a Xi :ma h/9702212 1 [ma h.FA] 13 Feb 1997
APPROXIMATION OF LIPSCHITZ FUNCTIONS BY ∆-CONVEX
FUNCTIONS IN BANACH SPACES
Manuel Cepedello Boiso
Feb ua y 13, 1997
Abs ac . In his pape we gi e some esul s abou he app oxima ion o a Lipschi z
unc ion on a Banach space by means o ∆-con ex unc ions. In pa icula , we
p o e ha he densi y o ∆-con ex unc ions in he se o Lipschi z unc ions o he
opology o uni o m con e gence on bounded se s cha ac e izes he supe e lexi i y
o he Banach space. We also show ha Lipschi z unc ions on supe e lexi e Banach
spaces a e uni o m limi s on he whole space o ∆-con ex unc ions.
0. In oduc ion and No a ions
A unc ion de ined on a Banach space Xis called ∆-con ex i i can be exp essed
as a di e ence o con inuous con ex unc ions o , equi alen ly, i i belongs o he
linea span o he con inuous con ex unc ions on X. The pu pose o his pape is
o gi e some necessa y and su icien condi ions o he app oxima ion o Lipschi z
unc ions by ∆-con ex unc ions. The ini ial mo i a ion o ou wo k comes om
wo ecen a icles o R. De ille, V. Fon and P. H´ajek ([DFH1] and [DFH2]). A
consequence o hei esul s is ha , unde ce ain condi ions on he Banach space X,
any con ex unc ion on Xwhich is bounded on bounded se s can be app oxima ed
by smoo h con ex unc ions. I is he e o e na u al o conside he class o Banach
spaces o which he ∆-con ex unc ions a e dense in he class o Lipschi z unc ions
in o de o ex end his p ope y o smoo h app oxima ions.
Ou main esul is he ollowing cha ac e iza ion o supe e lexi i y.
Theo em 0. Le Xbe a Banach space. Then Xis supe e lexi e i and only i
e e y Lipschi z unc ion on Xcan be app oxima ed uni o mly on bounded se s by
di e ences o con ex unc ions on Xwhich a e bounded on bounded se s.
A consequence o Theo em 0 is ha he abo e men ioned app oach does no
p o ide any new esul , because i wo ks only o supe e lexi e spaces ( o which
he smoo h app oxima ion p ope y is known using he exis ence o pa i ions o
uni y, see Ch. VIII o [DGZ]; o he analy ic app oxima ion case, see [K]).
Fo he supe e lexi e case, we gi e explici o mulas o he ∆-con ex app oxi-
ma ion o a Lipschi z unc ion. These o mulas, which a e simple han hose om
1991 Ma hema ics Subjec Classi ica ion. P ima y 46B20; Seconda y 46B10.
Key wo ds and ph ases. Con ex unc ions, supe e lexi i y in Banach spaces.
The au ho was suppo ed by a FPU G an o he Spanish Minis e io de Educaci´on y Ciencia.
Typese by A
M
S-T
EX
[S ], also p o ide uni o m con e gence on he whole space X. Speci ically, he de-
g ee o con e gence gi en by ou o mulas elies di ec ly on he o undi y o he
equi alen no m ha i is used. Simila ideas in his di ec ion can be ound in [A]
and [PVZ].
We hank P. H´ajek o b inging o ou a en ion he link be ween ou wo k and
he dis o ion heo em (see [OS] o de ini ions and de ails). Ou esul s p o ide
simple o mulas o deducing, on minimal supe e lexi e Banach spaces (such as ℓp,
1< p < ∞), he exis ence o a con ex unc ion which is no oscilla ion s able om
he exis ence o a Lipschi z unc ion which is no oscilla ion s able.
Le us ix some no a ion used in his pape . Fo a eal Banach space X, we
deno e an equi alen no m on Xby k · k and by BXi s closed uni ball unde his
no m. By con ex unc ion we will always mean con inuous con ex unc ion. We
will conside wo undamen al opologies on he se o con inuous unc ions de ined
on X:τκ( espec i ely τb) is he opology o uni o m con e gence on compac se s
o X( esp. uni o m con e gence on bounded se s o X).
