$C^k$-smooth approximations of LUR norms

0.25 MB·English

by  Petr Hajek

Ck-SMOOTH APPROXIMATIONS OF LUR NORMS
PETRHA´JEKANDANTON´INPROCHA´ZKA
Abstract. LetX beaWCGBanachspaceadmittingaCk-Fr´echetsmoothnorm. ThenX admitsan
equivalent norm which is simultaneously C1-Fr´echet smooth, LUR, and a uniform limitof Ck-Fr´echet
smoothnorms. IfX=C([0,α]),whereαisanordinal,thenthesameconclusionholdstruewithk=∞.
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0
2
1. Introduction
n
a The theory of Ck-Fr´echet smooth approximations of continuous functions on Banach spaces is well-
J
developed,thankstotheworkofmanymathematicians,whoseclassicalresultsandreferencescanbefound
3 in the authoritative monograph [2]. The known techniques rely on the use of Ck-Fr´echet smooth parti-
2
titons of unity, resp. certain coordinatewise smooth embeddings into the space c (Γ) (due to Torunczyk
0
] [23]). They are highly nonlinear, and even non-Lipschitz in nature. For example, if the given function
A
is Lipschitz or has some uniform continuity, trying to preserve the lipschitzness of the approximating
F smooth functions leads to considerable additional technical difficulties (e.g. [10], [11], [12], [13], [18],
. [19]).
h
t It is well-knownthat the (apparently harder,and less developed) paralleltheory of approximationsof
a
norms on a Banach space by Ck-Fr´echet smooth renormings requires different techniques.
m
Several open problems proposed in [2] are addressing these issues. In particular, if a Banach space
[ admits anequivalentCk-Fr´echetsmoothrenorming,is itpossibleto approximate(uniformly onbounded

1 sets) all norms by Ck-Fr´echet smooth norms? Even in the separable case, the answer is not known in
v full generality, although the positive results in [3] and [4] are quite strong, and apply to most classical
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Banachspaces. Inthenonseparablesetting,nogeneralresultsareknown,withasmallexceptionof[7]. In
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particular, one of the open problems in [2] is whether on a given WCG Banachspace with an equivalent
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3 Ck-Fr´echet smooth norm, there exists an equivalent locally uniformly rotund (LUR) norm which is a
. uniform limit on bounded sets of Ck-Fr´echet smooth norms. The notion of LUR is of fundamental
1
importance for renorming theory, and we refer to [2] and the more recent [20] for an extensive list of
0
9 authors and results.
0 Such a result is of interest for several reasons. It can be used to obtain rather directly the uniform
: approximationsofgeneralcontinuousoperators,byCk-Fr´echetsmoothones. Moreover,sinceLURnorms
v
i form a residualset in the metric space of allequivalent norms ona Banachspace,a positive answeris to
X
be expected. There is a closely related problem of obtaining a norm which shares simultaneously good
r rotundity and smoothness properties. By a famous result of Asplund [1], on every separable Asplund
a
space there exists an equivalent norm which is simultaneously C1-Fr´echet smooth and LUR. A clever
proof using Baire category, and disposing of the separability condition on the underlying Banach space,
wasdevisedin[6]([2], II.4.3). The theoremholdsinparticularinallWCGAsplund spaces(inparticular
all reflexive spaces). Its proof works under the assumption that the space admits an LUR norm, as well
as a norm whose dual is LUR. It is well-known that dual LUR implies that the original norm is C1-
Fr´echet smooth. However, using this approach one cannot in general handle norms with higher degree
Date:January2009.
2000 Mathematics Subject Classification. 46B20,46B03,46E15.
Key words and phrases. LUR,higherorderdifferentiability,renorming.
Supportedbygrants: Institutional ResearchPlanAV0Z10190503, A100190502, GACˇR201/07/0394.
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Ck-SMOOTH APPROXIMATIONS OF LUR NORMS 2
of differentiability, even in the separable case. Indeed, by [9] ([2], ProposititonV.1.3), a space admitting
a LUR and simultaneously C2-Fr´echet smooth norm is superreflexive. There is not even a rotund and
C2-Fr´echet smooth norm on c (Γ) ([14], [15]). In fact, one cannot even handle the proper case of LUR
0
and C1-Fr´echet smooth norms. Indeed, Talagrand [22] proved that C([0,ω ]) admits an equivalent C∞-
1
Fr´echet smooth norm, although it admits no dual LUR renorming. The existence of LUR renorming
of this space follows from Troyanski’s theorem [24]. In light of the previous results it is natural to ask
whether this space has a C1-Fr´echet smooth and simultaneously LUR renorming. This question was
posed on various occasions, e.g. in [8].
Our main result addresses both of the above mentioned open problems, namely higher smoothness
approximation and simultaneous LUR and C1-smoothness. Under reasonable assumptions (e.g. for
WLD, C(K) where K is Valdivia compact, or C([0,α]), i.e. the space of continuous functions on the
ordinal interval [0,α]), it gives a renorming which is simultaneously C1-Fr´echet smooth and LUR, and
admits a uniform approximation on bounded sets by Ck-Fr´echet norms. As a corollary we obtain a
positive solution to both of the mentioned problems. We should emphasize that it is unknown whether
C1-Fr´echet smooth norms are residual, or even dense, in the space of all equivalent norms on C([0,α]).
