The NSLUC property and Klee envelope

Transcription

1 Mathematische Annalen manuscript No. will be inserted by the editor The NSLUC property and Klee envelope A. Jourani L. Thibault D. Zagrodny Received: date / Accepted: date Abstract A notion called norm subdifferential local uniform convexity NSLUC is introduced and studied. It is shown that the property holds for all subsets of a Banach space whenever the norm is either locally uniformly convex or k-fully convex. The property is also valid for all subsets of the Banach space if the norm is Kadec-Klee and its dual norm is Gâteaux differentiable off zero. The NSLUC concept allows us to obtain new properties of the Klee envelope, for example a connection between attainment sets of the Klee envelope of a function and its convex hull. It is also proved that the Klee envelope with unit power plus an appropriate distance function is equal to some constant on an open convex subset as large as we need. As a consequence of obtained results, the subdifferential properties of the Klee envelope can be inherited from subdifferential properties of the opposite of the distance function to the complement of the bounded convex open set. Moreover the problem of singleton property of sets with unique farthest point is reduced to the problem of convexity of Chebyshev sets. Mathematics Subject Classification Primary 49J52, 46B20 Secondary 52A4, 4A50. This research was supported by Math-AmSud 3math-0 and ECOS projects Abderrahim Jourani Université de Bourgogne, Institut de Mathématiques de Bourgogne, UMR 5584 CNRS, BP , 2078 Dijon Cedex, France jourani@u-bourgogne.fr Lionel Thibault Université Montpellier II, Institut de Mathématiques, UMR 549 CNRS, Case Courier 05, Place Eugène Bataillon Montpellier Cedex, France and Centro de Modelamiento Matematico, Universidad de Chile, Chile lionel.thibault@univ-montp2.fr Dariusz Zagrodny Cardinal Stefan Wyszynski University, Faculty of Mathematics and Natural Sciences, College of Science, Institute of Mathematics, Dewajtis 5, Warsaw, Poland d.zagrodny@uksw.edu.pl

2 2 Jourani,Thibault,Zagrodny Keywords Klee envelope sup-convolution farthest point subdifferential Chebyshev sets NSLUC. Introduction Let X, be a real normed vector space. At the beginning of thirties of 20 century S. Mazur considered the following property: Every closed bounded convex subset of X, say D, is the intersection of closed balls containing it, see Mazur 933. There are several papers where this is investigated. We refer to Bandyopadhyay 992; Chen and Lin 998 for surveys on achievements and historical information of the property, which is called nowadays the Mazur Intersection Property, MIP for short. In other words, denoting by B[x, r] resp. Bx, r the closed resp. open ball centered at x with radius r > 0, we can express the Mazur Intersection Property as follows D = B[x, r]. x X,r>0,D B[x,r] Moreover we expect that D B[x, r] \ Bx, r for every ball with the smallest radius in the right-hand side of the equality. With this geometry we can relate an analytic reasoning. Namely, for every bounded subset S X we can define the farthest distance function from the set S also called antidistance function S x := sup x y y S and the set Q S x of farthest points from S to x, that is, Q S x := {y S : x y = S x}. Using the notions defined above the Mazur Intersection Property can be rewritten in the form D = B[x, D x]. x X The nonemptiness of the intersection D B[x, r]\bx, r with r = D x is in fact a question on nonemptiness of the set of farthest points. The above consideration can be embedded into a more general set up. Namely we can use the Klee envelope instead of the farthest distance function, that is, for λ > 0, p and an extended real-valued function f : X R {+ } we put κ λ,p fx := sup y X pλ x y p fy. Then the associated attainment set is denoted by Q λ,p fx, that is, Q λ,p fx := {y X : pλ x y p fy = κ λ,p fx}. 2

3 The NSLUC property and Klee envelope 3 Of course, the Klee function with p = and λ = coincides with the farthest distance function, that is, κ, ψ S x = S x and Q λ,p ψ S x coincides with Q S x, whenever f is the indicator function ψ S of a subset S of X, that is, ψ S x = 0 if x S and ψ S x = + otherwise. We shall see that results on the Klee envelope add new knowledge to the old problems recalled below concerning sets of farthest points. Let us also point out that using this set up we do not take care about the convexity of the set S, more generally about the convexity of the function f that we transform through the Klee envelope. However, it is worth considering whether the convexity is important or not in Definitions and 2. On the one hand, we show that it does not matter if we take f or cof in, we get the same value, see Proposition 7. On the other hand, in some spaces we have also the following implication d Q λ,p co f x = d co Q λ,p fx x + spand x, which entails the inclusions Q λ,p fx Q λ,p co f x co Qλ,p fx, 3 see Theorem 2. Thus the convex hull of the associated attainment set for co f can be recovered as the convex hull from the associated attainment set for f. When f is the indicator function ψ S, we obtain the equality Q S x = Q cos x, in several important cases increasing the range of its use, see Section 5 and Theorem 2. Of course, if Q S x is single-valued, then Q cos x is single-valued too in this case. So, natural is the question: Which sets and points have this property? There is also the old question posed by V. Klee see Klee 96 or Hiriart-Urruty 2007, closely related to this question, namely: Klee s question: Suppose that Q S x is a singleton for every x X. Must S consist of a single point? In fact, if we look carefully at Klee 96, Theorem.2 we notice that an affirmative answer to the above question by V. Klee implies the convexity of Chebyshev sets and vice versa, whenever X is a real Hilbert space; we recall that a set S is called Chebyshev provided that every point of the space X admits a unique nearest point in S. So far, both questions are unsolved in Hilbert spaces, see for example Hiriart-Urruty Let us also point out that not always investigating farthest points from a set S can be changed into finding farthest points from its closed convex hull. For example, if X := c 0, x :=, 2, 3,..., s i := i, 0,..., 0,, 0,... with being the i-th component for every i = 2, 3,..., S := {s 2, s 3,...}, then Q S x = but Q cos x = {0} since 0 cl wc0,l S co S. This example realizes the need of finding a large class of spaces and sets for which 3 holds true. For this reason we propose

