Exact asymptotic moduli in a Daugavet subspace of L1

Proof. Fix x and t. Clip a function y at the pointwise height |x|/t:

The clipped functions need not belong to E. Their norm closure K=\overline{C(B_E)}^{\,L^1} is compact. Indeed, every sequence in B_E has a subsequence Cauchy in measure. Clipping is pointwise 1-Lipschitz, so the clipped subsequence is also Cauchy in measure. For two of its terms b_i,b_j and any a>0,

Choose a small, then use absolute continuity of the integral and the measure-Cauchy property. This proves that the subsequence is Cauchy in L^1. Completeness proves the compactness assertion.

Fix \varepsilon>0 and take a finite \varepsilon-net b_1,\ldots,b_m of K. Set \sigma_i=\operatorname{sgn}(b_i) and

This is a closed subspace of finite codimension. For y\in S_F, choose i with \|C(y)-b_i\|_1<\varepsilon. Clipping preserves signs and reduces absolute values, so \|y-C(y)\|_1=1-\|C(y)\|_1. Since \int\sigma_i y=0, we obtain

Thus \|C(y)\|_1\le1/2+\varepsilon. Equation (2) now gives averaged increment at least t/2-t\varepsilon on S_F. The same bound holds for \|y\|_1\ge1 by radial monotonicity. Taking the supremum over F and letting \varepsilon\downarrow0 gives H_E(x,t)\ge t/2. The subspace is allowed to change with \varepsilon; this step does not assert that one subspace attains the limiting bound.

For every \|y\|_1\ge1, the triangle inequality also gives

Combining the two bounds proves the lemma.□

Proof. Suppose first that X has the Daugavet property. If the functional defining H is zero, then H=B_X and we may take y=w. Otherwise, after dividing the functional and threshold by its norm, write H=\{y\in B_X:\ell(y)>\alpha\} with \|\ell\|=1 and \alpha<1. Take 0<\eta<\min\{1-\alpha,\rho/2\}. The rank-one operator Ty=\ell(y)w has norm one, so there is y\in S_X with \|y+\ell(y)w\|>2-\eta. Replacing y by -y if necessary makes \ell(y)\ge0 without changing this inequality. It follows that \ell(y)>1-\eta>\alpha and

Conversely, assume the slice condition and let Tz=\ell(z)v be a nonzero rank-one operator with \|\ell\|=1. Set a=\|v\|=\|T\| and w=v/a. For 0<\eta<1, the slice condition provides y\in B_X with \ell(y)>1-\eta and \|w+y\|\ge2-\eta. A norm-one functional \varphi norming w+y takes value at least 1-\eta on each summand. Consequently,

Let \eta\downarrow0 and use \|I+T\|\le1+\|T\|. The identity is immediate when T=0.□

Proof. Choose u\in U, a continuous linear map T:X\to\mathbb R^k, and an open set O\subseteq\mathbb R^k such that u\in B_X\cap T^{-1}(O)\subseteq U. Put D=\overline{T(B_X)}. We want to place a convex combination of small slices of D inside O and then pull them back to B_X. For this it suffices to approximate T(u) by convex combinations of exposed points, that is, unique maximizers of linear functionals on D. We first verify that D is the closed convex hull of those points. Put M=\max_{q\in D}|q|, using a Euclidean norm. For a unit vector a and R>M, choose p_R\in D farthest from -Ra. Expanding the squares in |q+Ra|\le|p_R+Ra| gives

Thus p_R is the unique maximizer of the linear functional with coefficient p_R+Ra. If q_a maximizes a\cdot q on D, the same comparison gives

The exposed points therefore 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,\ldots,p_l and positive weights \lambda_i summing to one such that p=\sum_i\lambda_i p_i\in O. Choose r>0 with the Euclidean ball B(p,r)\subseteq O. If a_i\cdot q exposes p_i, compactness and uniqueness of its maximum give a threshold \beta_i<a_i\cdot p_i with

The sets H_i=\{y\in B_X:a_i\cdot T(y)>\beta_i\} are nonempty slices, since T(B_X) is dense in D. For y_i\in H_i,

Their convex combination consequently belongs to U.□

Proof. By Lemma 3.2, choose slices H_i and positive weights \lambda_i such that \sum_i\lambda_i=1 and \sum_i\lambda_iH_i\subseteq 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\|\ge2-\eta. A norm-one functional norming this sum takes value at least 1-\eta on each of its two summands. Therefore, with z_i=z_{i-1}+\lambda_i y_i,

The induction is well defined and yields

Now v=\sum_i\lambda_i y_i\in U and z_l=v-x. Let \eta\downarrow0; the reverse bound follows from v,x\in B_X.□

Proof of the formulas in Theorem 1.1. Lemma 2.1 gives the averaged lower bound. For the one-sided bound, choose a norm-one functional \varphi with \varphi(x)=1. Every y\in S_{\ker\varphi} satisfies

Hence D_E(x,t)\ge\max\{0,t-2\}.

