On linear selections of convex set-valued maps

гєжё R «мл c «Ё Ё ҐJ R ЇаЁ«R ҐЁп 2009, в. 43, ўлї. 2, б.???? Љ 517.988+514.172+517.982.256 c 2009. V. Yu. Protasov On linear selections of convex set-valued maps We study continuous subadditive set-valued maps taking points of a linear space X to convex compact subsets of a linear space Y. The subadditivity means that φ(x 1 + x 2 ) φ(x 1 ) + φ(x 2 ). We characterize all pairs of locally convex spaces (X, Y ), for which any such map has a linear selection, i.e., a linear operator A: X Y such that Ax φ(x), x X. The existence of linear selections for a class of subadditive maps generated by differences of a continuous function is proved. This result is applied to the problem of Lipschitz stability of linear operators in Banach spaces. 1. Introduction We consider set-valued maps taking points of a space X to convex compact sets of a space Y. In the sequel X and Y are linear locally convex spaces, M(Y ) is the set of nonempty convex compact subsets of the space Y. The set M(Y ) is partially ordered by inclusion: K 1 K 2, if K 1 K 2, and K 1 < K 2, if K 1 int K 2. We consider continuous maps φ: X M(Y ). The topology on the set M(Y ) is defined in a standard way by the support functions: the base of topology consists of all finite intersections of the sets O α,z = {K M(Y ) sup y K (z, y) < α}, z Y, α R. A map φ is called subadditive, if φ(x 1 + x 2 ) φ(x 1 ) + φ(x 2 ), x 1, x 2 X. In other words, the image of the sum is contained in the Minkowski sum of images. A map φ is convex, if φ(tx 1 + (1 t)x 2 ) tφ(x 1 ) + (1 t)φ(x 2 ), x 1, x 2 X, t [0, 1], and is positively homogeneous (in the sequel we call such maps homogeneous, skipping the word positively ), if φ(tx) = tφ(x), t [0, + ). Following [1] we use the word fan for a subadditive homogeneous map φ: X M(Y ). The simplest example of a subadditive map is φ(x) = {Ax}, where A is a linear continuous operator. If A is a compact convex set of linear operators, then the map φ(x) = {Ax A A} is subadditive as well. For any Lipschitz function f : R n R m the map φ(x) = Q(x), where Q(x) is the closure of convex hull of the set of differences The research is supported by the RFBR grants No 08-01-00208 and No 10-01- 00293, and by the grant of Dynasty foundation, 2010.

{f(z + x) f(z) z R n }, is subadditive. In contrast to the previous example, this map is not necessarily a fan. Another example of a fan is the prederivative of a Lipschitz function [1]. That notion has important applications in the theory of extremum. Subadditive convex maps have been intensively studied in the literature due to numerous applications in the theory of functions, nonlinear analysis, algebra, mathematical economics, etc. (see a review [2], the corresponding chapters in reviews [3], [4], the work [1], and references therein). One of the most important questions about a subadditive map: does it have a linear selection, i.e., a linear operator A (not necessarily continuous) such that Ax φ(x), x X? Let us remark that the existence of a continuous homogeneous selection, without linearity, follows from the classical work of Michael [5]. Is it true that any subadditive map possesses a linear selection? For all known examples from applications (including those mentioned above) the answer is affirmative. Besides, the existence of linear selections is proved for some other classes of maps. In particular, additive maps φ characterized by the property φ(x 1 +x 2 ) = φ(x 1 )+φ(x 2 ), x 1, x 2 X, always have linear selections [6]. A class of superadditive maps, which are, in some sense, opposite to the subadditive ones, has been studied in the literature in great detail. Such maps are characterized by the property φ(x 1 + x 2 ) φ(x 1 ) + φ(x 2 ), x 1, x 2 X. Every superadditive map possesses a linear selection. This was proved in [7] under the homogeneity assumption, that was omitted later in [8], [9]. For the subadditive maps the answer has not been obtained for the general case, although the existence of linear selections was established for several special cases important in applications (see [1], [10], see also [11], [12], where similar problems were analysed). In this paper we show that for general subadditive maps the answer is negative. Unless one of the spaces X or Y is one-dimensional, there always exists a fan φ: X M(Y ) that does not have a linear selection (Theorem 1). Only in the cases dim X = 1 or dim Y = 1 any subadditive map possesses a linear selection. An analogous fact takes place also for affine selections of convex maps (Theorem 2). These results are presented in 2 and proved in 3. Then we consider the class of subadditive maps generated by differences of uniformly continuous functions, and prove that for this class linear selections always exist (Theorem 3). This result is applied in 5 to solve the problem of the Lipschitz stability of linear operators in Banach spaces (Theorem 4). 2. The main results To begin with, we show that every subadditive map dominates a fan. Let us recall that we consider only continuous maps. Proposition 1. For every subadditive map φ: X M(Y ) there is a fan φ: X M(Y ) such that φ(x) φ(x), x X. 2

