How much more can a two-sided midpoint perturbation grow than a one-sided
one? For an infinite-dimensional real subspace of L^1 whose closed unit
ball is compact in measure and which has the Daugavet property, the answer
is exact: the averaged midpoint modulus is max{t/2, t-1} and the one-sided
modulus is max{0, t-2}, at every unit center and every positive radius.
The same weak-neighborhood geometry forbids every equivalent asymptotically
uniformly convex norm.
Write B_X and S_X for the closed unit ball and unit sphere of an
infinite-dimensional real Banach space X, and cof(X) for its closed
linear subspaces of finite codimension. For a norm N on X, a vector
x with N(x)=1, and t>0, put
H_N(x,t) = sup_{F in cof(X)} inf_{y in F, N(y)>=1}
( (N(x+ty)+N(x-ty))/2 - 1 ),
D_N(x,t) = sup_{F in cof(X)} inf_{y in F, N(y)=1}
( N(x+ty)-1 ).
The averaged midpoint modulus hat-delta_N(t) and the one-sided modulus
bar-delta_N(t) are the infima of these quantities over x in S_N. The
norm N is asymptotically uniformly convex (AUC) if
bar-delta_N(t)>0 for every t>0.
The averaged increment H_N and the one-sided increment D_N have
different convexity characters. Uniform convexity of Clarkson asks that
both ||x+ty|| and ||x-ty|| exceed one, on average, uniformly over
|x|=|y|=1. Milman obtained a linear-topology weakening, AUC, around
three decades later. The averaged version appeared only recently:
Dilworth, Kutzarova, Randrianarivony, Revalski and Zhivkov introduced
asymptotic midpoint uniform convexity (AMUC) and asked whether AMUC and
AUC are equivalent up to renorming.
J. A. Clarkson, Uniformly convex spaces, Trans. Amer. Math. Soc. 40
(1936), 396-414 (no open pdf).
V. D. Milman, Geometric theory of Banach spaces II, Russian Math.
Surveys 26:6 (1971), 79-163 (no open pdf).
./dilworth-kutzarova-randrianarivony-revalski-zhivkov-2016.pdf
J. Math. Anal. Appl. 436 (2016), 810-821, Definition 2.2:
"... the norm in a Banach space X is asymptotically midpoint
uniformly convex (in short AMUC) if alpha(lens(x, delta)) -> 0 as
delta -> 0 uniformly in x."
For subspaces of L^1 with compact unit ball in measure, Florent Baudier
answered the question negatively.
./baudier-2026.pdf
arXiv:2609.21283v1, 18 September 2026.
"Theorem 1. Let Y be a closed subspace of L^1 whose unit ball is
relatively compact with respect to the topology of convergence in
measure. Then Y is asymptotically midpoint uniformly convex."
"Corollary 1. The space X_KW is asymptotically midpoint uniformly
convex, fails the Point of Continuity Property, and thus does not
admit an equivalent norm that is asymptotically uniformly convex."
Baudier proved only a linear lower bound for the maximum-based modulus
bar-delta_{X_KW}(t)>=ct. The aim is to compute both moduli
exactly, for every unit center and every positive radius. Baudier's
argument passes through the L^1 splitting identity of Brezis and Lieb
along almost-everywhere convergent subsequences; the lower bound in
the clipping lemma instead uses a finite net of pointwise truncations
with sign functionals.
./brezis-lieb-1983.pdf
Proc. Amer. Math. Soc. 88 (1983), 486-490, Theorem 1 (L^1 splitting
identity).
The averaged and maximum-based moduli differ by at most a factor of two.
For each fixed y in S_X, both r -> N(x+ry) and r -> N(x-ry) are
even convex functions of r. Therefore the average and the maximum
are also even convex. Write m_N(t) for the modulus obtained by
replacing the average in H_N by the maximum and then taking the infimum
over unit centers. Positivity of hat-delta_N is AMUC, positivity of
bar-delta_N is AUC. The elementary inequality (a+b)/2 <= max{a,b}
gives hat-delta_N(t)<=m_N(t). For the reverse comparison, given a
finite-codimensional subspace F and a unit vector x, intersect F
with the kernel of a norm-one functional phi with phi(x)=1. Each
y in the intersection with N(y)=1 satisfies N(x+ty)>=1 and
N(x-ty)>=1, hence the average of the two excesses over one is at least
half of the maximum. Taking infima and suprema,
(1/2) m_N(t) <= hat-delta_N(t) <= m_N(t).
