When are Sums Closed?

Transcription

1 Division of the Humanities and Social Sciences Ec 181 KC Border Convex Analysis and Economic Theory Fall 2018 Winter 2019 Topic 20: When are Sums Closed? 20.1 Is a sum of closed sets closed? Example shows that the sum of two closed sets need not be closed. To state sufficient conditions for the sum to be closed we must make a fairly long digression Asymptotic cones A cone is a nonempty subset of R m closed under multiplication by nonnegative scalars. That is, C is a cone if whenever x C and λ R +, then λx C. A cone is nontrivial if it contains a point other than zero Definition Let E R m. The asymptotic cone of E, denoted AE is the set of all possible limits z of sequences of the form (λ n x n ) n, where each x n E, each λ n > 0, and λ n 0. Let us call such a sequence a defining sequence for z. This definition is equivalent to that in Debreu [1], and generalizes the notion of the recession cone of a convex set. This form of the definition was chosen because it makes most properties of asymptotic cones trivial consequences of the definition. The recession cone 0 + F of a closed convex set F is the set of all directions in which F is unbounded, that is, 0 + F = {z R m : ( x F ) ( α 0 ) [ x + αz F ]}. (See Rockafellar [3, Theorem 8.2].) Lemma (a) AE is indeed a cone. (b) If E F, then AE AF. (c) A(E + x) = AE for any x R m. (cc) 0 + E AE. (d) AE 1 A(E 1 + E 2 ). (e) A i I E i i I AE i. (f) AE is closed. KC Border: for Ec 181, src: AsymptoticCones v ::14.10

2 KC Border When are Sums Closed? 20 2 (g) If E is convex, then AE is convex. (h) If E is closed and convex, then AE = 0 + E. (The asymptotic cone really is a generalization of the recession cone.) (i) If C is a cone, then AC = C. (j) A i I E i i I AE i. The reverse inclusion need not hold. (k) If E + F is convex, then AE + AF A(E + F ). (l) A set E R m is bounded if and only if AE = {0}. Proof : Here are proofs of selected parts. treated as an exercise. The others are easy, and should be (cc) 0 + E AE. Let z 0 + E. Then for any n > 0 and any x E, we have x + nz E. But (x + nz) z, so z AE. 1 n (d) AE 1 A(E 1 + E 2 ). For x 2 E 2, by definition E 1 + x 2 E 1 + E 2, so by (b), A(E 1 + x 2 ) A(E 1 + E 2 ), so by (c), AE 1 A(E 1 + E 2 ). (f) AE is closed. Let x n be a sequence in AE with x n x. For each n there is a sequence λ n,m x n,m with lim m λ n,m x n,m = x n, λ n,m 0 as m, x n,m E, and each λ n,m > 0. Then for each k there is N k such that for all n N k, x n x < 1/k, and M k such that for all m M k, λ Nk,mx Nk,m x Nk < 1/k, and L k such that for all m L k, λ Nk,m < 1/k. Set P k = max{m k, L k }, y k = x Nk,P k, and λ k = λ Nk,P k. Then each λ k > 0, λ k 0 and λ k y k x < 2/k, so x AE. (g) If E is convex, then AE is convex. Let x, y AE and α [0, 1]. Since AE is a cone, αx AE and (1 α)y AE. Thus there are defining sequences λ n x n αx and γ n y n (1 α)y. Since E is convex, z n = λn γ n+λ n x n + γn γ n+λ n y n E for each n. Set δ n = γ n + λ n > 0. Then δ n 0 and δ n z n = λ n x n + γ n y n αx + (1 α)y. Thus αx + (1 α)y AE. (h) If E is closed and convex, then AE = 0 + E. In light of (cc), it suffices to prove that AE 0 + E, so let z AE, x E, and α 0. We wish to show that x + αz E. By definition of AE there is a sequence λ n z n z with z n E, λ n > 0, and λ n 0. Then for n large enough 0 αλ n < 1, so (1 αλ n )x + αλ n z n E as E is convex. But (1 αλ n )x + αλ n z n x + αz. Since E is closed, x + αz E. v ::14.10 src: AsymptoticCones KC Border: for Ec 181,

3 KC Border When are Sums Closed? 20 3 (i) If C is a cone, then AC = C. It is easy to show that C AC, as 1 nx x is a defining sequence. Since n AC is closed by (f), we have C AC. On the other hand if λ n 0 and x n C, then λ n x n C, as C is a cone, so AC C. (j) A i I E i i I AE i. The reverse inclusion need not hold. By (b), A i I E i AE j for each j, so A i I E i i I AE i. For a failure of the reverse inclusion, consider the even nonnegative integers E 1 = {0, 2, 4,...} and the odd nonnegative integers E 2 = {1, 3, 5,...}. Then E 1 E 2 =, so A(E 1 E 2 ) =, but AE 1 = AE 2 = AE 1 AE 2 = R +. But what if each E i is convex? (k) If E + F is convex, then AE + AF A(E + F ). Let z belong to AE + AF. Then there exist defining sequences (λ n x n ) E and (α n y n ) F with λ n x n + α n y n z. Let x E and y F. (If either E or F is empty, the result is trivial.) Then ( λ n (x n + y ) ) E + F and ( αn (x + y n ) ) E + F, so (λ n + α n ) is a defining sequence for z in E + F. ( λn (x n + y ) + α ) n (x + y n ) z, λ n + α n λ n + α n (l) A set E R m is bounded if and only if AE = {0}. If E is bounded, clearly AE = {0}. If E is not bounded, let {x n } be an unbounded sequence in E. Then λ n = x n 1 0 and (λ n x n ) is a sequence on the unit sphere, which is compact. Thus there is a subsequence converging to some x in the unit sphere. Such an x is a nonzero member of AE Example The asymptotic cone of a non-convex set need not be convex. Let E = {(x, y) R 2 : y = 1, x > 0}. This hyperbola is not convex and its x asymptotic cone is the union of the nonnegative x- and y-axes. But the asymptotic cone of a non-convex set may be convex. Just think of the integers in R Example It need not be the case that A(E + F ) AE + AF, even if E and F are closed and convex. For instance, let E be the set of points lying above a standard parabola: E = {(x, y) : y x 2 }. The asymptotic cone of E, which is the same as its recession cone, is just the positive y-axis: AE = {(0, y) : y 0}. KC Border: for Ec 181, src: AsymptoticCones v ::14.10

