SMOOTH LIPSCHITZ RETRACTIONS OF STARLIKE BODIES
ONTO THEIR BOUNDARIES IN INFINITE-DIMENSIONAL
BANACH SPACES
DANIEL AZAGRA AND MANUEL CEPEDELLO BOISO
Abs ac . Le Xbe an in ini e-dimensional Banach space and le Abe a Cp
Lipschi z bounded s a like body ( o ins ance he uni ball o a smoo h no m).
We p o e ha
(1) The bounda y ∂A is CpLipschi z con ac ible.
(2) The e is a CpLipschi z e ac ion om Aon o ∂A.
(3) The e is a CpLipschi z map T:A−→ Awi h no app oxima e ixed poin s.
1. In oduc ion and main esul s
The well known B ouwe ’s ixed poin heo em s a es ha e e y con inuous sel -
map o he uni ball o a ini e-dimensional Banach space admi s a ixed poin . This
is equi alen o saying ha he e is no con inuous e ac ion om he uni ball on o
he uni sphe e, o ha he uni sphe e is no con ac ible ( he iden i y map on he
sphe e is no homo opic o a cons an map). This esul is no longe ue in in ini e
dimensions (see [8]). In [14] B. Nowak showed ha o se e al in ini e-dimensional
Banach spaces B ouwe ’s heo em ails e en o Lipschi z mappings, and in [6] Y.
Benyamini and Y. S e n eld gene alized Nowak’s esul o all in ini e-dimensional
no med spaces, es ablishing ha o e e y in ini e-dimensional space (X, k·k) he e
exis s a Lipschi z e ac ion om he uni ball BX={x∈X:kxk ≤ 1}on o he
sphe e SX={x∈X:kxk= 1}, and ha SXis Lipschi z con ac ible.
In ecen yea s a lo o wo k has been done on smoo hness and Lipschi z p ope -
ies in Banach spaces (see [11, 5]). Following his end i is na u al o ask whe he
Nowak-Benyamini-S e n eld’s esul s can be sha pened so as o ge Cpsmoo h Lips-
chi z e ac ions o he uni ball on o he sphe e o e e y in ini e-dimensional Banach
space whose no m is Cpsmoo h. In his no e we will show ha his is indeed pos-
sible. In ac we gene alize hose esul s in wo ways. No only do hey hold o
he smoo h ca ego y bu also o a wide class o objec s han balls and sphe es: we
show ha o e e y in ini e-dimensional Banach space wi h a CpLipschi z bounded
s a like body A(whe e p= 0,1,2, . . . , ∞), he e is a CpLipschi z e ac ion o A
on o i s bounda y ∂A, and ∂A is also CpLipschi z con ac ible.
A his poin we need o in oduce some e minology. A closed subse Ao a
Banach space Xis said o be a s a like body p o ided Ahas a non-emp y in e io
and he e exis s a poin x0∈in Asuch ha each ay emana ing om x0mee s he
bounda y o Aa mos once. In his case we will say ha Ais s a like wi h espec o
x0. When dealing wi h s a like bodies, we can always assume ha hey a e s a like
1991 Ma hema ics Subjec Classi ica ion. P ima y: 46B20. Seconda y: 58B05, 46T05.
1
2 DANIEL AZAGRA AND MANUEL CEPEDELLO BOISO
wi h espec o he o igin (up o a sui able ansla ion), and we will do so unless
o he wise s a ed. Fo a s a like body A, we de ine he Minkowski unc ional o Aas
qA(x) = in {λ > 0|1
λx∈A}
o all x∈X. I is easily seen ha o e e y s a like body Ai s Minkowski unc-
ional qAis a con inuous unc ion which sa is ies qA( x) = qA(x) o e e y ≥0.
Mo eo e , A={x∈X|qA(x)≤1}, and ∂A ={x∈X|qA(x) = 1}, whe e ∂A
s ands o he bounda y o A. Con e sely, i ψ:X−→ [0,∞) is con inuous and
sa is ies ψ(λx) = λψ(x) o all λ≥0, hen Aψ={x∈X|ψ(x)≤1}is a s a like
body. Con ex bodies ( ha is, closed con ex se s wi h nonemp y in e io ) a e an
impo an kind o s a like bodies. We will say ha Ais a Cpsmoo h (Lipschi z)
s a like body p o ided i s Minkowski unc ional qAis Cpsmoo h (and Lipschi z) on
he se X q−1
A(0).