Recall that X has the Daugavet property if ||Id+T||=1+||T|| for every
rank-one bounded operator T on X. The identity was found by Daugavet
for compact T on C[0,1]; the general study of the property began two
decades later.
Daugavet, On a property of completely continuous operators in
the space C, Uspekhi Mat. Nauk 18:5 (1963), 157-158 (no open pdf;
mathnet paper 6415).
The property admits an internal characterization, due to Kadets, Shvidkoy,
Sirotkin and Werner and used by Kadets and Werner in the required form.
A nonzero real Banach space X has the Daugavet property if and
only if, for every w in S_X, every rho>0, and every nonempty slice
H = { y in B_X : l(y)>alpha } with l in X^* and alpha in R,
there is y in S_X cap H with ||w+y||>=2-rho. For the converse it
suffices that some y in H satisfy the same lower bound.
./kadets-shvidkoy-sirotkin-werner-2000.pdf
Trans. Amer. Math. Soc. 352 (2000), 855-873, Lemma 2.2.
./kadets-werner-2004.pdf
Proc. Amer. Math. Soc. 132 (2004), 1765-1773, Lemma 1.1.
Bourgain and Rosenthal built a closed subspace of L^1 whose unit ball
is compact in measure but which fails the Radon-Nikodym property.
Kadets and Werner refined its finite-dimensional enlargement into a
construction with both the Daugavet and the Schur property.
J. Bourgain, H. P. Rosenthal, Martingales valued in certain subspaces
of L^1, Israel J. Math. 37 (1980), 54-75. DOI 10.1007/BF02762868
(no open pdf).
./kadets-werner-2004.pdf
Summary: "a minor refinement of the Bourgain-Rosenthal
construction yields a Banach space with the Schur property and the
Daugavet property."
Both hypotheses hold simultaneously for this space: compactness in
measure, and the Daugavet slice property with unit witnesses.
Main theorem. Let E be an infinite-dimensional closed real subspace
of L^1(Omega,P) for a probability measure P. Suppose the closed unit
ball B_E is totally bounded in the metric
d(f,g) = inf { a>0 : P(|f-g|>a) < a }
of convergence in measure, and suppose E has the Daugavet property.
Then for every x in S_E and every t>0,
H_E(x,t) = max{t/2, t-1}, D_E(x,t) = max{0, t-2}.
Consequently the averaged and one-sided moduli hat-delta_E(t) and
bar-delta_E(t) equal the same two expressions at every positive
radius, and E admits no equivalent AUC norm.
Two ingredients drive the proof. The averaged lower bound comes from
pointwise clipping of the measure-precompact unit ball to a norm-compact
set, followed by finitely many sign constraints. The averaged upper
bound, the one-sided curve and the failure of AUC renorming all come
from a single geometric fact: every relatively weakly open neighborhood
of a unit vector meets the unit ball almost diametrically. A
quantitative realization of the Kadets-Werner enlargement, written down
with explicit constants, supplies a space with both hypotheses.
The clipping lemma. Let E be an infinite-dimensional closed real
subspace of L^1(Omega,P). If B_E is totally bounded in measure,
then H_E(x,t)>=max{t/2, t-1} for every x in S_E and t>0.
For real numbers a and b, the scalar identity
(|a+b| + |a-b|)/2 = max{|a|, |b|}
holds pointwise. Applying it with a=x and b=ty gives, almost
everywhere in Omega,
(1/2) ( ||x+ty||_1 + ||x-ty||_1 )
= int max{|x|, t|y|} dP
= 1 + int (t|y|-|x|)_+ dP
= t||y||_1 - int min{|x|, t|y|} dP.
Thus the averaged increment equals t||y|| minus the L^1 overlap of
|x| and t|y|. The lower bound reduces to an upper bound on this
overlap.
Proof of the clipping lemma. Fix x in S_E and t>0, and clip a
function y at the pointwise height |x|/t:
C(y) = sgn(y) min { |y|, |x|/t }.
