# Smoothness, Convexity, Porosity, and Separable Determination

# Smoothness, Convexity, Porosity, and Separable Determination

# Abstract and Keywords

This chapter shows how spaces with separable dual admit a Fréchet smooth norm. It first considers a criterion of the differentiability of continuous convex functions on Banach spaces before discussing Fréchet smooth and nonsmooth renormings and Fréchet differentiability of convex functions. It then describes the connection between porous sets and Fréchet differentiability, along with the set of points of Fréchet differentiability of maps between Banach spaces. It also examines the concept of separable determination, the relevance of the σ-porous sets for differentiability and proves the existence of a Fréchet smooth equivalent norm on a Banach space with separable dual. The chapter concludes by explaining how one can show that many differentiability type results hold in nonseparable spaces provided they hold in separable ones.

*Keywords:*
separable dual, Fréchet smooth norm, convex function, Banach space, renorming, Fréchet differentiability, porous sets, separable determination, nonseparable space

In this chapter we prove some results that will be crucial in what follows; in particular, we show that spaces with separable dual admit a Fréchet smooth norm. For the first time, we meet the *σ*-porous sets and see their relevance for differentiability: the set of points of Fréchet nondifferentiability of continuous convex functions forms a *σ*-porous set. We also give some basic facts about *σ*-porous sets and show that they are contained in sets of Fréchet nondifferentiability of real-valued Lipschitz functions. These results, and their proofs, are important in order to understand some of the development that follows. So, although much of this material is discussed in greater detail in [4], and we refer again to this book for the original references, we present here the most relevant results and arguments. Finally, we discuss a new way of treating separable reduction arguments that can be used to extend some of our results to nonseparable setting.

# 3.1 A Criterion of Differentiability of Convex Functions

Continuous convex functions on Banach spaces are easily seen to be locally Lipschitz (see [4, Proposition 4.6]). Thus the differentiability of such functions is naturally included, as a much simpler case, in the study of differentiability of Lipschitz functions. The key to the simplicity is that convexity implies “differentiability from below,” which is usually expressed more precisely as nonemptiness of its subdifferential *∂f*(*x*_{0}) of *f* at *x*_{0}.

Definition 3.1.1. Let

fbe a convex function defined on a convex setCand∈ C.x_{0}Thesubdifferential ∂f(x_{0})of f at x_{0}is the set of functionalsxsatisfying^{∗}∈ X^{∗}

Geometrically, this is the set of all linear functionals which support the convex epigraph of *f* at (*x*_{0}, *f*(*x*_{0})). Hence, by the Hahn-Banach theorem, in the interior points of *C* the subdifferential of a continuous convex function is nonempty.

Proposition 3.1.2.

Let f be a continuous convex function on an open convex subset C ⊂ X of a Banach space X, and let x ∈ C. Then for every x(^{∗}∈ ∂fx)and every u such that x ± u ∈ C,

(p.24)
*Proof.* The first inequality comes from the definition of the subdifferential. The same definition implies that and the second inequality follows by adding it to

An immediate corollary of this statement is a useful criterion for Fréchet or Gâteaux differentiability of convex functions on Banach spaces. Its main point is that differentiability may be easily proved without actually finding the derivative. Although one can prove criteria of differentiability avoiding the value of the derivative also in more general situations, they become too complicated and their usefulness seems to be limited to special tasks (see, for example, Section 3.6).

Proposition 3.1.3.

Let f be a continuous convex function on an open convex subset C of a Banach space X and let x ∈ C. Then

(i)

*f is Gâteaux differentiable at x if and only if for any u ∈ X*

(ii)

*f is Fréchet differentiable at x if and only if*

Proof.From Proposition 3.1.2 we immediately see that any element of the subdifferential offatxis its Gâteaux (resp. Fréchet) derivative.

We may notice that the above criterion did not use the full strength of convexity. For example, for the validity of (ii) we just needed nonemptiness of the “Fréchet subdifferential” of *f* at *x*, that is, existence of *x ^{∗} ∈ X ^{∗}* such that

We also remark that Proposition 3.1.2 shows that if a continuous convex function *f* is Gâteaux differentiable at *x* then its subdifferential at *x* has a single element, namely, its Gâteaux derivative at *x*. Conversely, if *∂f*(*x*) consists of a single point, *f* is Gâteaux differentiable at *x*. This may be seen by an application of the Hahn-Banach theorem. Details, and much additional information, may be found in [4, Chapter 4].

# 3.2 Fre´Chet Smooth and Nonsmooth Renormings

An important example of a continuous convex function on a Banach space is its norm. There are several known and for us very useful results on existence of interesting norms in rather general Banach space. The first result of this nature which we present here is due to Klee and Kadec independently.

Theorem 3.2.1.

Let X be a Banach space such that X= 0.^{∗}is separable. Then there exists an equivalent norm on X which is Fréchet differentiable at every point x

(p.25)
Such a norm is called a *Fréchet smooth* norm, or just *smooth* norm. Observe that no norm is (even Gâteaux) differentiable at the origin.

The proof of this theorem is based on the following simple yet important lemma. To state it, recall that the space *X* is said to be *locally uniformly convex* if *x _{n} − x →* 0 whenever

*x, x*are unit vectors and

_{n}*x*+

*x*2. Naturally, such norms are called locally uniformly convex.

_{n}→Lemma 3.2.2.

Assume that X^{∗}is locally uniformly convex. Then the norm in X is smooth.

