1 21 ISSN 1759-9008 1 Clarke s Generalized Gradien and Edala s L-derivaive PETER HERTLING Absrac: Clarke [2, 3, 4] inroduced a generalized gradien for real-valued Lipschiz coninuous funcions on Banach spaces. Using domain heoreic noions, Edala [5, 6] inroduced a so-called L-derivaive for real-valued funcions and showed ha for Lipschiz coninuous funcions Clarke s generalized gradien is always conained in his L-derivaive and ha hese wo noions coincide if he underlying Banach space is finie dimensional. He asked wheher hey coincide as well if he Banach space is infinie dimensional. We show ha his is he case. 2010 Mahemaics Subjec Classificaion 26E15, 46G05 (primary); 06A06, 03D78 (secondary) Keywords: Generalized gradien, L-derivaive, Banach space, direced complee parial order, bounded complee parial order 1 Inroducion Clarke [4, Page 1] observed ha nonsmooh phenomena in mahemaics and opimizaion occur naurally and frequenly, and here is a need o be able o deal wih hem. We are hus led o sudy differenial properies of nondiffereniable funcions. Moivaed by his observaion, Clarke [2, 3, 4] inroduced a generalized gradien f (x) for real-valued Lipschiz coninuous funcions f on Banach spaces. Using domain heoreic noions in he realm of compuable analysis, Edala [5, 6] inroduced a so-called L-derivaive for real-valued funcions on Banach spaces. Domain heory arose in he conex of compuer science and logic. While he primary applicaion of domain heory is in he semanics of programming languages, domain heoreic noions can also be applied successfully in compuable analysis, as shown in [5, 6] and in oher aricles by Edala. As Clarke s generalized gradien and Edala s L-derivaive are defined using raher differen mahemaical noions i is remarkable ha hey are closely conneced. Edala [5, 6] showed ha for Lipschiz coninuous funcions Clarke s generalized gradien is always conained in his L-derivaive and ha hese wo noions coincide if he underlying Banach space is finie dimensional. He asked wheher hey coincide as well if he Banach space is infinie dimensional. We show ha, indeed, for Lipschiz Published: April 2017 doi: 10.4115/jla.2017.9.c1
2 Peer Herling coninuous real funcions on arbirary Banach spaces hey coincide. In view of he fac ha boh noions are defined wihin differen mahemaical heories, he fac ha hey coincide for Lipschiz coninuous funcions may be aken as furher indicaion ha he resuling noion is a naural one. Tha hey coincide also shows ha he domain heoreic definiion by Edala can be used for a compuabiliy heoreic reamen of Clarke s generalized gradien. In he following secion we inroduce some basic noions and collec fundamenal facs abou a Banach space X, is dual space X and he weak opology on X. Then, in Secion 3, for a Lipschiz coninuous funcion on X, we inroduce Clarke s generalized direcional derivaive and Clarke s generalized gradien. In he following secion, for an open subse U of X, we inroduce a funcion ha may be considered as a global version of Clarke s generalized direcional derivaive. Then we prove a crucial propery of his funcion. Using his funcion, in Secion 5 we inroduce a cerain weak compac subse of X ha conains Clarke s generalized gradien. In Secion 6 we formulae several fundamenal facs concerning coninuous funcions f : X Y where X may be an arbirary opological space, eg a Banach space, and where Y is a direced complee parial order (dcpo). Furhermore, we consider some special bounded complee dcpo s. In Secion 7 we inroduce Edala s so-called ies of funcions ha are needed for he definiion of his L-derivaive. We also show how hey are relaed o he weak compac subses inroduced in Secion 5. Finally, in Secion 8 we inroduce Edala s L-derivaive and show ha for a Lipschiz coninuous funcion on an arbirary Banach space i coincides wih Clarke s generalized gradien. 2 Basic noions In his secion, we remind he reader of several well-known noions concerning a Banach space X, is dual space X and he weak opology on X. We will consider only vecor spaces over R, he field of real numbers. Le X be a real Banach space wih norm. For any x X and r > 0 le B(x, r) := {y X : y x < r} be he open ball wih radius r and midpoin x. For subses S, T X we wrie as usual S + T := {x X : ( s S) ( T) x = s + }. Le X be he linear vecor space of all coninuous linear funcions ζ : X R. Wih he norm defined by ζ := sup{ ζ(v) : v X, v 1}
