Clarke ’ s Generalized Gradient and Edalat ’ s L-derivative

Clarke [2, 3, 4] introduced a generalized gradient for real-valued Lipschitz continuous functions on Banach spaces. Using domain theoretic notions, Edalat [5, 6] introduced a so-called L-derivative for real-valued functions and showed that for Lipschitz continuous functions Clarke’s generalized gradient is always contained in this L-derivative and that these two notions coincide if the underlying Banach space is finite dimensional. He asked whether they coincide as well if the Banach space is infinite dimensional. We show that this is the case. 2010 Mathematics Subject Classification 26E15, 46G05 (primary); 06A06, 03D78 (secondary)


Introduction
Clarke [4, Page 1] observed that "nonsmooth phenomena in mathematics and optimization occur naturally and frequently, and there is a need to be able to deal with them.We are thus led to study differential properties of nondifferentiable functions."Motivated by this observation, Clarke [2,3,4] introduced a generalized gradient ∂f (x) for real-valued Lipschitz continuous functions f on Banach spaces.Using domain theoretic notions in the realm of computable analysis, Edalat [5,6] introduced a so-called L-derivative for real-valued functions on Banach spaces.Domain theory arose in the context of computer science and logic.While the primary application of domain theory is in the semantics of programming languages, domain theoretic notions can also be applied successfully in computable analysis, as shown in [5,6] and in other articles by Edalat.As Clarke's generalized gradient and Edalat's L-derivative are defined using rather different mathematical notions it is remarkable that they are closely connected.Edalat [5,6] showed that for Lipschitz continuous functions Clarke's generalized gradient is always contained in his L-derivative and that these two notions coincide if the underlying Banach space is finite dimensional.He asked whether they coincide as well if the Banach space is infinite dimensional.We show that, indeed, for Lipschitz The weak * topology is a topology on the set X * .It is defined to be the coarsest topology such that for any x ∈ X the function is continuous.Every subset U ⊆ X * that is open in the weak * topology is also open in the topology induced by the norm || • || * .The converse is in general not true; see, eg, Megginson [9, Theorem 2.6.2 and Corollary 2.6.3].It is well known that X * with the weak * topology is a topological vector space and a Hausdorff space; see, eg, Rudin [10, Page 66 and Theorem 1.12].Subsets of X * that are compact in the weak * topology will be called weak * compact.