*Proof.* Let *x ∈ X* with *x* = 1. Find *x ^{∗} ∈ X^{∗}* such that

*x*=

^{∗}*x*(

^{∗}*x*) = 1. For every

*u ∈ X*we choose a norm one functional

*y*in

^{∗}_{u}*X*such that Then

^{∗}Hence as *u →* 0. By the local uniform convexity of *X ^{∗}* we deduce that as

*u →*0. It follows that

*Proof of Theorem 3.2.1.* Since both *X* and *X ^{∗}* are separable there are a dense sequence in the unit sphere of

*X*and an increasing sequence of finite dimensional subspaces such that

Consider now the following maps from *X ^{∗}* into

_{2}:

The function on *X ^{∗}* given as

*x*is

^{∗}→ Tx^{∗}*w*-continuous on bounded sets and the function

^{∗}*x*is

^{∗}→ Sx^{∗}*w*-lower semicontinuous on bounded sets. Also

^{∗}Hence we can define the equivalent norm *||| · |||* on *X ^{∗}* by

Since all three summands are *w ^{∗}*-lower semicontinuous the unit ball in this norm is

*w*-closed. Thus

^{∗}*||| · |||*is norm dual to an equivalent norm on

*X*. By Lemma 3.2.2 it suffices to prove that (

*X*) is locally uniformly convex.

^{∗}, ||| · |||Since each of the expressions in (3.1) satisfies the triangle inequality, we obtain that

Hence (which we will not use) and, more important,

Since each of the expressions (3.2), (3.3), and (3.4) is positive,

By the (local) uniform convexity of _{2} it follows that

which implies that for each *n*,

Since the sequence (*x _{n}*) is dense in the unit sphere in

*X*and is bounded, we infer that in the

*w*-topology. For a contradiction, suppose that for some

^{∗}*ε >*0 there is a subsequence of (which we may assume to be the original sequence) such that

Pick *n* such that . Then for every *k* large enough there is a with . Since *F _{n}* is finite dimensional, a subsequence of converges to some . Thus

and we get the desired contradiction,

(p.27)
In the space _{1} it is easily seen that the norm is nowhere Fréchet differentiable. The next renorming result, whose first form is due to Leach and Whitfield, shows that this is a special case of a general phenomenon. We state and prove it in a rather strong form that will allow immediate applications to proving results showing that existence of suitable smooth functions implies separability of the dual.

Theorem 3.2.3.

Assume that(X, ·)is separable but X0^{∗}is nonseparable. Then for every< c <1there is an equivalent norm ||| · ||| on X such that ||| · ||| ≤ · , and for every x ∈ X, r >0,and every subspace Y of X such that X/Y has separable dual,

Proof.We choose 1> a > b > candε, τ >0 such that (1− τ)b −τ > c.LetEbe the family of norm separable subspaces ofX. For every^{∗}E ∈ Elet

Then

pis a pseudonorm on_{E}Xmajorized by·. Moreover, we notice that the pseudonormpsatisfies whenever_{E}E ⊃ E_{1}∪ E_{2}. Thus the infimum

is also a pseudonorm on *X* majorized by *·* . Consequently,

is an equivalent norm

Xmajorized by·.

Suppose now that *x ∈ X*, *r >* 0 and *Y* is a subspace of *X* such that *X/Y* has separable dual. We let and show that there is *y ∈ Y* with *y* = 1 such that for every *E ∈ E* one can find with

Indeed, if this were not the case, then for every

ybelonging to a countable dense subsetSof the unit sphere ofYwe would find∈ Esuch that(^{∗}y)≤for everyE_{y}yawith|y(^{∗}x)− p(x)| <2ε. ChooseE ∈ Econtaining such that|p(_{E}x)− p(x)| < ε. Then by the definition ofpthere is_{E}ysuch that^{∗}∈ M_{E}y(^{∗}x)> p(_{E}x)− ε, and so

Since we see that *y ^{∗}* belongs to all

*M*,

_{Ey}*y ∈ S*, which implies that

Hence

(p.28)
Having found *y ∈ Y* and such that *y* = 1 and (3.5) holds, we estimate

Hence *p*(*x* + *ry*) *− p*(*x*) *≥ −*2*ε* + *ar* = *br* and we are ready to finish the proof by concluding that

Corollary 3.2.4.

Assume that X is separable but X0^{∗}is nonseparable. Then for every< ε <1there is an equivalent norm ||| · ||| on X such that for every x,

Proof.Let||| · |||be the norm obtained in Theorem 3.2.3 with someε < c <1. Givenx ∈ X, we findxwith^{∗}∈ X^{∗}|||x= 1 and^{∗}|||x(^{∗}x) =|||x|||. Using the statement of Theorem 3.2.3 withYthe kernel ofx, we find for any given^{∗}r >0 a pointy ∈ Ysuch thaty ≤ rand

Since also we have

Remark 3.2.5. Recall that, for

ε >0, a functionf:X −→ Yis calledε-Fréchet differentiable atx_{0}if for someT ∈ L(X, Y) andδ >0,

whenever *x* is small enough.

We will study this notion in detail in the next chapter (where it is defined more formally in Definition 4.1.1), but we notice already that, given any the norm *||| · |||* from Corollary 3.2.4 (used with 2*ε* instead of *ε*) is not *ε*-Fréchet differentiable at any *x*_{0}. Indeed, if *x ^{∗} ∈ X^{∗}* andfor

*x < δ*, then

for all such *x*, which is incompatible with the conclusion of Corollary 3.2.4.

# 3.3 Fre´chet Differentiability of Convex Functions

It is clear from Corollary 3.2.4 that positive results on existence of Fréchet derivatives of continuous convex functions *f* : *X −→* R, with *X* separable, may hold only if *X ^{∗}*
(p.29)
is separable. The same is true of course for general Lipschitz maps from

*X*to any Banach space

*Y*.