The clipped functions need not lie in E. The norm closure
K = closure of C(B_E) in L^1 is nevertheless compact. Every
sequence b_1, b_2, ... in B_E has a subsequence Cauchy in measure,
because B_E is measure-total-bounded. The map C is pointwise
1-Lipschitz, so the clipped subsequence is also Cauchy in measure. If
c_i and c_j are two of its terms and a>0, then
||C b_i - C b_j||_1 <= a + int_{ |b_i-b_j|>a } (2|x|/t) dP.
Choose a small, and then late indices, using absolute continuity of
the integral of 2|x|/t and the measure-Cauchy property. The
subsequence is Cauchy in L^1. Compactness of K follows.
Fix eps>0 and take a finite eps-net b_1,...,b_m of K. Set
sigma_i = sgn(b_i) and
F = intersection_{i=1}^{m} ker ( y -> int sigma_i y dP ) in E.
This is a closed subspace of finite codimension. For y in S_F,
choose i with ||C(y)-b_i||_1<eps. Clipping preserves signs and
reduces absolute values, so ||y-C(y)||_1 = 1-||C(y)||_1. Since
int sigma_i y=0,
||C(y)||_1 - eps <= ||b_i||_1
= | int sigma_i (b_i-y) dP | <= 1 - ||C(y)||_1 + eps,
hence ||C(y)||_1 <= 1/2 + eps. Because int min{|x|,t|y|} dP equals
t||C(y)||_1, the averaged increment at y is t||y||_1 - t||C(y)||_1
>= t(1/2-eps). The estimate extends to every y in F with
||y||_1>=1 by the following monotonicity: for fixed y in F with
||y||_1=1, the function s -> s - int min{|x|, s|y|} dP has
nonnegative derivative at every s>=0, so the averaged increment at
s y exceeds its value at y. Taking the supremum over F and letting
eps decrease to zero gives H_E(x,t)>=t/2.
The same estimate supplies the second lower bound. Since
min{|x|,t|y|} <= |x|, the identity gives
(1/2)(||x+ty||_1+||x-ty||_1) - 1
>= t||y||_1 - 1 >= t - 1
for every ||y||_1>=1. Combining the two bounds,
H_E(x,t)>=max{t/2, t-1}. This proves the clipping lemma.
Almost diametral points in weak neighborhoods. A slice of B_X is a
nonempty set H = { y in B_X : l(y)>alpha } with l in X^* and
alpha in R.
Slice lemma. A nonzero real Banach space X has the Daugavet property
if and only if, for every w in S_X, every rho>0, and every slice
H of B_X, there is y in H with ||w+y||>=2-rho.
Proof. Suppose first that X has the Daugavet property. When the
defining functional is zero, H=B_X and y=w works. Otherwise divide
l and alpha by ||l|| so that ||l||=1 and alpha<1. Fix
0<eta<min{1-alpha, rho/2}. The rank-one operator T y = l(y) w has
norm one, so there is y in S_X with ||y+l(y) w||>2-eta. Replacing
y by -y leaves this inequality and the norm of y unchanged, and
makes l(y)>=0. Then l(y)>1-eta, because otherwise
||y+l(y)w||<=||y||+l(y)<=2-eta. Hence l(y)>alpha, and
||y+w|| >= ||y+l(y)w|| - (1-l(y)) > 2 - 2eta > 2 - rho.
Conversely, assume the slice condition. Let T z = l(z) v be a
nonzero rank-one operator with ||l||=1. Put a=||v||=||T|| and
w=v/a. For 0<eta<1 the slice condition provides y in B_X with
l(y)>1-eta and ||w+y||>=2-eta. A norm-one functional varphi
norming w+y takes value at least 1-eta on each summand. Therefore
||Id+T|| >= ||y+Ty|| = ||y + l(y) a w||
>= varphi(y) + a l(y) varphi(w)
>= (1-eta) + a (1-eta)^2.
Letting eta decrease to zero gives ||Id+T||>=1+||T||. The identity
||Id+T||<=1+||T|| is the triangle inequality, and the case T=0 is
immediate.
For the passage from slices to weak neighborhoods, the following
convexity fact is used. Its attribution to Bourgain, with a modern
discussion, appears in work of Abrahamsen, Nygaard and Poldvere.
