J. Eur. Math. Soc. 2, 199–216 (2000)
Digital Object Identifier (DOI) 10.1007/s100970000019
J. Lindenstrauss D. Preiss
A new proof of Fréchet differentiability
of Lipschitz functions
Received May 31, 1999 / final version
received February 16, 2000
Published online April 19, 2000 – c Springer-Verlag & EMS 2000
Abstract. We give a relatively simple (self-contained) proof that every real-valued Lipschitz
function on 2 (or more generally on an Asplund space) has points of Fréchet differentiability.
Somewhat more generally, we show that a real-valued Lipschitz function on a separable
Banach space has points of Fréchet differentiability provided that the closure of the set
of its points of Gâteaux differentiability is norm separable.
1. Introduction
The purpose of this paper is to give a simpler (self-contained) proof of the
main result of [4], namely that a Lipschitz function from 2 (or more generally from an Asplund space X) into has points of Fréchet differentiability.
The proof in [4] used an iterative method for finding a point of differentiability. Each stage of the iteration procedure involved a new renorming of
the space. The fact that X is Asplund was used to ensure the existence of
suitable good renormings. In this paper we also use an iterative procedure
but the fact that X is Asplund is used via slicing properties of bounded sets
in X .
Before we proceed we recall some background material (for a discussion
of it and further references see, e.g., Phelps [3]).
A function f from an open set in a Banach space X to Y is said to be
Gâteaux differentiable at x0 if there is a bounded linear operator T X Y
so that for every e X
lim f x0 te
t
0
f x0 t
Te If the limit above exists uniformly in e in the unit sphere of X we say that
f is Fréchet differentiable at x0 . Alternatively, f is Fréchet differentiable at
J. Lindenstrauss: Institute of Mathematics, The Hebrew University, Jerusalem 91904, Israel,
e-mail: [email protected]
D. Preiss: Department of Mathematics, University College London, London WC1E 6BT,
UK, e-mail: [email protected]
Mathematics Subject Classification (1991): 46G05, 58C20
200
J. Lindenstrauss, D. Preiss
x0 if and only if
f x0 y
f x0 Ty o y as y 0
If Y is the real line we denote the Gâteaux derivative T of f at the point x0
by f x0 X .
A Banach space X is called an Asplund space if every (real valued)
continuous convex function on X is Fréchet differentiable on a dense set.
A space X is Asplund if and only if every separable subspace of X has
a separable dual. If X is not Asplund there is an equivalent norm on X which
is nowhere Fréchet differentiable and thus in particular the assumption that
X is Asplund in the theorem of [4] quoted above cannot be weakened. If A
is a bounded set in X a -slice S of A is a set of the form S x A x x"! for some x X, $# 0 and where $ sup x x x A! .
We say in such a situation that S is a -slice defined by the vector x (to be
sure S depends also on ).
We shall consider here mainly separable Banach spaces X. It is well
known (see, e.g., [2]) that if f is a Lipschitz function from an open set G
in a separable X into then f is Gâteaux differentiable in a subset D of
G so that G % D is a Gauss null set. We shall work with the set A & X of Gâteaux derivatives of f at the points of D. This set is clearly bounded
in norm by the Lipschitz constant of f . In very crude terms our strategy is
to find a decreasing sequence Sn !'n( 1 of -slices of subsets of A whose
diameters tend to 0 and find points xn with f xn ) Sn so that xn x for
some vector x (this is easy to achieve) and so that f is Fréchet differentiable
at x with f x* limn f xn (this is delicate). What the proof gives in
particular is that starting with any non
empty slice S of A we find a point x
of Fréchet differentiability with f x + S. This also proves that the mean
value theorem holds for Fréchet derivatives.
The precise statement of the result we prove here is
Theorem. Let X be a separable Banach space and f a Lipschitz function
from an open subset G in X into . Let D & G be the set on which f
is Gâteaux differentiable and let A , f x- x D! . Assume that the
closure of A is norm separable. Then G contains a point of Fréchet
differentiability of f .
Moreover, for every segment . u /10324& G and for every m 5 f 067 f u there is a point x of Fréchet differentiability of f so that f x 08 u # m.
By the separation theorem (and the definition of Gâteaux derivatives) an
obvious equivalent way to state the theorem above is the following
Corollary. With the notations and assumptions as above we have that
cl conv f x9 x D!:; cl conv f x< x D= !
A new proof of Fréchet differentiability of Lipschitz functions
201
= is the subset of D consisting of all points at which f is Fréchet
where D
differentiable.
It should be pointed out that also the nonseparable case of the main
result of [4] follows from the theorem stated above. This will be shown in
Proposition 2 in the next section.
Besides being simpler than the proof of [4] (and this is the main point)
and giving a somewhat more general result (for separable X) there are also
some drawbacks to the proof presented here. In particular it does not seem
that the method of proof of this paper yields the following result (which
was obtained in [4] as a by-product): There is a subset N of the plane 2
with Lebesgue measure 0 so that every Lipschitz function from 2 into
is differentiable at some point of N.