Positive results on existence of points of Fréchet differentiability of continuous convex functions on spaces with separable dual are indeed valid, and are even valid in the sense of Baire category. Lindenstrauss proved that if *X* is separable and reflexive then every continuous convex *f* : *X −→* R is Fréchet differentiable on a dense *G _{δ}* set (hence outside a set of the first category). This result was extended by Asplund to every space with

*X*separable. In view of this result, spaces

^{∗}*X*with

*X*separable are called

^{∗}*Asplund spaces*. Notice that in the literature the term “Asplund space” often refers to Banach spaces

*X*such that for every separable subspace

*Z ⊂ X*the space

*Z*is separable. In this work, however, Asplund space will denote just spaces

^{∗}*X*with

*X*separable. Only in comments or remarks (especially when we treat separable determination in Section 3.6) we may occasionally mention the general case.

^{∗}A stronger result than Asplund’s with a much simpler proof was obtained by Preiss and Zaj´ıˇcek. It seems to be the first result connecting the question of Fréchet differentiability with the notion of porous set. (See Section 1.1 of the Introduction for the definition.) It turned out to play a central role in the whole subject and in particular in this volume.

Theorem 3.3.1.

A continuous convex function f on an Asplund space X is Fréchet differentiable outside a σ-porous set.

Proof.For each integerm ∈N cover the spaceXby a sequence of balls of radius 1^{∗}/(6m). Fork, m, n ∈N consider the sets

The set where *f* is not Fréchet differentiable is . We show that each is porous with constant 1*/*(6*mn*).

Fix *x ∈ A _{k,m,n}* and

*ε >*0. Diminishing

*ε*if needed, we achieve that

*f*is Lipschitz with constant 2

*n*on

*B*(

*x*, 2

*ε*). Indeed, let

*r, τ >*0 be such that

whenever *y ≤ r*. If *f* were not 2*n*-Lipschitz on a neighborhood of *x*, we could find points *y _{i}, z_{i} ∈ X* such that

*y*,

_{i}→ x*z*, and

_{i}→ xPut Then *w _{i} ∈ B*(

*x, r*) for large

*i*. By the convexity of

*f*we see that

(p.30)
On the other hand, for large *i*,

Combining the last two inequalities we get

which gives a contradiction for *i →∞*. By the definition of *A _{k},*we now find

*u*(

^{∗}∈ ∂f*x*)

*∩ B*and

_{k},*y ∈ X*with

*such that*

^{∗}_{m,n m}y < ε*u < n*and

We claim that

Assume otherwise, that there is a point *z* in this intersection. Let *v ^{∗}* be the element in

*∂f*(

*z*) witnessing that Then

*v*(

^{∗}*x − z*)

*≤ f*(

*x*)

*− f*(

*z*) and also

because both belong to the same ball *B _{k,m}*. Noticing that

we obtain

On the other hand, we have already found that and so using that *f* is 2*n*-Lipschitz we get

This contradiction finishes the proof.

# (p.31) 3.4 Porosity and Nondifferentiability

We first point out another important, although obvious, connection between porous sets and Fréchet differentiability.

Remark 3.4.1. If

Eis a porous set in a Banach spaceX, then thedistance function f(x) := dist(x,E) is nowhere Fréchet differentiable onE.

Proof.LetEbe a porous with constantcand letx ∈ E. Thenf(x) = 0, and sincefattains its minimum atxthe only possible derivative offatxis 0. Forε >0, letz ∈ Xsatisfyz ≤ εand

Then

which makes *f* (*x*) = 0 impossible.

A slightly more involved variant of this remark is that even a *σ*-porous set is contained in the set of points of Fréchet nondifferentiability of some real-valued Lipschitz function. This was proved in [40] (see also [4, Theorem 6.48]) for countable unions of closed porous sets, and Kirchheim observed that “closed” is in fact not needed.

Lemma 3.4.2.

Suppose that X has a Fréchet smooth norm and E ⊂ X is porous with constant c. Then there is f:X −→[0, 1]withLip(f)≤2such that

*Proof.* For every we define . By the 5*r*-covering theorem (see, e.g., [33, Theorem 2.1]), there are *x _{j} ∈ X \ E* such that the balls

*B*(

*x*), where

_{j}, r_{j}*r*=

_{j}*r*(

*x*), are disjoint, and for every

_{j}*x ∈ X \ E*there is

*j*such that

*B*(

*x, r*(

*x*))

*⊂ B*(

*x*, 5

_{j}*r*). Notice also that the balls

_{j}*B*(

*x*, 6

_{j}*r*) still do not meet

_{j}*E*. Define

Clearly, 0 *≤ f* (*x*) *≤* 1 and *f* has Lipschitz constant at most two on the closure of each ball *B*(*x _{j}, r_{j}*). Since on the boundary of these balls both formulas defining

*f*coincide, this implies that Lip(

*f*)

*≤*2.

The assertion (i) is obvious at the points where *f* (*x*) = 0. When *f* (*x*) *>* 0, *x* belongs to one of the balls *B*(*x _{j}, r_{j}*). In that case

*f*is Fréchet differentiable at

*x*and (i) follows.

(p.32)
To prove (ii), let *∈* and 0 1. It suffices to find *u* with such that *x E < ε < u u< ε* Indeed, since *f*(*x*) = 0 and *f*(*x − u*) *≥* 0, this will imply that

and so the statement.

By the porosity assumption on *E*, there is such that

Then

y /∈ Eand dist(y,E)≤ y x <1, and so Hence there isjsuch that In particular,Since

y ∈ B(x, 5_{j}r),_{j}x+ 5_{j}− x ≤ x − yr. Using also that_{j}x /∈ B(x, 6_{j}r), we get_{j}x6_{j}− x ≥rand so_{j}r. Hence_{j}≤ x − yu:=xsatisfies_{j}− xu ≤ x − y+ 5r6_{j}≤x − yand

Since *u ≤* 6*x − y < ε*, this shows exactly what we needed.