./abrahamsen-nygaard-poldvere-2019.pdf
Pacific. J. Math. 301 (2019), 385-394, attribution of Bourgain's
lemma on convex combinations of slices.
Bourgain's lemma. Every nonempty relatively weakly open subset U of
the closed unit ball of a real Banach space contains a convex
combination of nonempty slices of that ball.
Proof. Choose u in U, a continuous linear map T: X -> R^k, and
an open set O in R^k with u in B_X cap T^{-1}(O) contained in
U. Put D = closure of T(B_X) in R^k. A convex combination of
small slices of D inside O pulls back to one of small slices of
B_X inside U. It suffices to approximate T(u) by convex
combinations of exposed points of D, that is, unique maximizers of
linear functionals on D.
First, D is the closed convex hull of its exposed points. Put
M = max_{q in D} |q|, using the Euclidean norm. For a unit vector
a in R^k and R>M, choose p_R in D farthest from -R a.
Expanding the squares in |q+Ra| <= |p_R+Ra| gives
2 (p_R+Ra) . (q-p_R) <= -|q-p_R|^2 for all q in D.
So p_R uniquely maximizes (p_R+Ra).q on D. If q_a maximizes
a.q on D, the same comparison gives
0 <= a.q_a - a.p_R <= (|p_R|^2-|q_a|^2)/(2R) <= M^2/(2R).
The exposed points have the same supremum as D for every linear
functional. Separation proves the closed-convex-hull assertion, also
when D is a singleton.
Since T(u) in O, there are exposed points p_1,...,p_l and
positive weights lambda_i summing to one such that
p = sum_i lambda_i p_i lies in O. Take r>0 with the Euclidean
ball B(p,r) contained in O. If a_i.q exposes p_i, compactness
and uniqueness of the maximum produce a threshold beta_i<a_i.p_i
with
{ q in D : a_i.q > beta_i } contained in B(p_i,r).
The sets H_i = { y in B_X : a_i.T(y) > beta_i } are nonempty slices,
because T(B_X) is dense in D. For y_i in H_i,
| T( sum_i lambda_i y_i ) - p |
<= sum_i lambda_i |T(y_i)-p_i| < r.
The convex combination sum_i lambda_i y_i lies in U.
Diametral lemma. If X has the Daugavet slice property, then, for
every x in S_X and every relatively weakly open neighborhood U
of x in B_X,
sup_{v in U} ||v-x|| = 2.
./shvydkoy-2000.pdf
J. Funct. Anal. 176 (2000), 198-212. This statement is the
specialization used in the weak-open characterization, Lemma 3.
Proof. By Bourgain's lemma choose slices H_1,...,H_l and positive
weights lambda_i summing to one with sum_i lambda_i H_i contained
in U. Fix 0<eta<1 and set z_0=-x. At step i, apply the slice
property at z_{i-1}/||z_{i-1}|| to choose y_i in H_i with
|| z_{i-1}/||z_{i-1}|| + y_i || >= 2 - eta.
A norm-one functional varphi_i norming this sum takes value at least
1-eta on each summand. Therefore, with z_i = z_{i-1} + lambda_i y_i,
||z_i|| >= varphi_i(z_i)
>= varphi_i(z_{i-1}) + lambda_i varphi_i(y_i)
>= (1-eta)( ||z_{i-1}|| + lambda_i ).
The induction is well defined since ||z_0||=1, and yields
||z_l|| >= (1-eta)^l + sum_i lambda_i (1-eta)^{l-i+1}
>= 2 (1-eta)^l.
Now v = sum_i lambda_i y_i lies in U, and z_l = v - x. Letting
eta decrease to zero gives sup_{v in U}||v-x||>=2; the reverse bound
follows from v,x in B_X.
Completion of the two equality formulas. Fix any F in cof(E) and
choose a bounded finite-rank projection P of E with kernel F; any
linear section of the quotient map E -> E/F composed with the
quotient map gives one. For 0<eps<1/2 put
U_eps = { v in B_E : ||P(v-x)||<eps }.
The finite-dimensional range of P makes P weak-to-norm continuous,
so U_eps is relatively weakly open in B_E. By the diametral lemma
choose v_eps in U_eps with ||v_eps-x||>2-eps. Define
z_eps = (Id-P)(v_eps-x), L_eps = ||z_eps||,
y_eps = z_eps/L_eps.