The modulus o con exi y o he no m k · k de ined by
δk·k(ε) = in 1−
x+y
2
:x, y ∈BX;kx−yk ≥ ε(0 < ε < 2)
is called o powe ype p(p≥2) i δk·k(ε)≥Kεp, o some K > 0.
The concep o dyadic ee will play an impo an ole in he second pa o his
wo k. Ou ees a e geome ic ees con ained in Xand de ined as ollows. The
symbol αdeno e a mul i-index α= (α1
a
α2
a...a
αn)∈ {−1,1}<Nand |α|:= n. Fo
n∈N, a dyadic (n, θ)- ee Tin Xis a se o he o m xα∈X:α∈ {−1,1}<n
sa is ying he ollowing wo condi ions:
(1) xα=1
2xα
a
1+1
2xα
a
-1, o all |α|< n.
(2) kxα−xα′k ≥ θ > 0, o α6=α′.
The poin x∅will be called he oo o he ee T.
1. The posi i e esul s
Theo em 1.
Le (X, k · k)be a Banach space. The no m k · k is locally uni o mly con ex
( espec i ely uni o mly con ex) i and only i he ollowing p ope y holds: o e e y
Lipschi z unc ion on X, he sequence o unc ions ( n)n∈Nde ined by he o mula
n(x) := in
y∈Xn (y) + n2kxk2+ 2kyk2− kx+yk2o(n∈N, x ∈X)
is τκ-con e ging ( esp. τb-con e ging) o .
Rema k. Fo any unc ion on Xand n∈N, he unc ion nde ined as abo e is a
∆-con ex unc ion. This ollows immedia ely om he decomposi ion n=cn−dn,
wi h cn(x) := 2nkxk2and
dn(x) := sup
y∈Xnkx+yk2−2nkyk2− (y)(x∈X).
The unc ions cnand dna e clea ly con ex.
2
P oo o he Theo em 1.
Le us see i s ha he p ope y is necessa y. So, le be a Lipschi z unc ion
on X. We ha e o show ha he p e iously de ined sequence ( n)n∈N(τKo τb)-
con e ges o i he no m k · k sa is ies he co esponding o undi y condi ion.
We begin wi h he ollowing gene al esul :
Fac . Fo any poin x∈X, ( n(x))n∈Nis an inc easing sequence bounded abo e
by (x).
This Fac ollows immedia ely om aking y=xin he n’s in imum o mula
and om he inequali y
2kxk2+2kyk2−kx+yk2≥2kxk2+2kyk2−kxk+kyk2=kxk − kyk2≥0.(1)
Since o K∈Nwe ha e ha (K·n)=K
Kn, he p e ious Fac allows us o
suppose wi hou loss o gene ali y ha he Lipschi z cons an o is less han 1
(i.e. | (x)− (y)| ≤ kx−yk o all x, y ∈X).
We need o s udy o he in imum o mula de ining na a poin x∈X. Thus,
conside any poin y o which we ha e
(y) + n2kxk2+ 2kyk2− kx+yk2≤ (x).(2)
As is 1-Lipschi z, we deduce om (1) and (2) ha
nkxk − kyk2≤n2kxk2+ 2kyk2− kx+yk2≤ (x)− (y)≤ kx−yk.(3)
This las condi ion gi es a ela ion be ween he no ms o xand y. Ac ually, kyk
can be con olled by kxkin he ollowing way. Suppose ha kyk ≥ 1 + kxk, hen
by (3) we ge ha
1≤kxk − kyk≤kxk − kyk2≤1
nkx−yk ≤ 1
nkxk+1
nkyk.(4)
And we conclude o n≥3 ha kyk ≤ n+1
n−1kxk ≤ 2kxk.The e o e, we see ha i y
sa is ies (2) and n≥3 hen
kyk ≤ 2(1 + kxk).(5)
Then, o n≥3 we deduce ha
n(x) = in
kyk≤2(1+kxk) (y) + 2nkxk2+ 2nkyk2−nkx+yk2.(6)
In pa icula , he boundedness on bounded se s o n(and, consequen ly, o dn=
cn− n) ollows immedia ely.