The proof is in Appendix. Every selection of the map φ is a selection for φ as well. Therefore, proving the existence of linear selections we can restrict ourselves to homogeneous maps φ. Thus, it can always be assumed that a given subadditive map is homogeneous, and, consequently, is a fan. The next result answers the question: for which spaces X and Y every subadditive map φ: X M(Y ) possesses a linear selection? Theorem 1. If either dim X = 1 or dim Y = 1, then each subadditive map φ: X M(Y ) has a continuous linear selection. If the dimensions of the spaces are both bigger than 1, then there exists a fan φ: X M(Y ) that has no linear selection. Thus, in contrast to superadditive maps, the subadditive ones may have no linear selections. A similar problem is formulated for convex maps. In this case we analyze the existence of affine selections. If, for example, a convex map is univalued, i.e., all the sets φ(x) are oneelement, then φ is affine, and so it is an affine selection itself. For the the general convex maps, however, the following result similar to Theorem 1 holds: Theorem 2. If dim Y = 1, then every convex map φ: X M(Y ) possesses a continuous affine selection. If dim Y 2, then there exists a convex map φ: X M(Y ) that has no affine selection. Similar results can be established for maps φ: Ω M(Y ), where Ω X is an arbitrary convex domain (see Remark 2 below). 3. Proof of Theorems 1 and 2 The proofs of both theorems will be split into several steps and auxiliary results. To construct a map that does not possess a linear (or affine) selection, we first introduce so-called discretely convex maps defined on a finite set of points {x i } n i=1 X. One chooses the values of those maps so that the corresponding linear selections do not exist. Then one proves that piecewise linear extensions of such maps to the whole space X are convex maps. Proof of Theorem 1. The case dim X = 1. Let X = R. By Proposition 1 it may be assumed that φ is a fan. Whence, φ(0) = {0}. The subadditivity assumption yields 0 φ( 1) + φ(1), therefore there is b φ(1), for which b φ( 1). Then the map A: x xb is a linear selection for the map φ. The case dim Y = 1. For every x X the set φ(x) is a line segment, let us denote it as [φ (x), φ + (x)]. By Proposition 1 it may be assumed that φ is a fan. In particular, φ is convex. It follows that the function φ + is convex, and the function φ is concave. The epigraph P = {(x, t) X R t φ + (x)} of the function φ + and the hypograph Q = {(x, t) X R t φ (x)} of the function φ are both convex sets. The 3

continuity of the map φ implies that the set P has a nonempty interior. Hence, by the Hahn-Banach separation theorem [13, p. 38] there is a nonzero linear continuous functional (x, s) (X R) such that inf [(x, x) + st] sup [(x, x) + st]. (t,x) P (t,x) Q Let us denote the left-hand side of this inequality by a. The case s 0 is impossible, otherwise a = ; therefore, s > 0. For arbitrary x we have (x, x) + sφ + (x) a and (x, x) + sφ (x) a. Then for the affine function A φ (x) = a/s (x, x)/s for every x one has A φ (x) [φ (x), φ + (x)]. Whence, A φ is an affine selection. Since φ is a fan, it follows that φ(0) = {0}, and therefore A φ (0) = 0. Consequently, the operator A φ is linear. The case dim X 2, dim Y 2. It suffices to construct a fan φ: R 2 M(R 2 ) that does not have a linear selection. If such a fan exists, then for arbitrary spaces X, Y one can take their two-dimensional subspaces X 1 X, Y 1 Y, factorize X into a direct sum X = X 1 L, where L X is a closed subspace of codimension 2, and consider the corresponding projector P : X X 1. Then the map φ(x) = φ(p x) is a fan. Since φ X1 = φ, we see that the map φ has no linear selection. For constructing the map φ: R 2 M(R 2 ) we need the following auxiliary result proved in Appendix. Consider an arbitrary partition of the plane R 2 by rays r 1,..., r n starting from the origin. The rays are in consecutive order, two neighboring rays form an angle smaller than π/2. On each ray r i we chose an arbitrary direction vector x i and denote by α i, β i positive numbers such that x i = α i x i 1 + β i x i+1, i = 1,..., n, where we set x 0 = x n, x n+1 = x 1. An arbitrary function φ defined at the points x i, i = 1,..., n and taking values on the set M(Y ) is called discretely convex, if φ(x i ) α i φ(x i 1 ) + β i φ(x i+1 ), i = 1,..., n. The piecewise linear extension of a given function φ to the whole plane R 2, which will be denoted by the same symbol φ, is a homogeneous function φ: R 2 M(Y ) defined on each angle between neighboring rays r i and r i+1 by the formula φ(sx i + tx i+1 ) = sφ(x i ) + tφ(x i+1 ), s, t R +. Lemma 1. The piecewise linear extension of a discretely convex function φ defined at points x 1,..., x n R 2 and taking values on the set M(Y ) is a fan. Let x 1,..., x 6 R 2 be the vertices of a regular hexagon centered at the origin. They generate rays r i = {tx i t [0, + )} splitting the plane to six parts. Let also D i, i = 1,..., 6, be the consecutive main diagonals of a regular 12-gon centered at the origin. The function taking each point x i to the segment D i is discretely convex. Indeed, x i = x i 1 +x i+1, therefore α i = β i = 1; and the sum D i 1 +D i+1 is a rhombus that contains D i in its interior, which implies D i < D i 1 + D i+1. After 4