Although the proof given here is definitely simpler than that of [4] it
is still quite involved. The inherent reason for the difficulty of the proof
is that it provides an algorithm for producing a sequence of points xn !'n( 1
which converge to a point of Fréchet differentiability. Such an algorithm is
apparently not simple to find even in the classical case of maps from the line
into itself. The usual (very short and elegant) proofs of Lebesgue’s theorem
on the line prove automatically that the function is differentiable outside
a set of measure 0. Unfortunately, we are not aware of any definition of
a non trivial -ideal (or even just an ideal) of sets in an infinite dimensional
space X (even if X > 2 so that every Lipschitz function from X to
is Fréchet differentiable outside this -ideal (resp. ideal). Several natural
candidates for such a -ideal can be proved not to be suitable. The strongest
tool for proving this are the results of [6]. For example, it is not true that the
collection of Gauss null sets is such a -ideal. For this specific -ideal there
is now an even more striking counterexample. The Hilbert space 2 can be
renormed so that the new norm (which is of course a convex function) is
Fréchet differentiable only on a Gauss null set [1]. Thus if our algorithm
is applied to this specific norm in 2 we invariably end up in a fixed Gauss
null set.
The fact that no “almost everywhere” version of the theorem is known
leads to another complication. It is not known, for example, if two Lipschitz
functions f / g ? 2 have a common point of Fréchet differentiability,
or equivalently, whether a Lipschitz function from 2 to the plane has
a point of Fréchet differentiability. The second example in [6] shows that
the method used here to prove the theorem (i.e. via slices in the set of
Gâteaux derivatives) does not generalize in an obvious way to maps into k
with k # 1. In [6] it is shown that there is a Lipschitz mapping f from 2 to
3
so that
f 1 x e1 f 2 x e2 f 3 x e3 0
202
J. Lindenstrauss, D. Preiss
at every point of Fréchet differentiability of f , but so that f is Gâteaux
differentiable at the origin and
f 1 0 e1 f 2 0 e2 f 3 0 e3 1
(Here ei ! 3i( 1 are some vectors in 2 and f i x is the i’th component of the
derivative of f which is, by definition, a linear map from 2 to 3 ). A similar
example is given in [6] for mappings from p into 2 if 1 5 p 5 2.
(We mention here that in [6] there is a bad misprint which, unless
corrected, makes the proof unreadable. On page 227 in the statement of
Lemma 3 and in many places in pages 228 and 229 there is a meaningless
symbol g @ in the formulas. This symbol should be replaced everywhere
by g ACB rmmz A .)
The organization of this paper is as follows:
In Sect. 2 we prove some preliminary functional analytic results needed
to show the general form of our theorem. For the particular case of separable Asplund space one needs Proposition 1 and Remark 1 only under the
additional assumption that X is separable; under this assumption a simple
proof is provided by Remark 2.
In Sect. 3 we present the proof of a rather technical lemma concerning
Lipschitz functions from 2 to .
In Sect. 4 we present the algorithm for finding the promised point of
Fréchet differentiability of f .
In Sect. 5 we prove (using the result of Sect. 3) that the algorithm
presented in Sect. 4 really works.
2. Functional analytic preliminaries
The first result in this section shows how the assumption in the statement of
the theorem, that A has a norm separable closure, will be applied. This
result is related to several known results concerning the Radon-Nikodým
property, but we are not aware of a reference in which the same result
is proved. In any case, in order to make the proof of the theorem here
self-contained, we prove the proposition in detail.
Proposition 1. Let A a be a nonempty bounded subset of X so that the
closure
of
A is norm separable. Then for every x D 0 in X and every
0 5FEG5 x there is a slice of A which is determined by a vector y with
+H
y x
E and has a diameter at most E .
Proof. Assume that the assertion is false for some x and 0 5IEJ5
x and
assume also that A is contained in the unit ball of X .
Let K sup f xL f A! , let MNEO 2, PLQMRES 3 and put S N f A f xTIUUPV 2 MO! . By our assumption there are g1 / gW 1 X f A A new proof of Fréchet differentiability of Lipschitz functions
203
f x <#Y$Z PR! so that g1 gW 1 #FE ; i.e. there is a u in the unit sphere of
X so that g1 u [ gW 1 u <#\E .
Consider next the set f A f x M u ]#;YFP M g1 u ! . This
set is non empty (it contains g1 ) and is contained in S (since f x\#
^ZP M g1 f u <^ZP 2 M ). Also, for f in this set
M f u 9 f x M u 9#YZ^P M g1 u and hence
f u 9#
Let now 1 sup f x M 1 3P 1 aE , and
and put
1
g1 u [ZP_`M9
g1 u [ZES 3 M u f A! , let P 1 and M 1 be such that M 1 Y
5 EO 4,
^P 1 2 M 1 F
# Z^P M g1 u R/
b f A f x M u ZP 1 2 M 1 !O
From the discussion above it follows that the slice S1 satisfies S1 & S
and for f S1 we have f u 9# g1 u [^EO 3.
