
Delta(r,-theta) = Delta(r,theta) and it suffices to treat
0 <= theta <= pi/2, where Im F >= 0 and |Im F| = Im F.
Since F(r) is real, Im F(r) = 0, and by Lemma 4.1(d)
the function theta -> Im F( r e^{i theta} ) is strictly
increasing with derivative A(r,theta). Hence
Im F( r e^{i theta} ) = integral_0^{theta} A(r,phi) dphi,
T(r) = Im F( r e^{i pi/2} ) = integral_0^{pi/2} A(r,phi) dphi,
and therefore
Delta = integral_{theta}^{pi/2} A(r,phi) dphi
= integral_0^{s} A(r, pi/2 - xi) dxi.
Now A(r, pi/2 - xi) = (4/pi^2) arctan( 2 r sin(xi)/(1-r^2) ).
Using arctan(y) <= y and sin(xi) <= xi,
Delta <= (4/pi^2) (2r/(1-r^2)) integral_0^s xi dxi
= (4r/(pi^2 (1-r^2))) s^2
= (4r/(pi^2 (1-r)(1+r))) s^2
<= (2/pi^2) s^2/(1-r),
because 4r/(1+r) <= 2 for r >= 0. That is (2.3). Inequality
(2.4) is (2.3) solved for s. QED
Combining. Substitute (2.4) into (2.2):
(1-r)/s <= (1-r) / ( (pi/sqrt2) sqrt(Delta (1-r)) )
= (sqrt2/pi) sqrt( (1-r)/Delta ).
Hence
1/A(r,theta) <= pi/2 + C_0 sqrt( (1-r)/Delta(r,theta) ),
C_0 = (3 pi^3/8)(sqrt2/pi) = 3 sqrt2 pi^2 / 8. (2.5)
Finally specialise to a point on the positive horizontal slice
through height t. By definition the angle theta(r,t) satisfies
Im F( r e^{i theta(r,t)} ) = t, so by the definition of Delta,
Delta(r, theta(r,t)) = T(r) - |t|.
Thus, for every point of the slice with r >= 1/2 and r > r_0,
(2.6) 1/A(r,theta(r,t)) <= pi/2 + C_0 sqrt( (1-r)/(T(r)-|t|) ).
The constant C_0 is absolute; only its existence matters. What
matters for the argument is the shape of the right side: the
singularity at the tip is a square-root singularity in (1-r)
divided by the vertical gap T(r) - |t|, and that is precisely
what the next step can integrate against the exponentially small
weight r^{2k-1}.
4.7 Step 3: separate a compact part from the tip region.
Fix a cutoff radius r_* with 1/2 < r_* < 1. It will be chosen
at the very end, and it is chosen before k varies. The whole
point of the cutoff is this: below r_* the integrand is bounded
but the weight r^{2k-1} is tiny; above r_* the weight is not tiny
but we have the square-root estimate (2.6) and a positive lower
bound for the vertical gap.
Put
r_a = max(r_0, r_*),
so that the slice splits into the part with r_0 <= r <= r_a and
the part with r_a <= r <= R. One of the two may be empty.
Recall from Step 1 that
J_k(v,t)/k = (pi/4)(R^{2k} - r_0^{2k}) + E_k(v,t),
E_k(v,t) = k integral_{r_0}^{R} (1/A - pi/2) r^{2k-1} dr >= 0.
Since E_k >= 0 and the first term is at most (pi/4)R^{2k}, it is
enough to bound E_k. Trivially
E_k <= k integral_{r_0}^{R} (1/A) r^{2k-1} dr,
and we split that last integral at r_a.
The inner piece. Here 1/A can be large (near a tip), so we do
not use the A-estimate at all; we go back to the h variable,
where the integrand is manifestly bounded. Along the slice the
radius r = |g| increases strictly, so the set of points with
|g| <= r_* is exactly an initial segment of the slice. That
segment lies in the compact set F( {|w| <= r_*} ), which is
contained in D, so its horizontal length is at most
diam(D) = sup { |z_1 - z_2| : z_1, z_2 in D } < infinity.
On that segment, |g|^{2k-2} <= r_*^{2k-2} and
|g'|^2 = 1/|F'(g)|^2 <= max_{|w| <= r_*} 1/|F'(w)|^2, a finite
number because F' has no zeros in the disk. Therefore the
contribution of the inner piece to J_k/k, and hence to E_k, is at
most
(3.1) C_* k r_*^{2k-2},
C_* = diam(D) * max_{|w| <= r_*} |F'(w)|^{-2} < infinity.
The outer piece. Here r >= r_a >= r_* > 1/2, so (2.6) applies:
1/A - pi/2 <= C_0 sqrt( (1-r)/(T(r) - |t|) ).