Smoo h s a like bodies a e in e es ing because hey a e s ongly ela ed o bump
unc ions and o n-homogeneous polynomials in Banach spaces (see [2] and [3]),
he e o e hei geome ical p ope ies a e wo h s udying. I is wo h no ing ha
e e y Banach space ha ing a Cpsmoo h (Lipschi z) bump unc ion has a Cpsmoo h
(Lipschi z) bounded s a like body oo (and he con e se is also ue).
Be o e s a ing ou main esul we need a ew opological de ini ions. Le M,N
be closed subse s o a Banach space X. We will say ha wo maps , g :M−→ N
a e CpLipschi z homo opic p o ided he e exis an open subse Uo Xcon aining
M, an ε > 0, and a Cpsmoo h mapping H: (−ε, 1 + ε)×U−→ Xsuch ha
he es ic ion o H o [0,1] ×Mis a Lipschi z homo opy joining o g, ha is,
H: [0,1] ×M−→ Nis Lipschi z con inuous and sa is ies H(0, x) = (x) and
H(1, x) = g(x) o all x∈M. Mo eo e we will demand ha H( , x) = (x) o
≤0, x∈M, and H( , x) = g(x) o ≥1, x∈M.
I is no di icul o see ha , wi h his de ini ion, ‘being CpLipschi z homo-
opic’ endows he se o CpLipschi z mappings om Min o Nwi h an equi alence
ela ionship (one can join Cpsmoo h homo opies wi hou losing smoo hness o Lip-
schi zness).
A closed subse Mo Xis said o be CpLipschi z con ac ible i he iden i y
map on Mis CpLipschi z homo opic o a cons an map on M. Fo ins ance, i is
easy o check ha e e y CpLipschi z s a like body Ais CpLipschi z con ac ible.
I is also easy o see ha e e y wo maps on a (CpLipschi z) con ac ible se a e
always (CpLipschi z) homo opic ( hey a e bo h homo opic o a cons an ).
Finally, we will say ha :A−→ ∂A is a Cpsmoo h Lipschi z e ac ion om
he s a like body Aon o i s bounda y p o ided he e exis an open subse Uo X
con aining Aand a Cpsmoo h mapping R:U−→ Xsuch ha R ixes all he poin s
o ∂A, and he es ic ion o R o Ais Lipschi z con inuous and coincides wi h .
Ou main esul is he ollowing
Theo em 1.1. Le Xbe an in ini e-dimensional Banach space and le Abe a Cp
Lipschi z bounded s a like body. Then:
(1) The bounda y ∂A is CpLipschi z con ac ible.
(2) The e is a CpLipschi z e ac ion om Aon o ∂A.
SMOOTH LIPSCHITZ RETRACTIONS OF STARLIKE BODIES ONTO THEIR BOUNDARIES 3
(3) The e is a CpLipschi z map T:A−→ Awi h no app oxima e ixed poin s,
ha is, in {kx−T(x)k:x∈A}>0.
As a co olla y we ob ain he ollowing gene aliza ion o Benyamini-S e n eld’s
heo em:
Co olla y 1.2. Le (X, k·k)be an in ini e-dimensional Banach space wi h an equi -
alen no m k·kwhich is Cpsmoo h, and le BXand SXbe i s uni ball and uni
sphe e espec i ely. Then
(1) SXis CpLipschi z con ac ible.
(2) The e is a CpLipschi z e ac ion o BXon o SX.
(3) The e is a CpLipschi z map T:BX−→ BXwi h no app oxima e ixed
poin s.
I one is no in e es ed in he Lipschi z p ope y, i is a i ial consequence o he
main esul in [1] (see also [4]) ha he sphe e SXis Cpcon ac ible and he e a e Cp
smoo h e ac ions om BXon o SX. Un o una ely, he dele ing di eomo phisms
ob ained in [1, 4] a e no Lipschi z, and co olla y 1.2 canno be deduced using hose
esul s. As a ma e o ac , co olla y 1.2 p o ides a new esul e en in he case
X=`2wi h he usual hilbe ian no m.