Clarke s Generalized Gradien and Edala s L-derivaive 3 his space is a Banach space as well. As eiher ζ(v) = ζ(v) or ζ(v) = ζ(v) = ζ( v) and v = v we may wrie as well ζ = sup{ζ(v) : v X, v 1}. The weak opology is a opology on he se X. I is defined o be he coarses opology such ha for any x X he funcion l x : X R defined by l x (ζ) := ζ(x) is coninuous. Every subse U X ha is open in he weak opology is also open in he opology induced by he norm. The converse is in general no rue; see, eg, Megginson [9, Theorem 2.6.2 and Corollary 2.6.3]. I is well known ha X wih he weak opology is a opological vecor space and a Hausdorff space; see, eg, Rudin [10, Page 66 and Theorem 1.12]. Subses of X ha are compac in he weak opology will be called weak compac. 3 Clarke s Generalized Gradien The erminology in his secion is copied from Clarke [4, Chaper 2]. Le X be a Banach space wih norm. Le Y be a subse of X, and le c be a non-negaive real number. A funcion f : Y R is called Lipschiz coninuous wih Lipschiz consan c if for all x, y Y f (x) f (y) c x y. A funcion f : X R is called Lipschiz coninuous wih Lipschiz consan c near a poin x X if here is an ε > 0 such ha he resricion of f o he ball B(x, ε) is Lipschiz coninuous wih Lipschiz consan c. A funcion f : X R is called Lipschiz coninuous (near x) if here exiss a real number c 0 such ha f is Lipschiz coninuous wih Lipschiz consan c (near x). Le f : X R be Lipschiz coninuous near some poin x X. Fix an arbirary v X. The generalized direcional derivaive of f a x in he direcion v, denoed f (x; v), is defined by f (x; v) := lim sup z x, 0 f (z + v) f (z). In he following secion we spell his ou in more deail (Lemma 1(4)). For any x X he generalized gradien of f a x, denoed f (x), is defined as follows: f (x) = {ζ X : ( v X) ζ(v) f (x; v)}.
4 Peer Herling This is a nonempy, convex and weak compac subse of X ; see Clarke [4, Proposiion 2.1.2(a)]. 4 A Global Version of Clarke s Generalized Direcional Derivaive Le X be a Banach space wih norm. For a nonempy, open se U X and a funcion f : dom(f ) X R wih U dom(f ) ha is Lipschiz coninuous on U we define { f (z + v) f (z) f (U, v) := sup } : z U, > 0, z + v U for v X. We show in he following lemma ha his is well defined. This funcion can be considered as a global version of Clarke s generalized direcional derivaive. In he following lemma several elemenary asserions abou his funcion are colleced. Lemma 1 Le X be a Banach space, le U X be a nonempy open subse, and le f : dom(f ) X R wih U dom(f ) be a funcion Lipschiz coninuous on U. For he firs four of he following five asserions, fix an arbirary v X. (1) The value f (U, v) is well defined, and if c 0 is a Lipschiz consan for f on U hen f (U, v) c v. (2) If U U is a nonempy open subse of U hen f (U, v) f (U, v). (3) If x U hen f (x, v) f (U, v). (4) If x U hen lim n f (B(x, 2 n ), v) = f (x, v). (5) The funcion v f (U, v) is posiively homogeneous, i.e, for all r > 0 and v X, f (U, rv) = r f (U, v). Proof Fix some v X. (1) Le c 0 be a Lipschiz consan for f on U. Then for all z U and > 0 such ha z + v U we have f (z + v) f (z) f (z + v) f (z) c v = c v. This shows ha f (U, v) is well defined and saisfies f (U, v) c v. (2) This follows direcly from he definiion of f (U, v).
Clarke s Generalized Gradien and Edala s L-derivaive 5 (3) This is a consequence of he definiions of f (x, v) and of f (U, v). (4) Noe ha by he second saemen he sequence ( f (B(x, 2 n ), v)) n N is nonincreasing, and by he hird saemen i is bounded from below. Therefore is limi exiss. Tha is limi is equal o f (x, v) is simply a resaemen of he definiion of f (x, v). (5) We addiionally fix some real number r > 0. Then { } f (z + rv) f (z) f (U, rv) = sup : z U, > 0, z + rv U { } f (z + sv) f (z) = sup r : z U, s > 0, z + sv U s { } f (z + sv) f (z) = r sup : z U, > 0, z + sv U s = r f (U, v). Clarke s generalized direcional derivaive is posiively homogeneous and subaddiive [4, Proposiion 2.1.1]. In Lemma 1(5) we have seen ha also he global funcion v f (U, v) is posiively homogeneous. In he following lemma we show ha i is subaddiive as well if U is convex. Lemma 2 Le X be a Banach space, le U X be a convex open subse, and f : dom(f ) X R wih U dom(f ) be a funcion Lipschiz coninuous on U. Then he funcion v f (U, v) is subaddiive, ie, for all v, w X, f (U, v + w) f (U, v) + f (U, w). Proof We will fix some ε > 0 and show ha for any v, w X f (U, v) + f (U, w) f (U, v + w) ε. Once his is proved for any ε > 0, he asserion follows. f (U, v) + f (U, w) f (U, v + w) So, le us fix some ε > 0 and elemens v, w X. We choose a poin z U and a real number > 0 such ha z + (v + w) U and f (z + (v + w)) f (z) f (U, v + w) ε.