It remains to bound the vertical gap from below by a multiple of
r - r_a. Two facts. First, |t| = T(r_0) <= T(r_a) since T is
increasing and r_a >= r_0. Second, T' is increasing on (0,1),
because T'(r)/1 = (8/pi^2) arctanh(r)/r and the power series
arctanh(r)/r = 1 + r^2/3 + r^4/5 + r^6/7 + ...
has nonnegative coefficients. Hence, with
L_* = T'(r_*) = (8/pi^2) arctanh(r_*) / r_* > 0,
we get T'(r) >= L_* for all r >= r_*, and consequently
T(r) - |t| >= T(r) - T(r_a) >= L_*(r - r_a),
for all r >= r_a. (3.2)
Substituting (3.2) and (2.6) into the outer piece:
k integral_{r_a}^{R} (1/A - pi/2) r^{2k-1} dr
<= (C_0/sqrt(L_*)) * k integral_{r_a}^{R}
r^{2k-1} sqrt( (1-r)/(r - r_a) ) dr
<= (C_0/sqrt(L_*)) * I(k, r_a), (3.3)
where
I(k, r_a) = k integral_{r_a}^{1} r^{2k-1}
sqrt( (1-r)/(r - r_a) ) dr.
The extension of the upper limit from R to 1 is legitimate
because the integrand is nonnegative and R <= 1. This is the key
point of the whole section: it removes any dependence on how
close R is to the tip, in other words on how close the endpoint
v + it is to the boundary of D.
4.8 Step 4: the endpoint integral has an absolute bound.
Sub-lemma 4.5. For every integer k >= 1 and every
r_a with 0 <= r_a < 1,
I(k, r_a) <= 1 + pi/e.
Proof. Put H = k(1 - r_a) and substitute y = k(1 - r).
Then k dr = -dy, the limit r = r_a corresponds to y = H, the
limit r = 1 corresponds to y = 0, and
(1-r)/(r - r_a) = (y/k) / ((H - y)/k) = y/(H - y).
So
I(k, r_a) = integral_0^H (1 - y/k)^{2k-1}
sqrt( y/(H - y) ) dy.
Use log(1 - u) <= -u for 0 <= u < 1: it gives
(2k-1) log(1 - y/k) <= -(2k-1) y/k <= -y, so
(1 - y/k)^{2k-1} <= e^{-y}. Hence
I(k, r_a) <= integral_0^H e^{-y} sqrt( y/(H - y) ) dy.
Split the range into [0, H/2] and [H/2, H].
On [0, H/2] we have y <= H - y, so the square root is at most
1, and the integral there is at most
integral_0^{H/2} e^{-y} dy <= 1.
On [H/2, H] we have e^{-y} <= e^{-H/2}, and the substitution
y = H sin^2(phi) gives
integral_0^H sqrt( y/(H-y) ) dy
= integral_0^{pi/2} tan(phi) * 2H sin(phi) cos(phi) dphi
= 2H integral_0^{pi/2} sin^2(phi) dphi
= pi H / 2.
So the second part is at most (pi H/2) e^{-H/2}. Therefore
I(k, r_a) <= 1 + (pi H/2) e^{-H/2} <= 1 + pi/e,
because the maximum of H e^{-H/2} over H > 0 is 2/e. QED
The bound is uniform in H. That is exactly what is needed, since
H = k(1 - r_a) can tend to 0 (when the slice sits close to the
imaginary axis and the cutoff dominates) or to infinity (when
the slice approaches a tip), and both regimes occur as k grows
and the slice moves.
4.9 Assembling the proof of Proposition 4.2.
Combine (3.1) and (3.3). For every point v + it in D with v > 0
and every k >= 2,
J_k(v,t)/k <= (pi/4)|g(v+it)|^{2k}
+ C_* k r_*^{2k-2}
+ (C_0/sqrt(L_*)) (1 + pi/e). (4.1)
When the inner or outer piece is empty the corresponding term is
absent, so (4.1) remains valid in that case; and at v = 0 the
left side is 0. For v < 0 use oddness of J_k and evenness of
|g| (Step 1) to obtain the same bound for |J_k|/k.
Now define
eps_k = max { 0, sup_{v + it in D}
( |J_k(v,t)|/k - (pi/4)|g(v+it)|^{2k} ) }.
The supremum is finite for each k by (4.1), since the right side
is a constant independent of (v,t).
It remains to see that eps_k -> 0. Let epsilon > 0 be given.