Clarke's Generalized Gradient
The terminology in this section is copied from Clarke [4, Chapter 2].
Let X be a Banach space with norm || • ||.Let Y be a subset of X , and let c be a non-negative real number.A function f : Let f : X → R be Lipschitz continuous near some point x ∈ X .Fix an arbitrary v ∈ X .The generalized directional derivative of f at x in the direction v, denoted f • (x; v), is defined by In the following section we spell this out in more detail (Lemma 1(4)).
For any x ∈ X the generalized gradient of f at x, denoted ∂f (x), is defined as follows: We show in the following lemma that this is well defined.This function can be considered as a global version of Clarke's generalized directional derivative.In the following lemma several elementary assertions about this function are collected.
Lemma 1 Let X be a Banach space, let U ⊆ X be a nonempty open subset, and let f : dom(f ) ⊆ X → R with U ⊆ dom(f ) be a function Lipschitz continuous on U .For the first four of the following five assertions, fix an arbitrary v ∈ X .
(1) The value f (U, v) is well defined, and if c ≥ 0 is a Lipschitz constant for ( Proof Fix some v ∈ X . (1) Let c ≥ 0 be a Lipschitz constant for f on U .Then for all z ∈ U and t > 0 such that z + tv ∈ U we have This shows that f (U, v) is well defined and satisfies f (U, v) ≤ c • ||v||.
(2) This follows directly from the definition of f (U, v).
(3) This is a consequence of the definitions of f • (x, v) and of f (U, v).
(4) Note that by the second statement the sequence ( f (B(x, 2 −n ), v)) n∈N is nonincreasing, and by the third statement it is bounded from below.Therefore its limit exists.That its limit is equal to f (5) We additionally fix some real number r > 0. Then Clarke's generalized directional derivative is positively homogeneous and subadditive [4, Proposition 2.1.1].In Lemma 1 (5) we have seen that also the global function v → f (U, v) is positively homogeneous.In the following lemma we show that it is subadditive as well if U is convex.
Lemma 2 Let X be a Banach space, let U ⊆ X be a convex open subset, and Proof We will fix some ε > 0 and show that for any v, w ∈ X Once this is proved for any ε > 0, the assertion So, let us fix some ε > 0 and elements v, w ∈ X .We choose a point z ∈ U and a real number t > 0 such that z + t(v + w) ∈ U and Peter Hertling One might try to prove the assertion simply by replacing the quotient on the left hand side by the following term: If z + tv would happen to be an element of U , then this term would be a lower bound for f (U, w) + f (U, v), and we would be done.Unfortunately, the point z + tv may lie outside of the open set U ; see Figure 1.We will have to proceed differently.This is 4 where the convexity of U will be important.As U is convex, the line segment is a subset of U .Let k be a positive integer so large that t||v|| k < δ.
Then for i = 0, . . ., k the points are elements of the line segment L and, thus, of the open set U , and the points z i + t k v are elements of L + B(0, δ), and, hence, elements of the open set U as well; see Figure 1.
Using these points we can replace the quotient f (z+t(v+w))−f (z) t by a sum of quotients as follows: where ε > 0 was chosen arbitrarily.This ends the proof.
5 About Nonempty, Convex, Weak * Compact Sets Let X be a Banach space.In this section, first we observe that any nonempty, convex, weak * compact subset K of X * can be expressed with the help of its support function.
Then, using the global version of the generalized directional derivative, we introduce certain nonempty, convex, weak * star compact subsets of X * .
Let K be a nonempty, weak * compact subset of X * .It is well known (see, eg, Megginson [9, Corollary 2.6.9]) that any such set is bounded, ie, there is some non-negative real number B such that Peter Hertling This implies for all ζ ∈ K and v ∈ X The function s K : X → R is called the support function of K .
Lemma 3 Let X be a Banach space.Let K ⊆ X * be a nonempty, convex, and weak * compact subset of X * .Then Proof The inclusion "⊆" follows from the definition of s K (v).We wish to prove the inclusion "⊇".Therefore, let us consider some ξ ∈ X * with ξ ∈ K .By the second part of Beer [1, Theorem 1.4.2]there exist an element v ∈ X and a real number α such that either In the second case we obtain thus, (1) holds for −v and −α.Therefore, we can assume that (1) holds.Then and, hence, Let U ⊆ X be a nonempty, open subset of the Banach space X , and let f : dom(f ) ⊆ X → R with U ⊆ dom(f ) be a function Lipschitz continuous on U .We define Lemma 4 (1) K U,f is a nonempty, convex, and weak * compact subset of X * .
(3) By the second assertion we obtain ∂f (x) For the inverse inclusion consider some ζ ∈ n≥n 0 K B(x,2 −n ),f .Then, for all n ≥ n 0 and all v ∈ X , (compare Lemma 1(4)), and this implies ζ ∈ ∂f (x).That was to be shown.

Continuous Functions from an Arbitrary Topological Space to a Bounded Complete DCPO
It is the purpose of this section to provide several fundamental facts concerning continuous functions f : X → Y where X may be an arbitrary topological space, eg a Banach space, and where Y is a directed complete partial order (dcpo), in particular, where Y is a bounded complete dcpo.Furthermore, we consider some special bounded complete dcpo's.We will need all this later in order to define Edalat's L-derivative for arbitrary functions.In order to make the presentation self-contained we introduce basic notions about dcpo's as well.
A set Z with a binary relation ⊆ Z × Z satisfying the following three conditions Proof It is clear that (C(X, Z), C ) is a partial order.Now we show that C(X, Z) with C is even a dcpo.Let F ⊆ C(X, Z) be a C -directed set.Then for every x ∈ X , the set is -directed.We define a total function g : X → Z by g(x) := sup(F(x)).By the definition of g, for all f ∈ F we have f C g, thus g is an upper bound of F .Finally, if h is an arbitrary upper bound of F , then for all x ∈ X , g(x) = sup(F(x)) ≤ h(x).Hence, g is a least upper bound of F .
Let (Z, ) be a partial order.
• An element y ∈ Z is called a least element of Z if for all z ∈ Z , y z.Obviously, if a least element exists then it is unique.Usually, when a least element exists, it is denoted ⊥.
• A subset S ⊆ Z is called bounded if there exists an upper bound z ∈ Z for S.
• Z is called bounded complete if for any bounded subset S ⊆ Z there exists a supremum of S in Z .
Lemma 7 Any bounded complete partial order has a least element.
This is well known.For completeness sake we give the proof.
Proof In any partial order, every element is an upper bound of the empty set.Hence, in a bounded complete partial order sup(∅) exists.It is a least element.
The following proposition covers the cases of bounded complete dcpos that we will need.