2. The p oo s
The p oo o he main esul is a he echnical and will be spli in o h ee
p oposi ions and se e al lemmas. The gene al scheme o he p oo ollows ha o
[6], which in u n is a gene aliza ion wi h some modi ica ions o Nowak’s app oach
[14]. The p oo s in [6, 14] a e al eady in ol ed in hemsel es and he e hey will be
complica ed wi h he di icul ies peculia o smoo h maps and s a like bodies.
Fi s o all i should be no ed ha pa s (2) and (3) o heo em 1.1 a e s aigh -
o wa d consequences o (1). Indeed, assume ha ∂A is CpLipschi z con ac ible.
Then he e a e an open subse Uo Xcon aining ∂A and a Cpsmo h map H:
(−ε, 1 + ε)×U−→ Xsuch ha he es ic ion o H o [0,1] ×∂A is a Lipschi z
homo opy joining he iden i y o a cons an x0in ∂A. Wi hou loss o gene ali y
we may assume ha His de ined on (−∞,+∞)×Uand has he p ope y ha
H( , x) = x0 o all ≤0, x∈∂A, and H( , x) = x o all ≥1, x∈∂A. Then he
o mula
R(x) = H(2ψ(x)−1,x
ψ(x)),
whe e ψis he Minkowski unc ional o A, de ines a Cpsmoo h map on X {0}wi h
he p ope y ha R(x) = x/ψ(x) whene e ψ(x)≥1 and R(x) = x0i ψ(x)≤1
2.
Then one can ob iously ex end R(by pu ing R(0) = x0) o a Cpsmoo h map
R:X−→ Xsuch ha R(x) = xwhene e ψ(x) = 1 and R(x) = x0 o ψ(x)≤1
2.
The es ic ion o R o he se A={x∈X:ψ(x)≤1}gi es us a Cpsmoo h
e ac ion om Aon o i s bounda y. By using he ac ha H: [0,1]×∂A −→ ∂A
is Lipschi z, i is easily seen ha :A−→ ∂A is Lipschi z as well. This shows ha
pa (1) o he heo em implies (2). On he o he hand, once we ha e such a Cp
Lipschi z e ac ion one can easily ge a CpLipschi z map T:A−→ Awi h no
app oxima e ixed poin s: i is enough o ake T(x) = − (x).
4 DANIEL AZAGRA AND MANUEL CEPEDELLO BOISO
Le us now s a he p oo o pa (1) o 1.1. The ollowing lemma ells us ha
o e e y wo CpLipschi z bounded s a like bodies A1, A2 he pai (A1, ∂A1) is Cp
Lipschi z equi alen o he pai (A2, ∂A2). We omi he p oo o his esul since i
is an easy adap a ion o ha o P oposi ion 3 in [2].
Lemma 2.1. Le Xbe a Banach space, and le A1, A2be CpLipschi z bounded
s a like bodies. Then he e exis a Cpbi-Lipschi z di eomo phism g:X−→ Xsuch
ha g(A1) = A2,g(∂A1) = ∂A2, and g(0) = 0. Mo eo e , g(x) = µ(x)x, whe e
µ:X−→ [0,∞), and hence gp ese es he ays emana ing om he o igin.
The e o e, any CpLipschi z p ope y o a bounded s a like body o i s bounda y
is sha ed wi h all he bounded s a like bodies and hei bounda ies. In pa icula he
main heo em and all he auxilia y esul s which we will in oduce in his sec ion can
be p o ed o any pa icula CpLipschi z bounded s a like body in a Banach space
Xand hen, by using his lemma, ex ended o he es o CpLipschi z bounded
s a like bodies, which a e all equi alen . We will use his ac la e on wi hou
u he no ice.
We will also need he ollowing echnical de ini ion.
De ini ion 2.2. Le Xbe a Banach space wi h a Cpsmoo h Lipschi z bounded
s a like body A, and le ψbe i s Minkowski unc ional. Le Mbe a closed subse o
X,y0∈M, and ε > 0. Fo e e y y∈M,δ > 0, de ine he pseudoball Bψ
M(y, δ) =
{x∈M:ψ(x−y)≤δ}.