First choose r_* in (1/2, 1). The quantity L_* = T'(r_*) depends
on r_* only, and by Lemma 4.1(e),
L_* = (8/pi^2) arctanh(r_*)/r_* -> +infinity
as r_* -> 1.
So choose r_* close enough to 1 that
(C_0/sqrt(L_*)) (1 + pi/e) < epsilon/2.
Now freeze that r_*. The constant C_* is fixed, and
r_*^{2k-2} decays geometrically, so there is N such that
C_* k r_*^{2k-2} < epsilon/2 for every k >= N.
For such k, (4.1) gives eps_k < epsilon. Since epsilon was
arbitrary, eps_k -> 0, and Proposition 4.2 is proved.
Observe where the uniformity came from. The estimate (4.1) has
no (v,t)-dependent constant anywhere. The cutoff r_* was chosen
before k, the interval of integration was enlarged from R to 1,
and Sub-lemma 4.5 is uniform in the lower endpoint. All three
points are needed, and each of them corresponds to one way in
which the slice point can approach the boundary of D as k grows.
5. FINITE STRIP BODIES
----------------------
Now comes the assembly (C). We take a symmetric convex body
described by finitely many strips, build the holomorphic package
of Section 3 out of the planar map of Section 4, and check that
the resulting symplectic map lands where we want.
5.1 The statement.
PROPOSITION 5.1. Let n >= 2, let b_1,...,b_m be vectors in
R^n that span R^n, and set
K = { x in R^n : | b_j . x | <= 1 for j = 1,...,m }.
Then for every c with 0 < c < 4 there is a smooth symplectic
embedding
B^{2n}(c) -> int(K) x int(K^o).
The body K is a symmetric convex polytope with 0 in its
interior, and every symmetric convex polytope arises this way.
The general convex body follows in Section 6 by approximation.
5.2 The holomorphic package.
Extend each functional x -> b_j . x complex-linearly:
ell_j(z) = b_j . z, z in C^n.
These are complex-linear functions of z; if z = u + ix then
ell_j(z) = b_j . u + i b_j . x. Let F: D -> D and its inverse
g: D -> D be the maps of Lemma 4.1, and put
Omega = { z in C^n : ell_j(z) lies in D for every j }.
This is open and contains 0, since every ell_j(0) = 0 and 0 is in
D. Because the b_j span R^n, the linear map
z -> (ell_1(z),...,ell_m(z)) is injective, so there is a constant
C_b with
|z| <= C_b * max_j |ell_j(z)|.
Since D is bounded, Omega is bounded; and the same inequality
shows that K is compact. Also a small Euclidean ball around 0
lies inside every strip, so 0 is an interior point of K.
Fix an integer k >= 2 and set
f_j(z) = g(ell_j(z))^k,
f = (f_1,...,f_m),
tau = |f|^2 = sum_j |g(ell_j(z))|^{2k},
V = { z in Omega : tau(z) < 1 },
omega = dd^c( |z|^2 + tau ).
LEMMA 5.2. The package (Omega, f, k) satisfies hypotheses (H1)
and (H2) of Theorem 3.2.
Proof. (H1), first part. Since g(w) = 0 only at w = 0, the
common zero set of f on Omega is
{ z : ell_j(z) = 0 for all j } = {0},
by injectivity of z -> (ell_1(z),...,ell_m(z)).
(H1), second part. Near 0 we have g(w) = w h(w) with h
holomorphic and h(0) = 1/F'(0) nonzero. Hence
f_j(z) = ell_j(z)^k h(ell_j(z))^k, and since
|ell_j(z)| <= C |z| on a neighbourhood of 0,
|f(z)| <= C' |z|^k.
(H2). Fix s with 0 < s < 1 and consider the set
S = { z in Omega : tau(z) <= s }. If z is in S then for every
j,
|g(ell_j(z))|^k <= |f(z)| <= s^{1/2},
so ell_j(z) lies in F( { w : |w| <= s^{1/(2k)} } ), a compact
subset of D. By the linear bound, S is bounded. If z_n is a
sequence in S converging to some z, then each ell_j(z_n)
converges to ell_j(z), which therefore still lies in that
compact subset of D; so z is in Omega. And tau(z) <= s by
continuity. Hence S is closed and bounded in Omega, that is,
compact in Omega. QED
Consequently Theorem 3.2 applies, and for every k >= 2 it gives
a smooth symplectic embedding
(5.1) E: B^{2n}(c') -> (V, omega)
for every c' < pi k.
Here c' is a capacity in the same normalization as always,
B^{2n}(c') = { pi(|q|^2+|p|^2) < c' }; the number pi k in the
theorem is exactly what appears.