The paper is organized as follows. In Section 2, we introduce our notation and we present some
auxiliary lemmata. We include the easy proofs for reader’s convenience. The main result, its corollaries
and the frame of the proof of the main result are gathered in Section 3. Sections 4 and 5 then contain
the details of the construction.
2. Preliminaries
The closed unit ball of a Banachspace (X,k·k) is denoted by B , or B for short. Similarly, the
(X,k·k) X
open unit ball of X is BO =BO. By Γ we denote an index set. Smoothness and higher smoothness
(X,k·k) X
is meant in the Fr´echet sense.
Definition 2.1. Let A⊂ℓ∞(Γ). We say that a function f :ℓ∞(Γ)→R in A locally depends on finitely
manycoordinates (LFC)ifforeachx∈AthereexistsaneighborhoodU ofx,afiniteM ={γ ,...,γ }⊂
1 n
Γ and a function g :R|M| →R such that f(y)=g(y(γ ),...,y(γ )) for each y ∈U.
1 n
Definition 2.2. Let X be a vectorspace. A function g :X →ℓ∞(Γ) is saidto be coordinatewise convex
if, for eachγ ∈Γ, the function x7→g (x) is convex. We use the terms as coordinatewise non-negative or
γ
coordinatewise Ck-smooth in a similar way.
Lemma 2.3. Let X be a Banach space and let h : X → ℓ∞(Γ) be a continuous function which is
coordinatewise Ck-smooth, k ∈ N∪{∞}. Let f : ℓ∞(Γ) → R be a Ck-smooth function which locally
depends on finitely many coordinates. Then f ◦h is Ck-smooth.
Proof. Let x ∈ X be fixed. Since f is LFC, there is a neighborhood U of h(x), M = {γ ,...,γ } ⊂ Γ
1 n
and g : R|M| → R as in Definition 2.1. The function g is Ck-smooth, because f is Ck-smooth. As h is
continuous,thereexistsaneighborhoodV ofxsuchthath(V)⊂U. SincehiscoordinatewiseCk-smooth,
it follows that h(·)↾ :=(h(·)(γ ),...,h(·)(γ )) is Ck-smoothfrom X to R|M|. Finally, we havefor each
M 1 n
y ∈V that f(h(y))=g(h(y)↾ ) and the claim follows. (cid:3)
M
Lemma 2.4. Let Φ:ℓ∞(Γ)→R and let x∈ℓ∞(Γ) be such that
a) Φ is LFC at x,
b) Φ′(x)x6=0,
c) Φ(·) and Φ′(·) are continuous at x.
Then there is a neighborhood U of x and a unique function F : U → R which is continuous at x and
satisfies F(x)=1 and Φ( y )=1 for all y ∈U. Moreover F is LFC at x.
F(y)
Proof. The first part of the assertion follows immediately from the Implicit Function Theorem. We
will show that F is LFC at x. From the assumption a) we know that there is a neighborhood V of
Ck-SMOOTH APPROXIMATIONS OF LUR NORMS 3
x, M = {γ ,...,γ } ⊂ Γ, and g : Rn → R such that Φ(y) = g(y ↾ ) for all y ∈ V. It is obvious
1 n M
that g′(x ↾ )x ↾ = Φ′(x)x so it is possible to apply the Implicit Function Theorem to the equation
M M
g y = 1 to get h : V′ → R, where V′ is a neighborhood of x ↾ , such that h(x ↾ ) = 1 and h is
h(y) M M
co(cid:16)ntinu(cid:17)ous at x ↾M. There is a neighborhood U′ ⊂U ∩V of x such that we may define H :U′ →R by
H(y):=h(y ↾ )fory ∈U′. ThenH(x)=1andH iscontinuousatx. Also,Φ y =g y↾M =1.
M H(y) h(y↾M)
The uniqueness of F implies that F =H on U′ so F is LFC at x. (cid:16) (cid:17) (cid:16) (cid:17) (cid:3)
The following lemma is a variant of Fact II.2.3(i) in [2].
Lemma 2.5. Let ϕ:X →R be a convex non-negative function, x ,x∈X for r ∈N. Then the following
r
conditions are equivalent:
(i) ϕ2(xr)+ϕ2(x) −ϕ2(x+xr)→0,
2 2
(ii) limϕ(x )=limϕ(x+xr)=ϕ(x).
r 2
If ϕ is homogeneous, the above conditions are also equivalent to
(iii) 2ϕ2(x )+2ϕ2(x)−ϕ2(x+x )→0.
r r
Proof. Since ϕ is convex and non-negative, and y 7→y2 is increasing for y ∈[0,+∞), it holds
ϕ2(x )+ϕ2(x) x+x ϕ2(x )+ϕ2(x) ϕ(x)+ϕ(x ) 2 ϕ(x)−ϕ(x ) 2
r −ϕ2 r ≥ r − r = r
2 2 2 2 2
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
which proves (i) ⇒ (ii). The implication (ii) ⇒ (i) is trivial and so is the equivalence (i) ⇔ (iii). (cid:3)
Lemma 2.6. Let f,g be twice differentiable, convex, non-negative, real functions of one real variable.