4 4 Jourani,Thibault,Zagrodny a new notion, called norm subdifferential local uniform convexity NSLUC, see Definition, which allows us to get in particular 3 in several cases, for example: i the norm of X has the LUR property that is, the local uniform rotundity/convexity property; ii the norm of X is fully k convex; iii the norm of X is strictly convex and has the Kadec-Klee property, and S is relatively weakly sequentially compact; iv X is a Banach space whose norm has the Kadec-Klee property and the dual norm is Gâteaux differentiable off the origin; v the norm of X is strictly convex and S is relatively norm-compact, see Section 5 for details on relations among NSLUC properties and properties of the norm, and Theorem 3 where 3 is established in the listed cases. As an example of application of obtained results we relate the Klee envelope with a distance function, namely the sum of these two functions is constant on some open convex set, see Theorem 4. The two obvious consequence of the relationship between the Klee envelope and the distance functions are:. Subdifferential properties of one function can be inherited from subdifferential properties of the other one; 2. Another proof of the Klee idea that is, the Klee question is in fact a question on the convexity of Chebyshev sets is obtained, see Theorem 5 and Remark 0. For some results concerning differential properties of Klee envelopes we refer to the paper Wang 200 in the finite dimensional setting, and to Cibulka and Fabian 202 under the strong attainment. 2 Background Throughout we shall assume that X, is a real normed vector space, X is its topological dual and, is the pairing between X and X. We denote by B X, S X and B[x, r] resp. Bx, r the closed unit ball, the unit sphere and the closed resp. open ball of X centered at x with radius r > 0. By co and co, we denote the convex hull and the closed convex hull. The metric projection mapping on a subset S of X is defined by P S x := {s S : dx, S = x s }, x X, 4 where d, S is the distance function from the set S, that is, dx, S := inf x y. y S Let f : X R {+ } be an extended real-valued function. We recall that the Legendre-Fenchel conjugate of f is the function f : X R {, + } with f x := sup y X x, y fy for all x X.

5 The NSLUC property and Klee envelope 5 The subdifferential in the sense of convex analysis of f at a point x X where f is finite is defined by fx := {x X : x, y x fy fx, y X}. If f is finite at x, then x fx if and only if fx + f x = x, x ; so if the function f : X R {+ } is convex and lower semicontinuous, one has x fx x f x, 5 since under that condition f coincides with the restriction of f to X. If f is convex, the nonvacuity of fx at any point x where f is both finite and continuous is well known and this will be frequently used in the paper. The Fréchet subdifferential of f at the point x where f is finite is given by F fx := {x X : lim inf h 0 fx + h fx x, h h 0}. We adopt the convention F fx = when fx = +. Again with fx finite, the lower Dini directional derivative of f at x is given by d fx; h := lim inf w h;t 0 t fx + tw fx and when f is Lipschitz continuous near x we obviously have for all h X d fx; h = lim inf t fx + th fx. t 0 The Dini subdifferential of f at x is then the set fx := {x X : x, h d fx; h, h X} for x dom f and fx = if x / dom f, where dom f := {u X : fu < + } denotes the effective domain of f. When the set dom f is nonempty, one says that the function f is proper. Given any set-valued mapping M : X Y which can be P S, f, F f, f, it will be convenient as usual to denote by Dom M its effective domain and by Rge M its range, that is, Dom M := {x X : Mx } and Rge M := The graph of M is the set gph M := {x, y X Y : y Mx}. x X Mx.

6 6 Jourani,Thibault,Zagrodny 3 Properties of the Klee envelope Let f : X R {+ } be an extended real-valued function. We recall that for any reals λ > 0 and p, the Klee envelope of f with index λ and power p was defined in and the associated attainment set Q λ,p fx in 2. If κ λ,p f + the study of κ λ,p f and Q λ,p f is trivial. It is also worth pointing out that if κ λ,p fx 0 is finite for some x 0, then there is some real β such that Indeed, putting µ := κ λ,p fx 0 so, p2 p λ y p β fy for all y X. 6 pλ y x 0 p µ fy and noting that y p y x 0 + x 0 p 2 p y x 0 p + 2 p x 0 p, we see that for all y X p2 p λ y p pλ x 0 p µ pλ y x 0 p µ fy. The Klee envelope is a particular important case of supremal convolutions. Given functions g i : X R { } with i =,, n, the Moreau supremal convolution or sup-convolution of g,, g n is the function see Moreau 970 x φx := sup{g x + + g n x n : x + + x n = x}, where the supremum is taken over all n- tuples n-vectors x,, x n in X n such that x + + x n = x; so the Klee envelope κ λ,p f is the supremal convolution of f with the kernel function pλ p. If at least one of the functions g,, g n, say g, is convex, then φ is convex since the equality φx = sup g x u 2 u n + g 2 u g n u n u 2,,u n X n ensures that φ is the pointwise supremum of a family of convex functions. In particular, the Klee envelope κ λ,p f is always convex. For every real ε 0, it will be convenient to put Q ε λ,pfx := {y X : κ λ,p fx ε pλ x y p fy}, so Q ε λ,p fx whenever κ λ,pfx is finite and ε > 0. If f is finite at some point, say ȳ, then the inequality pλ x ȳ p fȳ κ λ,p fx tells us that κ λ,p f is coercive in the sense that κ λ,p fx + as x +. 7 In addition to that coercivity property, the next proposition establishes the local boundedness of the set-valued mapping Q ε λ,pf and some Lipschitz property of the function κ λ,p f. In view of the proof of the proposition, we must

7 The NSLUC property and Klee envelope 7 recall and this is not difficult to verify that the subdifferential of the convex function /p p is described for any x X by p p x = {x X : x, x = x x and x = x p } if p >, and with p = x = {x X : x and x, x = x }. 9 The latter means that 0 = B X and x = {x S X : x, x = x } for all x 0. Proposition Let X, be a normed vector space and f : X R {+ } be a proper function for which there exist two real numbers α, β with α > 0 such that α x p β fx for all x X. 0 The following hold: a With p = and λ /α the function κ λ, f is finite-valued and globally Lipschitz on X with /λ as a Lipschitz constant. b If p > and λ > /pα, the function κ λ,p f is finite-valued on X and Lipschitz continuous on each ball rb X of X with some Lipschitz contant L r p /λ therein. c With p and λ > /pα, for each pair of reals ε 0 and r > 0, the set-valued mapping Q ε λ,p is bounded over the ball rb X. Proof Let α, β be as given in the statement. First, assume that p = and λ /α. Fixing x X, we have for all y X, λ x y fy λ x + λ y α y + β x + β, λ so κ λ, fx is finite. Further, with x, x X taking the supremum over y X in the inequality λ x y fy λ x x + λ x y fy gives κ λ, fx κ λ, fx + λ x x, hence κ λ, f is Lipschitz on X with λ as a Lipschitz constant. Now assume that p and λ > /pα. Take any x X and write, for every y X, pλ x y p fy pλ x y p α y p + β. On the one hand, observing that pλ x y p α y p +β as y + since pλ α < 0, it results that pλ x p α p + β is bounded from above over X. From the function pλ x p f is also bounded from above 8