A poin y0is said o be an ε-escaping poin o ψin Mp o ided he e exis s a
Cpsmoo h Lipschi z mapping T:M−→ Msa is ying:
(1) Tis Lipschi z homo opic o he iden i y on M.
(2) in {ψ(Tny0−Tmy0) : n>m≥0} ≥ 10ε.
(3) Fo all n≥0,Tmaps Bψ
M(Tny0,2ε)isome ically on o Bψ
M(Tn+1y0,2ε)
and, mo eo e , Tis me ely a asla ion when es ic ed o hese se s.
(4) Fo all n≥0,T−1(Bψ
M(Tn+1y0,2ε)) = Bψ
M(Tny0,2ε).
Now we s a e he h ee auxilia y p oposi ions ha we will use in he p oo o he
main heo em.
P oposi ion 2.3. Le M,Nbe closed subse s o a Banach space Xwhich has a
CpLipschi z bounded s a like body Awi h Minkowski unc ional ψ. Suppose he e
is an ε-escaping poin y0in M. Le g: [−1,1] ×M−→ Nbe a CpLipschi z map
which cons an ly a ains he alue z0∈Nou side he se [1
4,3
4]×Bψ
M(y0, ε). Assume
mo eo e ha he e exis s an open subse Uo Xcon aining Mand an ex ension
g: (−1−ε, 1 + ε)×U−→ Xo gsuch ha gis Cpsmoo h and sa is ies g( , x) = z0
o all ∈(−1−ε, 1+ε)and x /∈Bψ
U(y0, ε). Then gis CpLipschi z homo opic o he
cons an unc ion z0in [−1,1] ×Mby means o a CpLipschi z homo opy Hτ( , x)
(0≤τ≤1,( , x)∈[−1,1] ×M) o which Hτ( , x) = z0whene e | | ≥ 3
4.
P oposi ion 2.4. Le Xbe an in ini e-dimensional Banach space wi h a CpLips-
chi z bounded s a like body. Then he e exis ε > 0and ano he CpLipschi z bounded
symme ic s a like body Wsuch ha i s bounda y ∂W has an ε-escaping poin wi h
espec o ψ=qW, he Minkowski unc ional o W.
SMOOTH LIPSCHITZ RETRACTIONS OF STARLIKE BODIES ONTO THEIR BOUNDARIES 5
P oposi ion 2.5. Le Xbe a Banach space and le Abe a CpLipschi z s a like body
which is bounded and symme ic, x0∈∂A,ε > 0. Then he iden i y map on ∂A is
CpLipschi z homo opic o a map :∂A −→ ∂A which cons an ly a ains he alue
−x0ou side he se {x∈∂A :ψ(x−x0)< ε}(whe e ψis he Minkowski unc ional
o A). Mo eo e , can be assumed o ha e a Cpsmoo h ex ension :U−→ X
(whe e Uis an open subse o Xcon aining ∂A) such ha (x) = −x0whene e
ψ(x−x0)≥ε,x∈U.
P oo o he heo em.
Le Ybe a closed hype plane o X. By P oposi ion 2.4 he e is a CpLipschi z
bounded symme ic s a like body Won Ysuch ha i s bounda y ∂W admi s an
ε-escaping poin y0, o some ε > 0 ha can be assumed o sa is y 0 < ε < 1
4. Le
qWbe he Minkowski unc ional o his s a like body. We may w i e X=R×Y.
Now, le Vbe a C∞smoo h Lipschi z bounded symme ic con ex body o he plane
R2such ha i s bounda y ∂V con ains he se
{( , s)∈R2:| | ≤ 1,|s|= 1},
and conside he Minkowski unc ional qVo V, which is a C∞smoo h equi alen
no m on R2. De ine now
ψ( , y) = qV( , qW(y))
o e e y ( , y)∈R×Y=X. I is clea ha ψis a CpLipschi z unc ion on X {0}
which is symme ic and posi i e homogeneous. Then
U={( , y)∈X:ψ( , y)≤1}
is a CpLipschi z bounded symme ic s a like body wi h he p ope y ha i s bound-
a y ∂U con ains he band [−1,1] ×∂W . Wi hou loss o gene ali y we can assume
ha U=A(see Lemma 2.1 and he p eceding ema ks), and i su ices o p o e he
heo em o his pa icula s a like body.