6 Peer Herling One migh ry o prove he asserion simply by replacing he quoien on he lef hand side by he following erm: f (z + (v + w)) f (z + v) f (z + v) f (z) + 5. a[1] a[2] a[3] a[4] a[5] If z + v would happen o be an elemen of U, hen his 4 erm would 1 5 4 1 3 2 be a lower bound for f (U, w) + f (U, v), and we would be done. Unforunaely, he poin z + v may lie 3 2 ouside of he open se U; see Figure 1. We will have o proceed differenly. This is U z + (v + w) U z k = z + (v + w) z + v z z 0 = z Figure 1: The diagram on he lef hand side illusraes a proof aemp for Lemma 2 ha does no work. The diagram on he righ hand side illusraes he consrucion in he proof for k = 5. where he convexiy of U will be imporan. As U is convex, he line segmen L := {z + r(v + w) : 0 r } is a subse of U. Since his line segmen L is compac and U is open here exiss some δ > 0 such ha even he se L + B(0, δ) is a subse of U. Le k be a posiive ineger so large ha Then for i = 0,..., k he poins v k z i := z + i (v + w) k are elemens of he line segmen L and, hus, of he open se U, and he poins z i + k v are elemens of L + B(0, δ), and, hence, elemens of he open se U as well; see Figure 1. < δ. 4
Clarke s Generalized Gradien and Edala s L-derivaive 7 Using hese poins we can replace he quoien follows: f (z + (v + w)) f (z) = f (z k) f (z 0 ) = = = = k 1 i=0 k 1 i=0 k 1 f (z i+1 ) f (z i ) f (z i + k (v + w)) f (z i) ( f (zi + k (v + w)) f (z i + k v) i=0 k 1 i=0 k 1 i=0 1 k f (z i + k (v + w)) f (z i + k v) /k 1 k k 1 1 f (U, w) + k f (U, v) i=0 = f (U, w) + f (U, v). f (z+(v+w)) f (z) + f (z i + k v) f (z ) i) k 1 + i=0 by a sum of quoiens as 1 k f (z i + k v) f (z i) /k We have shown f (U, v) + f (U, w) f (U, v + w) ε, where ε > 0 was chosen arbirarily. This ends he proof. 5 Abou Nonempy, Convex, Weak Compac Ses Le X be a Banach space. In his secion, firs we observe ha any nonempy, convex, weak compac subse K of X can be expressed wih he help of is suppor funcion. Then, using he global version of he generalized direcional derivaive, we inroduce cerain nonempy, convex, weak sar compac subses of X. Le K be a nonempy, weak compac subse of X. I is well known (see, eg, Megginson [9, Corollary 2.6.9]) ha any such se is bounded, ie, here is some non-negaive real number B such ha ( ζ K) ζ B.
8 Peer Herling This implies for all ζ K and v X Thus, for v X, by a real number is defined ha saisfies ζ(v) ζ v B v. s K (v) := sup{ζ(v) : ζ K} s K (v) B v. The funcion s K : X R is called he suppor funcion of K. Lemma 3 Le X be a Banach space. Le K X be a nonempy, convex, and weak compac subse of X. Then K = {ζ X : ( v X) ζ(v) s K (v)}. Proof The inclusion follows from he definiion of s K (v). We wish o prove he inclusion. Therefore, le us consider some ξ X wih ξ K. By he second par of Beer [1, Theorem 1.4.2] here exis an elemen v X and a real number α such ha eiher (1) ξ(v) > α and ( ζ K) ζ(v) < α or In he second case we obain ξ(v) < α and ( ζ K) ζ(v) > α. ξ( v) = ξ(v) > α and ( ζ K) ζ( v) = ζ(v) < α, hus, (1) holds for v and α. Therefore, we can assume ha (1) holds. Then and, hence, ξ {ζ X s K (v) α < ξ(v) : ( v X) ζ(v) s K (v)}. Le U X be a nonempy, open subse of he Banach space X, and le f : dom(f ) X R wih U dom(f ) be a funcion Lipschiz coninuous on U. We define K U,f := {ζ : X R : ζ is linear and ( v X) ζ(v) f (U, v)}. Lemma 4 (1) K U,f is a nonempy, convex, and weak compac subse of X. (2) For all x U, f (x) K U,f.