Let F : R2 → R be given as F(x,y) := f(x)g(y). For F to be convex in R2, it is sufficient that g is
convex and
(f′(x))2(g′(y))2 ≤f′′(x)f(x)g′′(y)g(y). (1)
for all (x,y)∈R2.
Proof. Let(x,y)∈R2 befixed. Sinceg isconvex,thefunctionF isconvexwhenrestrictedtothevertical
line going through (x,y). Let s = at+b (a,b ∈ R) be a line going through (x,y), i.e. y = ax+b. The
second derivative at a point (x,y) of F restricted to this line is given as:
f(x)g′′(y)a2+2f′(x)g′(y)a+f′′(x)g(y).
In order for the second derivative to be non-negative for all a ∈ R, it is sufficient that the discriminant
(2f′(x)g′(y))2−4f(x)g′′(y)f′′(x)g(y) of the above quadratic term be non-positive, which occurs exactly
when our condition (1) holds for (x,y). (cid:3)
Definition 2.7. We say that a function f : ℓ∞(Γ) → R is strongly lattice if f(x) ≤ f(y) whenever
|x(γ)|≤|y(γ)| for all γ ∈Γ.
Lemma 2.8. Let f : ℓ∞(Γ) → R be convex and strongly lattice. Let g : X → ℓ∞(Γ) be coordinatewise
convex and coordinatewise non-negative. Then f ◦g :X →R is convex.
Proof. Let a,b≥0 and a+b=1. Since g is coordinatewise convex and non-negative, we have
0≤g (ax+by)≤ag (x)+bg (y)
γ γ γ
for each γ ∈Γ. The strongly lattice property and the convexity of f yield
f(g(ax+by))≤f(ag(x)+bg(y))≤af(g(x))+bf(g(y))
so f ◦g is convex. (cid:3)
Definition 2.9. Let us define ⌈·⌉ : ℓ∞(Γ) → R by ⌈x⌉ = inf{t;{γ;|x(γ)|>t} is finite}. Then ⌈·⌉ is
1-Lipschitz, strongly lattice seminorm on (ℓ∞(Γ),k·k ).
∞
Ck-SMOOTH APPROXIMATIONS OF LUR NORMS 4
Proof. In fact ⌈x⌉ = kq(x)k , where q :ℓ∞(Γ) →ℓ∞(Γ)/c (Γ) is the quotient map and k·k the
ℓ∞/c0 0 ℓ∞/c0
canonical norm on the quotient ℓ∞(Γ)/c (Γ). Clearly, ⌈x⌉ = 0 if and only if x ∈ c (Γ). Let us assume
0 0
that ⌈x⌉ = t > 0. Then, for every 0 < s < t, there are infinitely many γ ∈ Γ such that |x(γ)| > s. It
follows that kx−yk >s for every y ∈c (Γ) and consequently kq(x)k ≥t. On the other hand, we
∞ 0 ℓ∞/c0
may define y ∈c (Γ) as
0
x(γ)−t if x(γ)>t,
y(γ):= x(γ)+t if x(γ)<−t,

0 otherwise.
Obviously kx−yk∞ ≤ t, so kq(x)kℓ∞/c0 ≤t. The strongly lattice property of ⌈·⌉ follows directly from
the definition. (cid:3)
Definition 2.10. Let (X,k·k) be a Banach space and let µ be the smallest ordinal such that |µ| =
dens(X). A system {P } of projections from X into X is called a projectional resolution of
α ω≤α≤µ
identity (PRI) provided that, for every α∈[ω,µ], the following conditions hold true
(a) kP k=1,
α
(b) P P =P P =P for ω ≤α≤β ≤µ,
α β β α α
(c) dens(P X)≤|α|,
α
(d) {P X :β <α} is norm-dense in P X,
β+1 α
(e) P =id .
µ X
S
If {P } is a PRI on a Banach space X, we use the following notation: Λ := {0} ∪ [ω,µ),
α ω≤α≤µ
Q :=P −P for all γ ∈[ω,µ) while Q :=P , and P := Q for any finite subset A of Λ.
γ γ+1 γ 0 ω A γ∈A γ
Lemma 2.11. Let X be a Banach space with a PRI {P } P. Then for each x∈X, ε>0, α∈[ω,µ]
α ω≤α≤µ
there is a finite set Aα(x)⊂Λ such that
ε
P x−P x <ε.
Aα(x) α
ε
We may choose A=Aα(x) in such a way(cid:13)that Q x6=0 f(cid:13)or β ∈A since P = Q .