In a similar way we define `W 1 sup f x IM u L f A! and choose
positive P W 1 and M W 1 so that M W 1 5FEO 4 /1M W 1 3P W 1 aE ,
W 1 ZP W 1 2 M W 1 #YZZPIM gW 1 u O/
S1
and put
1
c f A f x \M u U W 1 ZP W 1 2 M W 1 !R
Then SW 1 & S and for f SW 1, f u d5 gW 1 u EO 3. In particular if
f S1 and = f SW 1 then f u e = f u 9#FEO 3.
Since x M u x fM<bEO 2 5FE it follows from our assumptions that
there are g1g 1 and g1g W 1 in A so that g1g 1 x M u O/ g1g W 1 x M u #F 1 TP 1 and
g1g 1 u 1 h g1g W 1 u 1 <#YE for some vector u 1 in the unit sphere of X. Similarly,
we choose gW 1g 1 , and gW 1g W 1 in A and u W 1 in the unit sphere of X so that
gW 1g 1 x iM u O/ gW 1g W 1 x iM u #YjW 1 +P_W 1 and gW 1g 1 u W 1 k gW 1g W 1 u W 1 ?#*E .
We continue in an obvious way and define for every k-tuple l 1 /3l 2 / RR /3l k
of signs k 1 / 2 / RR elements gm 1 gon n npg m k A, vectors u m 1 gqn n nrg m k in the unit
W
sphere of X, numbers 0 5bMOm 1 gqn n nrg m k 5 2 k E and slices Sm 1gqn n nrg m k so that
SW
1
(i) gm 1 gqn n npg m k Sm 1 gqn n nrg m ks 1 ,
(ii) Sm 1gon n npg m kt 1 & Sm 1 gqn n npg m k ,
(iii) the slice Sm 1 gqn n nrg m k is determined by the vector x l 1 M u l 2 MRm 1 u m 1
RR l k MRm 1 gon n npg m ks 1 u m 1 gqn n npg m ks 1 ,
(iv) f Sm 1gon n npg m k g 1 , = f Sm 1 gqn n nrg m k g W 1 u
f u m 1 gqn n nrg m k [v= f u m 1 gon n npg m k 9#FES 3.
204
J. Lindenstrauss, D. Preiss
Note that the distance of the vector appearing in (iii) from x is smaller
than E . For every infinite sequence of signs l i ! i' ( 1 wl let gm be any H
limit point of the sequence gm 1 gon n npg m i ! i' ( 1 . If lx D l = i.e. if l i l = i for i
k
and l ky 1 D l = ky 1 then
#c| gm4 gm { u m 1 gqn n npg m k O|}xEO 3
and this contradicts the separability of the closure of A.
gmz gm {
Remark 1. If the closure of A is separable the same clearly holds for
every subset of A. It follows that any subset of A has slices of arbitrarily
small diameter which are determined by vectors arbitrarily close to any
given vector. This assertion is exactly what is used concerning A in the
proof of the theorem below.
Remark 2. If we make the stronger assumption that X itself is separ
able then
the
proof
of
Proposition
1
becomes
much
simpler.
Let
p
xG
sup f x < f A! . This is a continuous convex function and if X is separable, it is Fréchet differentiable at a dense set of vectors x. It is trivial to check
that whenever p is Fréchet differentiable at some point x, then this x determines slices of A of arbitrarily small diameter (cf. [3, Lemma 2.18]).
The second proposition we present in this section is a tool for reducing
differentiability problems to the separable case. In particular, it implies
immediately the main result of [4] in the non separable case, once we know
the theorem stated in Sect. 1. This reduction result appears already in [5,
Theorem 1]. We reproduce here its proof for the sake of completeness.
Proposition 2. Let X and Y be Banach spaces, G an open set in X and
f G Y a Lipschitz function. Then for every separable subspace V of X
there is a separable subspace W of X containing V which has the following
property: If f ~ G  W is Fréchet differentiable at some x G € W then f itself
if Fréchet differentiable at the same point.
Proof. We note first that f is Fréchet differentiable at x if (and obviously
only if) for every EJ# 0 there is a PG# 0 such that whenever u /10G X with
0 1 and whenever s and t satisfy 0 5b| s |/1| t |"5FP then
u t u 0‚ f x t H
(„ )
f x su f x s f x s 06 f x s
Eƒ
Indeed, by taking u f0 we deduce that Tu limt 0 f x tu 8 f x t
exists. It also follows from ( „ ) that T is linear and bounded. Using („ ) again
for u …0 , fixed s and t 0 we get
+H
Tu f x su f x s
EO 2
f x
A new proof of Fréchet differentiability of Lipschitz functions
205
for all u with the norm 1 and | s |}5FP . Thus T is the Fréchet derivative of f
at x.