4 KC Border When are Sums Closed? 20 4 So AE + A( E) is just the y-axis. A ( E + ( E) ) = R 2. Thus Now observe that E + ( E) = R 2, so AE + A( E) A ( E + ( E) ) When a sum of closed sets is closed We now turn to the question of when a sum of closed sets is closed. The following definition may be found in Debreu [1, 1.9. m., p. 22] Definition Let C 1,..., C n be cones in R m. We say that they are positively semi-independent if whenever x i C i for each i = 1,..., n, x x n = 0 = x 1 = = x n = 0. Clearly, any subcollection of a collection of semi-independent cones is also semi-independent. Note that in Example , A( E) = A(E), so these nontrivial asymptotic cones are not positively semi-independent Theorem (Closure of the sum of sets) Let E, F R m be closed and nonempty. Suppose that AE and AF are positively semi-independent. (That is, x AE, y AF and x + y = 0 together imply that x = y = 0.) Then E + F is closed, and A(E + F ) AE + AF. The proof relies on the following simple lemma, which is closely related to Lemma 1 in Gale and Rockwell [2] Lemma Under the hypotheses of Theorem , if (λ n ) is a bounded sequence of real numbers with each λ n > 0, (x n ) is a sequence in E, and (y n ) is a sequence in F, and if λ n (x n +y n ) converges to some point, then there is a common subsequence along which both (λ k x k ) and (λ k y k ) converge. Proof : It suffices to prove that both (λ n x n ) and (λ n y n ) are bounded sequences. Suppose by way of contradiction that λ n (x n + y n ) converges to some point, but say (λ n x n ) is unbounded. Since (λ n ) is bounded, it must be the case that both λ n x n and x n, so for large enough n we have λx n > 0. Thus for large n we may divide by λ n x n and define ˆx n = λ n λ n x n x n, ŷ n = λ n λ n x n y n, ẑ n = λ n λ n x n (x n + y n ), and observe that ẑ n = ˆx n + ŷ n. v ::14.10 src: AsymptoticCones KC Border: for Ec 181,

5 KC Border When are Sums Closed? 20 5 But ( λ n (x n + y n ) ) is convergent, and hence bounded, so ẑ n 0. In addition the sequence (ˆx n ) lies on the unit sphere, so it has a convergent subsequence, say ˆx k ˆx, where ˆx = 1. Then ŷ k = ẑ k ˆx k ˆx. But ŷ k = (λ k / λ k x k )y k, and λ k / λ k x k 0, so (λ k / λ k x k )y k is a defining sequence that puts ˆx AF. But a similar argument shows that ˆx AE. Since AE and AF are positively semi-independent, it follows that ˆx = 0, contradicting ˆx = 1. Thus (λ n x n ), is a bounded sequence, and by a similar argument so is (λ n y n ), so they have common subsequence on which they both converge. Proof of Theorem : First, E + F is closed: Let x n + y n z with {x n } E, {y n } F. By Lemma (with λ n = 1 for all n) there is a common subsequence with x k x and y k y. Since E and F are closed, x E and y F. Therefore z = x + y E + F, so E + F is closed. To see that A(E + F ) AE + AF, let z A(E + F ). That is, z is the limit of a defining sequence ( λ n (x n + y n ) ), where x n E and y n F. Since λ n 0, it is a bounded sequence. Thus by Lemma there is a common convergent subsequence, and by definition lim k λ k x k AE and lim k λ k y k AF, so z AE + AF Corollary Let E i R m, i = 1,..., n, be closed and nonempty. If AE i, i = 1,..., n, are positively semi-independent, then n i=1 E i is closed, and A n i=1 E i n i=1 AE i. Proof : This follows from Theorem by induction on n Corollary Let E, F R m be closed and let F be compact. Then E + F is closed. Proof : A compact set is bounded, so by Lemma (l) its asymptotic cone is {0}. Apply Theorem When is an intersection of closed sets bounded? Proposition Let E i R m, i = 1,..., n, be nonempty. If n i=1 AE i = {0}, then n i=1 E i is bounded. Proof : By Lemma (l), n i=1 E i is bounded if and only if A ( n i=1 E i ) = {0}. But by Lemma (j), A ( n i=1 E i ) n i=1 AE i, and the proposition follows. KC Border: for Ec 181, src: AsymptoticCones v ::14.10

6 KC Border When are Sums Closed? 20 6 References [1] G. Debreu Theory of value: An axiomatic analysis of economic equilibrium. Number 17 in Cowles Foundation Monographs. New Haven: Yale University Press. [2] D. Gale and R. Rockwell The Malinvaud eigenvalue lemma: Correction and amplification. Econometrica 44(6): [3] R. T. Rockafellar Convex analysis. Number 28 in Princeton Mathematical Series. Princeton: Princeton University Press. v ::14.10 src: AsymptoticCones KC Border: for Ec 181,