Nex pu x0= (1
2, y0)∈∂A and z0=−x0. By P oposi ion 2.5 he e exis s a
CpLipschi z map :∂A −→ ∂A which is CpLipschi z homo opic o he iden i y
on ∂A, and which has a Cpsmoo h ex ension :U−→ Xsuch ha (x) =
−x0whene e ψ(x−x0)≥ε,x∈U. No e ha i x= ( , y)∈∂A sa is ies
ψ(x−x0)< ε hen, since ε < 1
4, and aking in o accoun he pa icula shape o
∂A, we ha e ha ( , y)∈[1
4,3
4]×Bψ
∂W (y0, ε)⊂[−1,1] ×∂W. Then i is clea ha
g= |[−1,1]×∂W sa is ies he condi ions o P oposi ion 2.3 wi h M=∂W (bea in
mind ha g( , y) = ( , y) = z0whene e ψ(0, y −y0)≥εbecause he pseudoball
{x∈X:ψ(x−x0)< ε}is con ained in he cilynde {( , y)∈X:qW(y−y0)< ε}).
Since y0is an ε-escaping poin in ∂W, i ollows ha gis CpLipschi z homo opic,
as a map om [−1,1] ×∂W in o ∂A, o he cons an z0=−x0∈∂A, by a Cp
Lipschi z homo opy Hτ( , y) sa is ying Hτ( , y) = z0whene e | | ≥ 3
4.
Now, om he pa icula cons uc ion o ∂A, i is clea ha one can ex end Hτ
o a CpLipschi z homo opy Fτby de ining Fτ(x) = z0 o x∈∂A ([−1,1] ×∂W ),
and i is easily checked ha Fτis a CpLipschi z homo opy joining o he cons an
z0in ∂A. Since is i sel CpLipschi z homo opic o he iden i y on ∂A, we can
conclude ha ∂A is CpLipschi z con ac ible o a poin .

6 DANIEL AZAGRA AND MANUEL CEPEDELLO BOISO
Now we will gi e he p oo s o P oposi ions 2.3, 2.4 and 2.5.
P oo o P oposi ion 2.3.
Le Tbe he map associa ed o ψand he ε-escaping poin y0in De ini ion 2.2.
Le θ:R−→ [0,∞) be a CpLipschi z mapping such ha θis s ic ly inc easing
in (0,∞), θ(− ) = θ( ), θ(0) = 0, and θ( ) = | | o | | ≥ 1
8. Pick ano he non-
dec easing Cpmap ζ:R−→ Rsuch ha ζ( ) = 0 o ≤1
4and ζ( ) = 1 o ≥3
4.
Now le us de ine wo maps 0, 1: [−1,1] ×M−→ Nby
0( , x) = 


g(θ( ), T−n(x)) whene e ≥0, x ∈Bψ
M(Tn(y0), ε), and n≥0;
g(θ( ), T−n(x)) whene e ≤0, x ∈Bψ
M(Tn(y0), ε), and n≥1;
z0o he wise;
and
1( , x) = g(θ( ), T−n(x)) whene e ≥0, x ∈Bψ
M(Tn(y0), ε), and n≥0;
z0o he wise.
No e ha on each “ ec angle” [1
4,3
4]×Bψ
M(Tn(y0), ε) o [−3
4,−1
4]×Bψ
M(Tn(y0), ε),
n≥0, he mappings 0and 1a e de ined by he co esponding alue (wi h espec
o T−n) o gin he ec angle [1
4,3
4]×Bψ
M(y0, ε). All hese ec angles a e disjoin ,
by he de ini ion o ε-escaping poin . Since T−nis me ely an a ine asla ion o
Bψ
M(Tn(y0),2ε) on o Bψ
M(y0,2ε) and gis Cpsmoo h and Lipschi z, i is clea ha
he maps 0, 1a e Cpand Lipschi z as well.