8 8 Jourani,Thibault,Zagrodny over X and finite at some point according to the properness of f, hence κ λ,p f is finite-valued over X. On the other hand, fixing any reals ε 0 and r > 0, we note by that, for all x X and y Q ε λ,p fx, κ λ,p fx ε pλ x y p α y p + β. Choose some element y 0 X with fy 0 finite and consider any x rb X and y Q ε λ,p fx, that is, κ λ,pfx ε pλ x y p fy. We have fy 0 pλ x y 0 p fy 0 κ λ,p fx pλ x y p α y p + β + ε, so choosing some y p p y x we obtain by 8 and this entails or equivalently fy 0 pλ α y p + pλ y x p y p + β + ε pλ α y p + λ y, x + β + ε pλ α y p + λ y x p x + β + ε, fy 0 pλ α y p + r λ y + rp + β + ε, α pλ y p r λ y + rp β + fy 0 + ε, with α pλ > 0. It ensues that there exists some real γ > 0 depending only on λ, r, ε such that y γ for all y Q ε λ,p x since otherwise for a sequence y i i of elements of Q ε λ,p x with y i + as i +, the first member of the latter inequality would tend to, contradicting the inequality. This means that Q ε λ,p x γb X, and the set-valued mapping Q ε λ,pf is bounded on rb X as desired. Finally, assume that p > and λ > /pα. Fix ε > 0 and r > 0, and let γ > 0 given as above. Take any x, x rb X. Considering any y γb X and choosing x p p x y we see from 8 again that pλ x y p pλ x y p λ x, x x λ x x x = λ x y p x x.

9 The NSLUC property and Klee envelope 9 Consequently, then for every y γb X, pλ x y p pλ x y p pλ x y p fy r + γp x x, λ r + γp x x + λ pλ x y p fy. Taking the supremum over all y γb X and noting by what precedes that κ λ,p fu = sup y γb X pλ u y p fy for all u rb X, we obtain κ λ,p fx κ λ,p fx + r+γp λ x x. This finishes the proof. Remark a If κ λ0,pfx 0 is finite for some λ 0 > 0 and x 0 X, then by 6 there is some real β such that p2 p λ 0 y p β fy for all y X, so the inequality assumption 0 in the proposition is fulfilled. In particular, the Klee envelope κ λ, f is finite-valued and /λ-lipschitz on X if and only it is finite at some point in X. b If dom f is bounded and f is bounded from below, then for any real α > 0 there exists some β R such that 0 is satisfied. Indeed, considering a lower bound γ of f and putting µ := sup u dom f u, we see that α x p + γ αµ p γ + ψ dom f x fx for all x X. c For a nonempty subset S of the normed space X,, the existence of α > 0 and β R such that α x p β ψ S x for all x X amounts to requiring that the set S is bounded, which in turn is equivalent the property that, for each real α > 0, there exists some β R such that 0 holds true with f = ψ S. Indeed, under the minorization assumption 0 for f = ψ S with some α > 0 and β R, for all x S we have x p βα, hence S is bounded. On the other hand, by b the boundedness of S is equivalent to the fact that, for each real α > 0, there exists β R such that α x p β ψ S x for all x X. The set-valued mapping Q λ,pf satisfies a closedness property. Proposition 2 Let X, be a normed vector space and f : X R {+ } be a proper lower semicontinuous function for which there exist two real numbers α, β with α > 0 such that 0 is fulfilled. Assume either p = and λ /α or p > and λ > /pα. Then the graph gph Q λ,p f := {ε, x, y [0, + [ X X : y Qε λ,pfx} in [0, + [ X X of the set-valued mapping ε, x Q ε λ,p fx defined on [0, + [ X is closed in [0, + [ X X.

10 0 Jourani,Thibault,Zagrodny Proof Let ε i, x i, y i i be a sequence of elements of [0, + [ X X converging to ε, x, y with y i Q ε i λ,p fx i. Then for every i N, κ λ,p fx i ε i pλ x i y i p fy i, so from the lower semicontinuity of f and the continuity of κ λ,p f see the previous proposition we obtain κ λ,p fx ε pλ x y p fy. This means that y Q ε λ,pfx as required. The following lemma prepares the next result related to the behavior of Q λ,p f when both the norm and its dual norm are differentiable. Lemma Let X, be a normed space, let g i : X R { }, i =,, n, be extended real-valued functions and let φ be their supremal convolution, that is, φx = sup{g x + + g n x n : x + + x n = x}. Assume that φ x is finite and attained at x,, x n, that is, φ x = g x + + g n x n with x + + x n = x. Then and co D g x D g n x n D φ x, co F g x F g n x n F φ x, where co denotes the norm-closure in X of the convex hull. Proof Fix any x F g x and consider any real ε > 0. There exists some real δ > 0 such that x, u ε u g x + u g x, for all u δb X. Since φ x = g x + + g n x n, it follows that x, u ε u φ x + u φ x u δb X, thus x F φ x. It results that repeating the reasoning with i = 2,, n F g x F g n x n F φ x, and the latter entails the required inclusion since the Fréchet subdifferential of a function at any point of X is known to be convex and norm closed in X see, e.g., Mordukhovich 2006, Definition.83 and comments. The inclusion concerning the Dini subdifferential is obtained in a similar and easier way. Remark 2 With the same arguments, the first inclusion also holds with several other subdifferentials, for example with proximal subdifferential see, e.g., Schirotzek 2007, Definition 9.. b in place of the Dini subdifferential.