ε (cid:13) β (cid:13) A γ∈A γ
Proof. We will proceed by a transfinite induction on α. If α = ω, then Aω(x)P:= {0} for any ε > 0. If
ε
α = β +1 for some ordinal β, then Aα(x) := Aβ(x)∪{β} for all ε > 0. Finally, if α is a limit ordinal,
ε ε
we will use the continuity of the mapping γ 7→ P x at α [2, Lemma VI.1.2] to find β < α such that
γ
kP x−P xk<ε/2. Thus it is possible to set Aα(x):=Aβ (x). (cid:3)
β α ε ε/2
3. Main Result
Letusrecallthatanormk·kinaBanachspaceX islocallyuniformlyrotund (LUR)iflim kx −xk=0
r r
whenever lim 2kx k2+2kxk2−kx +xk2 =0.
r r r
(cid:16) (cid:17)
Theorem 3.1. Let k ∈N∪{∞}. Let (X,|·|) be a Banach space with a PRI {P } such that each
α ω≤α≤µ
Q X admits aC1-smooth, LURequivalent normwhich isalimit (uniformonboundedsets)ofCk-smooth
γ
norms. Let X admit an equivalent Ck-smooth norm k·k.
Then X admits an equivalent C1-smooth, LUR norm k|·|k which is a limit (uniform on bounded sets)
of Ck-smooth norms.
Our first corollary provides a positive solution of Problem 8.2 (c) in [8].
Corollary 3.2. Let α be an ordinal. Then the space C([0,α]) admits an equivalent norm which is
C1-smooth, LUR and a limit of C∞-smooth norms.
Ck-SMOOTH APPROXIMATIONS OF LUR NORMS 5
Proof of Corollary 3.2. By a result of Talagrand [22] and Haydon [17], C([0,α]) admits an equivalent
C∞-smooth norm. On the other hand, the natural PRI on C([0,α]) defined as
x(β) if β ≤γ,
(P x)(β)=
γ
(x(γ) if β ≥γ
has (P −P )X one-dimensional for each γ ∈[ω ,α). (cid:3)
γ+1 γ 0
Theorem 3.3. Let k ∈N∪{∞}. Let P be a class of Banach spaces such that every X in P
• admits a PRI {P } such that (P −P )X ∈P,
α ω≤α≤µ α+1 α
• admits a Ck-smooth equivalent norm.
Then each X in P admits an equivalent, LUR, C1-smooth norm which is a limit (uniform on bounded
sets) of Ck-smooth norms.
Proof. WewillcarryoutinductiononthedensityofX. LetX ∈P beseparable,i.e. dens(X)=ω. Then
we get the result from the theorem of McLaughlin, Poliquin, Vanderwerff and Zizler [21] or [2, Theorem
V.1.7].
Next, we assume for X ∈P that dens(X)=µ and that every Banach space Y ∈P with dens(Y)<µ
admits a C1-smooth, LUR norm which is a limit of Ck-smooth norms. Let {P } be a PRI on
α ω≤α≤µ
X such that Q X ∈ P for each α ∈ Λ. Then dens(Q X) ≤ |α+1| = |α| < µ. Thus the inductive
α α
hypothesis enables us to use Theorem 3.1. (cid:3)
The abovetheorem has immediate corollariesfor eachP-class(see [16] for this notion). The following
Corollary 3.4 solves in the affirmative Problem 8.8 (s) in [8] (see also Problem VIII.4 in [2]).
Corollary 3.4. Let X admit a Ck-smooth norm for some k ∈ N∪∞. If X is Vaˇsa´k (i.e. WCD) or
WLD or C(K) where K is a Valdivia compact, then X admits a C1-smooth, LUR equivalent norm which
is a limit (uniform on bounded sets) of Ck-smooth norms.
Proof of Theorem 3.1. Let 0 < c < 1. It follows from the hypothesis that, for each γ ∈ Λ, there are a
C1-smooth, LUR norm k·kγ on QγX and Ck-smooth norms (k·kγ,i)i∈N on QγX such that
ckxk≤kxk ≤kxk (2)
γ
for all x∈Q X and such that (1− 1)kxk ≤kxk ≤kxk for all x∈Q X.
γ i2 γ γ,i γ γ
We seek the new norm on X in the form
k|x|k2 :=N(x)2+J(x)2+kxk2.
We will insure during the construction that both N and J are C1-smooth and approximated by Ck-
smooth norms. In order to see that k|·|k is LUR, we are going to show that kx−x k → 0 provided
r
that
2k|x |k2+2k|x|k2−k|x+x |k2 →0 as r →∞. (3)
r r
Consider the following two statements:
a) kP x −P xk→0 for each finite A⊂Λ with 0∈/ {Q x:γ ∈A},
A r A γ
b) for every ε>0 there exists a finite A⊂Λ with 0∈/ {Q x:γ ∈A} and such that kP x−xk<ε and
γ A
kP x −x k<ε for all but finitely many r∈N.
A r r
Clearly, the simultaneous validity of a) and b) implies that kx−x k→0 as
r
kx−x k≤kP x −P xk+kP x−xk+kP x −x k.
r A r A A A r r
We construct N in such a way that we can prove in Lemma 4.1 that (3) implies a). Consequently, we
construct J in such a way that we can prove in Lemma 5.8 that (3) implies b). (cid:3)
Ck-SMOOTH APPROXIMATIONS OF LUR NORMS 6
4. About N
We may and do assume that the equivalent norms |·| and k·k satisfy
|·|≤k·k≤C|·|
for some C ≥1.