Theorem 3.4.3.

Let E be a σ-porous subset of a separable Banach space X. Then there is a Lipschitz function from X toRwhich is not Fréchet differentiable at any point of E.

Proof.IfXis nonseparable, the norm constructed in Corollary 3.2.4 provides the required example. Hence we assume that^{∗}Xis separable and so, by Theorem 3.2.1 that its norm is Fréchet smooth.^{∗}

Let where *E _{k}* is porous with constant

*c*, 0

_{k}*< c*1. Find the corresponding function

_{k}<*f*according to Lemma 3.4.2 and choose

_{k}*α*0 such that

_{k}>We show that the function

has the required property.

Obviously, *f* is well defined and Lipschitz. Suppose that *x ∈ E* and find *k* such that *x ∈ E _{k}*. By Lemma 3.4.2 (i)

(p.33) and a simple estimate by the Lipschitz constant says that

Hence, using for *f _{k}* the inequality 3.4.2 (ii), we have

which shows that *f* is not Fréchet differentiable at *x*.

Remark 3.4.4. If

Eis porous in the direction of a vectoru, the distance fromEis at any point ofEnondifferentiable in the direction ofu. IfXis separable, we may use the construction from the previous theorem with a Gâteaux smooth norm onXand so find, for any givenσ-directionally porous setE ⊂ X, a real-valued Lipschitz functionfsuch that for everyx ∈ Ethere is a direction in whichfis not differentiable. In particular,ffails to be Gâteaux differentiable at any point ofE.

The closure of a porous set is obviously nowhere dense, and thus *σ*-porous sets are of the first category. In a finite dimensional space porous sets are, by Lebesgue’s density theorem, sets of measure zero. We mention here in passing that even in the real line the *σ*-ideal of *σ*-porous sets is much smaller that the *σ*-ideal of sets which are of the first category and measure zero; see [49] or [51, p. 526] where it is pointed out that this result essentially goes back to Beurling and Ahlfors [5].

On the other hand, in infinite dimensional spaces porous sets need not be small in the sense of Gauss measure. In fact, Matouˇsek and Matouˇskov´a [34] (see also [4, Example 6.46]) proved that there is an equivalent norm on _{2} which is Fréchet differentiable only on a Gauss null set. This example was extended by Matouˇskov´a [35] to every separable uniformly convex space. (With similar reasoning, we will reuse this example in Chapter 5 to quickly see the difference between Gauss null sets and another *σ*-ideal of negligible sets, so-called Γ-null sets, which will be introduced there.) An earlier example in [40] (see also [4, Theorem 6.39]), which is, however, not related to differentiability, shows that porous sets are not small in the sense of Gauss measure in any infinite dimensional separable Banach space: any such space can be decomposed into two sets, one *σ*-porous and the other Gauss null.

# 3.5 Sets of Fre´Chet Differentiability Points

In this short section we show that for an arbitrary map between Banach spaces the set of its points of Fréchet differentiability is Borel; in fact it has type *F _{σδ}*. Such results are classical for real-valued functions of one real variable and have been extended to more dimensions by various authors. The particular extension we treat here is due to Zaj´ıˇcek [50]. Notice that in general the type of these sets cannot be improved even if the map is supposed to be Lipschitz. Indeed, by the result of Zahorski mentioned in
(p.34)
Chapter 2, for any

*F*subset

_{σδ}*E*of R whose complement has measure zero there is a Lipschitz function

*f*: ℝ

*−→*ℝ whose set of points of differentiability is precisely

*E*.

In this connection notice that for sets of points of Gâteaux differentiability the situation is quite different. For Lipschitz maps on separable spaces it is easy to check that these sets are Borel (see, for example, the beginning of the proof of [4, Theorem 6.42]), and an inspection of the argument reveals that they are *F _{σδ}*. But in general they may well be non-Borel. For continuous convex functions on separable spaces they are

*G*, but unlike the sets of their points of Fréchet differentiability, which are

_{δ}*G*in all Banach spaces, they may fail to be Borel in nonseparable spaces [44] and even in nonseparable Hilbert spaces [21].

_{δ}The main difficulty in proving that the set of points of differentiability is Borel is that, a priori, this set is obtained as a union over all possible values of the derivative, which is far from being a countable union. The usual way of overcoming this difficulty is via a criterion of Fréchet differentiability that does not need the value of the derivative. Here we use a different approach, which so far as we know is new in this context, via the canonical embedding of the target space into the second dual and compactness of balls in the *w ^{∗}*-topology.

For a map *f* : *X −→ Y* , *x ∈ X*, and *r >* 0 denote

. The reason for keeping *X* as a parameter in *ε*(*f, x, r,X*) will appear in the next section, where for *x* belonging to a subspace *U* of *X* we will look at the relation between differentiability of *f* : *X −→ Y* and of its restriction to *U*, hence at the relation between the quantities *ε*(*f, x, r,X*) and *ε*(*f, x, r,U*).

As the following simple observation says, the limit

measures how far

fis from being Fréchet differentiable atx.Observation 3.5.1.

A map f of a Banach space X to a Banach space Y is Fréchet differentiable at x ∈ X if and only if ε(f, x,X) = 0.Proof.Supposeε(f, x,X) = 0. Then there arer0 and_{k}>L(_{k}∈ LX, Y) such that for everyu ∈ Xwith This implies that

for

u <min{}{}{}{}{}{}{r. Hence_{j}, r_{k}}L2_{j}− L_{k}≤+ 2^{−j}, implying that the sequence^{−k}Lconverges to some_{k}L ∈ L(X, Y) andL − L2_{k}≤. Consequently,^{−k}

showing that *L* is the Fréchet derivative of *f* at *x*. The opposite implication is obvious.