For the upper bounds, fix any F\in\operatorname{cof}(E) and choose a bounded finite-rank projection P with kernel F. Such a projection is obtained by composing the quotient map with a linear section of E/F. For 0<\varepsilon<1/2,

is a relatively weakly open neighborhood of x. The finite-dimensional range of P makes it weak-to-norm continuous. By Lemma 3.3, choose v_\varepsilon\in U_\varepsilon with \|v_\varepsilon-x\|>2-\varepsilon, and put

Then

Thus the directions lie exactly in F, their unnormalized lengths tend to two, and the translated endpoints remain close to B_E.

For any t>0, set \lambda=\min\{t/L_\varepsilon,1\}. The identity

and the triangle inequality give the single bound

The right-hand side converges to 1+(t-2)_+=\max\{1,t-1\}. Since \|x-t y_\varepsilon\|\le1+t, the one-sided and averaged increments have upper limits at most

respectively. These are \max\{0,t-2\} and \max\{t/2,t-1\}. In particular the argument includes t=2, where the two branches meet. Each inner infimum for the fixed F is at most the corresponding limit. Taking the supremum over F proves the upper bounds. Both formulas hold for every unit center, so taking the infimum over centers leaves them unchanged.□

Proof of the renorming assertion in Theorem 1.1. Suppose N is an equivalent AUC norm. Choose a>0 with N(u)\ge a\|u\| for all u, and put R=\sup_{u\in B_E}N(u)<\infty. Fix, in this order,

The last choice is possible since R=\sup_{u\in S_E}N(u). The definition of \overline\delta_N(t_0) at x_0/N(x_0) supplies F\in\operatorname{cof}(E) such that

To justify the entire range of radii, convexity gives, for N(w)=1 and r\ge t_0,

Choose 0<\eta<\min\{(1+\gamma)N(x_0)-R,a/4\}. The set

is a relatively weakly open neighborhood of x_0: the infimum is the norm in the finite-dimensional quotient E/F. Equivalent norms have the same continuous functionals, hence the same weak topology. For u\in U, choose z\in F with N(u-x_0-z)<\eta. If N(z)\ge t_0N(x_0), Equation (3) gives

contrary to u\in B_E. Thus

This contradicts Lemma 3.3.□

Proof. Direct integration gives \E f=sp/(p-1)=1, and \PP(f>u)=(s/u)^p for u\ge s. For all u>0 the same expression is an upper bound. If Y=f-f' for independent copies, then Y is symmetric and \PP(\abs Y>u)\le2(s/u)^p. For fixed u>0, the variables X_j=a_jY_j\mathbf1_{\{\abs{a_jY_j}\le u\}} are independent and centered, and

Except on the event that some \abs{a_jY_j}>u, the sums Z and \sum_jX_j agree. The union bound and Chebyshev’s inequality therefore give

For the lower bound, when u\ge s, the event \{f>u+2s,\ f'\le2s\} gives

Here u+2s\le3u, 1-2^{-p}\ge1/2, and 3^p\le9. Suppose A>0 and omit zero coefficients. Since \abs{a_j}\le A, independence implies, for u\ge sA,

The last inequality uses 1-e^{-v}\ge v/2 for 0\le v\le1. Adjoin independent signs \varepsilon_j\in\{-1,1\} of equal probability. For each fixed realization of the terms, averaging first over the sign of a term of largest absolute value gives

because (\abs{b+c}+\abs{b-c})/2\ge\abs c. The signed sum has the same law as Z, by independence and symmetry. Consequently,

If A=0, both assertions are immediate.□

Proof of Lemma 6.2. We first fix the distribution of the new multipliers and prove the first and third conclusions uniformly in their number. Only then will we choose the number of multipliers to obtain the averaging conclusion.