ThebasicpropertiesofPRI[2,LemmaVI.1.2]andtheaboveequivalenceofnormsyield(kQ xk) ∈
γ γ∈Λ
c (Λ), and using kQ k ≤ 2C with the second inequality of (2), it follows that T : x ∈ (X,k·k) 7→
0 γ
(kQ xk ) ∈ (c (Λ),k·k ) is a 2C-Lipschitz mapping. Similarly for T : x ∈ X 7→ (kQ xk ) ∈
γ γ γ∈Λ 0 ∞ i γ γ,i γ∈Λ
c (Λ).
0
For each n∈N, we will consider an equivalent norm on c (Λ) given as
0
ζ (x):= sup x(γ)2
n
M∈Λnsγ∈M
X
where Λ := M ∈2Λ :|M|=n . It is easily seen that ζ is n-Lipschitz with respect to the usual norm
n n
on c (Λ). Also, ζ is obviously strongly lattice, so by Theorem 1 in [7], for each ε > 0 there is a C∞-
0 n
(cid:8) (cid:9)
smooth equivalent norm N on c (Λ) such that (1−ε)ζ (x) ≤ N (x) ≤ ζ (x) for all x ∈ c (Λ) with
n,ε 0 n n,ε n 0
kxk ≤1. Finally, we define
∞
1
N(x)2 := N2 (T(x))
2n+m n,m1
m,n∈N
X
Now the norm N(·) is C1-smooth since each Nn,1 ◦ T is 2nC-Lipschitz and C1-smooth. The latter
m
propertyfollowssinceeachNn,1 isnotonlyLFCbutitdependsonnonzerocoordinatesonly(cf. Remark
m
onpage 461in[17]). This fact is notexplicitely mentionedin[7]but follows fromthe proofthere (see [7,
p. 270]). We may define the approximating norms as
i
1
N (x)2 := N2 (T (x)).
i 2n+m n,m1 i
m,n=1
X
As a finite sum of Ck-smooth norms, N is Ck-smooth. Using kT (x)−T(x)k ≤ 2C for kxk ≤ 1, it is
i i ∞ i2
standard to check that N (x) → N(x) uniformly for kxk ≤ 1. We carry out some similar considerations
i
in more detail on page 14 when we demonstrate that J is approximated by Ck-smooth norms.
Lemma 4.1. Let us assume that (3) holds for x,x ∈X, r ∈N, and let A˜⊂Λ be a finite set such that
r
Q x6=0 for γ ∈A˜. Then kP x−P x k→0 as k →∞.
γ A˜ A˜ r
Proof. Let A := γ ∈Λ:kQ xk ≥min kQ xk . Let n := |A|. We may assume that k|x|k ≤ 1
γ γ α∈A˜ α α
whichimplieskTxnk∞ ≤2C. Using(3)andLemma2.5owemayassumethatk|xr|k≤2thuskTxk∞ ≤4C.
The convergence (3) and convexity (see Fact II.2.3 in [2]) imply that
2N2 (T(x ))+2N2 (T(x))−N2 (T(x+x ))→r 0
n,1 r n,1 n,1 r
m m m
for all m∈N. This further yields that
2ζ2(T(x ))+2ζ2(T(x))−ζ2(T(x+x ))→r 0
n r n n r
as well. Indeed, let ε>0 be given. We use that Nn,1 →ζn uniformly on bounded sets of c0(Λ) to find
m
m ∈N such that N2 (y)−ζ2(y) <ε/6 for all y ∈6CB and all m≥m . Now let r ∈N satisfy
0 n,1 n c0(Λ) 0 0
m
that for all r ≥r (cid:12)it holds 2N2 (T(cid:12) (x ))+2N2 (T(x))−N2 (T(x+x )) <ε/6. For each r ≥r
0(cid:12) n, 1 (cid:12) r n, 1 n, 1 r 0
(cid:12) m0 (cid:12) m0 m0
we obtain 2ζ2(T(x ))+2ζ2(T(x))−ζ2(T(x+x ))<ε.
n r n n r
Ck-SMOOTH APPROXIMATIONS OF LUR NORMS 7
Let B ∈Λ be arbitrary and let A ∈Λ such that
n r n
kQ (x+x )k2 =ζ (x+x ).
γ r γ n r
sγX∈Ar
Then
2ζ2(T(x ))+2ζ2(T(x))−ζ2(T(x+x ))≥2 kQ xk2 +2 kQ x k2 − kQ (x+x )k2
n r n n r γ γ γ r γ γ r γ
γX∈B γX∈Ar γX∈Ar
=2 kQ xk2 +2 kQ x k2 − kQ (x+x )k2
γ γ γ r γ γ r γ
(4)
γX∈Ar γX∈Ar γX∈Ar
+2 kQ xk2 − kQ xk2
 γ γ γ γ
γX∈B γX∈Ar
Since  
2 kQ xk2 +2 kQ x k2 − kQ (x+x )k2 ≥0
γ γ γ r γ γ r γ
γX∈Ar γX∈Ar γX∈Ar
we get from (4) that
liminf kQ xk2 ≥sup kQ xk2 :B ∈Λ =ζ (T(x))= kQ xk2 (5)
r γ γ  γ γ n n γ γ
γX∈Ar γX∈B  γX∈A
where the last equality follows from the definition of A. Equation (5) together with the definition of A
 
show that A = A for all r sufficiently large. We continue with such r and we choose B := A in (4) to
r
get that
2 kQ xk2 +2 kQ x k2 − kQ (x+x )k2 →r 0.