11 The NSLUC property and Klee envelope Below Lemma is applied to get relations between subdifferentials of the norm in power p and the Klee envelope at points where the Klee envelope is attained. The assertions b-c involve the Gâteaux derivative Dφ of a function, see, e.g., Deville, Godefroy and Zizler 993, page 2 for the definition and properties. The case p = 2 was first obtained in Cibulka and Fabian 202, Proposition 8. Proposition 3 Let X, be a normed vector space and f : X R {+ } be a proper function, and let p [, + [. For any x X such that κ λ,p fx is finite, the following hold: a For every y Q λ,p fx, one has λ p p x y κ λ,p fx. b If κ λ,p f is Gâteaux differentiable at x and Q λ,p fx, then for all y Q λ,p fx D κ λ,p f x, x y = λ x y p. c If κ λ, f is Gâteaux differentiable at x and Q λ, fx, then D κ λ, f x = /λ. d If p > and both the norm and its dual norm are Gâteaux differentiable off zero, then at any x X where κ λ,p f is Gâteaux differentiable, the set Q λ,p fx is at most a singleton. Proof The inclusion in a follows directly from the above lemma and b is a consequence of a and the descriptions 8 and 9. Concerning c with p =, the assumption on Q λ, fx allows us to choose some y Q λ, fx. It follows from 9 and the Gâteaux differentiability of κ λ, f at x that y x. Using this in the equality in b and noting that D κ λ, f x /λ because of the /λ-lipschitz property of κ λ, f, see Proposition and Remark, we see that D κ λ, f x = /λ as required. Now to justify d, suppose that both the norm and its dual norm are Gâteaux differentiable off zero. Then p p x is a singleton for every x X, say {J p x} = p p x, and the mapping J p : X X is one-toone. Consequently at any x X where κ λ,p f is Gâteaux differentiable, we have λ J p x y = D κ λ,p f x whenever y Q λ,p fx. Since J p is one-to-one, Q λ,p fx is at most a singleton. The result of Fitzpatrick 980, Theorem 2.9 says that the differentiability of the farthest distance function at a point x entails the same differentiability of the norm at an appropriate point. The next proposition extends the result to the function κ λ,p f. We refer also to Montesinos, Zizler and Zizler 20 for some results on differentiability of the farthest distance function.

12 2 Jourani,Thibault,Zagrodny Proposition 4 Let X, be a normed space and f : X R {+ } be a proper function with κ λ,p f finite at some point. Assume that x X is a point where κ λ,p f is Gâteaux resp. Fréchet differentiable. Then for any y Q λ,p fx, the function p is Gâteaux resp. Fréchet differentiable at x y with pλd κ λ,p f x as derivative at x y. Proof Suppose that κ λ,p f is Gâteaux resp. Fréchet differentiable at x and denote ζ := D κ λ,p f x. Suppose also that Q λ,p fx is nonempty. Put q := /pλ p. Fix any y Q λ,p fx and take x qx y. For any real t > 0 and any h X, we have qx y + th qx y = qx y + th fy qx y fy κ λ,p fx + th κ λ,p fx, hence we see through the inclusion x qx y that x, h t qx y + th qx y t κ λ,p fx + th κ λ,p fx. 2 Since the last member tends to ζ, h as t 0, it ensues that x, h ζ, h, for all h X, thus x = ζ. Replacing x by ζ in the first member of 2 yields 0 t qx y + th qx y t ζ, h t κ λ,p fx + th κ λ,p fx t ζ, h. Since the last member tends to zero resp. tends to zero uniformly with respect to h B X, we conclude that q is Gâteaux resp. Fréchet differentiable at x y with D κ λ,p f x as derivative there. Our next aim is to show that the set Dom Q λ, is quite large. We then start with the analysis of the nonemptiness of Q λ, fx, by noting that such a nonemptiness implies the equalities κ λ, fx = sup z X,y S X λ y, x z fz = λ x, x y fy for all y Q λ, fx and x x y. Moreover if 0 is fulfilled with α > λ and the space X is finite dimensional, then the emptiness of Q λ, fx implies the inequality κ λ, fx > sup z X,y S X λ y, x z fz, whenever f is proper and lower semicontinuous. Indeed, if the equality holds true, then κ λ, fx = lim λ x z i fz i i +

13 The NSLUC property and Klee envelope 3 for some sequence z i i in X, then it follows from Proposition c that the sequence z i i is bounded, thus we may assume that it converges to a certain ȳ. It results that κ λ, fx = lim i + λ x z i fz i = λ x ȳ fȳ κ λ, fx, so κ λ, fx = λ x ȳ fȳ, which means that ȳ Q λ, fx, a contradiction. This indicates the role of the set {x X : x κ λ, fx, sup λ x, x y fy < κ λ, fx} y X in investigating the set of points for which the attainment set is nonempty. Below it is shown that this set is not too large, namely it is of first category. Let us also recall that the G δ density property of Dom Q λ,2 f was studied in Cibulka and Fabian 202, Theorem 5 with p = 2, in fact the strong attainment on a dense G δ subset was proved. We consider in the foregoing theorem the case p =. The proof of the theorem as well as that of the following lemma use the main ideas of Lau 975, Lemma 2.2, Theorem 2.3. Lemma 2 Let X, be a normed space and f : X R {+ } be a proper function for which there are reals α > 0 and β R such that Then for any real λ > /α the set α x β fx for all x X. {x X : x κ λ, fx, sup y X x, x y fy < κ λ, fx} is a countable union of closed sets with empty interior, so it is a set of first category in the space X. Proof For each integer i N denote A i := {x X : x κ λ, fx, sup x, x y fy κ λ, fx y X i }, so clearly the set of the lemma coincides with i N A i. Let us first fix i N and show that A i is closed. Consider any sequence x n n of elements of A i converging to some x X, and for each n N choose by definition of A i some x n κ λ, fx n satisfying sup x n, x n y fy κ λ, fx n y X i. The /λ-lipschitz property of κ λ, f ensures that x n /λ, so extracting a subnet we may suppose that x n n converges weakly to some x in X. The norm weak closedness of gph κ λ, f guarantees that x κ λ, fx. Further, for every y X, since x n, x n y fy κ λ, fx n i

14 4 Jourani,Thibault,Zagrodny and κ λ, f is continuous, we also have x, x y fy κ λ, fx /i. It ensues that x A i, justifying the closedness of A i. It remains to prove that all A i have empty interior. Suppose, for some i N, that int A i and take some x X and r > 0 such that B[ x, r] A i. We know by Proposition that the set Q λ,f x is nonempty and bounded, hence we can define the real γ := sup{ x y : y Q λ, f x}. For ε := 2iγ+ r r, there exists by definition of κλ, f some ȳ Q ε λ, f x Q λ, f x dom f satisfying κ λ, f x ε < λ x ȳ fȳ κ λ,f x. 3 Define t := r/γ and u := x + t x ȳ B[ x, r], so u A i. From the definition of A i there is some u κ λ, fu such that sup u, u y fy κ λ, fu y X i. 4 On the other hand, by 3 and the equality u ȳ = + t x ȳ we also have κ λ, f x κ λ, fu < λ x ȳ fȳ + ε κ λ,fu = + tλ u ȳ fȳ + ε κ λ,fu u ȳ fȳ t + t λ + t fȳ + ε κ λ,fu, which ensures by the definition of κ λ, f and by 4 κ λ, f x κ λ, fu < + t κ λ,fu t + t fȳ + ε κ λ,fu = t + t κ λ,fu t + t fȳ + ε t + t u, ȳ u + fȳ = t r + t u, ȳ u + ε iγ + r, t + t fȳ + ε t i + t hence taking the equality ȳ u = +t t x u into account we obtain κ λ, f x κ λ, fu < u, x u + ε Since ε < iγ + r r, we deduce that κ λ, f x κ λ, fu < u, x u, r iγ + r. which contradicts the inclusion u κ λ, fu, completing the proof.