a slight perturbation of the segments D i we obtain segments D i. Since all the inequalities are strict, it follows that such perturbation preserves the discrete convexity of the function. Invoking Lemma 1, we see that the piecewise linear extension of the function φ: x i D i, i = 1,..., 6, is convex. Suppose that A is a linear selection of the map φ. Since x i = x i+3, we have Ax i = Ax i+3, i = 1, 2, 3. Therefore, for given i the point Ax i is uniquely defined as an intersection of the segments D i and D i+3. One can perturb the segment D 2 so that Ax 2 Ax 1 +Ax 3. Since x 2 = x 1 + x 3, this contradicts the linearity of the operator A. Remark 1. Let there be a finite set of points T = {x 1,..., x n } on the plane R 2 satisfying assumptions of Lemma 1 (all points x i are nonzero, the rays r i = { } tx i t [0, + ), i = 1,..., n, are consecutive, the angle between two neighboring rays is smaller than π/2), let also a convex homogeneous map F : R 2 M(Y ) be given. Then it is easily can be shown that the restriction of the function F to the set T is discretely convex. Lemma 1 implies that the converse is also true: every discretely convex map defined on T can be extended to the whole plane R 2 to a homogeneous map. Thus, discretely convex maps are precisely restrictions of all possible convex maps (from R 2 to M(Y )) to the set T. Proof of Theorem 2. The case dim Y = 1. The proof is literally the same as the proof of Theorem 1 for this case. The case dim Y 2. It suffices to construct a convex map φ: R M(R 2 ) that does not have a linear selection. The generalization to arbitrary spaces X, Y is realized in the same way as in Theorem 1. We need two lemmas proved in Appendix. The first one is an analogue of Lemma 1 for the straight line. Consider an arbitrary partition of a segment [a, b] with nodes a = x 1 < < x n = b and set T = {x i } n i=1. Let us write α i, β i for positive numbers such that x i = α i x i 1 + β i x i+1 and α i +β i = 1, i = 2,..., n 1. A function φ: T M(Y ) is discretely convex, if φ(x i ) α i φ(x i 1 )+β i φ(x i+1 ) for all i = 2,..., n 1. Piecewise linear extension of a given function to the whole segment [a, b] is defined on each segment [x i, x i+1 ], i = 1,..., n 1, as φ(tx i + (1 t)x i+1 ) = tφ(x i ) + (1 t)φ(x i+1 ), t [0, 1]. Lemma 2. The piecewise linear extension of a discretely convex function φ: T M(Y ) is a convex function on the segment [a, b]. Lemma 3. Let there be given five parallel planes in the space R 3 ; then for five straight lines in general position on those planes (each plane contains one line) there is no straight line meeting all of them. Consider 5 points on the plane a 1 = (1, 0), a 2 = (4/7, 1/7), a 3 = (1/3, 1/3), a 4 = (1/7, 4/7), a 5 = (0, 1) and the 5 segments A i = [ a i, a i ]. It is checked directly that A i < (A i 1 + A i+1 )/2, i = 2, 3, 4. 5