Then y_eps lies in S_F and
L_eps <= ||v_eps-x||+eps <= 2+eps,
L_eps >= ||v_eps-x||-eps > 2-2eps,
||x+z_eps|| = ||v_eps - P(v_eps-x)|| <= 1+eps.
So L_eps tends to two, 2-2eps<L_eps<=2+eps, and 1+eps bounds the
translated endpoint. For any t>0 put lambda = min{t/L_eps, 1}.
Since t - lambda L_eps = (t-L_eps)_+,
x + t y_eps = (1-lambda) x + lambda (x+z_eps) + (t-lambda L_eps) y_eps,
and the triangle inequality gives
||x+t y_eps|| <= (1-lambda) + lambda (1+eps) + (t-L_eps)_+
<= 1 + eps + (t-L_eps)_+.
Because L_eps>=2-2eps, the right-hand side is at most
1+eps+(t-2+2eps)_+, which tends to 1+(t-2)_+ as eps decreases to
zero. Also ||x-t y_eps||<=1+t. The one-sided increment at direction
y_eps in S_F is bounded by 2eps+(t-2+2eps)_+; the averaged
increment is bounded by (t+(t-2+2eps)_+)/2 up to O(eps). The inner
infimum over S_F is at most the value at y_eps, and F was
arbitrary. Taking eps to zero,
H_E(x,t) <= max{t/2, t-1}, D_E(x,t) <= max{0, t-2}.
For the one-sided lower bound, choose a norm-one functional varphi
with varphi(x)=1 and take F=ker varphi. Every y in S_F
satisfies varphi(x+ty)=1 and ||x+ty||>=t-1, hence
||x+ty||-1>=max{0,t-2}. So D_E(x,t)>=max{0,t-2}. Together with the
clipping lemma, both lower bounds H_E(x,t)>=max{t/2,t-1} and
D_E(x,t)>=max{0,t-2} hold at every unit center, and the four
inequalities assemble into the two equalities of the main theorem.
Failure of AUC renorming. The averaged modulus grows linearly near zero,
whereas the one-sided modulus stays zero up to t=2. So the inherited
norm is not AUC. Ruling out every equivalent AUC norm uses only the
weak-neighborhood geometry of the diametral lemma, not the exact curves.
Proof. Suppose N is an equivalent AUC norm on E. Choose a>0 with
N(u)>=a||u|| for all u, and put R=sup_{u in B_E} N(u)<infty. Fix, in
this order,
t_0 = a/(4R), 0<gamma<bar-delta_N(t_0),
x_0 in S_E with (1+gamma) N(x_0) > R.
Such x_0 exists since R=sup_{u in S_E} N(u). The definition of
bar-delta_N(t_0) at x_0/N(x_0) provides F in cof(E) with
N(x_0+z) >= (1+gamma) N(x_0) whenever z in F and N(z)>=t_0 N(x_0). (*)
To extend (*) to a larger scale, use the convexity of
r -> N(x_0/N(x_0)+r w) for N(w)=1: for every r>=t_0,
N(x_0/N(x_0)+r w) - 1
>= (r/t_0) ( N(x_0/N(x_0)+t_0 w) - 1 ).
Choose 0<eta<min{(1+gamma)N(x_0)-R, a/4} and set
U = { u in B_E : inf_{z in F} N(u-x_0-z) < eta }.
The infimum is the norm in the finite-dimensional quotient E/F, so U is
a relatively weakly open neighborhood of x_0 in B_E: equivalent norms
have the same continuous functionals and therefore the same weak
topology. For u in U pick z in F with N(u-x_0-z)<eta. If
N(z)>=t_0 N(x_0), then (*) gives N(u)>(1+gamma)N(x_0)-eta>R, which
contradicts u in B_E. Hence N(z)<t_0 N(x_0) and
N(u-x_0) < t_0 N(x_0)+eta <= t_0 R+eta < a/2,
||u-x_0|| < 1/2.
Every u in U lies at distance less than 1/2 from x_0, contradicting the
diametral lemma.
Construction of a space with both properties. The construction
multiplies selected old unit vectors by independent positive random
variables that have mean one but are small with high probability. Each
product has small L^1 overlap with every old unit vector, so its sum
with any old unit vector has norm close to two. Averages of many
products recover the selected old vectors in norm. A uniform estimate
keeps the entire enlarged unit ball close in measure to the old one,
which permits the iteration to preserve total boundedness in measure.