Clarke s Generalized Gradien and Edala s L-derivaive 9 (3) Le n 0 be a nonnegaive ineger wih B(x, 2 n 0) U. Then f (x) = n n 0 K B(x,2 n ),f. Proof (1) Le c 0 be a Lipschiz consan for f on U. According o Lemma 1(1) for any v X we have f (U, v) c v. Hence, for ζ K U,f, ζ(v) c v and ζ(v) = ζ( v) c v = c v, hus ζ(v) c v. This implies ζ X, and K U,f is conained in he se {ζ X : ζ c}. By Alaoglu s heorem (see, eg, Megginson [9, Theorem 2.6.18]) his se is weak compac. As K U,f is a weak closed subse of his se, K U,f is iself weak compac; see, eg, Engelking [7, Theorem 3.1.2]. I is also clear ha K U,f is a convex se. Tha K U,f is nonempy follows from he second asserion and he fac ha f (x) is nonempy. (2) This follows from Lemma 1(3). (3) By he second asserion we obain f (x) K B(x,2 n ),f for any n n 0, hence f (x) n n 0 K B(x,2 n ),f. For he inverse inclusion consider some ζ n n 0 K B(x,2 n ),f. Then, for all n n 0 and all v X, ζ(v) f (B(x, 2 n ), v), hence, ζ(v) lim f (B(x, 2 n ), v) = f (x, v) n (compare Lemma 1(4)), and his implies ζ f (x). Tha was o be shown. 6 Coninuous Funcions from an Arbirary Topological Space o a Bounded Complee DCPO I is he purpose of his secion o provide several fundamenal facs concerning coninuous funcions f : X Y where X may be an arbirary opological space, eg a Banach space, and where Y is a direced complee parial order (dcpo), in paricular, where Y is a bounded complee dcpo. Furhermore, we consider some special bounded complee dcpo s. We will need all his laer in order o define Edala s L-derivaive for arbirary funcions. In order o make he presenaion self-conained we inroduce basic noions abou dcpo s as well. A se Z wih a binary relaion Z Z saisfying he following hree condiions
10 Peer Herling (1) ( z) z z (reflexiviy), (2) ( x, y, z) (x y y z) x z (ransiiviy), (3) ( y, z) (y z z y) y = z (anisymmery), is called a parial order. Le (Z, ) be a parial order. An elemen z Z is called an upper bound of a subse S Z if ( s S) s z. An elemen z Z is called a supremum or leas upper bound of a subse S Z if i is an upper bound of S and if for all upper bounds y of S one has z y. Obviously, if a se S has a supremum, hen his supremum is unique. Then we denoe i by sup(s). A subse S Z is called direced if i is nonempy and for any wo elemens x, y S here exiss an upper bound z S of he se {x, y}. Z is called a dcpo if for any direced subse S Z here exiss a supremum of S in Z. A subse S Z is called upwards closed if for any elemens s, z Z: if s S and s z hen z S. The following lemma is well known. Lemma 5 (See, eg, Goubaul-Larrecq [8, Proposiion 4.2.18]) Le (Z, ) be a parial order. The se of all subses O Z saisfying he following wo condiions: (1) O is upwards closed, (2) if S Z is a direced subse wih sup(s) O hen S O, is a opology on Z, called he Sco opology. Consider now some parial order (Z, ) and an arbirary opological space X. We call a oal funcion f : X Z Sco coninuous if i is coninuous wih respec o he given opology on X and he Sco opology on Z. Le C(X, Z) denoe he se of all Sco coninuous funcions f : X Z. On his se we define a binary relaion C by f C g : ( x X) f (x) g(x). Proposiion 6 Le (Z, ) be a dcpo, and le X be an arbirary opological space. Then C(X, Z) wih C is a dcpo. Furhermore, if F C(X, Z) is a C -direced se hen he funcion g : X Z defined by g(x) := sup({f (x) : f F}) is Sco coninuous and he leas upper bound of F.