γ γ γ r γ γ r γ
γ∈A γ∈A γ∈A
X X X
Since x7→ kQ xk2 is an equivalent LUR norm on P X, it follows that kP (x−x )k→0 and,
γ∈A γ γ A A r
by continuiqtyPof PA˜, we obtain the claim of the lemma. (cid:3)
5. About J
Let {φ } be a system of functions satisfying
η 0<η<1
(i) φ : [0,+∞) → [0,+∞), for 0 < η < 1, is a convex C∞-smooth function such that φ is
η η
strictly convex on [1−η,+∞), φ ([0,1−η])={0} and φ (1)=1.
η η
(ii) If 0<η ≤η <1 then φ (x)≤φ (x) for any x∈[0,1].
1 2 η1 η2
One example of such a system can be constructed as follows: let φ : R → R be C∞-smooth such
that φ(x) = 0 if x ≤ 0, φ(1) = 1 and φ is increasing and strictly convex on [0,+∞). We define
φ (x) :=φ(x−(1−η)) for all x∈[0,1]. Now the system {φ } satisfies (ii) since η 7→ x−(1−η) is increasing
η η η η
for every x∈[0,1) while the validity of (i) follows from properties of φ.
We define a function Φ :ℓ∞(Γ)→(−∞,+∞] by
η
Φ (x)= φ (|x(γ)|).
η η
γ∈Γ
X
Let us define Z :ℓ∞(Γ)→R as the Minkowski functional of the set C ={x∈ℓ∞(Γ);Φ (x)<1/2}.
η η
Lemma 5.1. Let 0 < η < 1 be fixed. Then Z is a strongly lattice seminorm such that (1−η)Z (x) ≤
η η
kxk and Z is LFC, C∞-smooth and strictly positive in the set
∞ η
A (Γ):={x∈ℓ∞(Γ):⌈x⌉<(1−η)kxk }.
η ∞
Moreover (1−η)Z (x)<kxk for all x∈A (Γ).
η ∞ η
Ck-SMOOTH APPROXIMATIONS OF LUR NORMS 8
Proof. The set C is symmetric convex with zero as interior point (indeed, (1−η)Bℓ∞(Γ) ⊂ C) so Zη is
1 -Lipschitz and convex.
1−η
Let A′(Γ) := {x∈ℓ∞(Γ):⌈x⌉<1−η}. This set is convex and open since ⌈·⌉ is a continuous and
η
convex (seminorm). The function Φ is in A′(Γ) a locally finite sum of convex C∞-smooth functions,
η η
thus it is a convex function which is LCF and C∞-smooth in A′(Γ).
η
Let us fix x ∈ A′(Γ) such that Φ (x ) = 1/2. Then, since φ is increasing at the points where it is
0 η η 0 η
not zero, we get Z (x ) = 1 and Φ′(x )x > 0. As is usual, we consider the equation Φ x = 1.
η 0 η 0 0 η Zη(x) 2
By the Implicit Function Theorem, this equation locally redefines Zη and proves that Zη is(cid:16)C∞-s(cid:17)mooth
onsomeneighborhoodU ofx since Φ is. MoreoverbyapplicationofLemma2.4wegetthatZ isLFC
0 η η
at x .
0
To provethat Z is LFC,strictly positiveandC∞-smoothin A (Γ) itis enoughto showthatfor each
η η
x∈A (Γ) there is λ>0 suchthat λx∈A′(Γ) andΦ (λ·x)=1/2and then use the homogeneity of Z .
η η η η
Let x ∈ A (Γ). Then x < 1−η and since A′(Γ) is convex, it follows that [0, x ] ⊂ A′(Γ).
η kxk∞ η kxk∞ η
We have for such x that lΦη(kxxkm∞) ≥ 1, Φη(0·x) = 0 and the mapping λ 7→ Φη(λx) is continuous for
λ∈[0, 1 ]. Hence there must exist λ∈(0, 1 ) such that λx∈A′(Γ) and Φ (λ·x)=1/2.
kxk∞ kxk∞ η η
We continue showing that Z is strongly lattice. First observe that Φ is strongly lattice as φ is
η η η
nondecreasing. Let|x|≤|y|andZ (x)=1. Thenx∈∂C whichimpliesthat⌈x⌉=1−η orΦ (x)=1/2.
η η
Since both functions ⌈·⌉ and Φ are strongly lattice, we conclude that ⌈y⌉≥1−η or Φ (y)≥1/2 which
η η
in turn implies that Z (y) ≥ 1. For a general x we employ the homogeneity of Z , so Z is strongly
η η η
lattice.