Peter Hertling
Proposition 8 Let Y be a nonempty topological vector space whose topology is a Hausdorff topology.Let (1) Then Z with defined as reverse inclusion is a bounded complete dcpo with least element Y .
(2) If S ⊆ Z is bounded or directed then Proof First, we observe that (Z, ) is a partial order.
Next, we prove the following claim.
Claim 1: For any S ⊆ Z , the set K∈S K is either empty or an element of Z .
For the proof, we distinguish three cases.
Case I: S = ∅.Then K∈S K = Y , and this is an element of Z .
Case II: S = {Y}.Then again K∈S K = Y , and this is an element of Z .
Case III: S contains at least one element K 0 that is different from Y , ie, that is nonempty, convex and compact.As all elements of S are closed (any compact subset of a Hausdorff space is closed; see, eg, Engelking [7,Theorem 3.1.8])and the intersection of arbitrarily many closed sets is closed as well, the intersection K∈S K is a closed subset of Y .In fact, it is a closed subset of the compact set K 0 , and hence (see, eg, [7, Theorem 3.1.2])compact itself.Note also that all elements of S are convex and that the intersection of arbitrarily many convex sets is again a convex set.Thus, K∈S K is a convex and compact subset of Y .If this intersection is not empty then it is an element of Z .We have proved Claim 1.
We continue with the following claim.
Claim 2: If S ⊆ Z is bounded or directed then K∈S K is not empty.
Let S ⊆ Z be a bounded set, and let K 0 ∈ Z be an upper bound of S. This means K 0 ⊆ K for all K ∈ S, hence, K 0 ⊆ K∈S K .And as all elements of Z are nonempty, K 0 is nonempty.Hence, K∈S K is nonempty.Now let S ⊆ Z be directed.If S = {Y} then K∈S K = Y , and this set is nonempty.Let us assume that S contains at least one element K 0 that is different from Y , hence, K 0 is a nonempty, convex, compact subset of Y .Using the assumption that S is a directed set, one shows by induction that the set Clarke's Generalized Gradient and Edalat's L-derivative 13 has the finite intersection property, ie, the intersection of any finite subset of S is nonempty.As the elements of S are closed subsets of the compact set K 0 , this implies that We have proved Claim 2.
Now let S ⊆ Z be bounded or directed.Claims 1 and 2 imply that K∈S K is an element of Z .It is clear that K∈S K is an upper bound of S. On the other hand, any upper bound of S must be a subset of any K in S, hence, a subset of K∈S K .Thus, this intersection is the least upper bound of S.
Example 9 Let us apply Proposition 8 to Y = R.Hence, let be the set whose elements are all nonempty, compact intervals and the whole set of real numbers.We define as reverse inclusion: Then (I, ) is a bounded complete dcpo with least element R. The supremum of a bounded or directed set S ⊆ I is the intersection I∈S I .
Example 10 Let X be a Banach space.We apply Proposition 8 to Y := X * with the weak * topology.It is well known that X * with the weak * topology is a topological vector space and a Hausdorff space; see, eg, Rudin [10,Page 66].We define Z convex by On Z convex we define again as reverse inclusion.Then (Z convex , ) is a bounded complete dcpo with least element X * .The supremum of a bounded or directed set S ⊆ I is the intersection K∈S K .
Edalat [5, Section 3], [6, Section 4] defined his L-derivate as the supremum of a certain class of "elementary step functions" from a Banach space X to Z convex from Example 10.These elementary step functions are Scott continuous.We will see that under suitable conditions the supremum of a certain class of such functions exists in the dcpo of Scott continuous functions.
Definition 11 Let X be an arbitrary topological space and (Z, ) be a partial order with least element ⊥.For any open subset U ⊆ X and any z ∈ Z we define the (total) function (U z) : X → Z by We call such a function an elementary step function.