For every x G we choose a countable subset A x with the following property: Whenever s and t are non zero rational numbers with
2 max | s |/1| t |†5 dist x / X % G there are u /10L A x so that u 0 1
and
06[ f x t
f x su f x s f x s 06 f x s
‡L| s | sup f x t z e[ f x t f x sz[ f x s
f x s [ f x s z /1ˆ X / z 1 f x t u
We now define inductively an increasing sequence of separable subspaces of X. Put V0 V . Once Vn has been defined let Cn be a dense sequence
in it and let Vny 1 be the closed linear span of Cn ‰
A x) x Cn € G ! .
'
We claim that W r ( 0 Vn has the desired property. Let x G € W and
assume that f ~ W is Fréchet differentiable at x with T= as its derivative there.
Then for every E-# 0 there is a 0 5‡P EOL5 min Eƒ/ dist x / X % G so that
for every z W with z x 5FP ES
f z [
f x eŠT= z
x
5YE z x 28 If f is not Fréchet differentiable at x it would follow from the observation
in the beginning of the proof that there are u /10- X of norm 1, an EX# 0
and rational s / t, with 0 5f| s |/1| t |e5P EO 4 so that ( „ ) fails to hold. From
'n( 1 Cn so that
the definition of W it follows that there is a y x 5
1‹| t | 2`| s | 1`| t | 2‹| s |†
y
min | s |/1| t |†5FP ES 4 /
L y x T= 5FEO 4 /
f y[ f x 5FEO 4
and
1 06S f y t f y su S f y s f y s 06S f y s #FEƒ
Since A y& W and by the definition of A y , there are u /10 W of norm
f y t u 1 so that
0 [ f y t f y su [
f y s0 [
f y s #FET\| s |?# 3EO
f y t u
f y s
4
On the other hand the left-hand side of the preceding inequality does not
exceed
206
J. Lindenstrauss, D. Preiss
0 [ f x t f y su f x s
f y s 0 [ f x s 1‹| t | 2`| s | f y
5 T= y t u 0 [ x t y su x s y E y t u 0 [ x 28 | t |
E y su x y s 0 x 28 | s | EO 4
e 5 1‹| t | 2`| s | y x T= E 4 y x `| t | EO 4 5 3EO 4 f y t u
f x
s0
x s
2 y
x ` | s | 28
Thus we are led to a contradiction and the proposition is proved.
3. Measure theoretic lemmas
The purpose of this section is to prove Lemma 3 below. We start with the
well known result that the so called Hardy-Littlewood maximal operator is
of weak type (1,1). We will denote by m the Lebesgue measure on the line.
Lemma 1. Let g
L 1 then for every Œd# 0
H
m t Mg t9Œ‹!
2aŒ
| g tO| dt
where
Mg te
| g sO| ds m I W
sup
I
Consequently, if | g tR|
H
H
1 give
0 Mg t
Mg t 2 dt
1
H
H
0
I is an interval containing t 1 for every t then Fubini’s Theorem and the bound
m t Mg t
0
1
1
2
 s
2
1
# s! ds | g tO| dtds 4
0
m t Mg t#Y s ! ds
| g tO| dt The next lemma is a slight variant of a lemma used already in [4, Lemma 3.3].
Lemma 2. Suppose a 5 b are real numbers and h . a / b2Ž
H
function whose Lipschitz constant is 1. Assume that
b
| h tR| dt 2 | h b[ h aO|
a
Then there is a measurable subset A
&
a / b so that
a Lipschitz
A new proof of Fréchet differentiability of Lipschitz functions
207
| h tO| dt,
| h tR| dt 8 b a for every s A,
(ii) h s<
H
8 h sO| t s | for every s A and t . a / b2 .
(iii) | h te h sR|
Proof. Let g ti h t whenever the derivative exists (and take it as 0
otherwise). Put gy max g / 0 and
b
| g u O| du 8 b a
t . a / b2z g t#
D
a
B c t D g t5 Mg t 2 64!
A D % B
It is evident that every s D (and thus s A) satisfies (ii). For s A
and t d. a / b2
H
H
‘ g u du Mg sO| t s | 8 h sS| t s |
| h t[ h sO|
tg s ’
m A
(i)
b
a
1
16
b
a
and thus also (iii) holds. It remains to verify (i),
b
a
| g tO| dt 2
H
b
a
b
2
a
gy t dt
b
g t dt
a
gy t dt 1
2
a
gy t dt
a
| g tR| dt b
b
2
h b[ h a
Thus
b
a
H
| g tO| dt 4
H
b
a
gy t dt
4
D
and consequently
b
g t dt 4
D
| g tO| dt
b
1
2
g t dt 4
‘
g ’q“
ab D
gy t dt
a
H
| g tR| dt 8
g t dt a
D
In view of Lemma 1,
B
H
g t dt
B
H
Mg t 2 dt 64
b
a
| g tR| dt 16
and hence
b
a
H
| g tO| dt 8
A
or
b
a
g t dt 8
g t dt
H
8
B
A
H
| g tO| dt 16
g t dt
A
H
g t dt 16m AO
1
2
b
a
| g tO| dt
208
J. Lindenstrauss, D. Preiss
We come now to the lemma which is the purpose of the present section.