15 The NSLUC property and Klee envelope 5 Theorem Let X, be a Banach space and f : X R {+ } be a proper function for which there are reals α > 0 and β R such that α x β fx for all x X, and let λ > /α. Assume that, for each x λ B X, the infimum of the function f + x, is attained. Then the set Dom Q λ, f contains a dense G δ subset of X. Proof Denote by M the set of first category in the statement of Lemma 2 above. Fix any x X \ M and by Lemma 2 choose some x κ λ, fx such that sup x, x y fy κ λ, fx. y X Since κ λ, fx λ B X according to the /λ-lipschitz property of κ λ, f, we can take thanks to the attainement assumption some ȳ X satisfying x, ȳ + fȳ = inf x, y + fy, y X or equivalently It results that x, x ȳ fȳ = sup x, x y fy. y Y κ λ, fx x, x ȳ fȳ λ x ȳ fȳ κ λ,fx, and this justifies the inclusion ȳ Q λ, fx, concluding the proof of the theorem. Let us notice that, assuming that the effective domain of f is a compact set and f is lower semicontinuous, we guarantee the assumption that the infimum of the function f + x, is attained for every x λ B X. It is also worth observing that if f is constant on its effective domain dom f which is additionally assumed to be weakly closed and bounded, then by James theorem, see Holmes 975 the attainment assumption in Theorem means that dom f is weakly compact. In particular, if the effective domain of f is B X, then X has to be a reflexive Banach space, whenever the attainement assumption in Theorem is satisfied. The next corollary is a direct consequence of Theorem. Before stating the corollary, let us recall that, given a topology θ on X, a function φ : X R {+ } is θ-inf-compact provided that all lower sections {x : φx r} are θ-compact, for all r R.

16 6 Jourani,Thibault,Zagrodny Corollary Let X, be a Banach space and f : X R {+ } be a proper function for which there are reals α > 0 and β R such that α x β fx for all x X, and let λ > /α. Assume, for some topology θ on X, that the function f+ x, is θ-inf-compact for all x λ B X. Then the set Dom Q λ, f contains a dense G δ subset of X. The second corollary assumes the weak inf-compactness of f. Corollary 2 Let X, be a Banach space and f : X R {+ } be a proper function for which there are reals α > 0 and β R such that α x β fx for all x X, and let λ > /α. Assume that f is weakly inf-compact. Then the set Dom Q λ, f contains a dense G δ subset of X. Proof Fix any x λ B X and note that fx + x, x α λ x β. This entails that µ := inf X f + x, is finite and coincides with inf ρbx f + x, for some real ρ > 0. Putting r := + µ and C := {x X : fx + x, x r} ρb X, we see that µ = inf C f + x,. Further, for any x C we have fx r x, x r + ρ x, hence C {x X : fx r + ρ x }. The latter set being weakly compact by the weak inf-compactness of f, the weakly closed subset C is weakly compact too. Consequently, the infimum µ is attained on the set C according to the weak lower semicontinuity of f + x,. The corollary then follows from Theorem. Taking for f the indicator function of a nonempty weakly compact subset of X in the latter corollary yields the following result: Corollary 3 Theorem 2.3 in Lau 975 Let X, be a Banach space and S be a nonempty weakly compact subset of X. Then Dom Q S contains a dense G δ subset of X. The next corollary makes use of the Mackey topology on the topological dual space X ; we refer to Schaefer 97, IV Duality, 3. Locally Convex Topologies Consistent with a Given Duality. The Mackey-Arens Theorem for the definition and properties of that topology.

17 The NSLUC property and Klee envelope 7 Corollary 4 Let X, be a Banach space and f : X R {+ } be a proper lower semicontinuous convex function for which there are reals α > 0 and β R such that α x β fx for all x X, and let λ > /α. Assume that at every x λ B X the Legendre-Fenchel conjugate f is finite and continuous with respect to the Mackey topology τx, X. Then the set Dom Q λ, f contains a dense G δ set of X. Proof The function f being proper, lower semicontinuous and convex, the continuity of f at an element x dom f with respect to the Mackey topology τx, X is known to entail that f x, is weakly inf-compact see Moreau 966, Corollary 8.2. The assertion of the corollary then follows from Corollary. Remark 3 For the proper lower semicontinuous convex function f : X R {+ }, we know that there always exist reals α, β satisfying the inequality α x β fx for every x X. The assumption of the latter corollary requires the positiveness of such a real α. Since κ λ, f is Lipschitz with /λ as Lipschitz constant, we know that Rge κ λ, f λ B X. Adapting the proofs of Lemma 4. and Theorem 4.2 in Whestphal and Schwartz 998, we can show more. Lemma 3 Let X, be a normed space and f : X R {+ } be a proper function with κ λ, f finite at some point. Let also x i i be a sequence of elements of X with lim x i = +. Then for any x dom f and any i + sequence x i i with x i κ λ,fx i, one has lim x x i x i, = lim i + x i x i + x i = λ. Proof Since the function κ λ, f is assumed to be finite at some point, we know that it is finite on X and Lipschitz with /λ as a Lipschitz constant. Fix any x dom f. Observe that on the one hand by definition of κ λ, fx i λ x i x fx κ λ, fx κ λ, fx i κ λ, fx, and on the other hand by the inclusion x i κ λ,fx i It follows that, for large i N, x i, x x i κ λ, fx κ λ, fx i. λ fx + κ λ,fx κ λ,fx i κ λ, fx x i x x i x x i, x i x x i x i x λ,