Mo eo e , he uppe bound on kykgi en by (5) oge he wi h he condi ion (3)
gi es
0≤2kxk2+ 2kyk2− kx+yk2≤1
nkx−yk ≤ 1
nkxk+1
nkyk ≤ 3
n(1 + kxk).(7)
The condi ion (7) gi es he c ucial s ep o he p oo . In ac , i he no m k · k
e i ies one o he o undi y p ope ies s a ed in heo em hen by (7) ncan be
3
chosen big enough o necessa ily en o ce y o be close o x. The e o e, n(x) mus
be close o (x). Le us jus i y his asse ion.
Suppose ha he sequence o unc ions nis no compac ly con e ging o .
Since he sequence o ∆-con ex unc ions ( n)nis inc easing and is con inuous,
Dini’s heo em ells us ha τκ-con e gence o n o is equi alen o he poin wise
con e gence. Then, he e exis s a poin x0∈Xsuch ha ( n(x0))ndoes no
con e ge o (x0).
As ( n(x0))nis inc easing, he e exis s some ε0>0 such ha o any n∈Nwe
ha e n(x0) + ε0< (x0). By de ini ion o n, we can ind a sequence o (yn)nin
Xso ha
n(x0) + ε0≤ (yn) + 2nkx0k2+ 2nkynk2−nkx0+ynk2+ε0≤ (x0).(8)
Since is 1-Lipschi z, we ge om (8) ha
kx0−ynk ≥ (x0)− (yn)≥ε0>0.(9)
Since yn e i ies he condi ion (8), i ollows om (7) ha we ha e
0≤2kx0k2+ 2kynk2− kx0+ynk2≤3
n(1 + kx0k)−−−→
n→∞ 0.(10)
Bu (9) and (10) show ha he no m k · k can no be locally uni o mly con ex a
x0(c .Ch. II. P op. 1.2 o [DGZ]). Tha p o es he compac con e gence o n o
i he no m k · k is locally uni o mly con ex.
A simple p oo o he uni o m con e gence o he sequence nin he uni o mly
con ex case ollows he same lines. We will gi e la e a di ec quan i a i e app oach
(see he p oo o Theo em 3). I he sequence ndoes no con e ge uni o mly on
a bounded se o X, he e is an ε0>0 and a bounded sequence (xn)nso ha
n(xn) + ε0< (xn), n∈N. Then, o each xn(n∈N) we can choose yn e i ying
(yn) + 2nkxnk2+ 2nkynk2−nkxn+ynk2+ε0≤ (xn).(11)
Simila easonings as be o e imply om (11) ha he nex wo s a emen s a e
ul iled:
kxn−ynk ≥ (xn)− (yn)≥ε0>0 (12)
0≤2kxnk2+ 2kynk2− kxn+ynk2≤3
n(1 + kxnk)−−−→
n→∞ 0 (13)
We ha e limn→∞ 3
n(1+kxnk) = 0 since he sequence (xn)nis bounded. Mo eo e ,
(5) implies ha he sequence (yn) is also bounded. The e o e, (12) and (13) show
ha he no m k · k canno be uni o mly con ex (c . Ch. IV Lemma 1.5 [DGZ]).
The necessi y o he p ope y is p o ed.
Con e sely, he p ope y s a ed in he heo em is su icien . We i s show ha
he p ope y o compac con e gence implies he ollowing claim.
4
Claim 1.1. Fo any sequence (xn)n∈N⊂Xand any poin x0∈X, he condi ion
limn→∞ 2kx0k2+ 2kxnk2− kx0+xnk2= 0 implies dx0,(xn)n= 0.