By assump ion, Tis CpLipschi z homo opic o he iden i y; le Gτ, τ ∈[0,1],
be a CpLipschi z homo opy joining he iden i y o Tin M. Then
Fτ( , x) =  0( , x) o ≥0;
0( , Gτ(x)) o ≤0,
is a CpLipschi z homo opy joining 0 o 1in [−1,1]×M. Now, he map F1
τ( , x) =
1(θ( )(1−ζ(τ))+ζ(τ), x) de ines a CpLipschi z homo opy joining 1 o he cons an
z0, and i is no di icul o see ha he map
F0
τ( , x) = ( 0(θ( )ζ(τ) + (1 −ζ(τ)), x) o x /∈Bψ
M(y0, ε);
g( , x) o x∈Bψ
M(y0, ε)
is a CpLipschi z homo opy joining g o 0(he e we use he ac ha g( , x) = z0
whene e ψ(x−x0)≥ε,x∈U). We can hen ob ain he desi ed homo opy Hτ
by applying successi ely F0
τ,Fτand F1
τ. Since all o hese homo opies ha e he
cons an alue z0 o | | ≥ 3
4, he same is ue o Hτ.
In o de o p o e P oposi ion 2.4 a numbe o a he echnical lemmas and ac s
will be equi ed. Le us ix some s anda d no a ion used h oughou hese s a e-
men s. I Kis a subse o Xand x∈X, we deno e by dψ(x, K) := in {ψ(x−y) :
y∈K}. Also, he closed s a like body Bψ
X(x, ) = {y∈X:ψ(y−x)≤ }will be
simply w i en as Bψ(x, ).
The i s echnical ool we need is somehow a smoo h e sion o U yshon’s lemma.
SMOOTH LIPSCHITZ RETRACTIONS OF STARLIKE BODIES ONTO THEIR BOUNDARIES 7
Lemma 2.6. Le Xbe a Banach space, and le Abe a CpLipschi z bounded sym-
me ic s a like body wi h Minkowski unc ional ψand Kbe a compac subse o X.
Then, o e e y > 0 he e exis s a CpLipschi z unc ion = ψ, ,K :X−→ [0,1]
such ha
(1) (x) = 1 whene e dψ(x, K)≤ /2, and
(2) (x) = 0 whene e dψ(x, K)≥ .
P oo . Le Lψbe he Lipschi z cons an o ψ(i.e.,ψ(x)−ψ(y)≤Lψkx−yk, o all
x, y ∈X). Since Kis compac he e exis x1, . . . , xl∈Ksuch ha
K⊂
l
[
j=1
Bk·kxj,
4Lψ.(∗)
Then pick a non-dec easing C∞ unc ion g:R−→ [0,1] such ha g−1(0) = (−∞,3
4 ]
and g−1(1) = [7
8 , ∞). Pu
h(x) =
l
Y
j=1
g(ψ(x−xj))
o all x∈X. Since he unc ions x7→ g(ψ(x−xj)) a e all bounded, Lipschi z and
Cp, he unc ion h, being a ini e p oduc o such unc ions, is also Cpsmoo h and
Lipschi z. Mo eo e , no e ha he Lipschi z cons an o honly depends on ψ, and
he numbe o elemen s o he co e ing (∗).
By he cons uc ion o hi is qui e clea ha h(x) = 1 i x /∈ ∪l
j=1Bψ(xj,7
8 ), and
he e o e h(x) = 1 whene e dψ(x, K)≥ . Mo eo e , is easy o see ha h(x) = 0 i
x∈G:= ∪l
j=1Bψ(xj,3
4 ). Le us check ha G⊇ {x∈X:dψ(x, K)≤ /2}. In ac ,
i x∈Xis such ha dψ(x, K) = /2, ake y∈Kin such a way ha ψ(x−y) = /2
and xjso ha y∈Bk·k(xj,
4Lψ). Then i ollows
ψ(x−xj)≤ψ(x−y) + Lψky−xjk ≤ 3
4 .
In o de o conclude he p oo i su ices o ake (x) = 1 −h(x). 