(p.35)
Notice that the notion of *ε*-Fréchet differentiability, which we have already mentioned in Remark 3.2.5, is, since we are normally interested in small *ε >* 0, for all practical purposes equivalent to the requirement that *ε*(*f, x,X*) *≤ ε.* More precisely, *ε*(*f, x,X*) *< ε* implies that *f* is *ε*-Fréchet differentiable at *x*, and this implies that *ε*(*f, x,X*) *≤ ε*. In particular, the *ε*-Fréchet differentiability of *f* at *x* for every *ε >* 0 is equivalent to the fact that *ε*(*f, x,X*) = 0, and so Observation 3.5.1 translates to the often quoted

Observation 3.5.2.

Any map f:X −→ Y which is ε-Fréchet differentiable at a point x ∈ X for every ε >0is Fréchet differentiable at x.

In general, the infimum defining *ε*(*f, x, r, X*) is not attained, but it is attained in the presence of compactness. This also allows precise description of *ε*-differentiability using the quantities *ε*(*f, x, r,X*).

Observation 3.5.3.

Suppose that f:X −→ Y , where Y is a dual Banach space and ε(f, x, r,X)< ∞. Then there is L ∈ L(X, Y)such that for every u with u < r.

*Consequently, f is ε-Fréchet differentiable at x if and only if ε*(*f, x, r, X*) *≤ ε for some r >* 0.

Proof.For everyi ≥1 pickL(_{i}∈ LX, Y) such that

for every *u* with *u < r*. Since then *L _{i}*(

*u*)

*− L*

_{1}(

*u*)

*≤*2(

*ε*(

*f, x, r,X*) + 1)

*u*for

*u < r*, the operators

*L*form a bounded sequence in

_{i}*L*(

*X, Y*). Let

*K >*0 be such that

*L*. Choose a free ultrafilter U on N. Assuming that

_{i}≤ K*Y*=

*Z*, we consider for every

^{∗}*u ∈ X*and

*z ∈ Z*the limit

Since the absolute value of this limit is dominated by *K u z*, it defines for every fixed *u* an element of *Z ^{∗}* =

*Y*and, consequently, it also defines a bounded linear operator

*L ∈ L*(

*X, Y*) satisfying

It follows that for every *z ∈ Z*, *u ∈ X*, and *η >* 0 there are arbitrarily large *i ∈* N such that *|*(*L _{i}u*)

*z −*(

*Lu*)

*z| ≤ ηz*. Then for every

*u ∈ X*with

*u < r*we find a norm one

*z ∈ Z*such that

and infer that

(p.36)
Given *η >* 0, *i* can be arbitrarily large, and then letting *η →* 0 we obtain the statement.

After this digression we return to the real theme of this section.

Proposition 3.5.4.

Suppose that f:X −→ Y , where Y is a dual Banach space. Then for every c >0the set {x ∈ X | ε(f, x,X)< c} is F_{σ}.

Proof.Let

We show that

for every This will establish that the set in question is hence *F _{σ}*.

Let and find *x _{i} ∈ E_{j,k}* such that

*x*. (Note that this implicitly means that

_{i}→ x*c >*3

*/k*.) For each

*i*we choose

*L*(

_{i}∈ L*X, Y*) such that

If *i*_{0} is such that for *i ≥ i*_{0} then for every *u <* _{2}^{1} * _{j}* ,

Hence and so (*L _{i}*) is a bounded sequence. Arguing in the same way as in the proof of Observation 3.5.2, we may find

*L ∈ L*(

*X, Y*) such that for every

*z ∈ Z*,

*u ∈ X*, and

*η >*0 there is

*i ∈*N such that (

*L*)

_{i}u*z −*(

*Lu*)

*z ≤ ηz*.

Suppose that We find a unit vector *z ∈ Z* such that

Then we find *i ∈* N such that and conclude that

Hence

Corollary 3.5.5.

The set of points of X at which a map f from X to a Banach space Y is Fréchet differentiable is F_{σδ}.(p.37)

Proof.We may considerfas a mapping intoY, as this does not change the notion of Frechet ´differentiability. Then we see from Observation 3.5.1 that its set of points of Frechet ´differentiability is which is^{∗∗}Fby Lemma 3.5.4._{σδ}

# 3.6 Separable Determination

The key idea behind separable determination (or separable reduction) is that some notions with which we wish to work in nonseparable Banach spaces may in fact use countability strongly enough so that statements about them hold in a nonseparable space provided they hold in its separable subspaces. There are several approaches to this. Here we follow the approach that started in [38], as modified by various authors. Although in all applications one makes the final deduction using just one separable subspace, it is convenient to know that the family of subspaces that can be used is so large that one easily join countably many arguments together. We will therefore use the concept of rich families of subspaces introduced in [6] by Borwein, Moors, and an unnamed mathematician (about whom the authors say “whose incisive comments formed the genesis of this paper”).

Definition 3.6.1. Let

Xbe a Banach space. A familyRof separable subspaces ofXis calledrichif

(i) for every increasing sequence *R _{i}* in

*R*, belongs to

*R*, and

(ii) each separable subspace of

*X*is contained in an element of*R*.

Recall that for us “a subspace” means “a closed subspace,” so this is indeed the definition from [6], from which also comes the following statement.

Proposition 3.6.2.

The intersection of countably many rich families is a rich family.

*Proof.* The requirement (i) of the definition is obvious. To show (ii), let (*R _{n}*) be a sequence of rich families and

*Y*a separable subspace of

*X*. Let (

*n*) be a sequence of natural numbers in which each number occurs infinitely often. Denote

_{i}*R*

_{0}=

*Y*and for

*k*= 1, 2,

*…*use the property (i) recursively to choose such that Observing that the subspace satisfies, for each

*n*,

we infer from the property (i) of *R _{n}* that

*R ∈ R*. Hence

_{n}*R*belongs to the intersection of the

*R*, and it obviously contains

_{n}*Y*.