Finally, if x ∈ A (Γ), then the above considerations imply that Φ x = 1/2. This is possible
η η Zη(x)
only if there is some γ ∈Γ such that x(γ) >1−η, and the moreover c(cid:16)laim fo(cid:17)llows. (cid:3)
Zη(x)
Lemma 5.2. Let 0<η ≤η <1. Then Z (x)≤Z (x) for every x∈A (Γ).
1 2 η1 η2 η2
Proof. First of all, if x ∈ A (Γ), then x ∈ A (Γ). So the equivalence Z (λx) = 1 ⇔ Φ (λx) = 1/2
η2 η1 ηi ηi
holds for both i=1,2. Let us assume that Z (λx) =1 for some λ>0. Then the ordering of functions
η1
φ yields 1/2=Φ (λx)≤Φ (λx) which results in Z (λx)≥1. (cid:3)
η η1 η2 η2
Lemma 5.3. Let 0<η <1 be given and let x ,x∈A (Γ) (r ∈N) be non-negative (in the lattice ℓ∞(Γ))
r η
such that
2Z2(x)+2Z2(x )−Z2(x+x )→0 as r →∞.
η η r η r
Then x (γ)→x(γ) for any γ ∈Γ such that x(γ)>Z (x)(1−η).
r η
Proof. The assumption and Lemma 2.5 yield
x+x
r
Z (x )→Z (x) and Z →Z (x). (6)
η r η η η
2
(cid:18) (cid:19)
Let us put x˜:= x and x˜ := xr . We get from (6) that
Zη(x) r Zη(xr)
2Z2(x˜)+2Z2(x˜ )−Z2(x˜+x˜ )→0.
η η r η r
Since Z (x˜)=Z (x˜ )=1, the above implies that
η η r
λ :=Z (x˜+x˜ )→2.
r η r
Wemaydeducefromx,x ∈A (Γ)thatΦ (x˜)=1/2=Φ (x˜ )forallk ∈N. Also,Φ (λ−1(x˜+x˜ ))=1/2
r η η η r η r r
for all but finitely many k ∈ N. Indeed, if Φ (λ−1(x˜+x˜ )) 6= 1/2, then λ−1(x˜+x˜ ) ∈ ∂A′(Γ). Then
η r r r r η
in fact λ−1(x˜+x˜ ) = 1−η. As x˜ ∈ A′(Γ), there is ξ > 0 such that ⌈x˜⌉+ξ < 1−η. By the same
r r η
reasoning ⌈x˜ ⌉<1−η. By the convexity (subaditivity) of ⌈·⌉ and these estimates one has
(cid:6) r (cid:7)
⌈x˜+x˜ ⌉≤⌈x˜⌉+⌈x˜ ⌉<2(1−η)−ξ.
r r
Ck-SMOOTH APPROXIMATIONS OF LUR NORMS 9
Finally, λ < 2(1−η)−ξ which can happen only for finitely many r as λ →2.
r 1−η r
As Φ is continuous at x˜ and λ →2, it follows
η r
Φ ((λ −1)−1x˜)→1/2.
η r
Consequently
(1−λ−1)Φ (λ −1)−1x˜ +λ−1Φ (x˜ )−Φ λ−1(x˜+x˜ ) →0. (7)
r η r r η r η r r
Let a > 1−η. The definition of φ and a compactness argument imply that for each ε > 0 there exists
(cid:0) η (cid:1) (cid:0) (cid:1)
∆>0 such that if for reals r,s,α it holds
• 0≤r ≤4max{kx˜k ,sup kx˜ k },
∞ r r ∞
• a≤s≤4max{kx˜k ,sup kx˜ k },
∞ r r ∞
• 1 ≤α≤ 3, and
4 4
• αφ (r)+(1−α)φ (s)−φ (αr+(1−α)s)<∆,
η η η
then |r−s|<ε.
Inparticular,leta>1−ηbesuchthat{γ ∈Γ;x˜(γ)>1−η}={γ ∈Γ;x˜(γ)>a}andletγ ∈Γbesuch
that x˜(γ) > a. Then for r large enough we have (λ −1)−1x˜(γ) > a so we may substitute r := x˜ (γ),
r r
s := (λ −1)−1x˜(γ) and α := λ−1. It follows from (7) that one has (λ −1)−1x˜(γ)−x˜ (γ) → 0 as
r r r r
k →∞. Since λ →2 and using (7), we finally get that x (γ)→x(γ) as k →∞. (cid:3)
r r (cid:12) (cid:12)
(cid:12) (cid:12)
The following system of convex functions is at the heart of our construction. We recall that C ≥1 is
the constant of equivalence between the norms |·| and k·k, which was introduced in Section 4.