By Lemma 3 there are segments Ãi arbitrarily close to the segments A i and such that the 5 segments (Ãi, (i 1)/4) R 3 cannot be intersected by one line. Then for the points T = {x i } i=1,...,5, x i = (i 1)/4, of the segment [0, 1] the function φ(x i ) = Ãi is discretely convex. Consequently, its piecewise linear extension φ: [0, 1] M(R 2 ) is a convex function (Lemma 2), for which there is no affine selection. Extending the function φ from the segment [0, 1] to R by formulas φ(x) = (1 4x)φ(0) for x 0 and φ(x) = (4x 3)φ(1) for x 1, we get a convex map φ: R M(R 2 ) that does not have an affine selection. Remark 2. The proof of Theorem 2 is applicable for maps φ: Ω M(Y ) as well, where Ω X is an arbitrary open convex domain. Indeed, taking a pair of points a, b Ω and setting x i = 5 i a + i 1 b, 4 4 i = 1,..., 5, we define a discretely convex map φ at points {x i }, then we extend it to the segment [a, b] and further to Ω. We see that in case dim Y 2 there is a convex map φ: Ω M(Y ) without affine selections. Similarly Theorem 1 is generalized to maps φ: C M(Y ), where C X is an arbitrary convex cone. 4. On selections of maps generated by continuous functions In this section we study one special class of subadditive maps. Those maps are generated by differences of a continuous function on a Banach space. Linear selections of such maps are applied in problems of stability of linear operators. This issue will be discussed in 5. Consider a uniformly continuous map f : X Y between Banach spaces X and Y. Suppose that the space Y is dual to some normed space (for spaces possessing this property we shall use the short notation dual space ). In the sequel we denote the norm in X by, and the norm in Y by. For arbitrary u X we write Q(u) = Q f (u) for the closure (in the *-weak topology of Y ) of the convex hull of the set {f(x + u) f(x) x X}. Thus, Q f (u) is the *-weak closure of convex hull of all possible changes of the function f corresponding to the change u of the argument. In what follows we omit the index f. The uniform continuity of the map f implies that all the sets Q(u) are bounded and, consequently, compact in the *-weak topology. If f is an affine map, then all Q(x) are one-point sets, and so the map φ(x) = Q(x) is linear. For arbitrary f the map φ is set-valued and subadditive. It appears that this map always has a continuous linear selection. 6

Theorem 3. Suppose f : X Y is a uniformly continuous map of Banach spaces X and Y, and the space Y is dual; then there exists a continuous linear operator A: X Y, for which Ax Q(x), x X. None of the assumptions of this theorem can be relaxed, as it is shown in examples below. To prove the theorem we consider first the case of finite-dimensional X. Then we extend it to an arbitrary Banach space X applying the following result from [15] (we slightly change notation in the statement): Theorem A. Let Y be a topological space, X be a set, and T be a directed family of subsets of X, i.e. for every L 1, L 2 T there exists a set L T that contains both L 1 and L 2. Let also for each L T a nonempty compact set P L of maps from L to Y be given. We assume that for every L 1, L 2 T the relation L 1 L 2, F P L2 implies that F L1 P L1. Then, if L T L = X, then there exists a map F : X Y, such that F L P L for every L T. Proof of Theorem 3. Consider first the case dim X <. For every n 3 we denote by B n and B n the balls in X centered at the origin, of radii n 2 and n 2 2n respectively. Let f n (u) = 1 [f(x + u) f(x)] dx, ν(b n ) B n h n (a, u) = 1 [f(x + a + u) f(x + a)] dx ν(b n ) B n (we use the Riemann-Bochner integral [14] and write ν(m) for the volume of a set M ). Obviously, f n (u) Q(u). For arbitrary v, u X such that u, v < 1 and for every x B n all the three points x + u, x+v, and x+u+v lie in B n. Let ω(t) = sup x y t f(x) f(y) be the modulus of continuity of the map f. We have [f(x+z) f(x)] dx Ω ν(ω)ω( z ). Applying this estimate first for Ω = B n \ B n, z = u, then for Ω = B n \ (u + B n ), z = v, and finally for Ω = B n \ B n, z = u + v, we obtain f n (u) h n (0, u) f n (v) h n (u, v) f n (u + v) h n (0, u + v) 7 ( 1 ν( B n ) ν(b n ) ( 1 ν( B n ) ν(b n ) ( 1 ν( B n ) ν(b n ) ) ω( u ), ) ω( v ), ) ω( u + v ).