Clarke s Generalized Gradien and Edala s L-derivaive 11 Proof I is clear ha (C(X, Z), C ) is a parial order. Now we show ha C(X, Z) wih C is even a dcpo. Le F C(X, Z) be a C -direced se. Then for every x X, he se F(x) := {f (x) : f F} = {z Z : ( f F) z = f (x)} is -direced. We define a oal funcion g : X Z by g(x) := sup(f(x)). Firs, we claim ha his funcion g is Sco coninuous. Fix some poin x X, and le O Z be a Sco open se wih g(x) O. We have o show ha here is an open se U X wih x U and g(u) O. From sup(f(x)) = g(x) O and from he assumpion ha F(x) is a direced se we conclude F(x) O, hence, here is some f F wih f (x) O. As f is Sco coninuous, here is some open se U X wih x U and f (U) O. Thus, for all y U, on he one hand we have f (y) O, and on he oher hand, by definiion of g, f (y) g(y). As O is Sco open and, hus, upwards closed, we obain g(y) O. This shows g(u) O. Thus, we have shown ha g is Sco coninuous. By he definiion of g, for all f F we have f C g, hus g is an upper bound of F. Finally, if h is an arbirary upper bound of F, hen for all x X, g(x) = sup(f(x)) h(x). Hence, g is a leas upper bound of F. Le (Z, ) be a parial order. An elemen y Z is called a leas elemen of Z if for all z Z, y z. Obviously, if a leas elemen exiss hen i is unique. Usually, when a leas elemen exiss, i is denoed. A subse S Z is called bounded if here exiss an upper bound z Z for S. Z is called bounded complee if for any bounded subse S Z here exiss a supremum of S in Z. Lemma 7 Any bounded complee parial order has a leas elemen. This is well known. For compleeness sake we give he proof. Proof In any parial order, every elemen is an upper bound of he empy se. Hence, in a bounded complee parial order sup( ) exiss. I is a leas elemen. The following proposiion covers he cases of bounded complee dcpos ha we will need.
12 Peer Herling Proposiion 8 Le Y be a nonempy opological vecor space whose opology is a Hausdorff opology. Le Z := {Y} {K : K Y is a nonempy, convex, compac se}. (1) Then Z wih defined as reverse inclusion is a bounded complee dcpo wih leas elemen Y. (2) If S Z is bounded or direced hen sup(s) = K S K. Proof Firs, we observe ha (Z, ) is a parial order. Nex, we prove he following claim. Claim 1: For any S Z, he se K S K is eiher empy or an elemen of Z. For he proof, we disinguish hree cases. Case I: S =. Then K S K = Y, and his is an elemen of Z. Case II: S = {Y}. Then again K S K = Y, and his is an elemen of Z. Case III: S conains a leas one elemen K 0 ha is differen from Y, ie, ha is nonempy, convex and compac. As all elemens of S are closed (any compac subse of a Hausdorff space is closed; see, eg, Engelking [7, Theorem 3.1.8]) and he inersecion of arbirarily many closed ses is closed as well, he inersecion K S K is a closed subse of Y. In fac, i is a closed subse of he compac se K 0, and hence (see, eg, [7, Theorem 3.1.2]) compac iself. Noe also ha all elemens of S are convex and ha he inersecion of arbirarily many convex ses is again a convex se. Thus, K S K is a convex and compac subse of Y. If his inersecion is no empy hen i is an elemen of Z. We have proved Claim 1. We coninue wih he following claim. Claim 2: If S Z is bounded or direced hen K S K is no empy. Le S Z be a bounded se, and le 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 elemens of Z are nonempy, K 0 is nonempy. Hence, K S K is nonempy. Now le S Z be direced. If S = {Y} hen K S K = Y, and his se is nonempy. Le us assume ha S conains a leas one elemen K 0 ha is differen from Y, hence, K 0 is a nonempy, convex, compac subse of Y. Using he assumpion ha S is a direced se, one shows by inducion ha he se S := {K K 0 : K S}
Clarke s Generalized Gradien and Edala s L-derivaive 13 has he finie inersecion propery, ie, he inersecion of any finie subse of S is nonempy. As he elemens of S are closed subses of he compac se K 0, his implies ha K S K ; see, eg, [7, Theorem 3.1.1]. I is clear ha K S K = K S K. We have proved Claim 2. Now le S Z be bounded or direced. Claims 1 and 2 imply ha K S K is an elemen of Z. I is clear ha K S K is an upper bound of S. On he oher hand, any upper bound of S mus be a subse of any K in S, hence, a subse of K S K. Thus, his inersecion is he leas upper bound of S. Example 9 Le us apply Proposiion 8 o Y = R. Hence, le I := {R} {[a, b] : a, b R, a b} be he se whose elemens are all nonempy, compac inervals and he whole se of real numbers. We define as reverse inclusion: A B : B A. Then (I, ) is a bounded complee dcpo wih leas elemen R. The supremum of a bounded or direced se S I is he inersecion I S I. Example 10 Le X be a Banach space. We apply Proposiion 8 o Y := X wih he weak opology. I is well known ha X wih he weak opology is a opological vecor space and a Hausdorff space; see, eg, Rudin [10, Page 66]. We define Z convex by Z convex := {X } {K : K X is a nonempy, convex, weak compac se}. On Z convex we define again as reverse inclusion. Then (Z convex, ) is a bounded complee dcpo wih leas elemen X. The supremum of a bounded or direced se S I is he inersecion K S K. Edala [5, Secion 3], [6, Secion 4] defined his L-derivae as he supremum of a cerain class of elemenary sep funcions from a Banach space X o Z convex from Example 10. These elemenary sep funcions are Sco coninuous. We will see ha under suiable condiions he supremum of a cerain class of such funcions exiss in he dcpo of Sco coninuous funcions. Definiion 11 Le X be an arbirary opological space and (Z, ) be a parial order wih leas elemen. For any open subse U X and any z Z we define he (oal) funcion (U z) : X Z by { z if x U, (U z)(x) := if x U. We call such a funcion an elemenary sep funcion.