The lemma has a rather special and quite involved statement. It is tailormade to its application in Sect. 5. The proof of the lemma is however not
very complicated. It uses just the previous two lemmas and Fubini’s theorem
in the plane.
H
W
Lemma 3. Let 0 5F”•/1M–/3—5 1 with MGU— , let r / s # 0 with s 10 6 ”•MO— r.
be a Lipschitz function which satisfies
Let g .˜ 2r / 2r 2™\. 0 / s27
| g 0 / sR|}#F” s
H
| g u / 0O| ” s 32 for all u d.˜ 2r / 2r 2
and is such that both partial derivatives g1 u /10‚ and g2 u /106 are a.e. of
absolute value at most 1. Finally, let N &š.˜ r / r 2™›. 0 / s 2 be a set of
(Lebesgue) measure 0.
Then there is a u /106d.˜ r / r 2Ž™I. 0 / s2?% N so that
(i) g1 u /106#bz—
(ii) g1 u /106#FE 1 b” s 128r or | g2 u /106O|•#FE 2 c”1 16
H‡
H
M 8 g1 u /106 y R| t | , if | t | r.
(iii) | g u t /106[ g u /106O|
and
Proof. We may clearly assume that N contains all points of non differentiability of g in .˜ r / r 2:™\. 0 / s2 . Note that by the assumptions on g
H
| g u /106O| 2s for all u /1069$.˜ 2r / 2r 2:™\. 0 / s2‚
Let I be the set of all 0^c. 0 / s2 for which Nœ- u u /106d N ! has
r
measure 0, let C1 4‹M and k 32. If W r | g1 u /10‚O| du # kC1 s for some
0X I then by Lemma 2 there is a u I.˜ r C1 s / r C1 s2% Nœ so that
g1 u /106 C1 s r (which is clearly #\E 1 and
H
H
| g u t /106 g u /106O| 8 g1 u /10‚S| t | whenever | u t | r H
If | t |
r but | u t |} r then by the choice of u, | t |• C1 s and
H
H
H
| g u t /106 g u /106O| 4s 4 | t | C1 Mƒ| t |
and therefore the point u /106 we found has the desired properties. Thus we
may assume from now on that
r
W
r
H
| g1 u /106R| du kC1 s for all 0 I Let M be the set of points u /10‚V.˜ r / r 2™ I such that at least one of the
following happens
H
z— ,
(a) g1 u /10‚
(b) There is a t with | t |
H
r so that | g t /106 g u /106O|•#bMƒ| t
u| .
A new proof of Fréchet differentiability of Lipschitz functions
By Lemma 1 and Chebyshev’s inequality for every
m u
209
0+ I ,
H
u /1069 M ! kC1 sa— 2kC1 s`M C2 s r r
By Fubini’s theorem
there is an r0 / so that the set I1 of those
4 2
0+ I such that r0 /1069 M ‰ N or r0 /1069 M ‰ N has measure at most
c2 s 8C2 s2 r. Note that by the definition of C2 and the assumption on s
in the statement of the lemma c2 5YE 2 2.
Every u /106V D ž.˜ r0 / r0 2:™ˆ. 0 / s2•% M ‰ N clearly satisfies (i) in
the statement of the Lemma. It also satisfies (iii), even in the stronger form
H
H
| g u t /106 g u /106O| M–| t | if | t | r (Ÿ )
H
H
Indeed, this is clear by the definition of M if | u t |
r. If | t |
r and
H
H
| u t |•# r then since | u | r0 r 2 we deduce that | t |•# r 2. But then
H
H H
| g u t /106 g u /10‚O| 4s 8s r r 2 Mƒ| t |
as desired. Consequently, if for some u /106) D, | g2 u /106O|•#\E 2 this point
will satisfy all the requirements in the lemma and the proof will be finished.