The scheme and its enlargement are due to Kadets and Werner.
./kadets-werner-2004.pdf
Proc. Amer. Math. Soc. 132 (2004), 1765-1773, Lemma 2.4 and
Theorem 2.5. Constants 10, 54, 108 are the explicit values
extracted from that argument.
Ambient space and measure metric. Take Omega=[0,1]^N with the product
Lebesgue measure P. The metric
d(f,g) = inf { a>0 : P(|f-g|>a)<a }
induces convergence in measure on measurable functions modulo equality
almost everywhere. The union bound gives the triangle inequality:
P(|f-h|>a+b) <= P(|f-g|>a)+P(|g-h|>b) <= a+b. The identity d(f,g)=0
holds exactly when f=g almost everywhere. Markov's inequality gives
d(f,g)<=||f-g||^{1/2}, so norm convergence implies convergence for d.
A subspace depends on finitely many coordinates if one finite coordinate
set determines all its elements. Write B(G) and S(G) for the closed
unit ball and sphere of a subspace G.
Heavy-tail lemma. Let 1<p<=3/2, s=(p-1)/p, and f=s V^{-1/p}, with V
uniform on (0,1). Let Y_1,...,Y_n be independent variables, each
distributed as f-f' for an independent copy f' of f. For real scalars
a_1,...,a_n, put A=(sum_j |a_j|^p)^{1/p} and Z=sum_j a_j Y_j. Then
E f=1 and
P(|Z|>u) <= 10 (s A/u)^p (u>0),
E|Z| >= A/54.
Proof. Direct integration gives E f=s p/(p-1)=1 and
P(f>u)=(s/u)^p for u>=s; the same formula is an upper bound for every
u>0. If Y=f-f', then Y is symmetric and P(|Y|>u)<=2(s/u)^p. Fix
u>0 and put X_j=a_j Y_j 1_{|a_j Y_j|<=u}. The X_j are independent and
centered, and by layer cake,
E X_j^2 <= int_0^u 2v P(|a_j Y_j|>v) dv
<= 4(s|a_j|)^p u^{2-p}/(2-p).
Outside the event that some |a_j Y_j|>u, the sums Z and sum_j X_j
agree. The union bound gives P(some |a_j Y_j|>u)<=2(sA/u)^p; Chebyshev
gives P(|sum_j X_j|>u)<=4(sA/u)^p/(2-p). Adding,
P(|Z|>u) <= (2+4/(2-p))(s A/u)^p <= 10(s A/u)^p,
since p<=3/2 gives 2-p>=1/2 and hence 4/(2-p)<=8.
For the lower bound, when u>=s the event {f>u+2s, f'<=2s} gives
P(|Y|>u) >= (s/(u+2s))^p (1-2^{-p}) >= (1/18)(s/u)^p,
using u+2s<=3u, 1-2^{-p}>=1/2 and 3^p<=9. Suppose A>0 and omit zero
coefficients. Since |a_j|<=A, independence gives, for u>=sA,
P(max_j |a_j Y_j|>u)
= 1 - prod_j P(|a_j Y_j|<=u)
>= 1 - exp( - sum_j P(|a_j Y_j|>u) )
>= (1/36)(s A/u)^p,
where the middle step uses prod(1-t_j)<=exp(-sum t_j) and the last uses
1-exp(-v)>=v/2 for 0<=v<=1, applied with v=(1/18)(sA/u)^p<=1.
Adjoin independent signs epsilon_j in {-1,1} of equal probability. For
each fixed realization of the terms, averaging over the sign of a term
of largest absolute value gives
E_epsilon | sum_j epsilon_j a_j Y_j | >= max_j |a_j Y_j|,
because (|b+c|+|b-c|)/2>=max{|b|,|c|}. The signed sum has the same law
as Z by symmetry and independence. Hence
E|Z| >= E max_j |a_j Y_j|
>= (1/36) int_{sA}^infty (sA/u)^p du
= (sA)/(36(p-1)) = A/(36p) >= A/54,
using s=(p-1)/p in the second-to-last identity and p<=3/2 at the end.