Parameter choices. Choose c\in(0,h/4), and then r,q>0 sufficiently small that

With s=(p-1)/p, choose 1<p\le3/2 close enough to 1 that

These choices are possible because s\to0 as p\downarrow1. We fix p now; the integer n will be chosen last.

Near-additivity with the old space. For any prospective n, take mn independent copies f_{kj} of the variable f in Lemma 6.3, each on a fresh coordinate, and put

Independence from the old coordinates gives \norm{v_{kj}}=1. For every w\in S(G),

Indeed, on \{f_{kj}>c\} bound the minimum by \abs w, and on its complement by c\abs{u_k}; independence then gives the last bound. Thus (5) holds for every n.

Approximation of the whole unit ball. We next prove (7), also uniformly in n. Fix g\in B(G'_n) and choose any representation

Fix k. On an auxiliary product probability space resample all the variables in the kth group, leaving the old coordinates and all 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

The first equality uses \norm{u_k}=1 and independence of the old coordinates. Lemma 6.3, applied to the differences of the paired copies, therefore gives

In particular, the bound does not depend on n or on the chosen representation of g.

The estimate for z_k-z'_k says that z_k is concentrated near some constant. Indeed, averaging the conditional probability over z'_k gives a deterministic c_k\in\R with \PP(\abs{z_k-c_k}>r)<q. These constants may depend on g; no simultaneous choice is needed. Set

The event D depends only on the fresh coordinates, has probability at least 1-mq, and on D we have \abs{g-\psi}\le r\sum_k\abs{u_k}. Its independence from the old coordinates gives

At this point \psi need not belong to the old unit ball. Set \psi_0=\psi/\max\{1,\norm\psi\}\in B(G). The cost of this normalization is bounded by

Markov’s inequality restricted to D and the bound on its complement now yield

This proves (7) for every n.

We have now proved both near-additivity and whole-ball approximation without placing an upper bound on n.

Recovering the old vectors by averages. It remains to choose n for (6). For independent copies f_1,\dots,f_n of the fixed integrable variable f, write f^M=\min\{f,M\}. Truncating, centering, and applying the variance bound to the bounded independent copies gives

The first term tends to zero as M\to\infty; after M is fixed the second tends to zero as n\to\infty. Choose n so that the left side is at most h. All groups have the same law, and independence gives

Taking G'=G'_n completes the proof. This space uses just finitely many additional coordinates; the earlier two conclusions persist because their estimates held for every finite n.□

Proof of Theorem 6.1. Start with G_1=\operatorname{span}\{1\} and put h_b=2^{-b}. Given G_b, choose a finite h_b-net \{u_{b,k}:1\le k\le m_b\} of S(G_b) in norm and apply Lemma 6.2 with h=h_b. Denote the resulting space by G_{b+1} and its new unit vectors by v_{b,kj}, 1\le j\le n_b. All stages depend on finitely many coordinates, so the construction continues on \Omega. Set

Each inclusion G_b\subseteq G_{b+1} is strict: if any v_{b,kj} belonged to G_b, substituting w=-v_{b,kj} in (5) would give 0\ge2-h_b. Thus E is infinite-dimensional.

The entire unit ball is totally bounded in measure. The union \bigcup_bB(G_b) is norm dense in B(E): if x_j in the algebraic union tends to x\in B(E), then x_j/\max\{1,\norm{x_j}\} also tends to x. Repeated use of (7) gives, for a>b and g\in B(G_a),

The bound is automatic for a\le b. Distance to a fixed nonempty set is a 1-Lipschitz function for d, so norm density and (4) extend it to every g\in B(E). The set 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 is an \eps-net of B(E).

The Daugavet slice property. Now fix w,H,\ell,\alpha,\rho as in Theorem 6.1. There is u\in S(E) with \ell(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)}\to u and w^{(b)}\to w in norm. Choose a net point u_{b,k_b} with \norm{u_{b,k_b}-u^{(b)}}\le h_b and set

By (6), \norm{a_b-u}\le2h_b+\norm{u^{(b)}-u}\to0. Thus \ell(a_b)>\alpha for all sufficiently large b, and for each such b at least one summand v_{b,k_bj} lies in H. For every summand, (5) also gives

Take b large enough that the last two error terms sum to less than \rho. The summand in H is the required unit vector.

Finally, Lemma 3.1 turns this slice property into \norm{I+T}=1+\norm T for every rank-one operator T on E.□