18 8 Jourani,Thibault,Zagrodny hence passing to the limit as i + keeping in mind that x i + as i + gives lim x x i x i, = lim i + x i x i + x i = λ. Proposition 5 Let X, be a reflexive Banach space and f : X R {+ } be a proper function with κ λ,p f finite-valued. a If p >, then Rge κ λ,p f = X. b If the norm is smooth and strictly convex or equivalently both norms and its dual norm are smooth, that is, Gâteaux differentiable off zero, then Rge κ λ, f is convex and B0, /λ Rge κ λ, B[0, /λ] = cl Rge κ λ,. Proof a The assertion a follows from a classical result. We sketch the arguments. Fixing any x X, since κ λ,p fx x, x + as x + see 6, the weakly lower semicontinuous function κ λ,p f x, has a global minimizer x X, hence the Moreau-Rockafellar sum rule yields x κ λ,p f x. b The Gâteaux differentiability assumption on 2 entails that the duality mapping J := 2 2 is single-valued and norm-weak continuous. Let x X be a point where f is finite. Fix any x X with x = /λ. Choose some x S X with x, x = /λ. By 8 we have x = λ Jx. For each integer i N, choosing x i κ λ,fix + x the above lemma ensures that lim i x i, x = lim i x i = /λ. Consider a subsequence that we do not relabel of x i i converging weakly to some z, it is clear that z /λ and z, x = /λ, so z = /λ. It results that z = λ Jx, hence z = x. On the other hand, denoting by φ the Legendre-Fenchel conjugate of κ λ, f, by 5 one has cl Rge κ λ, f = cldom φ, thus cl Rge κ λ, f is a weakly closed convex set in X. Since x i Rge κ λ,f, it results that x cl Rge κ λ, f. Consequently, λ S X is included in the closed convex set cl Rge κ λ, f, which entails B[0, /λ] cl Rge κ λ, f, and this inclusion is an equality since the converse follows from the Lipschitz property of κ λ, f with constant /λ. Now let us show that B0, /λ Rge κ λ, f. Fix any x B0, /λ. Since B[0, /λ] = cl Rge κ λ, f as seen above, we can take a sequence x i, x i i of elements of gph κ λ, f such that lim i + x i x = 0. The above lemma ensures that the sequence x i i is bounded since x < λ in this case, hence a subsequence converges weakly to some x. By the weak-norm closedness of the graph of the subdifferential of the continuous convex function κ λ, f, we obtain x κ λ, fx, which justifies the desired inclusion B0, /λ Rge κ λ, f.

19 The NSLUC property and Klee envelope 9 Finally, the strict convexity of and the inclusions B0, /λ Rge κ λ, f B[0, /λ] guarantee the convexity of Rge κ λ, f. The statement a is in fact a simple consequence of Zagrodny 994, Proposition 3.5, where an equivalent condition to the reflexivity of the space was given. In view of Proposition 3c the Gâteaux differentiablity of κ λ, f at x implies that Dκ λ, fx λ S X whenever Q λ, fx. Thus it is natural to investigate the set C λ f := {x X : κ λ, fx λ S X }. It follows from Proposition 5 b that there exists at least one point from the domain of κ λ, f which is not in C λ f. As a consequence, the Klee envelope κ λ, f is not a smooth function. In other words, there must exist a point in its domain where it is not Gâteaux differentiable, whenever X is a reflexive Banach space with both norm and its dual norm being smooth. Additional properties of the Klee envelope can be obtained in the case when X, is a Hilbert space and p = 2. Indeed, writing in that case 2λ x y 2 fy = 2λ x 2 + λ x, y fy + 2λ y 2, and taking the supremum over y X gives κ λ,2 fx = 2λ x 2 + f 2λ 2 λ x. Further, given λx, y gph Q λ,2 f, we have κ λ,2 f λx 2λ λx 2 = 2λ λx y 2 fy 2λ λx 2 = 2λ y 2 + y, x fy. Putting φu := f 2λ 2 u for all u X, we deduce on the one hand that κ λ,2 f λx as well as φx are finite, and on the other hand that φx + y, u x = u, y + 2λ y 2 fy φu, for all u X. This says that y φx = f 2λ 2 x. We have then proved the following:

20 20 Jourani,Thibault,Zagrodny Proposition 6 Assume that X, is a Hilbert space and f : X R {+ } is a proper function. Then for all x X and κ λ,2 fx = 2λ x 2 + f 2λ 2 λ x Q λ,2 f λx f 2λ 2 x. Corollary 5 Assume that X, is a Hilbert space and f : X R {+ } is a proper function with κ λ,2 f finite at some point in X. Then the inverse subdifferential κ λ,2 f is single-valued and λ-lipschitz on X. Proof By the first equality in Proposition 6 the lower semicontinuous convex function g, defined by gx := f 2λ 2 λ x for all x X, is proper, hence the resolvent mapping J := Id X + λ g is single-valued and -Lipschitz on X see, e.g., Moreau 965, Propositions 5.b and 6.a or Bauschke and Combettes 20, Corollary This means by the first equality in Proposition 6 again that λκ λ,2 f is single-valued and - Lipschitz on X or equivalently κ λ,2 f is single-valued and λ-lipschitz on X, as easily seen. Finishing this Section we would like to emphasize, that there are other papers concerning subdifferential or differential properties of the Klee envelope. For example Proposition 4.4 and Theorem 4.7 in Wang 200 provide such some results in the finite dimensional setting. In Proposition 3 and Theorem 7 of Cibulka and Fabian 202, these properties were investigated under the strong attainment assumption with p = 2. 4 Klee envelope of the lower semicontinuous convex hull function For a set S of the normed space X,, according to the Mazur Intersection Property, we define the Mazur hull Maz S of S as the intersection of all closed balls containing S, so a closed set fulfills the Mazur intersection property if and only if it coincides with its Mazur hull. Clearly, from the very definition Maz S = B[x, S x], x X so Maz S x S x for any x X. This combined with the inclusions S co S co S Maz S entails that for any bounded set S S x = co S x = co S x = Maz S x, for all x X. 5

21 The NSLUC property and Klee envelope 2 Given a function f : X R {+ }, its convex hull co f : X R {, + } is defined by co fx = inf{r R : x, r co epi f}, where epif denotes the epigraph of f, that is, epi f = {y, r X R : fy r}. Clearly, it is the greatest convex function majorized by f and m co fx = inf{ t i fy i : y i X, t i > 0, m t i y i = x, m t i = }. Similarly, the lower semicontinuous convex hull or closed convex hull co f : X R {, + } of f is defined by co fx = inf{r R : x, r co epi f}. It follows from the construction that co f is convex and lower semicontinuous and it is the greatest lower semicontinuous convex function less or equal to f. It also satisfies the following properties co [epif] = epi co f, co [dom f] dom co f. This allows us to express this closed convex hull function in the following manner in the case where f is lower semicontinuous: For all x X there exist m n sequences of elements m n n in N, t n n,, t n m n n in ]0, ], with t n i =, and y n n,, y n m n n in dom f such that lim n + m n t n i yi n = x, cofx = lim n + m n t n i fyi n. The foregoing proposition extends the two first equalities in 5 to Klee envelopes of functions. Proposition 7 Let X, be a normed space and let f : X R {, + } be an extended real-valued function. For all x X, κ λ,p fx = κ λ,p co f x = κλ,p cof x. 6 Proof Since co f co f f, we see that κ λ,p f κ λ,p co f κ λ,p co f. Now fix x X and take any y X. Consider any convex combination m t iy i = y, that is, t i > 0 and m t i =. For every i =,..., m, we have pλ x y i p fy i κ λ,p fx,