The notion of separable determination could be defined formally in the following way. A property *P* of pairs (*x,U*), where *x ∈ X* and *U* is a subspace of *X*, is *separably determined* if there is a rich family *R* of separable subspaces of *X* such that for every *R ∈ R*and *x ∈ R*, the pair (*x,X*) has property *P* if and only if (*x,R*) does. Usually, however, one does not feel that a given property is of this form, and so we state and
(p.38)
prove the separable reduction statements without using this notion. To illustrate this point, separable determination of Fréchet differentiability would, according to this definition, be expressed as: for any *f* : *X −→ Y* , the property “*f* is Fréchet differentiable at the point *x* in direction *U* ” of pairs (*x,U*) is separably determined. In our opinion, it is more revealing (although slightly longer) to say that there is a rich family *R* of separable subspaces of *X* such that for every *R ∈ R*and *x ∈ R*, the function *f* is Fréchet differentiable at *x* if and only if its restriction to *R* is Fréchet differentiable at *x*.

We will now present a new general setup for a number of separable determination questions. For simplicity, let us concentrate only on Fréchet differentiability. This is a problem of linear approximation, so one may imagine that the real statement is that even the error of the best linear approximation of our function on balls around points of *R* is the same for the original problem and for its restriction to *R*. We may measure the error of this approximation inside a subspace *U* by

but a technically more convenient expression is

To allow more applications, we will write the fraction as

where *F* is a function on *U* ^{2} *× Y* ^{2}. We also allow an arbitrary number of variables instead of just two. A more delicate point is that we have not a single function, but a collection of them, and this collection depends on the subspace *U* of the ambient space *X*. The last point is important, as it leads to the idea of measuring the error as a supremum not over the whole space but over its separable subspaces, and thereby it leads to the question of separable determination of a somewhat strange looking minimax type problem.

The general setup we arrived at is as follows. For every separable subspace *U* of *X*, let F(*U*) be a collection of non-negative functions on *U ^{p} ×Y ^{p}* such that the restriction of functions from F(

*U*) to

*V ⊂ U*belongs to F(

*V*). By F we denote the collection of all F(

*U*). (These collections depend also on

*p*and

*Y*, but since

*p*and

*Y*are always clear from the context, we do not indicate this dependence.)

To simplify the notation, we denote for *x ∈ X* and *u ∈ X ^{p}*,

Also, for *u ∈ X ^{p}* we put

(p.39)
The quantities measuring the approximability of *f* are now defined in the natural way. First, for *F ∈* F(*U*) we let

and then define

Notice that for separable *W* we have a simpler formula,

This formula holds for any subspace of *X* provided we have a family F of functions on *X ^{p} ×Y ^{p}* and each F(

*U*) is the collection of their restrictions to

*U*. This simpler formula will be true in some but not all of our applications. In the opposite direction, it would be natural to restrict the domain of F(

^{p}×Y^{p}*U*) only to finite dimensional

*U*, but we do not have any application of this more general approach.

For the validity of separable determination, we still need another assumption. For that recall that in (3.6) the linear operators *L* are (as long as we bound their norms) uniformly equicontinuous. So, up to a countable decomposition (which we can handle using Proposition 3.6.2), we assume that our collections are uniformly equicontinuous. This is enough for our purposes, but let us note that we will actually use only the following consequence of uniform equicontinuity. For every *ε >* 0 there is *δ >* 0 such that whenever *F ∈* F(*U*), *u, v ∈ U ^{p}*,

*y ∈ Y*, and

^{p}*u − v < δ*, then

*|F*(

*u, y*)

*− F*(

*v, y*)

*| < ε*.

Observation 3.6.3.

If ε and δ are as above and W is a subspace of X, then for every x, ˜x ∈ W with x −˜x < δ,

*Proof.* Fix any 0 *< c < β*(*f, x,W*, F). By definition of *β*(*f, x,W*, F) there is a separable subspace *U ⊂ W* such that *β*(*f, x, U, F*) *> c* for every *F ∈* F(*U*). Let *V* be the linear span of *U ∪ {x −* ˜*x}*. Then, given any *F ∈* F(*V* ), the restriction of *F* to *U* belongs to F(*U*), and so we may find *u ∈ U ^{p}* such that

*F*(

*u, f*(

*x*+

*u*))

*> c*. It follows that, with ˜

*u*=

*u*+

*x −*˜

*x ∈ V*,

^{p}Hence proving the statement.

Lemma 3.6.4.

Suppose thatFis uniformly equicontinuous. Then for every function f:X −→ Y the family R of separable subspaces R of X such that for every x ∈ R,

*is rich on X.*

(p.40)

Proof.We first show a slightly stronger version of the first requirement from the definition of richness. If (W) is an increasing sequence of separable subspaces of_{i}Xsuch that for eachithe set of ˜x ∈ Wwith_{i}

is dense in *W _{i}*, then belongs to

*R*. To this aim we observe that the inequality

*β*(

*f, x,W*, F)

*≤ β*(

*f, x,X*, F) is obvious. To prove the opposite, suppose that

*x ∈ W*,

*η >*0, and

*β*(

*f, x,W*, F)

*< ∞*. By uniform equicontinuity of F there is

*δ >*0 such that

*|F*(

*u, y*)

*− F*(

*v, y*)

*| < η*whenever

*F ∈*F(

*W*),

*u, v ∈ W*, and

*u − v < δ*. Choose

*i*large enough that

*B*(

*x, δ*)

*∩ W*=

_{i}*∅*. By assumption, there is ˜

*x ∈ B*(

*x, δ*)

*∩W*for which (3.7) holds. Hence, applying Observation 3.6.3 twice, we get

_{i}To prove the second requirement from the definition of richness we show that there is a way to assign to every separable subspace *V* of *X* a separable subspace *V ⊃ V* of *X* such that *β*(*f, x, V* , F) *≥ β*(*f, x,X*, F) for every *x* from a dense subset of *V* . This will finish the proof since, defining = *V* and = *W _{k}*, we infer from

*W*

_{1}

*W*

_{k}_{+1}what we have proved above that belongs to

*R*.