Fac 2.7. Le Xbe a Banach space which has a CpLipschi z bounded symme ic
s a like body Awi h Minkowski unc ional ψand > 0. Then o some M > 0
one has ha o e e y a, b wi h kak=kbk=1
4 he e exis s a CpLipschi z unc ion
a,b :X−→ [0,1] whose Lipschi z cons an is less han o equal o M, and which
sa is ies ha
(1) a,b(x) = 1 whene e dψ(x, [a, b]) ≤ /2, and
(2) a,b(x) = 0 whene e dψ(x, [a, b]) ≥ .
P oo . Fix > 0. Fo e e y wo a bi a y poin s a, b o Xsa is ying kak=kbk=1
4,
conside he compac se K= [a, b]. F om Lemma 2.6, he e exis s a unc ion a,b
ha e i ies condi ions (1) and (2). We only ha e o ensu e ha he unc ion a,b
cons uc ed in he p oo o Lemma 2.6 can be chosen wi h a Lipschi z cons an
ha does no depend on he segmen [a, b]. As we ema ked be o e, he Lipschi z
cons an o a,b only depends on he numbe o elemen s o he ini e co e ing chosen
in (∗). Bu , since he diame e o any segmen [a, b] is uni o mly bounded, o e e y
8 DANIEL AZAGRA AND MANUEL CEPEDELLO BOISO
pai aand bi is easy o ind an app op ia e co e ing o [a, b] wi h a ixed numbe
o elemen s. 
Lemma 2.8. Le Xbe a Banach space which has a CpLipschi z bounded symme ic
s a like body Awi h Minkowski unc ional ψ. Then o e e y > 0 he e exis s a
cons an L > 0so ha o e e y a, b ∈Xwi h kak=kbk=1
4 he e is a map
F=Fa,b :A−→ Asa is ying
(1) Fis CpLipschi z, and he Lipschi z cons an o Fis less han o equal o L
(and he e o e only depends on ψand , bu no on a, b).
(2) Fmaps Bψ(a, /2) isome ically on o Bψ(b, /2); in ac Fis me ely a ans-
la ion when es ic ed o hese se s, and F(a) = b.
(3) F−1(Bψ(b, /2)) = Bψ(a, /2).
(4) F(x) = xwhene e dψ(x, [a, b]) ≥ .
(5) Fmaps lines pa allel o he segmen [a, b]in o hemsel es.
P oo . Fo e e y such a, b le us de ine F=Fa,b :X−→ Xby
F(x) = x+ a,b(x)(b−a)
o all x∈X, whe e a,b is he co esponding unc ion ob ained om ac 2.7. I is
clea ha Fis Cpsmoo h and Lipschi z on X, wi h a Lipschi z cons an no g ea e
han L=M+ 1. The e o e Fsa is ies condi ion (1) o he lemma.
I is e iden om he de ini ions o Fand a,b ha Fsa is ies p ope ies (2), (4)
and (5) as well. Le us see ha Fsa is ies p ope y (3). I F(x)∈Bψ(b, /2) hen
2≥ψ(b−F(x)) = ψ( a,b(x)a+ (1 − a,b(x))b−x).
Since 0 ≤ a,b(x)≤1 we ha e a,b(x)a+ (1 − a,b(x))b∈[a, b] and, he e o e, i
ollows ha dψ(x, [a, b]) ≤ /2 and a,b(x) = 1. Hence o h, we ha e
2≥ψ(b−F(x)) = ψ(b−(x+ (b−a)) = ψ(a−x),
which means ha x∈Bψ(a, /2). This shows ha F−1(Bψ(b, /2)) = Bψ(a, /2).

Lemma 2.9. Le Xbe an in ini e-dimensional Banach space which has a CpLip-
schi z bounded symme ic s a like body Awi h Minkowski unc ional ψ. Then he e
exis some ε > 0and a poin x0in he in e io o Awhich is an ε-escaping poin
in Awi h espec o a map T:A−→ Awhich in addi ion o p ope ies (1)–(4) o
De ini ion 2.2 sa is ies T(x) = xwhene e ψ(x)≥3
4.
P oo . Wi hou loss o gene ali y we can assume ha BX⊆A. Since Ais a bounded
s a like body we know ha he e exis s some α > 0 such ha αkxk ≤ ψ(x)≤ kxk o
all x∈X. No e ha no ma e how Tis de ined, Twill be CpLipschi z homo opic
o he iden i y on Abecause Ais s a like and hence CpLipschi z con ac ible (so
ha bo h Tand he iden i y a e homo opic o a cons an on A).