22 22 Jourani,Thibault,Zagrodny hence m t i pλ x y i p m t i fy i κ λ,p fx, which combined with the convexity of x p yields pλ x y p m t i fy i κ λ,p fx. Taking the supremum of both members over all convex combinations of y, we obtain m m m pλ x y p inf{ t i fy i : t i y i = y, t i > 0, t i = } κ λ,p fx, in other words pλ x y p co fy κ λ,p fx. For every u X, taking the limit superior of both members of the latter inequality as y u gives pλ x u p co fu κ λ,p fx, hence taking now the supremum over all u X guarantees that κ λ,p co f x κ λ,p fx. This and what precedes justifies the desired equalities in 6. 5 The NSLUC property We begin this section by defining a class of closed sets which will be involved in the next section for the study of relationship between the attainment sets of κ λ,p f and κ λ,p co f. Definition Let S be a subset of the normed space X,. We say that S has norm subdifferential local uniform convexity property, NSLUC in short, if for every bounded subset S S with 0 cl S and every u S X for which there is a continuous linear functional u u satisfying one can find a real β > 0 such that inf s S s u, s u > 0, 7 s S, s u, s + β s u, s u. 8 Let us point out that if we omit the restriction that S must be bounded, then even in Hilbert spaces or simply in the Euclidean space R 2, we cannot ensure that 8 holds true. As an example, in a Hilbert space H endowed with the inner product let us take the subset S = {v + nu : n 0}, where u = v = and u v = 0. Then, u := u u and inf s S ds, span u =, thus 7 is satisfied. So, if there exists β > 0 for which

23 The NSLUC property and Klee envelope 23 relation 8 is fulfilled, otherwise stated, for all n N, β + n v + nu or equivalently β 2 + 2βn + n 2 n 2 +, that is, β 2 + 2βn, then we arrive at a contradiction. Below we provide another characterization of the NSLUC property by means of distances from the kernel of the functional u u and the space generated by u, whenever u S X. For this reason, fix any u S X and u u. Since u, u = 0, we know that the space X is the algebraic even topological direct sum of the vector spaces ker u and span u, that is, X = ker u span u. Further, for every x X, noticing by the well-known distance formula from a closed hyperplane see, e.g., Fabian et al. 20, Exercice 2.2, page 7 that dx, ker u = u, x u = u, x, we see that p x := x u, x u is a nearest point with respect to of x in ker u. Putting p 2 x := u, x u span u, it is known and obvious that p 2 is the projector onto span u, that is, x = p x + p 2 x for all x X. Let S S be a bounded subset, with 0 cl S such that S keeps the space span u uniformly far, that is, inf s S ds, span u > 0, 9 then assuming NSLUC property, there exists β > 0, such that s S, s s p s + β s p 2 s or equivalently for the proof see Proposition 8 below s S, s ds, ker u + γds, span u, 20 for some γ > 0. When X, is a Hilbert space and S keeps the set span u uniformly far, such a real γ can be precised. Let such a bounded subset S of a Hilbert space X, be given. Put M := sup s and r := inf s S s S ds, span u. Note that, for all s S, s 2 = s p s 2 + p s 2 = ds, ker u 2 + s u, s u 2 ds, ker u 2 + ds, span u 2. So, to ensure the desired inequality it suffices to choose γ so that for all s S ds, ker u 2 + 2γds, ker u ds, span u + γ 2 ds, span u 2 ds, ker u 2 + ds, span u 2,

24 24 Jourani,Thibault,Zagrodny and by definitions of M and r this holds true if in particular 2γM 2 +γ 2 M 2 r 2 M 2 γ 2 +2γ + M 2 +r 2 M 2 γ + 2 M 2 +r 2, which in turn is satisfied in particular for γ := M+ M 2 +r 2 M. In fact, as it will be established hereafter, in several normed vector spaces the implication 9 20 is satisfied. This can be easily seen whenever the characterization of the NSLUC property given below is used. Proposition 8 Let S be a subset in the normed space X, and S S with 0 cl S be a bounded subset. Then for every u S X and u u relation 7 holds if and only if 9 holds. Moreover, the following assertions are equivalent: i S has NSLUC property, ii for every bounded subset S S with 0 cl S and every u S X satisfying relation 9 and every continuous linear functional u u one can find a real γ > 0 such that relation 20 holds true. Proof We start our proof by establishing the following equivalence: relation 9 is equivalent to relation 7 for every u u. Since for every u u, ds, span u s u, s u, then the implication 9 7 is trivial. Now, assume that relation 7 holds for every u u and suppose that relation 9 does not hold. Then there exists a bounded sequence s i i of elements of S such that lim ds i, span u = 0. Since the space i + span u is finite dimensional, extracting subsequence, we may assume that lim s i = λu and hence lim i + i + u, s i = λ and lim s i u, s i u = 0, i + which contradicts relation 7. The implication i ii is easy to observe, because of the previous part of the proof and for all s X, ds, ker u = u, s and ds, span u s u, s u. Now, we prove the implication ii i. Fix any bounded set S S and u S X for which there is u u fulfilling 7 and for each i N there exists s i S such that s i < u, s i + i s i u, s i u. 2 By ii, there exists γ > 0 depending only on S and not on s i i such that s i ds i, ker u + γds i, span u, and hence, using the equality d, ker u = u,, we see that relation 2 ensures that γds i, span u < i s i u, s i u. Since the sequence s i i is bounded, we obtain that which contradicts relation 9. lim ds i, span u = 0, i +