5.3 The phase-space map.
Split z = u + ix with u and x in R^n. Write
v_j = b_j . u = Re ell_j(z),
t_j = b_j . x = Im ell_j(z),
and recall the slice primitives J_k from Proposition 4.2, which
are evaluated at (v_j, t_j). Define
Phi_k: V -> R^n_q x R^n_p,
q = x,
p = -u - sum_{j=1}^m J_k(v_j, t_j) b_j. (5.2)
This is smooth because each J_k is smooth on D and each ell_j
maps V into D.
LEMMA 5.3. Phi_k^* omega_0 = omega.
Proof. We compute term by term. By definition
omega_0 = sum_{l=1}^n dq_l ^ dp_l,
and dq_l = dx_l while
dp_l = -du_l - sum_{j=1}^m b_{j,l}
( d_v J_k(v_j,t_j) dv_j
+ d_t J_k(v_j,t_j) dt_j ).
Also dv_j = sum_l b_{j,l} du_l and dt_j = sum_l b_{j,l} dx_l,
by definition of v_j and t_j. Hence, summing over l,
Phi_k^* omega_0
= sum_l du_l ^ dx_l
- sum_{j,l} b_{j,l} d_v J_k dx_l ^ dv_j
- sum_{j,l} b_{j,l} d_t J_k dx_l ^ dt_j.
The third sum vanishes: for each fixed j it equals
-d_t J_k dt_j ^ dt_j = 0. In the second sum,
sum_l b_{j,l} dx_l = dt_j, so it equals
- sum_j d_v J_k dt_j ^ dv_j = + sum_j d_v J_k dv_j ^ dt_j.
Therefore
Phi_k^* omega_0 = sum_l du_l ^ dx_l
+ sum_j (d_v J_k)(v_j,t_j) dv_j ^ dt_j.
(5.3)
Now compare with omega = dd^c(|z|^2 + tau). By Section 3.1,
dd^c|z|^2 = sum_l du_l ^ dx_l.
It remains to identify dd^c tau. Each summand of tau is
|g(ell_j(z))|^{2k}, which depends on z only through the pair
(v_j, t_j), and that pair is related to z holomorphically.
So it suffices to do the computation in the plane. There,
for a holomorphic G and the function |G|^{2k} on the plane,
dd^c( |G(v+it)|^{2k} )
= k^2 |G(v+it)|^{2k-2} |G'(v+it)|^2 dv ^ dt.
This is the calculation at the end of Section 3.1 with
|zeta|^2-coefficient: writing rho = |G|^2, one has
Delta(rho^k) = 4k^2 rho^{k-1}|G'|^2, and dd^c of anything is
one quarter of the Laplacian on the plane, so the coefficient
is k^2|G|^{2k-2}|G'|^2.
Apply this with G = g, whose derivative is g' and whose
derivative along the slice is what appears in J_k. By
Proposition 4.2,
d_v J_k(v,t) = k^2 |g(v+it)|^{2k-2} |g'(v+it)|^2.
Comparing with the plane formula gives
dd^c( |g(ell_j(z))|^{2k} )
= (d_v J_k)(v_j,t_j) dv_j ^ dt_j.
Summing over j and adding the |z|^2 part turns (5.3) into
Phi_k^*omega_0 = dd^c(|z|^2 + tau) = omega. QED
5.4 Phi_k is injective.
LEMMA 5.4. Phi_k: V -> R^{2n} is injective.
Proof. Suppose Phi_k(z_1) = Phi_k(z_2). The first n
coordinates of Phi_k are q = x, so x_1 = x_2; call that common
value x. Consequently every t_j = b_j . x agrees, and only
the real parts u_1, u_2 remain to be compared.
Put
Du = u_2 - u_1, v_{ji} = b_j . u_i,
p_i = the momentum of Phi_k(z_i), i = 1,2.
By (5.2),
p_2 - p_1 = -Du
- sum_j ( J_k(v_{j2},t_j) - J_k(v_{j1},t_j) ) b_j.
Take the dot product with Du. Since b_j . Du = v_{j2} - v_{j1},
(p_2 - p_1) . Du
= - |Du|^2
- sum_j ( J_k(v_{j2},t_j) - J_k(v_{j1},t_j) )
( v_{j2} - v_{j1} ).
Now fix j. Both v_{j1} and v_{j2} lie in the horizontal slice
of D at height t_j, which by Lemma 4.1(f) is the interval
(-V(t_j), V(t_j)). So the whole segment between them lies in
D at that height, and by Proposition 4.2 the function
v -> J_k(v,t_j) is nondecreasing there. Hence each product
( J_k(v_{j2},t_j) - J_k(v_{j1},t_j) ) ( v_{j2} - v_{j1} )
is nonnegative. Therefore
(p_2 - p_1) . Du <= -|Du|^2 <= 0.