Since he Claim 1.1 is also alid o any subsequence o he gi en sequence
(xn)n, we can s eng hen he conclusion o he claim o limn→∞ kxn−x0k= 0 and
he locally con exi y o he no m k · k is e i ied (c .Ch. II. P op. 1.2 o [DGZ]).
P oo o he Claim 1.1. Conside he Lipschi z unc ion g(x) = dx, (xn)n,x∈X.
Fo x0∈Xand ε > 0, he poin wise con e gence o he sequence (gn)n o ga x0
ells us ha he e exis s Nε∈Nso ha
dx0,(xn)n=g(x0)≤in
y∈Xng(y) + Nε2kx0k2+ 2kyk2− kx0+yk2o+ε. (14)
Using g(xn) = 0, we can e alua e he inequali y (14) a y=xn(n≥Nε) and
deduce ha
dx0,(xn)n≤Nεlim
n→∞ 2kx0k2+ 2kxnk2− kx0+xnk2+ε=ε.
The e o e, dx0,(xn)n= 0 and he claim is p o ed.
I he p ope y o uni o m con e gence on bounded se s o Xholds, he nex
claim, analogous o he p e ious one, also does.
Claim 1.2. Le (xn)n∈Nand (yn)n∈Nbe wo bounded sequences in Xsuch ha
limn→∞ 2kxnk2+ 2kynk2− kxn+ynk2= 0. Then limm→∞ dym,(xn)n= 0.
P oo o he Claim 1.2. As be o e, conside he Lipschi z unc ion d·,(xn)nand
ε > 0. This ime, he p ope y o uni o m con e gence gi es a posi i e in ege Nε
so ha o all m∈N he nex inequali y is sa is ied.
dym,(xn)n≤in
z∈Xndz, (xn)n+Nε2kzk2+ 2kymk2− kz+ymk2o+ε. (15)
Taking z=xmin he in imum o (15) we ob ain
dym,(xn)n≤Nε2kxmk2+ 2kymk2− kxm+ymk2+ε≤2ε,
o mla ge enough, and he claim is p o ed.
In o de o inish wi h he p oo , we shall show ha he alidi y o Claim 1.2
implies ha he no m k · k is uni o mly con ex. I no , (c . Ch. IV Lemma 1.5
[DGZ]) he e exis s wo bounded sequences (xn)n∈Nand (yn)n∈Nin Xsa is ying
ha
lim
n→∞ 2kxnk2+2kynk2−kxn+ynk2= 0 and kxn−ynk ≥ 1 ( o all n∈N). (16)
Fi s , he no m k · k is locally uni o mly con ex (since Claim 1.1 clea ly holds).
I ollows ha he sequence (xn)nhas no no m clus e poin . Indeed, i o some
x0∈X he e exis s (xnk)k−→
kx0, hen
lim
k→∞ 2kx0k2+ 2kynkk2− kx0+ynkk2= lim
k→∞ 2kx0k2+ 2kxnkk2− kx0+xnkk2= 0
and he e o e (ynk)k−→
kx0. A con adic ion wi h he ac kxn−ynk ≥ 1, o all
n∈N, o (16).
Hence, passing o a subsequence, we can suppose ha o some 1 > α > 0 we
ha e ha kxn−xmk ≥ α, o all n6=m. Now, we need he ollowing echnical
lemma, whose p oo will be gi en la e .
5
Lemma 1.3. Le be (λn)n∈N⊂[0,1] and (xn)n∈N,(yn)n∈N wo bounded sequences
in Xso ha 2kxnk2+ 2kynk2− kxn+ynk2−−−→
n→∞ 0. Then he sequence o con ex
combina ions zn=λnxn+ (1 −λn)yn(n∈N) sa is ies ha 2kxnk2+ 2kznk2−
kxn+znk2−−−→
n→∞ 0.
Fo any n∈N, ake 0 ≤λn≤1 such ha o zn:= λnxn+ (1 −λn)ynwe
ha e kxn−znk=α
2. Then, he Lemma 1.3 and he Claim 1.2 used join ly
imply ha limm→∞ d(zm,(xn)) = 0. Bu , kxn−znk=α
2>0 and also o n6=m
kxn−zmk ≥ kxn−xmk − kxm−zmk ≥ α−α
2=α
2>0, a con adic ion.