Lemma 5.4. There exist
• a decreasing sequence of positive numbers δ ց0; δ <2C;
n 1
• a decreasing sequence of positive numbers ρ ց0;
n
• positive numbers κ > 0 such that for each n ∈ N the sequence (κ ) is decreasing and
n,m n,m m
m
κ →0; for each n,m∈N one has ρ >2κ ;
n,m n n,m
• an equi-Lipschitz system of non-negative, C∞-smooth, 1-bounded, convex functions
{g :D →R:n,m∈N,l=1,...,n},
n,m,l n,l
where D :=[0,2nC−δ (n−l)]×[0,1+2nC], satisfying (with n,m,l ∈N, l ≤n, resp. l <n
n,l n
in (A2),(A5))
(A1) g (t,s)=0 iff (t,s)∈[0,lδ ]×[0,1+2nC]=:N ;
n,m,l n n,l
(A2) g (t,s)≥g (t,s)+ρ whenever (t,s)∈D \N ;
n,m,l n,m,l+1 n n,l n,l+1
(A3) if (t,0)∈D \N , then s7→g (t,s) is increasing on [0,1+2nC] and
n,l n,l n,m,l
g (t,1+2nC)−g (t,0)≤κ ;
n,m,l n,m,l n,m
(A4) if (t,0)∈D , then g (t,0)=g (t,0);
n,l n,m,l n,m+1,l
(A5) for all (t,s)∈D \N it holds g (t,s)<g (t+r,s) provided r >δ .
n,l n,l n,m,l n,m,l+1 n
(A6) Let (t,s)∈D \N . If (t ,s )∈D and t →t and g (t ,s )→g (t,s) as
n,l n,l r r n,l r n,m,l r r n,m,l
r →∞, then s →s.
r
(A7) The mapping (t,s)7→g (|t|,|s|) is strongly lattice in D .
n,m,l n,l
Proof. Let f :R→[0,+∞) be defined as
0, for t≤0,
f(t):=
(exp(−t12) for t>0.
Itiselementary(onemayuseLemma2.6)tocheckthatf(t)·(s2+s+1)isconvexinthestrip(−∞,10−1]×
[0,10−1] so the function
g(t,s):=f(10−1t)·((10−1s)2+10−1s+1)
Ck-SMOOTH APPROXIMATIONS OF LUR NORMS 10
is convex in the strip (−∞,1]×[0,1]. We take for (δ ) just any decreasing null sequence of positive
n n
numbers such that δ <2C, and we define
1
t−δ l s
n
g (t,s):=g ,θ .
n,m,l n,m
(2C−δ )n 1+2nC
(cid:18) n (cid:19)
whereθ ∈(0,1)willbechosenlater. Nowsinceourfunctionsg arejustshiftsandstretchesofone
n,m n,m,l
non-negative, C∞-smooth, 1-bounded, Lipschitz, convex function, it follows that all g share these
n,m,l
properties (with the same Lipschitz constant).
Properties (A1), (A4) and (A5) are straightforward, see also Figure 5. Notice that, when t > 0, the
κ
n,m
1+2nC
ρ
n
0 δnl δn(l+1) 2nC−(n−l)δn
function s 7→ g(t,s) is increasing on [0,1]. This implies the first part of (A3). In order to satisfy (A2),
we may define ρ as
n
ρ :=inf{g (t,s)−g (t,s):l,m∈N,l<n,(t,s)∈D \N }
n n,m,l n,m,l+1 n,l n,l+1
whichevaluatesasρ =g (2δ ,0)=f δn ց0asn→∞. Noticethatthisρ doesnotdepend
n n,1,1 n (2−δn)n n
on the choice of θn,m. On the other hand,(cid:16)in order(cid:17)to fulfill (A3), κn,m may be defined as
κ :=sup{g (t,1+2nC)−g (t,0):l≤n,(t,0)∈D }
n,m n,m,l n,m,l n,l
whichevaluatesasκ =g (2nC,1+nC)−g (2nC,0). Weseethat,byanappropriatechoiceof
n,m n,m,n n,m,n
θ (in particular, for each n∈N, the sequence (θ ) should be decreasing to zero), one may satisfy
n,m n,m m
the requirements ρ >2κ and κ ց0 as m→∞.
n n,m n,m
For the proof of (A6) let us assume that s 9s. The fact that g (t ,·)→g (t,·) uniformly on
r n,m,l r n,m,l
[0,1+2nC] leads quickly to a contradiction.
Finally (A7) follows since g is non-decreasing in D in each variable. (cid:3)
n,l
Letusfix,foreachδ >0,someC∞-smooth,convexmappingξ from[0,+∞)to[0,+∞)whichsatisfies
δ
ξ ([0,δ])={0}, ξ (t)>0 for t>δ and ξ (t)=t−2δ for t≥3δ. Such a mapping can be constructede.g.
δ δ δ
by integrating twice a C∞-smooth, non-negative bump.
Lemma 5.5. Let n,m ∈ N be fixed and let us define a mapping H : BO → ℓ∞(F ) where
n,m (X,k·k) n
F = (A,B)∈2Λ×2Λ :|A|≤n,B ⊂A,A6=∅6=B by
n
(cid:8) (cid:9)
H x(A,B):=g ξ (kQ xk ),ξ (kP x−xk) .
n,m n,m,|A| δn γ γ δn B 
γ∈A
X
 