H
E 2 for all u /106G D. Hence, in
We can thus assume that | g2 u /106O|
particular, for all 0+d. 0 / s 2 ,
s
r¡
H
r¡
p¡
| g r0 /106O| | g r0 / 0O| | g2 r0 / tO| dt
0
H
r¡
r¡
” s 32 | g2 r0 / tO| dt ‘
| g2 r0 / tO| dt
I1
0g s ’o“ I1
H
H
” s 32 c2 s E 2 s 2E 2 s H
Since m I1 c2 s there is a 0+ I % I1 –€¢. 1 2c2 s / s2 which we fix from
now on. We have
H
” s 5b| g 0 / sO| | g 0 /106O| 2c2 s
r0
H
| g r0 /106O| | g1 t /106R| dt 2c2 s
W r0
r0
H
| g1 t /106O| dt 2 E 2 c2 s
W r0
r0
H
| g t /106O| dt ” s 2 W r0 1
Hence
W
r0
r0
| g1 t 1/ 06O| dt #Y” s 2. Since also
H
H
| g r0 /106[ g r0 1/ 06O| 4E 2 s ” s 4
210
J. Lindenstrauss, D. Preiss
we deduce
from Lemma 2 that there is a u
” s 32r0 9UE 1 and
d.˜ r0 / r0 2£% Nœ so that g1 u /106#
H
| g u t /10‚ g u /106R| 8 g1 u /106S| t |/ whenever u t d.˜ r0 / r0 2‚
In order to verify that for this choice of u /106 assertion (iii) also holds, we
H
have to consider only points t with | t |
r and | u t |8# r0 . Let =t be of
the same sign as t so that | u t= |: r0 . Since 0Z I % I1 it follows that
u t= /106 D and hence, using (Ÿ ),
H
| g u t /106 g u /106O| | g u t /106¤ g u t= /106O| | g u t= /106¤ g u /106O|
H
H‡
M–| t t= | 8 g1 u /106S| t= |
M 8 g1 u /106R| t |
4. The algorithm
In this section we describe the algorithm we use for proving the theorem
stated in the introduction. We use the notation appearing in the statement
of the theorem. We just add now for convenience the assumption that the
xH 1
for all
Lipschitz constant of f is at most 21 (and thus also f x
2
x D).
0*
We denote e0 u 0* u , 0 sup f x e0 F x D! ,
— 0 0 m u c0 3 and c0 ¥ 0 2— 0 . Since f is Gâteaux differentiable outside a set of Gaussian measure 0 we can replace the segment . u /1032 by a very close parallel segment . u r/103o2 so that f is Gâteaux
differentiable a.e. on . u /10 2 and infer that there is an x¦ D such that
f x¦ e0 *# m u §0 . Hence — 0 # 0 and we may find x0 D so that
W
f x0 e0 #b 0 10 10 — 0 . We let 0 5§P 0 5 1 be such that the open ball
B x0 /3P 0 (with center x0 and radius P 0 ) belongs to G. We let
D0
x x
D € B x0 /3P
2
0
and f x e0 <# c0
All the further points we choose will be in D0 and note that f x x D0 !
is a subset of a slice of A defined by the vector e0 .
W
We let ” k 2 k for k 0 / 1 / 2 /1RR and will define inductively (in
this order) numbers k /3¨ k /1M k , unit vectors ek , numbers k / ck /3— k , points xk ,
numbers © k /3P k and sets Dk . We will do this choice so that in particular the
following hold:
(1)
(2)
(3)
D j and D j & D j W 1 for all j 0 /
H
2— j
f x j e j e c j for j 0 /
for every x D j there is ª«bª x / j95Y j such that
| f x te j [ f xe t f x j e j O|•5Yª| t | for all | t |}5FP j xj
A new proof of Fréchet differentiability of Lipschitz functions
211
0 we define the values we have not defined as yet by 0 2,
© 0 1 2 and ¨ 0 ,M 0 1. Obviously, (1) and (2) hold for j 0 (where
DW 1 means D), and (3) is satisfied with ª« 3 2.
Let k 1 and assume that all the above have been defined for values
For k
smaller than k. We put
(4)
k
10W 8 min ” k2W 1 — kW 1 M kW 1 /3 kW
¨ k 10W 8 min — kW 1 /3 k2 3/ ¨ kW
(5)
1
1
M k 10W 8 min ” k /3 k O
(6)
By Proposition 1 there is a unit vector ek X with ek ekW 1 5¨ k
which defines a slice of the set f x9 x DkW 1 ! of diameter less than M k .