Since h n (0, u + v) = h n (0, u) + h n (u, v), from the triangle inequality it follows that ( f n (u) + f n (v) f n (u + v) 1 ν( B ) n ) 2(ω( u ) + ω( v )). ν(b n ) Since ν( B n )/ν(b n ) 1 as n, it follows that for every u and v we have f n (u)+f n (v) f n (u+v) 0 as n. Applying the continuity we derive λf n (u) f n (λu) 0 as n for every λ R. Using now the inductive argument, we conclude that for every basis {e i } of the space X one has ( ) x i f n (e i ) f n x i e i 0 as n. i i Therefore, for any element z Z, where Z = Y, it follows that x i (z, f n (e i )) (z, f n (x)) 0 as n, (1) i where x = i x ie i. Since f n (e i ) Q(e i ) for all n and the set Q(e i ) is *-weak compact, we see that the sequence {f n (e i )} n N has *-weak limit point g i Q(e i ). Whence, for every z Z and for each i = 1,..., dim X there is a subsequence {n k } N, for which (z, f nk (e i )) (z, g i ), k. The set of indices i is finite, hence for every z there exists a common subsequence for all i. Let us now define a linear operator A: X Y by the formula Ax = i x ig i, where x = i x ie i. Applying (1) to the subsequence {n k }, we deduce that (z, Ax) (z, f nk (x)) 0, k. Since for every z Z there is such a subsequence, it follows that the element Ax is a *-weak limit point of the sequence {f n (x)} Q(x). Consequently, Ax Q(x), which completes the proof in case dim X <. Let now X be an arbitrary Banach space. Consider the set T of all its finite-dimensional subspaces. For every L T we write P L for the set of linear operators A L : L Y such that Ax Q(x), x L. We have shown that the set P L is nonempty for every L T. Furthermore, P L is compact in the *-weak topology, because every operator A L P L is defined in a unique way by the finite set {A L e i }, where {e i } is a basis of the space L. Finally we observe that if L 1 L 2, then A L2 L1 P L1 for every A L2 P L2. Therefore, the sets P L satisfy the assumptions of Theorem A, applying which we get an operator A: X Y such that A L P L for all L T. This operator is linear, since it is on every finite-dimensional subspace. It is continuous, which follows from the uniform continuity of the map f. Since f(x + u) f(x) ω( u ), 8

u X, we see that the set Q(u) lies in a ball of radius ω( u ), and hence Au ω( u ). For u = 1 we obtain A ω(1). Remark 3. Comparing Theorems 1 and 3 shows, in particular, that not every subadditive map φ: X M(Y ) can be represented in the form φ(x) = Q f (x) for a suitable function f. Remark 4. The uniform continuity of the map f is essential in Theorem 3. For instance, for the scalar function f(x) = x 3 one has Q(x) = [x 3 /4, + ), therefore any linear selection has to satisfy Ax x 3 /4, which is impossible. Convexity of all the sets Q is also essential. As a simple example one can take the function f : R R 2, f(x) = (cos x, sin x). The corresponding set-valued map φ(x) = {f(a + x) f(a), a R}, is subadditive; each image φ(x) is a circle. If there is a linear selection A, then A 0, because all the sets φ(x) are contained in the disc of radius 2 centered at the origin. Whence, 0 φ(x) for every x, which is impossible, since, for example, 0 / φ(π). We see that in the definition of the sets Q(x) one cannot avoid taking convex hull. The duality condition for the space Y cannot be omitted either, as the following example shows. Example 1. Theorem 3 may fail if the space Y is not dual, or if the *-weak closure is replaced by the weak closure in the definition of the sets Q(u). Note that for convex sets the weak closure coincides with the usual closure. Let us show that there is a Lipschitz function f : R l 1, for which the corresponding set-valued map φ(x) = Q(x) (defined with the usual closure) has no linear selection. Consider a system of univariate functions: {g k ( )} k N defined as follows: { 1 x 1, x [0, 2], g 1 (x) = 0, x / [0, 2], 2x 2 k 1, x [2 k 2, 2 k 1 ], g k (x) = 2 k x, x [2 k 1, 2 k ], k 2. 0, x / [2 k 2, 2 k ], All the functions g k are compactly supported, piecewise linear, and Lipschitz continuous with constant 2. Furthermore, the sum k N g k(x) equals to x for x 0, and to 0 for x 0. For every x this sum contains at most two nonzero terms. Consider the map f : R l 1 defined as follows: f(x) = y = (y 1, y 2,... ) l 1, where y i = g i ( x) + g i (x). For every x the element f(x) has at most two nonzero coordinates, hence the map f is Lipschitz with constant 4. Consequently, if there is a linear selection of the map φ(x) = Q(x), then it is continuous: A 4. For 9