25 The NSLUC property and Klee envelope 25 Remark 4 Let S be a subset of the normed space X, having the NSLUC property. Let u, u gph with u S X and let α > 0. If s i i is a sequence of elements of S, then the following implication holds true. lim s i = i + lim i + u, s i = α = lim s i αu = 0 i + Let us also observe that, if X has the NSLUC property, then any subset S of X has the NSLUC property too. Below it is proved that if the unit sphere has the NSLUC property, then the whole set X has the property too. Remark 5 The sphere S X of the normed space X, has the NSLUC property if and only if the whole set X has the NSLUC property. Proof Suppose that S X has the NSLUC property. Let us fix a bounded set S X with 0 cl S and u, u gph such that Put and notice that A S X. Since inf s S s u, s u > A := { s s : s S } S 0 inf a A a u, a u inf s S s u, s u, it follows from 22 and the NSLUC property for S X that there is β > 0 such that a A, a u, a + β a u, a u. 23 It is a simple consequence of 23 that s S, s u, s + β s u, s u, which gives the NSLUC property of the whole set X. The converse implication is a direct consequence of the observation preceding the remark. In the following proposition, we shall show that if a subset S of X has the NSLUC property, then the norm of the space is strictly convex on S and the set S has the Kadec-Klee property with respect to the norm. Let us recall that the norm is strictly convex on a subset S of X if the following implication holds true: x, y S, x + y = x + y, x 0, y 0 x = x y. 24 y If 24 holds true for S = S X, then one just says that the norm is strictly convex or the space X, is stricly convex.

26 26 Jourani,Thibault,Zagrodny We say that a set S X has Kadec-Klee property with respect to the norm whenever any sequence x i i of elements of S converging weakly to x X along with lim x i = x converges strongly to x that is, x i x 0 as i + i +. So, the norm has the Kadec-Klee property if and only if the whole set X has the Kadec-Klee property with respect to. Proposition 9 Let S be a set in the normed space X, having the NSLUC property. Then the norm is strictly convex on S and S has the Kadec-Klee property with respect to. Proof Strict convexity: Let x, y S, with x + y = x + y, x 0 and y 0. We shall show that x = λy, where λ := x y. Suppose that x λy > Put S = {x, y} and u = x+y x + y. Then u =. We claim that, for all u u y u, y u > 0, and x u, x u > 0. Indeed, if y = u, y u, then u, y = y. If u, y = y, then y = λ x, and this contradicts relation 25. If u, y = y, then y = y x +2 y x, and hence y = 0, and this contradicts y 0 and hence the claim is justified. Then, by the NSLUC property of S, there exists β > 0 such that x u, x + β x u, x u, and y u, y + β y u, y u, and adding these two inequalities gives x + y u, x + y + β x u, x u + y u, y u. Since x + y = x + y and u, x + y = x + y, it ensues that x + y x + y + β x u, x u + y u, y u, which contradicts the inequality x u, x u + y u, y u > 0. Kadec-Klee property: Let x i i be a sequence of elements of S and x X be such that x i i converges weakly to x and lim x i = x. We may i + assume that x 0, otherwise we are done. Put α := x, u := α x and take u u. It follows from Remark 4 that lim x i x = 0. i + Now we give examples of sets satisfying the NSLUC property. Let us recall first that the norm of X is locally uniformly rotund LUR, or simply X, is LUR see, e.g., Deville, Godefroy and Zizler 993, if the following condition holds: lim x i =, i + lim x i + x = 2 = i + lim x i x = 0. i +

27 The NSLUC property and Klee envelope 27 Proposition 0 If the normed space X, is LUR, then X has the NSLUC property. Proof Fix any bounded set S X with 0 cl S and any u S X for which there is u u fulfilling 7. Suppose that there is no real β > 0 satisfying relation 8. Then for each i N there exists s i S such that s i < u, s i + i s i u, s i u. 26 Further by 7, for δ := inf s S s u, s u > 0, we have s i u, s i u δ, for all i N. 27 As S is bounded, the sequence s i i is bounded too and hence lim i + i s i u, s i u = 0. Note that, because of 26 and the relation 0 cl S, each convergent subsequence which exists because of the boundedness of S of the sequence u, s i i has a nonzero limit. Put α i = u, s i for all i N. We may suppose that α i 0, i and lim i + α i = α 0. Using 26, we get lim Now, note that i + α i 2 = u + u, α i and hence lim i + lim u α i i + following inequality s i = and u, α i s i =, for all i N. 28 s i = u, u + α i s i u + α i s i u + α i s i 29 u + α i s i = 2. So the LUR property of X implies that s i = 0. Because of 7, the last equality contradicts the α i s i u α i inf s S s u, s u. For any fixed integer k 2, a normed space X, is called fully k-convex see, e.g., Fan and Glicksberg 955 and Fan and Glicksberg 958, if every sequence x n n of elements of X satisfying lim x n =, and k n + k x νi for any k indices ν ν k is a Cauchy sequence. Fan and Glicksberg

28 28 Jourani,Thibault,Zagrodny proved that X, is fully k-convex if and only if every sequence x n n of elements of X satisfying k lim ν,,ν k k x νi =, is a Cauchy sequence. It is also shown in Fan and Glicksberg 958, that for any integer k 2, every fully k-convex normed space is strictly convex and has the Kadec-Klee property. Further, Polak and Sims exhibited in Polak and Sims 983 an example of a Banach space which is fully 2 convex but not locally uniformly rotund. In the following proposition, we shall show that all subsets of a fully k- convex Banach space have the NSLUC property. This allows us to say that there is no equivalence between the LUR property of a Banach space and the NSLUC property of the space. Proposition If X, is a fully k-convex Banach space, for some integer k 2, then X has the NSLUC property. Proof Fix any bounded subset S X with 0 cl S and any u S X for which there is u u fulfilling 7. We repeat the proof of Proposition 0, to get the existence of δ > 0 and a sequence s i i of elements of the bounded subset S such that s i u, s i u δ i N, 30 the sequence u s i, s i i converging to α 0 and lim i + u = as well as, s i 26. Putting ε α := signα, there is some i 0 such that u, s i = ε α u, s i for all i i 0. Let ν ν k be any k indices greater than i 0. Using relation 26 gives ε α u, s ν + + s νk s ν + + s νk < ε α u, s ν + + s νk + k j= ν j s νj u, s νj u, and hence noting that k j= ν j s νj u, s νj u 0 as ν j for j =,, k we get k lim ν,,ν k k s νi α =. Since X is fully k-convex, the sequence s i i converges to some s, and s 0 because 0 cl S. Consequently, u, ε α s = ε α s or equivalently u ε α s. Since u u, we obtain u+ εαs ε αs = 2. The strict convexity of X because of the full k-convexity of X, ensures that u = ε αs ε α s. Thus s = u, s u, and this contradicts relation 30 by passing to the limit in 30.