14 Peer Herling I is easy o see ha any such funcion is Sco coninuous. In fac, we will need he following sronger saemen, which is an immediae consequence of Goubaul- Larrecq [8, Lemma 5.7.10]. Lemma 12 Le X be an arbirary opological space and (Z, ) be a parial order wih leas elemen. Le F be a finie se of elemenary sep funcions from X o Z such ha for each x X he se {f (x) : f F} has a leas upper bound. Then he funcion g : X Z defined by g(x) := sup({f (x) : f F}) is Sco coninuous. Noe ha his lemma implies ha every elemenary sep funcion is Sco coninuous. Proof This is an immediae consequence (almos a reformulaion) of [8, Lemma 5.7.10]. For he formulaion of he desired resul he following noion is useful. We call a se F of oal funcions from X o Z poinwise bounded if for every x X, he se is -bounded. F(x) := {f (x) : f F} = {z Z : ( f F) z = f (x)} Theorem 13 Le X be an arbirary opological space, and le (Z, Z ) be a bounded complee dcpo. Le F be a se of elemenary sep funcions from X o Z. If F is poinwise bounded hen he oal funcion g : X Z defined by g(x) := sup{f (x) : f F} is Sco coninuous and a leas upper bound of F. Proof Le F be a poinwise bounded se of elemenary sep funcions. Remember ha we have already seen ha every elemenary sep funcion is coninuous, ie, F C(X, Z). Le he oal funcion g : X Z be defined by g(x) := sup{f (x) : f F}. This funcion is well defined because we assume ha F is poinwise bounded and Z is bounded complee. We have o show ha g is Sco coninuous and a leas upper bound of F. In fac, once we have shown ha g is Sco coninuous i is clear ha g is a leas upper bound of F. Thus, we are now going o show ha g is Sco coninuous.
Clarke s Generalized Gradien and Edala s L-derivaive 15 For any finie subse E F, he se {f (x) : f E} is bounded because F is poinwise bounded. As Z is bounded complee, sup{f (x) : f E} exiss. According o Lemma 12, he funcion g E : X Z defined by g E (x) := sup{f (x) : f E} is Sco coninuous. I is clear ha i is a leas upper bound of E, ie, g E = sup(e). I is sraighforward o see ha he se D := {g E : E is a finie subse of F} is a direced subse of C(X, Z). By Proposiion 6 sup(d) exiss and is a Sco coninuous funcion. Finally, i is also sraighforward o see ha sup(d) C g and g C sup(d), hus, sup(d) = g. As sup(d) is Sco coninuous, he proof is finished. 7 Ties of Funcions Le X be a Banach space. As in Example 10 we define Z convex by Z convex := {X } {K : K X is a nonempy, convex, weak compac se}. On Z convex we define again as reverse inclusion. In Example 10 we already menioned ha (Z convex, ) is a bounded complee dcpo wih leas elemen X. Noe ha for any K Z convex and any x X he se K(x) := {r R : ( ζ K) r = ζ(x)} is an elemen of I as defined in Example 9. Le V X be a nonempy open subse of X. Following Edala [5, Definiion 1], [6, Definiion 3.1], for a nonempy open se U V and an elemen K Z convex, we call he se δ V (U, K) := {f : f : dom(f ) R is a funcion wih U dom(f ) V and ( x, y U) K(x y) f (x) f (y)} he single ie of O wih K. Here, he formula K(x y) f (x) f (y) has o be undersood wih respec o he dcpo of Example 9. We can rewrie his definiion wihou using domain heoreic language as follows: δ V (U, K) = {f : f : dom(f ) R is a funcion wih U dom(f ) V and ( x, y U) ( ζ K) f (x) f (y) = ζ(x y)}. Remark 14 Acually, Edala [5, Definiion 1], [6, Definiion 3.1] considered only convex open ses U. Convexiy of U will be imporan laer. Bu since he definiion of δ(u, K) makes sense for arbirary open U and since we wish o show where convexiy of U will be imporan, righ now we do no resric ourselves o convex U.