P oo o he Lemma 1.3. By he inequali y
0≤kxnk − kynk2≤2kxnk2+ 2kynk2− kxn+ynk2−−−→
n→∞ 0,(17)
we ha e ha
lim
n→∞ kxnk − kynk= 0.
Since he sequences (xn)nand (yn)na e bounded we also ha e ha
lim
n→∞ kxnk2− kynk2= 0.(18)
Bu hen we deduce om (17) and (18) ha
lim
n→∞ kxnk2−
xn+yn
2
2= lim
n→∞ kynk2−
xn+yn
2
2= 0.(19)
On he o he hand, using he con exi y o he unc ion k · k2we ge he ollowing
gene al lowe es ima e o 0 ≤λ≤1
kλx+(1−λ)yk2≥
x+y
2
2
−1−2λmax (
x+y
2
2
−kxk2
,
x+y
2
2
−kyk2).
Pu ing (17),(18),(19) and he las inequali y oge he , we ob ain ha
0≤2kxnk2+ 2kznk2− kxn+znk2
= 2kxnk2+ 2kλnxn+ (1 −λn)ynk2− k(1 + λn)xn+ (1 −λn)ynk2
≤2kxnk2+ 2λnkxnk2+ 2(1 −λn)kynk2−4
1 + λn
2xn+1−λn
2yn
2
≤2kxnk2+ 2kynk2− kxn+ynk2+λnkxnk2− kynk2
+ 4λnmax (
xn+yn
2
2
− kxnk2
,
xn+yn
2
2
− kynk2)−−−→
n→∞ 0.
The lemma is p o ed and his concludes he p oo o Theo em 1.
Rema k. The in imum o mula used o he de ini ion o ( n) in he Theo em 1 is
closely ela ed o he well-known in -con olu ion o mula o by nk · k2:
(nk · k2)(x) := in
y∈X (y) + nkx−yk2(n∈N,x∈X).(20)
6
In ac , hese wo in imum o mulas a e iden ical i he no m k · k is a Hilbe ian
no m, because o he pa allelog am iden i y. Howe e , o a non-Hilbe ian no m
k · k he unc ions gi en by he in -con olu ion o mula can no be exp essed in
gene al as ∆-con ex unc ions.
As a co olla y o he abo e ema k, we ha e ha he o mula o Theo em 1
con e ges uni o mly on X o case o a Hilbe ian no m k · k (since i is well-known
ha he in -con olu ion o mula o (20) con e ges uni o mly on he whole space X,
see [LL]). The ques ion ha na u ally a ises is whe he his emains ue o no o
a gene al uni o mly con ex no m. The answe is gi en in he ollowing p oposi ion.
P oposi ion 2. Le (X, k · k)be a Banach space. I o e e y Lipschi z unc ion
on X he sequence o unc ions
n(x) := in
y∈Xn (y) + n2kxk2+ 2kyk2− kx+yk2o
con e ges o uni o mly on X, hen he modulus o con exi y o he no m k · k is
o powe ype 2.
P oo o he P oposi ion 2. This uni o m con e gence p ope y has he ollowing
consequence analogous o Claim 1.1 and Claim 1.2 and whose p oo is iden ical
o ha o Claim 1.2.
Claim 2.1. I (xn)n∈Nand (yn)n∈Na e wo sequences (no necessa ily bounded)
in Xsa is ying ha limn→∞ 2kxnk2+ 2kynk2− kxn+ynk2= 0, hen hey also
e i y ha he limm→∞ dym,(xn)n= 0.