each i the absolute value of ith coordinate of the element f(a+x) f(a) is uniformly bounded for all a R by some constant m i. Whence sup z Q(x) z i m i, and therefore (Ax) i m i for all x. Consequently, (Ax) i = 0, i N, and thus A 0. On the other hand, if we set e = (1, 1,... ) l, we obtain (e, f(x)) = i N y i(x) = x for all x R. Therefore, (e, z) = x for every z Q(x). In particular, (e, Ax) = x, which is impossible for x 0. 5. Lipschitz stability of linear operators Problems on stability of linear maps originated with Ulam and Hyers ([16], [17]). They have been thoroughly studied in the literature (see a detailed review [18]). Let us briefly describe the issue. There are many criteria of linearity of a map f : X Y. For example, the so-called Cauchy equation f(x 1 + x 2 ) = f(x 1 ) + f(x 2 ), x 1, x 2 X, characterizes linearity of continuous maps. Assume that conditions of some criterion are satisfied approximately, with a certain error. Does it imply that there is a linear map close enough to f? The stability property means that every almost linear map can be approximated by a linear one. We involve Theorem 3 on linear selections to solve the Lipschitz stability problem. Let f : X Y be a map of Banach spaces X and Y. For an arbitrary pair of distinct points x 1, x 2 X we denote {x 1, x 2 } f = (f(x 2 ) f(x 1 ))/ x 2 x 1. For a given number ε 0 consider the following property of the map f : {a, b} f {b, c} f ε for every points a, c X and b [a, c]. (2) Thus, for given ε 0 inequality (2) has to be satisfied for every triple of points a, b, c X such that b lies on the line segment [a, c]. In case ε = 0 only affine operators satisfy condition (2). For arbitrary ε > 0 Lipschitz ε/2-perturbations of affine operators, i.e., f = A+h, where A is an affine map and h is a Lipschitz map with the constant ε/2, possess property (2). Is the converse true: every map f possessing property (2) is a Lipschitz perturbation of an affine operator? Definition 1. A pair of Banach spaces (X, Y ) possesses the property of Lipschitz linear stability (LLS), if there exists a function C(ε) such that lim ε 0 C(ε) = 0 and for every continuous map f : X Y satisfying (2) for some ε 0 there is a continuous linear operator A: X Y, for which {x 1, x 2 } f A C(ε), x 1, x 2 X. An addition of a constant to the operator A does not change the value {x 1, x 2 } f A. Therefore, if there is an affine approximating operator A, then there is also a linear one. Whence, LLS means that for every map f satisfying (2) there exists a linear operator A, for which the map 10

f A is Lipschitz continuous with constant C(ε). The LLS property was introduced by Páles in [19], where he posed the following problem: for which spaces X does the pair (X, R) possess the LLS property? This problem is closely related to the problem of Ulam-Hyers stability for p = 1 [20], [21]. Let us recall that for a given p 0 the Ulam- Hyers stability property means that every continuous map f : X Y, satisfying the condition f(x 1 +x 2 ) f(x 1 ) f(x 2 ) δ( x 1 p + x 2 p ) can be approximated by a linear operator A so that f(x) Ax C(δ) x p, where C(δ) 0, δ 0. It is known that all pairs of Banach spaces X, Y possess this property for each p 1. Moreover, for any map f the approximating linear operator is unique. The case p = 1 is exceptional: already for X = Y = R there is no stability (see [18] for more details). The LLS property is a natural analogue of the Ulam-Hyers stability for p = 1. Let us emphasize that in this case the approximating operator is not necessarily unique. On other problems of stability related to LLS see [15], [22] [24]. Theorem 3 makes it possible to obtain a comprehensive solution, even for a more general problem: if X is an arbitrary space, and Y is a space dual to a normed space, then the LLS property holds with C(ε) = 2ε. This is true, in particular, whenever Y is reflexive, for example, for Y = R, which answers the question of Z. Páles. Theorem 4. If X is an arbitrary Banach space, and Y is a space dual to a normed space, then the LLS property holds with C(ε) = 2ε. Proof. Let us first show that condition (2) implies that the value {x 1, x 2 } f is almost the same for all parallel line segments [x 1, x 2 ] (in the sequel we omit the index f ). More precisely, if vectors x 2 x 1 and y 2 y 1 are codirectional, then {x 1, x 2 } {y 1, y 2 } 2ε. Take an arbitrary point z on the extension of the line segment [x 1, x 2 ] through the point x 2 and denote by y 2 the closest point of the segment [y 1, z] to the point y 2. As z we have y 2 y 2 and z y 2 = z x 2 + O(1), therefore {y 2, z} {x 2, z} 0. On the other hand, {x 1, x 2 } {x 2, z} ε and {y 2, z} {y 1, y 2} ε, hence, invoking the triangle inequality, we get {x 1, x 2 } {y 1, y 2} 2ε + {y 2, z} {x 2, z}. Taking limit as z we obtain {x 1, x 2 } {y 1, y 2 } 2ε. Since the map f is continuous, it follows that for all sufficiently small δ the value sup u =δ f(u) f(0) is bounded by some constant C. Consequently, {x 1, x 2 } C/δ + 2ε for all x 1, x 2 X. This means that f is Lipschitz continuous. Furthermore, (f(x + u) f(x)) (f(y + u) f(y)) 2ε u for every x, y X. 11