If p_1 = p_2 then the left side is 0, so |Du|^2 <= 0, that is
u_1 = u_2, and then z_1 = z_2. QED
Injectivity together with Lemma 5.3 gives more than injectivity.
Since Phi_k^*omega_0 = omega is nondegenerate, the differential
DPhi_k has trivial kernel at every point, so Phi_k is a local
diffeomorphism; an injective local diffeomorphism is a
diffeomorphism onto its image, with a smooth inverse. Hence
Phi_k: (V, omega) -> (image, omega_0) is a symplectic
embedding. (5.4)
5.5 The image lies in int(K) x S int(K^o), for S close to pi k/4.
The position coordinate. For z in V we have ell_j(z) in D, so
t_j = Im ell_j(z) lies in (-1,1). Thus |b_j . x| < 1 for every
j, which is exactly q = x in int(K).
The momentum coordinate. Introduce the support function
h_K(p) = max_{y in K} <p, y>, p in R^n.
It is subadditive, positively homogeneous, and determines
interiors by int(K^o) = { p : h_K(p) < 1 }. The last fact is the
definition of the polar together with the fact that a strict
inequality holds at an interior point. Since |b_j . y| <= 1 for
every y in K, we have
h_K(+- b_j) <= 1, j = 1,...,m.
Also, because Omega is bounded, so is its projection to the
u-coordinate, and hence
C_K := max { h_K(-u) : z = u + ix in Omega } < infinity.
This constant does not depend on k. Now estimate, using
subadditivity and positive homogeneity of h_K together with
h_K(+-b_j) <= 1:
h_K(p) / k
<= h_K(-u)/k + sum_j |J_k(v_j,t_j)| h_K(b_j) / k
<= C_K/k + sum_j |J_k(v_j,t_j)|/k. (5.5)
Apply Proposition 4.2 to each summand:
sum_j |J_k(v_j,t_j)|/k
<= (pi/4) sum_j |g(ell_j(z))|^{2k} + m*eps_k
= (pi/4) tau(z) + m*eps_k.
Substituting into (5.5) and using tau(z) < 1 on V,
h_K(p)/k < C_K/k + pi/4 + m*eps_k. (5.6)
Now let eta > 0. Since eps_k -> 0 and C_K, m are fixed before k
varies, there is k_0 such that for all k >= k_0
C_K/k + m*eps_k < eta*pi/4.
For such k and every z in V, (5.6) gives h_K(p) < S * 1 with
S = (1 + eta) pi k / 4,
that is, p lies in S int(K^o) = { p : h_K(p) < S }. Together
with the position bound,
(5.7) Phi_k(V) is contained in int(K) x S int(K^o).
Note carefully the order of the choices, because it is easy to
get it backwards. The number eta is chosen first. Then k is
chosen large. The cutoff radius r_* inside the proof of
Proposition 4.2 is chosen LATER than eta but EARLIER than k, and
it is forced to be close to 1; that is why eps_k can be made
small without the cutoff depending on k.
5.6 Rescale and finish.
Fix c with 0 < c < 4. Choose eta > 0 so small that
(1 + eta) c < 4,
and then, by (5.7), fix an integer k such that the inclusion
holds. With S = (1+eta)pi k/4 we have
S c = (1 + eta) pi k c / 4 < pi k,
so by (5.1) there is a smooth symplectic embedding
E: B^{2n}(S c) -> (V, omega).
Introduce the two rescaling maps
D_{sqrt S}: B^{2n}(c) -> B^{2n}(S c), D_{sqrt S}(z) = sqrt(S) z,
L_S: R^{2n} -> R^{2n}, L_S(q,p) = (q, p/S).
Then:
(i) D_{sqrt S} maps B^{2n}(c) onto B^{2n}(S c), because
pi|sqrt(S) z|^2 < S c exactly when pi|z|^2 < c.
(ii) D_{sqrt S}^* omega_0 = S * omega_0, since scaling all
2n coordinates by sqrt(S) multiplies every area by S.
(iii) L_S^* omega_0 = (1/S) * omega_0, since only the p
coordinates are scaled, by the factor 1/S.
(iv) L_S maps int(K) x S int(K^o) into int(K) x int(K^o).
Now define
Psi = L_S o Phi_k o E o D_{sqrt S}:
B^{2n}(c) -> int(K) x int(K^o).
It is a composition of embeddings, hence an embedding, and
Psi^* omega_0
= D_{sqrt S}^* ( E^* ( Phi_k^* ( L_S^* omega_0 ) ) )
= S * (1/S) * omega_0
= omega_0,
using (ii), (iii) and Lemma 5.3. Therefore Psi is a smooth
symplectic embedding of B^{2n}(c) into int(K) x int(K^o), as
required. Since c < 4 was arbitrary, Proposition 5.1 is proved.