Le (wn)n∈Nbe a no malized basic sequence in Xwi h bio hogonal unc ionals
(w∗
n)n∈N⊂X∗sa is ying kw∗
nk ≤ 4 (one can always ake such a sequence, see [10],
p. 93), and pu zn=1
4wn o all n∈N. Le us deno e by Ln,k he s aigh line
{ zn+ (1 − )zk: ∈R}passing h ough znand zk( o n6=k). I is easy o see
SMOOTH LIPSCHITZ RETRACTIONS OF STARLIKE BODIES ONTO THEIR BOUNDARIES 9
ha , i {n, k}∩{m, l}=∅ hen kx−yk ≥ 1
32 o all x∈Ln,k, y ∈Lm,l. This implies
ha ψ(x−y)≥α
32 o all x∈Ln,k, y ∈Lm,l; ha is, dψ(Ln,k, Lm,l)≥α
32 .
Now ake =α
320 , and o e e y n, k ∈N,n6=k, pick a unc ion Fn,k :A−→ A
sa is ying he condi ions o Lemma 2.8 o a=znand b=zk, and pu ε= /4. Fo
his choice o εand we ha e
dψ(Ln,k, Lm,l)≥α
32 ≥10 > 20ε. (∗∗)
No e ha , by his inequali y and he cons uc ion o F, i {n, k}∩{m, l}=∅ hen
Fn,k(x) = xwhene e dψ(x, Lm,l)≤ = 4ε, and in pa icula whene e Fm,l(x)6=x
o x=Fm,l(y) o some y6=x. Then he in ini e composi ion
V1(x) = (· · · ◦ F2n−1,2n◦ · · · ◦ F3,4◦F1,2)(x)
is well de ined and sa is ies
(1) V1is Lipschi z. Indeed, ake in o accoun ha all he maps Fn,k in ol ed in
he de ini ion o V1ha e a Lipschi z cons an which is less han o equal o a
ixed cons an L, and he in ini e composi ion de ining V1is uni o mly locally
ini e. In ac o e e y x∈A he e exis s a neighbou hood o xin Asuch
ha V1coincides wi h one o he Fn,k when es ic ed o his neighbou hood.
F om hese p ope ies and om he ac s ha Acon ains he uni ball BX,
which is a con ex se , and V1ob iously es ic s o he iden i y ou side BX,
one can easily deduce ha V1is Lipschi z (wi h a Lipschi z cons an less
han o equal o L) on A.
(2) V1is Cpsmoo h ( his is again a consequence o he ac ha he in ini e
composi ion de ining V1is locally ini e and all he unc ions Fn,k a e Cp
smoo h).
(3) V1maps Bψ(z2n−1,2ε) isome ically (in ac i is me ely a ansla ion when
es ic ed o his se ) on o Bψ(z2n,2ε).
(4) V−1
1(Bψ(z2n,2ε)) = Bψ(z2n−1,2ε).
(5) V1(x) = xwhene e ψ(x)≥3
4.
Le us de ine as well
V2(x) = (· · · ◦ F2n,2n+1 ◦ · · · ◦ F4,5◦F2,3)(x).
Then V2is also a CpLipschi z map ha sa is ies
(3’) V2maps Bψ(z2n,2ε) isome ically (in ac i is a ansla ion when es ic ed
o his se ) on o Bψ(z2n+1,2ε).
(4’) V−1
2(Bψ(z2n+1,2ε)) = Bψ(z2n,2ε).
(5’) V2(x) = xwhene e ψ(x)≥3
4.
Now le us de ine T=V2◦V1. I is clea ha Tis a CpLipschi z map. I only emains
o check ha z1is an ε-escaping poin o T. Indeed, as said abo e, condi ion (1) o
De ini ion 2.2 is i ially sa is ied. I is also clea ha Tnz1=z2n+1, and condi ion
(2) o 2.2 ollows om (∗∗) abo e. Finally, condi ions (3) and (4) o 2.2 ollow
espec i ely om (3, 3’) and (4, 4’) abo e. 
P oo o P oposi ion 2.4.