The case A=0 is immediate.
Enlargement lemma. Suppose G is a nonzero finite-dimensional subspace
of L^1 depending on finitely many coordinates, m>=1, u_1,...,u_m lie
in S(G), and 0<h<1. Then there exist a finite-dimensional G'
containing G and depending on finitely many coordinates, an integer
n>=1, and vectors v_kj in S(G') with 1<=k<=m and 1<=j<=n, such that
(a) ||w+v_kj|| >= 2-h for every w in S(G), every k, every j;
(b) ||u_k - (1/n) sum_{j=1}^n v_kj|| <= h for each k;
(c) for every g in B(G') there is psi_0 in B(G) with d(g,psi_0)<=h.
Proof. Choose c in (0,h/4), then r,q>0 with mq<1 and
mq + h^{-1}( 2mr + (1-mq)^{-1} - 1 ) < h.
With s=(p-1)/p, choose 1<p<=3/2 close enough to one that
s<c, 2((s/c)^p + c) < h, 10(108 s/r)^p < q.
The choices are possible because s decreases to zero as p decreases to
one. Fix p; the integer n is chosen last.
Near-additivity. For any prospective n, take mn independent copies
f_kj of f from the heavy-tail lemma, each on a fresh coordinate. Put
v_kj = u_k f_kj, G'_n = span( G union {v_kj} ).
Independence from the old coordinates gives ||v_kj||=1. For w in S(G)
and each k,j, the scalar identity applied pointwise to w and u_k f_kj
gives
||w+v_kj|| = 2 (1/2)(||w+v_kj||+||w-v_kj||) - ||w-v_kj||
>= 2(2 - int min{|w|, |u_k| f_kj} dP) - 2
= 2 - 2 int min{|w|, |u_k| f_kj} dP.
Split the last integral over {f_kj>c} and its complement. On {f_kj>c}
the minimum is at most |w| in absolute value, so this part integrates
to at most P(f_kj>c). On the complement the minimum is at most c|u_k|,
so this part integrates to at most c. Independence of f_kj from the
old coordinates is used in both bounds. Therefore
||w+v_kj|| >= 2 - 2(P(f_kj>c)+c) = 2 - 2((s/c)^p + c) > 2 - h,
proving (a) for every n.
Approximation of the whole unit ball. Fix g in B(G'_n) and choose a
representation
g = x + sum_{k=1}^m u_k z_k, x in G, z_k = sum_{j=1}^n a_kj f_kj.
Fix k. On an auxiliary product probability space, resample all the
variables in the kth group, leaving the old coordinates and the other
groups unchanged. Write z'_k for the resampled sum and g^{(k)} for the
resulting function. The new variables are independent copies, so
g^{(k)} has the same law as g, and
E|z_k-z'_k| = E|u_k (z_k-z'_k)| = E||g-g^{(k)}|| <= 2.
The first equality uses ||u_k||=1 and independence from the old
coordinates. Applying the heavy-tail lemma to the differences of the
paired copies yields
A_k = (sum_j |a_kj|^p)^{1/p} <= 108,
P(|z_k-z'_k|>r) <= 10(108 s/r)^p < q,
where the first bound uses E|Z|>=A/54 and the second uses the upper
bound of the heavy-tail lemma with A_k<=108. Neither bound depends on
n or on the chosen representation.
The estimate on z_k-z'_k concentrates z_k near a deterministic c_k in
R, with P(|z_k-c_k|>r)<q. The constant may depend on g. Put
psi = x + sum_k c_k u_k in G,
D = intersection_k { |z_k-c_k| <= r }.
The event D depends only on the fresh coordinates, has probability at
least 1-mq, and satisfies |g-psi|<=r sum_k |u_k| pointwise on D.
Independence of D from the old coordinates gives
1 >= E(|g| 1_D) >= (1-mq)( ||psi|| - mr ),
so ||psi||<=mr+(1-mq)^{-1}. Put psi_0=psi/max{1,||psi||} in B(G).
The cost of normalization is ||psi-psi_0||<=mr+(1-mq)^{-1}-1. Markov's
inequality restricted to D and the bound on its complement give
P(|g-psi_0|>h) <= mq + h^{-1}( mr + ||psi-psi_0|| )
<= mq + h^{-1}( 2mr + (1-mq)^{-1} - 1 ) < h.