The factor 4 appears here as the product of the leading constant
pi/4 from the slice estimate with the rescaling S ~ pi k/4.
6. ARBITRARY BODIES, UPPER BOUND, CONSEQUENCES
----------------------------------------------
Three things remain. First, pass from finite strip bodies to
arbitrary symmetric convex bodies; this is an approximation, and
it costs an arbitrarily small enlargement of the position factor.
Second, prove the matching upper bound c_G(U_K) <= 4. Third,
extract the volume-product inequality and its consequences.
6.1 Approximating a convex body by finitely many strips.
LEMMA 6.1. Let K be an origin-symmetric convex body in R^n
and let alpha > 0. Then there are finitely many vectors
b_1,...,b_m in R^n spanning R^n such that, with
K' = { x in R^n : | b_j . x | <= 1, j = 1,...,m },
one has K subset K' subset (1+alpha) K and (K')^o subset K^o.
Proof. Preliminary: 0 is an interior point of K. Since K
has nonempty interior it contains some open ball B(y,r); by
symmetry it contains B(-y,r); and by convexity, for |z| < r,
z = (1/2)(y+z) + (1/2)(-y+z), y+z in B(y,r), -y+z in B(-y,r),
so z lies in K. Hence B(0,r) subset K. Since K is also
bounded, there exist r, R > 0 with
B(0,r) subset K subset B(0,R).
Then K^o satisfies B(0,1/R) subset K^o subset B(0,1/r): the
left inclusion because |<p,y>| <= R|p| <= 1 for |p| <= 1/R,
the right one because <p, y> <= r|p| <= 1 for every y in K
forces |p| <= 1/r. So K^o is a compact symmetric convex body
with 0 in its interior, and in particular
rho := min { h_{K^o}(v) : |v| = 1 } >= 1/R > 0.
Bipolar identity. We need K^{oo} = K. The inclusion
K subset K^{oo} is immediate: if x in K and p in K^o then
<p,x> <= 1. For the reverse, let x be outside K. Since K is
closed and convex, a separating hyperplane gives a nonzero v
and a number c with
<v,x> > c >= <v,y> for every y in K,
that is <v,x> > h_K(v). Here h_K(v) > 0, because 0 is an
interior point of K. Hence p = v/h_K(v) lies in K^o and
<p,x> = <v,x>/h_K(v) > 1, so x is not in K^{oo}. This proves
K^{oo} = K.
The net. Put lambda = 1 + alpha and choose d with
0 < d <= alpha * rho / (1 + alpha).
Since K^o is compact, it contains a finite d-net: points
b_1,...,b_m in K^o such that every point of K^o is within
distance d of some b_j. Let P = conv{ +-b_1, ..., +-b_m }.
Symmetry of K^o gives P subset K^o.
The support function estimate. Fix a unit vector v and let
p* be a maximizer of <v, .> over K^o, so <v,p*> =
h_{K^o}(v). Choose a net point b_j with |b_j - p*| <= d.
Then
h_P(v) >= <v, b_j> = <v,p*> - <v, p* - b_j>
>= h_{K^o}(v) - |p* - b_j|
>= h_{K^o}(v) - d
>= h_{K^o}(v) - alpha*h_{K^o}(v)/(1+alpha)
= lambda^{-1} h_{K^o}(v),
using h_{K^o}(v) >= rho and the choice of d in the third
line.
Now use the standard fact that for closed convex sets with 0
in their interiors, A subset B if and only if h_A <= h_B.
The estimate above holds for every unit v and both sides are
positively homogeneous, so it holds for all v; hence
lambda^{-1} K^o subset P subset K^o.
Polars reverse inclusions, so, using (P)^o = {x : <p,x> <= 1
for all p in P} and the bipolar identity applied to P and to
K^o,
K = (K^o)^{oo} subset P^o subset (lambda^{-1} K^o)^o
= lambda K,
and
(P^o)^o = P subset K^o.
Finally, P^o is exactly the finite strip body of the
statement: p in P^o means <+-b_j, x> <= 1 for every j, that
is |b_j . x| <= 1. Since P contains lambda^{-1}K^o, which
has interior, P has interior, which forces b_1,...,b_m to
span R^n. Setting K' = P^o finishes the proof. QED
6.2 The upper bound.
We prove c_G(U_K) <= 4 by exhibiting a linear symplectic change
of coordinates that places U_K inside a product of a strip with
a plane, and then applying Gromov's nonsqueezing theorem.