Whence, the distance between two arbitrary points from the set Q(u) does not exceed 2ε u. By Theorem 3 there is a linear selection A for the map φ(u) = Q(u). Since Au Q(u) and f(x + u) f(x) Q(u), we see that (f(x + u) f(x)) Au 2ε u for every x, u X, which yields {x + u, x} f A 2ε. Appendix. Proofs of auxiliary results Proof of Proposition 1. Since the set φ(x) is convex, it follows that the sum of n sets φ(x) equals to nφ(x). The subadditivity of the map φ yields φ(nv) nφ(v), n N, v X. (3) Consider the following maps φ (x) = t 1 φ(tx) and φ (x) = 2 k φ(2 k x). t (0,+ ) k N {0} Let us show that actually φ = φ. The inequality φ (x) φ (x) is obvious. To prove the opposite inequality we observe that {2 k φ(2 k x)} k N {0} is a system of embedded compact sets (this follows from (3)), hence it has a nonempty intersection, i.e., φ (x). For every positive t there exists a sequence t j = 2 j /p j, where p j N, converging to t. The continuity of the map φ implies that t 1 j φ(t j x) t 1 φ(tx). Applying (3) for n = p j, v = t j x, we obtain φ (x) 2 j φ(2 j x) = 2 j φ(p j t j x) p j 2 j φ(t j x) = t 1 j φ(t j x). Thus, φ (x) t 1 j φ(t j x) for all j N, from which we conclude that φ (x) t 1 φ(tx). This inequality is valid for each t (0, + ), therefore φ (x) φ (x). Thus, φ = φ. In particular, for every x the image φ (x) is a nonempty convex compact set. The homogeneity of the map φ is obvious, let us establish the subadditivity. For an arbitrary y φ (x 1 + x 2 ) and for every k 0 one has y 2 k φ(2 k x 1 + 2 k x 2 ) 2 k (φ(2 k x 1 ) + φ(2 k x 2 )). Consequently, there exists y i (k) 2 k φ(2 k x i ), i = 1, 2, such that y = y 1 (k) + y 2 (k). Since 2 k φ(2 k x i ) φ(x i ), it follows that y i (k) φ(x i ). In particular, all the points y 1 (k), k 0, are in the compact set φ(x 1 ). Hence, the sequence {y 1 (k)} has a limit point ỹ 1. All terms of this sequence, except for finitely many of them, lie in the compact set 2 n φ(2 n x 1 ), whence ỹ 1 2 n φ(2 n x 1 ) for each n, and therefore ỹ 1 φ (x 1 ) = φ (x 1 ). Since ỹ 2 = y ỹ 1 is a limit point of the sequence {y y 1 (k)} k 0, and y y 1 (k) = y 2 (k) 2 k φ(2 k x 2 ), we see that ỹ 2 φ (x 2 ). Thus, for every y φ (x 1 + x 2 ) there exists 12

ỹ 1 φ (x 1 ), ỹ 2 φ (x 2 ) such that y = ỹ 1 + ỹ 2. Consequently, φ (x 1 + x 2 ) φ (x 1 ) + φ (x 2 ). This means that φ is subadditive. The proof of continuity of the map φ is elementary, and we skip it. Proof of Lemma 1. Let us first establish the lemma for a usual scalar function φ taking values in R. The piecewise linear extension of such a function is convex on every sufficiently small neighborhood of an arbitrary point x R 2 \{0}. Indeed, if x does not belong to the rays r i, then φ is linear on a small neighborhood of that point x; if, otherwise, x r i, then the convexity of that function on some small neighborhood follows from the inequality φ(x i ) α i φ(x i 1 ) + β i φ(x i+1 ). Thus, the function φ is locally convex, and hence it is convex globally on R 2. For an arbitrary map φ taking values in M(Y ) and for every u Y we consider the scalar function g u (t) = s φ(t) (u), where s K (u) = sup y K (u, y) is the support function of a convex compact set K M(Y ). For every u the support function is additive and homogeneous in K. Furthermore, by the convex strict separation theorem [13, с. 38] we have K 1 K 2 if and only if s K1 (u) s K2 (u), u Y. That is why for every u the scalar function g u (t) is discretely convex, and hence its piecewise linear extension is convex. For that piecewise linear extension φ we have s φ((1 λ)α+λβ) (u) (1 λ)s φ(α) (u) + λs φ(β) (u) for every λ [0, 1] and u Y, and therefore φ((1 λ)α + λβ) (1 λ)φ(α) + λφ(β). Thus, the map φ is convex. Proof of Lemma 2 coincides with the proof of Lemma 1. Proof of Lemma 3. For the sake of simplicity we assume that distances between those planes are equal. Points of the ith line are defined as a i x i + b i, where x i, b i R 3 and a i R is a scalar parameter, i = 1,..., 5. If five points are collinear, then the following three equalities hold: a i x i + b i = (a i 1 x i 1 + b i 1 + a i+1 x i+1 + b i+1 )/2 for i = 2, 3, 4. Each vector x i is of dimension 2, hence the numbers a 1,..., a 5 satisfy a system of 6 linear equations. Coefficients of those equations are polynomials in x i, b i. Since there exists at least one position of five lines, when they cannot be intersected by one straight line, it follows that there are values x i, b i, for which the linear system in nondegenerate. Consequently, in general position it is nondegenerate as well. In this case the five lines are not intersected by one line. The author is grateful to two anonymous referees for their attentive reading and many valuable comments. <ЁвҐа вга [1] A. D. Ioffe, Nonsmooth analysis: differential calculus of nondifferentiable mappings, Trans. Amer. Math. Soc., 266:1 (1981), 1 56. [2] А. М. Рубинов, Сублинейные операторы и их приложения, УМН, 32:4 (1977), 113 174. 13