Put
k sup f x ek 9 x DkW 1 !O
+H
Since f x
1 2 for all x
D,
kW 1 ek ekW 1 2
and, since xkW 1 DkW 1 ((1) for j k 1),
k f xkW 1 ek f xkW 1 ekW 1 [ ek ekW 1 2 (8)
(7)
H
k
k 1) and (5),
H
ek ekW 1 5Y¨ k — kW 1 5 f xkW 1 ekW 1 [ ckW 1
and thus (8) implies that k #
f xkW 1 ekW 1 ckW 1 2. Hence we may
By (2) (for j
choose ck so that
(9)
k
# ck #
1
ekW 1 ckW 1 2 /
k ck 5 10W
(10)
and so that
(11)
f x kW
the diameter of f x x
3 2
k
/
DkW 1 / f x ek 9# ck ! is at most M k For future reference we record that by (9) and by (2) (for j
(12)
ckW
1
#
f x kW
1
ekW 1 [
ckW 1 2 U— kW 1 min k ck 4 /1M k and a point xk DkW 1 such that
W
f xk ek 9# max k 10 10 ” k2 — k M k / k ck 2 We now take —
(13)
ck
k 1
k
212
J. Lindenstrauss, D. Preiss
These two choices ensure in particular that (2) holds for j
(13) with (7) and (8), we infer that
k. Combining
| f xk ek e f xkW 1 ekW 1 R|•5 10W 9 ” k2W 1 — kW 1 M kW 1 By (3) (for j 1 /1RRO/ k 1 there is a © k # 0 so that
Hˆ
| f xk teH j [ H f xk [ t f x jH e j O|
j Z© k R| t |
(15)
for all 1 j k 1 and | t |
P j
choose P k # 0 so that P k 5 min © k 4 /3P k2W 1 , B xk / 2P k2 Z&
Next we
B xkW 1 /3P k2W 1 and
H
H
(16)
| f xk tek f xk [ t f xk ek O| 10W 3 k | t | for | t | P k Finally we let Dk be those points x in DkW 1 € B xk /3P k2 which satisfy
f x ek G# ck and for which there is a ªUfª x / k with 0 5cª\5 k so
that
H
H
| f x tek [ f x[ t f xk ek O| ª| t | if | t | P k Clearly xk Dk so that (1) holds also for j k. By the definition of Dk
also (3) holds for j k. Moreover, (11) implies that
(17)
the diameter of f x x Dk ! is at most M k (14)
Note that we do not claim at this stage that the set Dk is in any sense
large. It may very well consist just of one point xk . Still our inductive choice
of all parameters works. Of course once a set D j reduces to a single point
x j then xk x j for k j and all the sets Dk with k j consist of the
single point xk .
Since P k 0 as k ­¬ it is clear that x limk
' xk exists. To prove
that this point has the desired property we shall prove (in the next section)
the following lemma.
Lemma 4. There are numbers with j 0 so that for every k #
# 0 and ® j # 0 for j 0 / 1 / 2 /1RR
0
H
+H
(18)
® j
| f xk y f xk [ f x j yR| j y whenever y
where the xk ! are the points constructed above.
j
j
Let us show now the simple fact that Lemma 4 implies the theorem. First
note that it follows from (18) that for any two positive integers j1 and j2 ,
dH
j1 j2 and hence x lim j f x j H exists and
xH j2 f x j1 < f L
x f x j -H j . Consequently, | f xk y¯ f xk x yO| 2 j y
whenever y
® j and k # j. Keeping j fixed and letting k ¬ , we
infer that
H
+H
| f x ye f x[
x yO| 2 j y if y
® j
A new proof of Fréchet differentiability of Lipschitz functions
213
Thus f is Fréchet differentiable at x and f x e x . Since all the points x j
belong to the set D0 we get that f x : lim j f x j satisfies f x e0 < c0 ,
hence f x 09 u 9# m as desired.
5. The proof of Lemma 4
In this section we prove Lemma 4 and this will conclude the proof of the
theorem. We keep the notation of the previous section and claim that suitable
j and ® j ( j 0 / 1 /1RR ) for which (18) holds are given by
103 ” j j ® j 10W 9 ” j — j M j P j y 1 © j y 1 Indeed, assume that (18) is false for some k # j 0. Since +H
clear that j 1. Let y X with y
® j be such that
| f xk y[ f xk [ f x j yO|‹#b j y j
0
1 it is
By moving y slightly we may assume that at almost every point of the
intersection of G and the affine span of xk / xk y / xk e j the function f is
Gâteaux differentiable.
We next pass into a setting where we can apply Lemma 3. Let s y ,
by
e y s, r 106 sa” j — j M j and define g .˜ 2r / 2r 2™I. 0 / s 2¯
0 e f xk [ f x j ue j 0 eO
We intend to apply Lemma 3 with the given choices of r / s / g, with ”°b” j /
—*c— j /1M)fM j and N the set of u /106 such that f is not Gâteaux differentiable at xk ue j 0 e. We check first that all the assumptions of Lemma 3
are satisfied.
To estimate g u / 0 we cannot use (3) directly since this would give an
estimate
in terms of r and
not in terms of s. We therefore use (14) to replace
f x j e j by f x j y 1 e j y 1 . We proceed as follows:
g u /106e
f xk ue j
Since
H
h” j — j M j P j y 1
H
we deduce, using (3), (4), (5) and (14), that for | u |
2r
H
| g u / 0O| | f xk ue j y 1[ f xk [ f x j y 1 ue j y 1 O|
e j y 1 e j 2 | f x j y 1 e j y 1 [ f x j e j R|†R| u |
H
W 9 2
j y 1 ¨ j y 1 2 10 ” j — j M j | u |
H
W
W
2r 10 8 ” 2j — j M j 2 10 9 ” 2j — j M j
H
2s 10W 2 2 10W 3 ” j ” j s 32 2r
2 106 sh” j — j M
j
H
2 106 ®
j
214
J. Lindenstrauss, D. Preiss
It remains to note that
| g 0 / sO|1;| f xk y f xk [ f x j yR|•#b j y ” j s Having verified that all the assumptions of Lemma 3 hold we use it to find
a point u /106.˜ r / r 2Ž™I. 0 / s2 so that
(19)
xk ue j 0 e
f is Gâteaux differentiable at x
g1 u /106#bz—
(20)
j
g1 u /106#F” j s 128r or | g2 u /106O|•#F”
(21)
j
16
H‡
H
| g u t /106 g u /106O|
M j 8 g1 u /106 y O| t | for | t | r (22)
We now make the following
Claim. x
D j W 1 and f x e j # cj.