This proves (c) for every n.
Recovering the old vectors by averages. Choose n for (b). For
independent copies f_1,...,f_n of f, write f^M=min{f,M}. Truncating,
centering and applying the variance bound to the bounded independent
copies gives
E| (1/n) sum_j f_j - 1 | <= 2 E(f-f^M) + ( Var(f^M)/n )^{1/2}.
The first term tends to zero as M increases; with M fixed, the second
tends to zero as n increases. Choose n so that the left side is at
most h. All groups have the same law, and independence gives
||u_k - (1/n) sum_j v_kj|| = E|1 - (1/n) sum_j f_kj| <= h
for each k. This proves (b). Taking G'=G'_n completes the enlargement.
The space G' uses finitely many additional coordinates, and the
estimates in (a) and (c) held for every n.
Quantitative Kadets-Werner space. Start with G_1=span{1} and put
h_b=2^{-b}. Given G_b, choose a finite h_b-net {u_{b,k} :
1<=k<=m_b} of S(G_b) in norm and apply the enlargement lemma with
h=h_b. Let G_{b+1} and the vectors v_{b,kj} (1<=j<=n_b) be the result.
Every stage depends on finitely many coordinates, so the construction
continues on Omega. Set E= closed norm closure of the union over b of
G_b in L^1.
Strict inclusions. If some v_{b,kj} belonged to G_b, substituting
w=-v_{b,kj} in (a) would give 0>=2-h_b, a contradiction. Thus every
inclusion G_b contained in G_{b+1} is strict, and E is
infinite-dimensional.
Total boundedness in measure. The union over b of B(G_b) is norm dense
in B(E): if x_j in the algebraic union tends in norm to x in B(E),
then x_j/max{1,||x_j||} also tends to x. Repeated use of (c) gives,
for a>b and g in B(G_a),
inf_{psi in B(G_b)} d(g,psi) <= sum_{l=b}^{a-1} h_l <= 2^{1-b}.
The bound is automatic for a<=b. Distance to a fixed nonempty set is
a 1-Lipschitz function for d, so norm density extends the estimate to
every g in B(E). The ball B(G_b) is norm compact and therefore
d-compact. Given eps>0, choose b with 2^{1-b}<eps/2 and a finite
eps/2-net of B(G_b) for d. The preceding distance estimate shows that
this net is an eps-net of B(E). So B(E) is totally bounded in d.
Daugavet slice property. Fix w in S(E), a slice
H = { y in B(E) : l(y)>alpha } with l in E^*, and rho>0. There is u
in S(E) with l(u)>alpha: the supremum of a linear functional over the
ball equals its supremum over the sphere, including when the functional
is zero. Normalizing approximants from the increasing union gives
u^{(b)},w^{(b)} in S(G_b) with u^{(b)}->u and w^{(b)}->w in norm.
Choose a net point u_{b,k_b} with ||u_{b,k_b}-u^{(b)}||<=h_b and set
a_b = (1/n_b) sum_{j=1}^{n_b} v_{b,k_b j}.
By (b), ||a_b-u||<=2h_b+||u^{(b)}-u||->0. Thus l(a_b)>alpha for all
sufficiently large b, and for each such b at least one summand
v_{b,k_b j} lies in H. For every summand, (a) gives
||w+v_{b,k_b j}|| >= 2 - h_b - ||w-w^{(b)}||.
Take b large enough that h_b+||w-w^{(b)}||<rho. The summand in H is
the required unit vector. The slice lemma turns this slice property
into ||Id+T||=1+||T|| for every rank-one bounded operator T on E.
End of proof. The Kadets-Werner space E is an infinite-dimensional
closed real subspace of L^1(Omega,P), its unit ball is totally bounded
in d, and E has the Daugavet property. The clipping lemma gives
H_E(x,t)>=max{t/2,t-1}; the finite-rank projection argument gives
H_E(x,t)<=max{t/2,t-1} and D_E(x,t)<=max{0,t-2}; a norming functional
gives D_E(x,t)>=max{0,t-2}. Taking infima over unit centers preserves
the identities. The weak-neighborhood argument at a near-maximizer of
an arbitrary equivalent norm bars every equivalent AUC norm. The
existence theorem supplies at least one such E.