By Claim 1.2, any no m which sa is ies he conclusion o Claim 2.1 is uni o mly
con ex. In ac , Claim 2.1 insu es ha he modulus o con exi y o he no m k · k
is o powe ype 2. I no (see [H]), he e exis s wo sequences (xn)n∈Nand (yn)n∈N
such ha
2kxnk2+ 2kynk2<kxn+ynk2+1
nkxn−ynk2xn6=yn,n∈N.(21)
I we ake un:= xn
kxn−ynkand n:= yn
kxn−ynkin (21) we ob ain ha
0≤2kunk2+ 2k nk2− kun+ nk2<1
n−−−→
n→∞ 0,(22)
kun− nk= 1.(23)
Since he no m k·k is uni o mly con ex and condi ions (22) and (23) holds, he
sequences (un)nand ( n)ncan no be bounded. No ice ha we ha e also om (22)
ha limn→∞ kunk − k nk= 0. Thus, passing o a subsequence we can suppose
ha
sup{kunk,k nk} + 1 <min{kun+1k,k n+1k} o all n∈N
=⇒ kun− mk ≥ kun|| − k mk≥1 (n6=m). (24)
Assembling (23) and (24) we deduce ha d m,(un)n= 1 (m∈N). Bu Claim
2.1 and (22) imply ha limmd m,(un)n= 0 which is a con adic ion.
Theo em 3 below p o ides a con e se o P oposi ion 2 and an explici o mula
o uni o m app oxima ion o Lipschi z unc ions by ∆-con ex unc ions on supe -
e lexi e Banach spaces. The alidi y o his o mula depends upon he exis ence
in he supe e lexi e space o an equi alen no m which is enough o und.
7
Theo em 3. Le (X, k · k)be a Banach space whose no m k · k has i s modulus o
con exi y o powe ype p(p≥2). Then o e e y Lipschi z unc ion on X he
ollowing sequence o ∆-con ex unc ions
p
n(x) = in
y∈Xn (y) + n2p−1kxkp+ 2p−1kykp− kx+ykpo(n∈N, x ∈X)
con e ges o uni o mly on X.
P oo o he Theo em 3. We essen ially need he ollowing lemma o p o e he
esul .
Lemma 3.1. I he modulus o con exi y o he no m k·k is o powe ype p(p≥2)
hen he e exis s a posi i e cons an Ck·k ≤1such ha o e e y pai x, y ∈X he
ollowing inequali y holds:
Ck·kkx−ykp≤2p−1kxkp+ 2p−1kykp− kx+ykp.
P oo o he Lemma 3.1. Unde he assump ion o he lemma (see [H]), he e exis s
a posi i e cons an C′
k·k ≤2 such ha o e e y pai u, ∈Xwe ha e ha
ku+ kp+ku− kp≥2kukp+C′
k·kk kp.(25)
The lemma ollows om he change o a iables x=u+
2and y=u−
2in (25).
Gi en a Lipschi z unc ion on X, we use he p e ious Lemma 3.1 and he
ac ha p
n(x)≤ (x) ( o all x∈Xand n∈N) o ob ain he ollowing chain o
inequali ies:
(nCk·kk · kp)≤ p
n≤ (n∈N).(26)
Since p≥2, he p e ious in -con olu ion o mula in (26) con e ges o uni o mly
on X(see [LL]). The e o e, he same is ue o ( p
n)n. Fo ins ance, i is a
1-Lipschi z unc ion we ge he ollowing es ima e
k − p
nk∞≤ k − (nCk·kk · kp)k∞≤1
nCk·k 1
p−1
.
The i s consequence ha s ems om he Theo em 1 is ano he p oo o a
esul due o G.A. Edga ([E]).
Co olla y 4. Le XBanach space wi h an equi alen locally uni o mly con ex
no m. Then he σ- ields o Bo el se s o he no m and weak opologies a e he
same.
P oo o he Co olla y 4. The i s pa o he Theo em 1 a i ms ha he exis ence
o a locally uni o mly con ex no m on Ximplies he ollowing:
∀ :X→RLipschi z ∃(cn)n,(dn)n⊂Con (X) such ha =τκ- lim
n→∞(cn−dn).