16 Peer Herling Lemma 15 Le V X be a nonempy open subse of X, and fix some K Z convex. (1) For any nonempy open ses U 1, U 2 V U 1 U 2 = δ V (U 1, K) δ V (U 2, K). (2) Le U be a nonempy open subse of V, and le K Z convex be differen from X, ie, le K be a nonempy, convex, weak compac subse of X. If f δ V (U, K) hen (a) f is Lipschiz coninuous on U and (b) K U,f K. Proof (1) This is clear. (2) (a) We already menioned ha any nonempy, convex, weak compac subse of X is bounded, ha is, here exiss some c 0 such ha for all ζ K, ζ c. Consider arbirary x, y U. Due o he assumpion f δ V (U, K) here exiss some ζ K wih ζ(x y) = f (x) f (y). We obain f (x) f (y) = ζ(x y) ζ x y c x y. (b) This shows ha f is Lipschiz coninuous on U. Due o he assumpion f δ V (U, K), for any z U and > 0 wih z + v U here exiss some ζ K wih This implies f (z + v) f (z) = ζ(z + v z) f (U, v) s K (v). = ζ(v). By Lemma 3 we obain K U,f K. Noe ha in he following lemma we consider convex U. Lemma 16 Le V X be a nonempy open subse of X. Le U be a nonempy convex open subse of V. Le f : dom(f ) R wih U dom(f ) V be a funcion ha is Lipschiz coninuous on U. Then f δ V (U, K U,f ).
Clarke s Generalized Gradien and Edala s L-derivaive 17 Proof We have o show ha for every x, y U here exiss some ζ K U,f f (x) f (y) = ζ(x y). Consider some x, y U. wih If x = y hen we choose an arbirary ζ K U,f (remember ha by Lemma 4(1) K U,f is no empy) and obain f (x) f (y) = 0 = ζ(0) = ζ(x y). Now we consider he case x y. We se w := x y. Le W be he one-dimensional subspace of X generaed by w, and le he linear funcion ζ 0 : W R be defined by ζ 0 (α w) := α (f (x) f (y)), for any α R. Then for all α 0 and for α < 0 ζ 0 (αw) = α (f (x) f (y)) f (y + 1 w) f (y) = α 1 α f (U, w) = f (U, αw), ζ 0 (αw) = α (f (x) f (y)) = ( α) (f (y) f (x)) f (x + 1 ( w)) f (x) = ( α) 1 ( α) f (U, w) = f (U, ( α) ( w)) = f (U, αw). Hence, for all v W we have ζ 0 (v) f (U, v). Now remember ha he funcion v f (U, v), mapping X o R, is posiively homogeneous (Lemma 1(5)) and subaddiive (Lemma 2; his is where he convexiy of U is used). By he Hahn Banach Exension Theorem (see, eg, Megginson [9, Theorem 1.9.5]) here exiss a linear funcion ζ : X R saisfying (1) ζ(v) = ζ 0 (v) for all v W and (2) ζ(v) f (U, v) for all v X. The firs equaion implies ζ(x y) = ζ(w) = ζ 0 (w) = f (x) f (y). The second equaion implies ha ζ is an elemen of K U,f.