To define *V* , let *S* be a countable dense subset of *V* . For every *x ∈ S* and rational *c < β*(*f, x,X*, F) find a separable subspace *U _{x,c}* of

*X*such that

*β*(

*f, x,U*)

_{x,c}, F*> c*for every

*F ∈*F(

*U*). Then the space

_{x,c}*V*given as the closed linear span of

*V*together with all these

*U*has the required property.

_{x,c}In the following simple applications, we have a family F of functions on *X ^{p} ×Y ^{p}*, and F(

*U*) is the collection of their restrictions to a subspace

*U*. Hence the simpler formula for

*β*(

*f, x,W*, F) applies to all subspaces of

*X*.

Corollary 3.6.5.

Let f:X −→ Y . Then there is a rich family R on X such that for every R ∈ R, x ∈ R, and r >0,

*Proof.* For rational *s, C >* 0 let F* _{s,C}* (

*U*) consist of the single function

By Lemma 3.6.4 there is a rich family *R _{s,C}* such that for every

*R ∈ R*and

_{s,C}*x ∈ R*,

Let *r >* 0. By Proposition 3.6.2 the family

(p.41)
is rich on *X*. Let *R ∈ R* and *x ∈ R*. By taking the supremum in (3.8) over rational 0 *< s < r* and*C >* 0 we get

This simple result is all that we need to show that porosity is separably determined. To give a detailed statement, we define for *E ⊂ X* the porosity of *E* at a point *x ∈ X* as

If for some *r >* 0 there is no such *c*, we let *p*(*E, x*) = 0. If *Y* is a subspace of *X*, we define the porosity of *E* at *x* in the direction of *Y* by the same formula in which *y* is restricted to belong to *x* + *Y* .

Corollary 3.6.6.

Let E ⊂ X. Then there is a rich family R on X such that for every R ∈ R and every x ∈ R, the porosity of E at x in the direction of R is equal to its porosity (in the direction of X).

*Proof.* Observe that

and use the previous result for *f* (*x*) = dist(*x,E*).

By Proposition 3.6.2 we have an immediate corollary.

Corollary 3.6.7.

Let E ⊂ X be a σ-porous subset of X. Then there is a rich family R on X such that for every R ∈ R, the set E ∩ R is σ-porous in R.

Before discussing differentiability, it is natural to have a brief look at continuity.

Corollary 3.6.8.

For any function f:X −→ Y there is a rich family R such that for every R ∈ R and every x ∈ R, the function f is continuous at x if and only if its restriction to R is.

Proof.Use Lemma 3.6.4 with, for example, F(U) consisting of functions

It suffices to observe that *f* is continuous at *x* if and only if *β*(*f, x,X*, F) = 0.

We now turn our attention to linear approximations related to Fréchet differentiability. For that, we recall the quantities *ε*(*f, x, r,U*) from the previous section. As we could have already seen there, constructing a *Y* -valued linear operator is a nontrivial task, and so more delicate separable determination results need a compactness assumption in the target space *Y* . We will therefore consider functions with values in dual spaces (weaker assumptions may be also used). Incidentally, one may notice that in this situation the infimum defining *ε*(*f, x, r,U*) is actually attained.

(p.42) Theorem 3.6.9.

For every function f:X −→ Y , where Y is a dual Banach space, there is a rich family R on X such that ε(f, x, r,R) =ε(f, x, r,X)whenever R ∈ R, x ∈ R, and r >0.

*Proof.* Since the inequality *ε*(*f, x, r,R*) *≤ ε*(*f, x, r,X*) is obvious, we show the converse.

Let *ψ _{k}* : [0,

*∞*)

^{4}

*−→*[0,

*∞*),

*k ≥*1, be continuous functions with compact support having the following property. For every

*r >*0 there is a sequence (

*k*)

_{j}*⊂*N such that

For *k, l ∈* N and a separable subspace *U* of *X*, define F* _{k,l}*(

*U*) as the collection of functions

where *L ∈ L*(*U, Y* ) and *L ≤ l*. It is easy to see that each F* _{k,l}* satisfies the assumptions of Lemma 3.6.4. Using also Proposition 3.6.2, we find a rich family

*R*on

*X*such that

for every *x ∈ R* and *k, l ∈* N.

Suppose now that *R ∈ R*, *x ∈ R*, and *r, ε >* 0. Find a sequence (*k _{j}*)

*⊂*N such that (3.9) holds. By definition of

*ε*(

*f, x, r,R*) there is

*T ∈ L*(

*R, Y*) such that

Hence

which shows that

for every *j ∈* N. Thus, fixing *l ∈* N such that *l > T*, we see

for each *j*. This means that for every *j ∈* N and every separable subspace *U* of *X*, there is with and

Consider on the set I = *{*(*j,U*) *| j ∈* N, *U ⊂ X* separable subspace*}* a partial order defined as (*j,U*) (*k, V* ) if and only if *j ≤ k* and *U ⊂ V* . Then the sets *{*(*k, V* ) *|* (*j,U*) (*k, V* )*}*, (*j,U*) *∈* I, form a filter base on I. Let U be an ultrafilter extending this filter base. Assuming that *Y* = *Z ^{∗}* we put for any