Nonsqueezing, stated exactly as used. Gromov's theorem says
that if there is a symplectic embedding
B^{2n}(c) -> B^2(C) x R^{2n-2},
where the target carries the standard form sum dq_j ^ dp_j with
the first factor being the (q_1,p_1)-plane, then c <= C. We
quote only this. Two remarks: it is the only external input in
the whole paper, and it is the standard formulation with the
same capacity normalization as everywhere else (a ball of radius
R has capacity pi R^2, so B^2(C) has radius sqrt(C/pi)).
Step 1: a dual pair of coordinates. Maximize any nonzero
linear functional over K. Let q_0 in K be a maximizer, put
H = h_K(v) = <v,q_0> > 0 for that functional, and set
p_0 = v / H.
Then p_0 lies in K^o, because <p_0, y> = <v,y>/h_K(v) <= 1 for
every y in K; and
<q_0, p_0> = 1.
Step 2: a symplectic linear change. Since <q_0,p_0> = 1, the
hyperplane ker p_0 does not contain q_0, so R^n is the direct sum
of ker p_0 and the span of q_0. Choose a basis e_2,...,e_n of
ker p_0 and let A be the matrix with columns q_0, e_2, ..., e_n.
Define the linear map
(q,p) -> (Q, P) = ( A^{-1} q , A^T p ).
This is symplectic: for any (q,p), summing over the index i,
sum_i dQ_i ^ dP_i
= sum_i (A^{-1})_{i k} (A^T)_{i l} dq_k ^ dp_l
= sum_{k,l} ((A^{-1})^T A^T)_{k l} dq_k ^ dp_l
= sum_k dq_k ^ dp_k,
because (A^{-1})^T A^T = (A A^{-1})^T is the identity. Its first
coordinates are the pairing of q with p_0 and of p with q_0:
Q_1 = (A^{-1} q)_1 = <p_0, q>, P_1 = (A^T p)_1 = <q_0, p>.
Indeed, <p_0, A e_1> = <p_0,q_0> = 1 and <p_0, A e_i> = 0 for
i >= 2; so the coefficient of q_0 in the expansion of q is
<p_0,q>, which is the first entry of A^{-1}q.
Step 3: the strips. On U_K:
- q lies in int(K), so <p_0,q> <= h_K(p_0) = 1 with equality
only on the boundary of K; hence Q_1 = <p_0,q> < 1. By
symmetry -p_0 also lies in K^o, giving Q_1 > -1.
- p lies in int(K^o), so <q_0,p> <= h_{K^o}(q_0) = 1 with
equality only on the boundary; hence P_1 < 1, and P_1 > -1
by symmetry.
Therefore the symplectic change of coordinates maps U_K into
(-1,1) x (-1,1) x R^{2n-2},
with the square lying in the (Q_1,P_1)-plane.
Step 4: the square is an area-4 disk. The open square has area
4, and any two open disks or squares of equal area in the plane
are symplectomorphic. Here is an explicit map, for completeness.
Put R = 2/sqrt(pi), so that the disk of radius R has area 4, and
let
h(X) = sqrt(R^2 - X^2), -R < X < R,
H(X) = -1 + integral_{-R}^{X} h(s) ds.
Since integral_{-R}^{R} h = (area of the disk of radius R)/2 = 2,
we get H(R) = 1, and H(-R) = -1; and H' = h > 0. So H is a
smooth diffeomorphism of (-R,R) onto (-1,1). Define
(u,v) |-> ( X = H^{-1}(u), Y = v * h(H^{-1}(u)) ).
The image is the open disk X^2 + Y^2 < R^2, and the Jacobian is
dX/du * dY/dv = (1/h(X)) * h(X) = 1,
so the map is area preserving. Applying it to the (Q_1,P_1)
pair gives a symplectic embedding
U_K -> B^2(4) x R^{2n-2}.
Step 5: nonsqueezing. If B^{2n}(c) embedded symplectically into
U_K, then composing with the above would give a symplectic
embedding of B^{2n}(c) into B^2(4) x R^{2n-2}, and Gromov's
theorem would force c <= 4. Hence
(6.1) c_G(U_K) <= 4.
6.3 Proof of Theorem 1.1.
Fix n >= 2 and an origin-symmetric convex body K in R^n, and
fix c with 0 < c < 4.
Choose alpha > 0 with
lambda * c < 4, lambda = 1 + alpha.
Such alpha exists because c < 4. Apply Lemma 6.1 to obtain a
finite strip body K' with
K subset K' subset lambda K, (K')^o subset K^o.