18 Peer Herling 8 Edala s L-derivaive In his secion we consider he seing of Example 10, ha is, X is a Banach space, and on Z convex := {X } {K : K X is a nonempy, convex, weak compac se} we define as reverse inclusion. Then, (Z convex, ) is a bounded complee dcpo wih leas elemen X ; see Example 10. Noe ha in he following lemma we do no assume ha he open se U considered here is convex. Lemma 17 Le X be a Banach space, V X a nonempy open subse of X, U V a nonempy open subse of V, K Z convex, and f : V R a funcion wih f δ V (U, K). If f is Lipschiz coninuous near some poin x V hen f (x) (U K)(x). Proof If x U hen (U K)(x) = X, and f (x) X is clear. Le us consider he case x U. Then (U K)(x) = K. Thus, we have o show f (x) K. This is clear if K = X. Le us assume ha K is no equal o X, hence, ha K is a nonempy, convex, weak compac subse of X. According o Lemma 15(2), f is Lipschiz coninuous on U and K U,f K. According o Lemma 4(2) we have f (x) K U,f. Togeher we obain f (x) K. Le X be a Banach space, V X a nonempy open subse of X, and f : V R an arbirary funcion. We define: D(f ) := {(U, K) : U V is nonempy, open, and convex and K Z convex and f δ V (U, K)}. Firs, we noe ha D(f ) is no empy. For example, if U V is an open ball (balls are convex) hen (U, X ) is an elemen of D(f ) (his is shown by an applicaion of he Hahn Banach Exension Theorem in a similar manner as in he proof of Lemma 16). Lemma 18 The se F := {(U K) : (U, K) D(f )} is poinwise bounded.
Clarke s Generalized Gradien and Edala s L-derivaive 19 Proof I is sufficien o show ha for each x X he se F(x) is a bounded subse of Z convex. We disinguish wo cases. Case I: There is no pair (U, K) D(f ) wih x U and K X. Then F(x) = {X }. This se is bounded by X iself, he leas elemen of Z convex. Case II: There is some pair (U 0, K 0 ) wih x U 0 and K 0 X. Then K 0 is a nonempy, convex, weak compac subse of X. According o Lemma 15(2), f is Lipschiz coninuous on U 0 and, hence, Lipschiz coninuous near x. By Lemma 17 f (x) (U K)(x) for any (U, K) D(f ). This shows ha F(x) is bounded by f (x). Now we can define Edala s L-derivaive [5, Secion 3], [6, Secion 4]. Definiion 19 Le X be a Banach space, V X be a nonempy open subse of X, and f : V R be an arbirary funcion. The funcion Lf defined by L(f ) := sup{(u K) : (U, K) D(f )} is called L-derivaive of f. According o Lemma 18 and Theorem 13 his funcion Lf is well defined, Sco coninuous and a leas upper bound of he se {(U K) : (U, K) D(f )}. Noe ha we do no make any assumpion abou he funcion f. The following resul describes he L-derivaive via Clarke s generalized gradien. Theorem 20 Le X be a Banach space, and le V X be a nonempy open subse of X. Le f : V R be an arbirary funcion. Fix some poin x V. (1) If f is no Lipschiz coninuous near x hen Lf (x) = X. (2) If f is Lipschiz coninuous near x hen Lf (x) = f (x). Since we have already done mos of he work, he proof is fairly shor. Proof Le D(f ) be defined as before Lemma 18. We have already seen ha D(f ) is no empy. According o Theorem 13, Lemma 18 and he las asserion in Example 10 Lf (x) = sup{(u K)(x) : (U, K) D(f )} = (U K)(x). (U,K) D(f )
20 Peer Herling (1) Le us assume ha f is no Lipschiz coninuous near x. Then here canno exis a pair (U, K) D(f ) wih nonempy, convex, compac K X because oherwise, according o Lemma 15(2), f were Lipschiz coninuous on U and, hence, Lipschiz coninuous near x. We conclude ha (U K)(x) = X for all (U, K) D(f ). Hence, also L(f )(x) = X. (2) Le us assume ha f is Lipschiz coninuous near x. On he one hand, according o Lemma 17, f (x) (U K)(x) = Lf (x). (U,K) D(f ) For he oher direcion, noe ha here is some n 0 such ha B(x, 2 n 0) V and f is Lipschiz coninuous on B(x, 2 n 0). Then, according o Lemma 4(3), f (x) = n n 0 K B(x,2 n ), f. And by Lemma 16 (remember ha balls in X are convex), for all n n 0 we have f δ V (B(x, 2 n ), K B(x,2 n ),f ). This implies (B(x, 2 n ), K B(x,2 n ),f ) D(f ), hence, Lf (x) n n 0 K B(x,2 n ), f. We have shown Lf (x) f (x) as well. Acknowledgemens The auhor was suppored by he EU gran FP7-PEOPLE-2011-IRSES No. 294962: COMPUTAL. Noe added in proof The auhor has recenly learned ha Proposiion 6 is conained in K. Keimel and J.D. Lawson, Coninuous and compleely disribuive laices, Laice Theory: Special Topics and Applicaions, Volume 1 (G. Gräzer and F. Wehrung, ediors), Birkhäuser/Springer (2014) pages 5 53.
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