*u ∈ X*and

*z ∈ Z*,

Since the absolute value of this limit is bounded by *lu z*, it defines for every fixed *u ∈ X* an element of *Y* = *Z ^{∗}* and, consequently, it also defines a bounded linear operator

*L ∈ L*(

*X, Y*) satisfying

It follows that for every separable subspace *U*_{0}, any *u ∈ U*_{0}, every unit vector *z ∈ Z*, any *η >* 0, and *j*_{0} *∈* N, there are *j >* and a subspace *U ⊃ U*_{0} such that

- Suppose now that
u ∈ U_{0}and 0< u < r. Find a unit vectorz ∈ Zsuch that

Also, find *j*_{0} *∈* N such that for every *j > j*_{0},

Finally, we find *j >* and *U ⊃ U*_{0} such that *|*(*L*_{(j,U)}*u*)*z −* (*Lu*)*z| < εu*, and estimate

where the last step follows from (3.10). Hence *ε*(*f, x, r,X*) *≤ ε*(*f, x, r,R*) + 4*ε*, and so, since *ε >* 0 is arbitrary, *ε*(*f, x, r,X*) *≤ ε*(*f, x, r,R*).

It is now immediate to deduce the separable determination statement for Fréchet differentiability. It is a variant of Zaj ´ıˇcek’s [50] strengthening of the separable reduction statement originating in [38] (see also [27]).

Theorem 3.6.10.

For every f:X −→ Y there is a rich family R on X such that for every R ∈ R, f is Fréchet differentiable (as a function on X) at every x ∈ R at which its restriction to R is Fréchet differentiable (as a function on R).(p.44)

Proof.We may considerfas a mapping intoY; this does not change the notion of Fréchet differentiability. Hence the statement follows from Theorem 3.6.9 and Observation 3.5.1.^{∗∗}Remark 3.6.11. The proof of the previous theorem cannot be repeated verbatim for

ε-Fréchet differentiability, whereε >0 is fixed. But if the target spaceYis a dual space, we may combine Theorem 3.6.9 and Observation 3.5.3 to get that for everyf:X −→ Ythere is a rich familyRonXsuch that for everyR ∈ R,fisε-Fréchet differentiable (as a function onX) at everyx ∈ Rat which its restriction toRisε-Fréchet differentiable (as a function onR).

Notice that Corollary 3.6.5 applied to the distance from *E* shows that a set *E* in *X* is nowhere dense if and only if there is a rich family in each element of which *E* is nowhere dense. Hence first category sets are separably determined in the same way as *σ*-porous sets are in Corollary 3.6.7. More interestingly, this statement may be “lifted” to the Borel Γ* _{n}*- and Γ-null sets that will be introduced in Definition 5.1.1. In Corollary 5.6.2 we will prove that Γ

*- and Γ-nullness of Borel sets are separably determined.*

_{n}By a straightforward use of separable reduction, a number of results from this book may be easily extended to a nonseparable setting. Since this would be rather mechanical, we will not state the results that may be obtained, but just give several examples illustrating the technique of proving nonseparable versions of some of the theorems that we prove later.

(1)

*Real-valued Lipschitz, and even cone-monotone functions on (nonseparable) Asplund spaces (i.e., spaces such that for every separable subspace Z ⊂ X the space Z*The Lipschitz statement is an immediate consequence of Theorems 12.1.1 and 3.6.10. (Notice that this is how the Fréchet differentiability result for Lipschitz functions was extended to nonseparable Asplund spaces already in [39], but that the slicing technique of [27] can prove it without the use of separable reduction.) In the case of functions monotone with respect to a cone^{∗}is separable) have points of Fréchet differentiability.*C*we find the rich family*R*from Theorem 3.6.10 and choose*x*_{0}*∈ C*. Further we observe that the family of*R ∈ R*that contain*x*_{0}is rich in*X*and the restriction of*f*to such*R*is cone-monotone. Hence we conclude the argument by a reference to Theorem 12.1.3. We also notice that this argument shows that the additional (mean value) statements of Theorems 12.1.1 and 12.1.3 also hold in nonseparable Asplund spaces.(2)

*σ-porous sets in (nonseparable) Asplund spaces are*Γ_{1}*-null.*For this, just combine Theorem 10.4.1 with Corollaries 3.6.7 and 5.6.2.(3)

*For any (possibly uncountable) set*Δ,*every Lipschitz map of**to*R2^{n}, where*≤ n ≤ p < ∞, has points of Fréchet differentiability.*To see this, notice first that the family of sets where*C*runs through countable subsets of Δ, is rich in(Δ). Suppose now that_{p}*f*:(Δ)_{p}*−→*R, where 2^{n}*≤ n ≤ p < ∞*, is Lipschitz. By Theorem 3.6.10 combined with Proposition 3.6.2, we can find a rich subfamily*R*of the above family such that for every (p.45)*R ∈ R*,*x ∈ R*, the function*f*is Fréchet differentiable at*x*iff its restriction to*R*is. By Theorem 13.1.1 the latter happens at some points of*R*, and so*f*is Fréchet differentiable at these points.(4)

*For any (uncountable) set*Δ,*every Lipschitz map of c*_{0}(Δ)*to a Banach space with the RNP is Fréchet differentiable*Γ*-almost everywhere.*Arguing as in the previous point, we find a rich subfamily*R*consisting of spaces isomorphic to*c*_{0}such that for every*R ∈ R*,*x ∈ R*, the given Lipschitz map is Fréchet differentiable at*x*iff its restriction to*R*is. By Theorem 6.4.3 the latter occurs Γ-almost everywhere in*R*. Since, by Corollary 3.5.5, the set of points of Fréchet nondifferentiability of*f*is a Borel subset of*c*_{0}(Δ), we conclude from Lemma 5.6.1 that*f*is Fréchet differentiable at Γ-almost every