We show first that this will conclude the proof. Indeed, by (13) and (21)
we have either
f x e j f x j e j g1 u /106
f x j e j ” j s 128r f x j e j 10W 6 ” 2j — j M j 128 #Y
j
j
which is a contradiction, or by (11), (13) and (21)
H‡ LH
f x j [ f x
” j 16 5c| g2 u /106O|
M
j
and this again is a contradiction in view of (6).
It remains to prove the claim. Note first that by (2), f x j e j 9# c j H
H
H H
and by (5) and (12), ¨ i — iW 1 ci ciW 1 for all i 1. Hence, for 0 i
f x e j e
j
j
e j f x j e j g1 u /106 ei e j
jW 1
f x j e j Z— j ¨ ky 1 f x j e j e^— j c j ci # ci k( i
f x ei 9
—
ei
This proves the second assertion in the claim and also that x D0 , since
LH
H
x xk
r s P j y 1 / xk x j 5YP 2j and B x j / 2P 2j 9& B x0 /3P 02 .
It remains to prove that x Di for i 1 / 2 /1RRh/ j 1 and we do this
H
by induction on i. Let i
j 1 and assume that x DiW 1 . What has to be
A new proof of Fréchet differentiability of Lipschitz functions
215
Di is that there is a ª i 5F i so that
H
H
| f x tei [ f x t f xi ei O| ª i | t | for all | t | P i shown in order to ensure that x
We shall verify this by treating separately three different ranges of t.
H
r. Since x
Case I, | t |
(12)) we have
g1 u /106
f x e j H
DiW 1 and since f x j e j # c j # ci (by (2) and
f x j ej
1
H
f x ei e
1
2 e j ei
f x j ej
j
i ci ¨ i 5F i ci ¨ iy 1 2 k( i y 1
LH
Hence, using that xi / x j Di and so (17) implies that f x j R f xi M i,
we infer from (22) that
| f x tei [ f x t f xi ei O|
H]
e j ei f x j [ f xi O| t | | f x te j [ f x[ t f x j e j R|
H
2¨ iy 1 M i M j 8 g1 u /106 y | t |
H
2¨ iy 1 2 M i 8 ¨ iy 1 i ci | t |
In view of (4), (5), (6) and (10) the number in front of | t | is strictly smaller
than i .
Case II, r
we get
5b| t |
H
P j y 1 2. In this case, since xk D j y 1 /1|p01|
H
s and | u |
H
r
| f x tei [ f x[ t f xi ei O|
H§
e j ei f xi [ f x j y 1 | f x te j [ f x t f x j y 1 e j O| O| t |
H§
2¨ iy 1 M i O| t | | f xk t u e j ¤ f xk ue j ¤ t f x j y 1 e j R| s
H§
4¨ iy 1 M i O| t |
| f xk t u e j y 1 e f xk ue j y 1 t f x j y 1 e j y 1 O| s
H§
4¨ iy 1 M i O| t |
| f xk t u e j y 1 e f xk t u f x j y 1 e j y 1 O|
| f xk ue j y 1 e f xk [ u f x j y 1 e j y 1 O| s
H§
4¨ iy 1 M i 3 j y 1 s rO| t |
4¨ iy 1 M i 3 j y 1 10W 6 ” j — j M j R| t |
and again the number in front of | t | is strictly less than
i.
216
J. Lindenstrauss, D. Preiss
Case III, P
by i),
H
y 2 5| t |
j 1
H
P i . By (15) (in which k is replaced by j 1 and j
| f x tei e f x t f xi ei R|
H
| f x tei f x j y 1 tei O| | f x f x j y 1O |
| f x j y 1 tei e f x j y 1 [ t f xi ei O|
H x x jy 1 i Z© j y 1 O| t |
H x xk xk x j y 1 i Z© j y 1 R| t |
H
2
2 2r P j y 1 aP j y 1 i Z© j y 1 | t |
Since r 106 ® j h” j — j M j it follows from the definitions of ® j and P j y 1 that
also here the number in front of | t | is strictly less than i .
This, finally, finishes the proof of Lemma 4 and thus of the theorem.
Acknowledgements. We thank Professor Bernd Kirchheim for suggesting several significant
improvements of our arguments.
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