Passing to interiors and using that the interior of lambda K is
lambda int(K),
int(K) subset int(K') subset lambda int(K),
int((K')^o) subset int(K^o). (6.2)
Proposition 5.1, applied to the finite strip body K', gives a
smooth symplectic embedding
e: B^{2n}(lambda c) -> int(K') x int((K')^o).
By (6.2) the target sits inside lambda int(K) x int(K^o), so e
may be regarded as taking values there. Introduce
Q_lambda(q,p) = ( q/lambda, p ),
D_{sqrt lambda}(z) = sqrt(lambda) z,
and consider
Phi = Q_lambda o e o D_{sqrt lambda}:
B^{2n}(c) -> U_K. (6.3)
This is well defined: D_{sqrt lambda} maps B^{2n}(c) onto
B^{2n}(lambda c), the image of e lies in lambda int(K) x int(K^o),
and Q_lambda maps that into int(K) x int(K^o). It is a smooth
embedding, being a composition of embeddings. Finally
D_{sqrt lambda}^* omega_0 = lambda * omega_0,
Q_lambda^* omega_0 = lambda^{-1} * omega_0,
e^* omega_0 = omega_0,
so
Phi^* omega_0 = lambda * lambda^{-1} * omega_0 = omega_0.
Therefore c_G(U_K) >= 4. Combined with (6.1), c_G(U_K) = 4, and
every c < 4 is attained. This proves Theorem 1.1.
One bookkeeping point about the order of choices. For each
prescribed c the construction is finite: choose alpha, then K'
and its m defining vectors, then (inside Proposition 5.1) choose
eta with (1+eta)lambda c < 4, then choose k large enough that
C_{K'}/k + m*eps_k < eta*pi/4. Each choice depends on data
fixed before it, and no limit of embeddings is taken.
6.4 The symmetric Mahler inequality.
COROLLARY 6.2. For every integer n >= 1 and every
origin-symmetric convex body K in R^n,
vol_n(K) vol_n(K^o) >= 4^n / n!.
Proof. The case n = 1. Here K = [-a,a] with a > 0, and
K^o = [-1/a, 1/a]. The product of the lengths is
(2a)(2/a) = 4 = 4^1/1!.
The case n >= 2. First, interiors do not change the volume.
Let C be a convex body containing B(0,r). For 0 < t < 1 and
every z in C,
tz + (1-t)B(0,r) subset tC + (1-t)B(0,r) subset C
by convexity, so tC is contained in int(C). Hence
t^n vol_n(C) = vol_n(tC) <= vol_n(int C) <= vol_n(C),
and letting t increase to 1 gives vol_n(int C) = vol_n(C).
Apply this to K and to K^o, both of which contain a
neighbourhood of 0.
By Fubini,
vol_{2n}(U_K) = vol_n(K) * vol_n(K^o). (6.4)
Now let c < 4. Theorem 1.1 gives a smooth symplectic
embedding Phi: B^{2n}(c) -> U_K. Since Phi^*omega_0 =
omega_0, taking n-th exterior powers gives
Phi^*( omega_0^n ) = omega_0^n,
and omega_0^n is n! times the volume form, so
|det DPhi| = 1 and Phi preserves volume. Therefore, using
vol_{2n}(B^{2n}(c)) = c^n/n!,
c^n/n! = vol_{2n}(B^{2n}(c))
= vol_{2n}(Phi(B^{2n}(c)))
<= vol_{2n}(U_K)
= vol_n(K) vol_n(K^o).
Letting c increase to 4 gives 4^n/n! <= vol_n(K) vol_n(K^o).
QED
6.5 Sharpness of the constant.
Take K = [-1,1]^n, the cube. Its volume is 2^n. Its polar is
K^o = { p : sum_j |p_j| <= 1 }.
Intersecting K^o with the closed positive orthant gives the
simplex { p_j >= 0, sum p_j <= 1 }, of volume 1/n!, and the 2^n
orthants give 2^n such pieces meeting only along faces.
Therefore
vol_n(K^o) = 2^n / n!, vol_n(K) vol_n(K^o) = 4^n / n!.
So the constant in Corollary 6.2 cannot be improved, and by
Theorem 1.1 the ball of every capacity below 4 sits inside
U_K = (-1,1)^n x (int of the crosspolytope).
6.6 An add-on, not used elsewhere.
Theorem 1.1 also transports symplectic packings: if
sum_i R_i^n < 4^n and R_i + R_j < 4 for all i < j, then some
c < 4 dominates all these inequalities strictly, the ball-packing
theorem of the companion paper packs the closed balls of
capacities R_i into B^{2n}(c), and Theorem 1.1 carries the
packing into U_K. That packing theorem is an external result,
not used anywhere else in this document.