Le us check ha his p ope y implies he equi alence o he wo Bo el amilies,
Bo (X, k · k) and Bo (X, w). Ob iously, Bo (X, w)⊆Bo (X, k · k). To see he
o he inclusion, ake Fak · k-closed se o X. Then conside he Lipschi z unc ion
(·) = dis (·, F). No e ha he p e ious p ope y implies ha is he poin wise
limi o a sequence o w-Bo el unc ions (because e e y con inuous con ex unc ion
is w-lowe -semicon inuous). The e o e, is w-Bo el and so Fis w-Bo el.
Now, we p oceed o s a e wo applica ions o Theo em 1 and Theo em 3
o he s udy o supe e lexi e Banach spaces. Ac ually, we will show in he nex
sec ion ha bo h o hem cha ac e ize supe e lexi i y.
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Co olla y 5. Le Xbe a supe e lexi e Banach space. Deno e by Con (X) he se
o con inuous con ex unc ions on X,Con b(X) he subse o Con (X)consis ing o
he unc ions which a e bounded on bounded se s and UCb(X) he class o unc ions
on Xwhich a e uni o mly con inuous on bounded se s o X. Then
spanτbCon b(X)=UCb(X).
P oo o he Co olla y 5. F om he P op. 1.6 o [Ph] ollows immedia ely ha
Con b(X) = Con (X)∩ UCb(X)⊆ UCb(X). Le us show he τb-densi y o he
∆-con ex unc ions in he se UCb(X).
No ice ha he se o Lipschi z unc ions on Xis dense in UCb(X) unde he
opology τb. This ac can be p o ed using again he in -con olu ion o mula. Mo e
p ecisely, gi en ∈ UCb(X) and bounded on X, (nk · k)(x) is a sequence
o Lipschi z unc ions τb-con e ging o . Since he bounded unc ions o UCb(X)
a e clea ly τb-dense in UCb(X), he densi y o he Lipschi z unc ions is he e o e
deduced.
Thus, he co olla y holds i we show ha he con ex unc ions {cn, dn}n∈Nob-
ained in he p oo o Theo em 1 a e in Con b(X). Clea ly, cn(·) = 2nk · k2∈
Con b(X) and, as we ema ked du ing he p oo o his heo em, dn=cn− nis
also bounded on bounded se s o X.
Co olla y 6. Le Xbe a supe e lexi e Banach space. Wi h he same no a ions
o he p e ious co olla y, one has
spanτuCon b(X)⊃ UC(X)
(whe e τuis he opology o uni o m con e gence on X).
P oo o he Co olla y 6. As we ema ked du ing he p oo o he p e ious Co ol-
la y 5, he esul is p o ed i we show he τu-densi y o he subse o ∆-con ex
unc ions Con b(X) in he se o uni o mly con inuous unc ions UC(X).
Bu his ollows om Theo em 3 and Pisie ’s eno ming heo em ([P]) ha
gi es an equi alen no m wi h modulus o con exi y o powe ype p( o some
p≥2) on e e y supe e lexi e Banach space.
Rema k. Fo a simple and mo e geome ical p oo o Pisie ’s heo em we e e o
[L].
2. The nega i e esul s
In his pa , we will show ha he o undi y condi ions on he no m needed in
Theo em 1 can no be d opped. Fo ins ance, e en he poin wise con e gence ails
o some Banach spaces, as he ollowing coun e -example shows.
Example 7. The e exis s a Lipschi z unc ion on ℓ∞which can no be a poin wise
limi o a sequence o ∆-con ex unc ions.
In he a icle [T], M. Talag and p o ed ha Bo (ℓ∞, w)&Bo (ℓ∞,k · k∞). Tak-
ing a k · k∞-closed, non w-Bo el se B he unc ion d(·, B) is a Lipschi z unc ion
which can no be he poin wise limi o any sequence o ∆-con ex unc ions (since
∆-con ex unc ions a e w-Bo el).
On he o he hand, he τb-densi y p ope y o he span o Con b(X) in UCbis
a cha ac e iza ion o supe e lexi i y. This conclusion comes om he ollowing
heo em.
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