[3] Ю. Г. Борисович, Б. Д. Гельман, А. Д. Мышкис, В. В. Обуховский, О новых результатах в теории многозначных отображений. I. Топологические характеристики и разрешимость операторных соотношений, ў Є.: Итоги науки и техники. Математический анализ, в. 25, ВИНИТИ, М., 1987, 123 197. [4] Ю. Г. Борисович, Б. Д. Гельман, А. Д. Мышкис, В. В. Обуховский, Многозначные отображения, ў Є.: Итоги науки и техники. Математический анализ, в. 19, ВИНИТИ, М., 1982, 127 230. [5] E. Michael, Continuous selections. I, Ann. of Math., 63:2 (1956), 361 382. [6] V. V. Gorokhovik, Representations of affine multifunctions by affine selections, Set-Valued Anal., 16:2 3 (2008), 185 198. [7] А. Я. Заславский, О существовании линейного селектора суперлинейного многозначного отображения, Матем. заметки, 29:4 (1981), 557 566. [8] A. Smajdor, Additive selections of superadditive set-valued functions, Aequationes Math., 39:2 3 (1990), 121 128. [9] A. Smajdor, W. Smajdor, Affine selections of convex set-valued functions, Aequationes Mathematicae, 51 (1996), 12 20. [10] D. Popa, Additive selections of (α, β)-subadditive set-valued maps, Glas. Mat. Ser. III, 36:1 (2001), 11 16. [11] Z. Páles, Linear selections for set-valued functions and extension of bilinear forms, Arch. Math., 62:5 (1994), 427 432. [12] K. Nikodem, Z. Páles, Sz. W asowicz, Multifunctions with selections of convex and concave type, Math. Pannon., 11:2 (2000), 249 260. [13] Г. Г. Магарил-Ильяев, В. М. Тихомиров, Выпуклый анализ и его приложения, УРСС, М., 2000. [14] J. Diestel, J. J. Uhl, Vector Measures, Amer. Math. Soc., Providence, RI, 1977. [15] J. Tabor, D. Yost, Applications of inverse limits to extensions of operators and approximation of Lipschitz functions, J. Approx. Theory, 116:2 (2002), 257 267. [16] S. M. Ulam, A collection of mathematical problems, Interscience Publ., New York London, 1960. [17] D. H. Hyers, On the stability of the linear functional equation, Proc. Nat. Acd. Sci. USA, 27 (1941), 222 224. [18] Th. M. Rassias, On the stability of functional equations and a problem of Ulam, Acta Appl. Math., 62:1 (2000), 23 130. [19] J. Chmielinski, Report on the 12th ICFEI, Ann. Acad. Pedagog. Crac. Stud. Math., 7 (2008), 125 159. [20] S.-M. Jung, On the Hyers-Ulam-Rassias stability of approximately additive mappings, J. Math. Anal. Appl., 204:1 (1996), 221 226. [21] P. Gǎvruta, A generalization of the Hyers Ulam Rassias stability of approximately additive mappings, J. Math. Anal. Appl., 184:3 (1994), 431 436. [22] B. E. Johnson, Approximately multiplicative maps between Banach algebras, J. London Math. Soc., Ser. 2, 37:2 (1988), 294 316. 14

[23] Z. Gajda, On stability of additive mappings, Internat. J. Math. Sci., 14:3 (1991), 431 434. [24] P. Šemrl, The stability of approximately additive functions, in: Stability of Mappings of Hyers Ulam Type (eds. Th. M. Rassias, J. Tabor), Hadronic Press, Palm Harbor, FL, 1994, 135 140. Moscow State University Поступило в редакцию department of Mechanics and Mathematics 12 апреля 2010 г. e-mail: v-protassov@yandex.ru Key words set-valued map, linear selection, subadditivity, Lipschitz function, stability of linear operators 15