Ties of Functions
Let X be a Banach space.As in Example 10 we define Z convex by Z convex := {X * } ∪ {K : K ⊆ X * is a nonempty, convex, weak * compact set}.
On Z convex we define again as reverse inclusion.In Example 10 we already mentioned that (Z convex , ) is a bounded complete dcpo with least element X * .Note that for any K ∈ Z convex and any x ∈ X the set is an element of I as defined in Example 9. Let V ⊆ X be a nonempty open subset of X .Following Edalat [5, Definition 1], [6, Definition 3.1], for a nonempty open set U ⊆ V and an element K ∈ Z convex , we call the set the single tie of O with K .Here, the formula K(x − y) f (x) − f (y) has to be understood with respect to the dcpo of Example 9. We can rewrite this definition without using domain theoretic language as follows:

Peter Hertling
Lemma 15 Let V ⊆ X be a nonempty open subset of X , and fix some K ∈ Z convex .
(1) For any nonempty open sets (2) Let U be a nonempty open subset of V , and let K ∈ Z convex be different from X * , ie, let K be a nonempty, convex, weak * compact subset of ) This is clear.
( This shows that f is Lipschitz continuous on U . (b) Due to the assumption f ∈ δ V (U, K), for any z ∈ U and t > 0 with z + tv ∈ U there exists some ζ ∈ K with This implies f (U, v) ≤ s K (v).
By Lemma 3 we obtain K U,f ⊆ K.
Note that in the following lemma we consider convex U .
Lemma 16 Let V ⊆ X be a nonempty open subset of X .Let U be a nonempty convex open subset of V .Let f : dom(f ) → R with U ⊆ dom(f ) ⊆ V be a function that is Lipschitz continuous on U .Then f ∈ δ V (U, K U,f ).
Proof We have to show that for every x, y ∈ U there exists some Hence, for all v ∈ W we have ζ 0 (v) ≤ f (U, v).Now remember that the function v → f (U, v), mapping X to R, is positively homogeneous (Lemma 1(5)) and subadditive (Lemma 2; this is where the convexity of U is used).By the Hahn Banach Extension Theorem (see, eg, Megginson [9, Theorem 1.9.5]) there exists a linear function ζ : X → R satisfying The second equation implies that ζ is an element of K U,f .

Figure 1 :
Figure 1: The diagram on the left hand side illustrates a proof attempt for Lemma 2 that does not work.The diagram on the right hand side illustrates the construction in the proof for k = 5.

First, we claim
that this function g is Scott continuous.Fix some point x ∈ X , and let O ⊆ Z be a Scott open set with g(x) ∈ O.We have to show that there is an open set U ⊆ X with x ∈ U and g(U) ⊆ O. From sup(F(x)) = g(x) ∈ O and from the assumption that F(x) is a directed set we conclude F(x) ∩ O = ∅, hence, there is some f ∈ F with f (x) ∈ O.As f is Scott continuous, there is some open set U ⊆ X with x ∈ U and f (U) ⊆ O. Thus, for all y ∈ U , on the one hand we have f (y) ∈ O, and on the other hand, by definition of g, f (y) g(y).As O is Scott open and, thus, upwards closed, we obtain g(y) ∈ O.This shows g(U) ⊆ O. Thus, we have shown that g is Scott continuous.

Remark 14 Actually, Edalat [ 5 ,
Definition 1],[6, Definition 3.1]  considered only convex open sets U .Convexity of U will be important later.But since the definition of δ(U, K) makes sense for arbitrary open U and since we wish to show where convexity of U will be important, right now we do not restrict ourselves to convex U .
[7, K U,f is contained in the set {ζ ∈ X * : ||ζ|| * ≤ c}.By Alaoglu's theorem (see, eg, Megginson [9, Theorem 2.6.18])thisset is weak * compact.As K U,f is a weak * closed subset of this set, K U,f is itself weak * compact; see, eg, Engelking[7, Theorem 3.1.2].It is also clear that K U,f is a convex set.That K U,f is nonempty follows from the second assertion and the fact that ∂f (x) is nonempty.
For any finite subset E ⊆ F , the set {f (x) : f ∈ E} is bounded because F is pointwise bounded.As Z is bounded complete, sup{f (x) : f ∈ E} exists.According to Lemma 12, the function g E : X → Z defined byg E (x) := sup{f (x) : f ∈ E} is Scott continuous.It is clear that it is a least upper bound of E, ie, g E = sup(E).It is straightforward to see that the set D := {g E : E is a finite subset of F}is a directed subset of C(X, Z).By Proposition 6 sup(D) exists and is a Scott continuous function.Finally, it is also straightforward to see that sup(D) C g and g C sup(D), thus, sup(D) = g.As sup(D) is Scott continuous, the proof is finished.