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Home , Algebraic interior, Convex set, Interior (topology)

Some basic topological and algebraic properties of convex sets

In what follows, C is a in a X.

Topological . Let us start with the following easy observation. Observation 0.1. Let C be a convex set in a normed space X. If C contains an 0 open B (z0, r), then 0¡ ¢ B (1 − t)z0 + ty, (1 − t)r ⊂ C whenever y ∈ C, t ∈ [0, 1). 0 Proof. Fix y ∈ C, t ∈ [0, 1). Since C is convex, it contains the set (1 − t)B (z0, r) + ty = B0((1 − t)x + ty, (1 − t)r). ¤ Corollary 0.2. Let C be a convex set in a normed space X. If x ∈ int(C) and y ∈ C, then [x, y) ⊂ int(C). In particular, int(C) is a convex set. Theorem 0.3. Let C be a convex set in a normed space X. (a) If either int(C) 6= ∅ or C is finite-dimensional, then int(C) = int(C). (b) If int(C) 6= ∅, then C = int(C). Proof. First part of (a). Suppose that int(C) 6= ∅. Thus C contains an open ball 0 B (z0, r). One inclusion is obvious. Let x ∈ int(C). There exists % > 0 such 0 0 that B (x, %) ⊂ C. If x = z0, we are done. If x 6= z0, fix y ∈ B (x, %) such that x ∈ (z0, y), that is, there exists t ∈ (0, 1) with x = (1 − t)z0 + ty. Since y ∈ C, there 0 0 0 0 exists y ∈ C be such that ky −y k < r. For x = (1−t)z0 +ty , Observation 0.1 says that B0(x0, (1−t)r) ⊂ C. Then x ∈ int(C), since kx−x0k = (1−t)ky−y0k < (1−t)r. Second part of (a). Let C be finite-dimensional. If aff(C) = X, then int(C) 6= ∅ by the “relative interior theorem”; hence we can apply the first part of (a). If aff(C) 6= X, the affine hull of C is closed (since it is a translate of a finite-dimensional linear set) and has no interior points. Since int(C) ⊂ int(C) ⊂ int(aff(C)) = int(aff(C)) = ∅, we have int(C) = int(C) = ∅.

(b) Obviously, C ⊃ int(C). If x ∈ C, fix z0 ∈ int(C). Arbitrarily near to x we can find some y ∈ C; but since [z0, y) ⊂ int(C) (Corollary 0.2), arbitrarily near to x we can find points of int(C). Thus x ∈ int(C). ¤

Algebraic interior. Let C be a convex set in a vector space X, x ∈ C. We say that x is an algebraically interior point of C if for every v ∈ X there exists t > 0 such that x + tv ∈ C. In this case, we shall write x ∈ a-int(C). (For a nonconvex set E ⊂ X, an algebraically interior point is a point x ∈ E such that for each v ∈ X there exists t > 0 such that [x, x + tv] ⊂ E.) In other words, x ∈ a-int(C) if and only if x ∈ intL(C ∩ L) for each L ⊂ X passing through x. The following proposition is an analogue of Corollary 0.2. Proposition 0.4. Let C be a convex set in a vector space X. If x ∈ a-int(C) and y ∈ C, then [x, y) ∈ a-int(C). In particular, the set a-int(C) is convex. 1 2

Proof. Let 0 < λ < 1, z = (1 − λ)x + λy, v ∈ X. Take t > 0 such that x + tv ∈ C. Then z +(1−λ)tv ∈ C since z +(1−λ)tv = (1−λ)(x+tv)+λy. Now, the assertion easily follows. ¤ If X is a normed space, then obviously int(C) ⊂ a-int(C). The following theo- rem states some sufficient conditions for equality of the two sets. Before stating it we recall the following fundamental theorem of R.L. Baire, which we state in the following form. Theorem 0.5 (Baire Category Theorem). Let T be either a complete or a Hausdorff locally compact topologicalS space. If Fn (n ∈ N) are closed sets without interior points, then also the n∈N Fn has no interior points. Exercise 0.6. Use the Baire Category Theorem to prove the following two facts. 1. Every compact countable set in a Hausdorff has at least one isolated point. 2. The algebraic dimension of a is either finite or uncountable. Theorem 0.7. Let C be a convex set in a normed linear space X. Then int(C) = a-int(C) provided at least one of the following conditions is satisfied. (a) int(C) 6= ∅; (b) C is finite-dimensional; (c) X is Banach and C is an Fσ-set.

Proof. (a) Let x ∈ a-int(C). Fix z0 ∈ int(C). If x 6= z0, there exists y ∈ C such that x ∈ (z0, y). Then x ∈ int(C) by Corollary 0.2. (b) Let x ∈ a-int(C). This implies that aff(C) = X and hence X is finite- dimensional. By the “relative interior theorem”, int(C) = ri(C) is nonempty. Apply the part (a). S (c) Let x ∈ a-int(C). We can suppose that x = 0. Write C = k Fk where each Fk is a . Then [ [ [ X = tC = nC = nFk . t>0 n n,k By the Baire Category Theorem, there exist n, k such that nFk has a nonempty interior. Thus int(Fk) 6= ∅ and hence int(C) 6= ∅. Apply (a). ¤

Directions from a point. Let C be a convex set in a vector space X, x0 ∈ C. Let us consider two sets of directions: the set c(x0,C) of directions of the half-lines from x0 to the points of C, and the set i(x0,C) of directions of the lines L such that L ∩ C is a nondegenerate segment containing x0 in its relative interior. More precisely,

c(x0,C) = {v ∈ V : x0 + tv ∈ C for some t > 0} ,

i(x0,C) = {v ∈ V : x0 ± tv ∈ C for some t > 0} . Let us state main properties of these two sets. 3

Proposition 0.8. Let X, C, x0 be as above. (a) x0 ∈ x0 + i(x0,C) ⊂ x0 + c(x0,C) ⊂ aff(C). (b) c(x0,C) = (C − x0). (c) c(x0,C) is a ,£ and i(¤x0,C) is a linear set. (d) i(x0,C) = c(x0,C) ∩ −c(x0,C) . (e) x0 + i(x0,C) is£ the largest affine set¤ A such that x0 ∈ intA(C ∩ A). (f) aff(C) = x0 + c(x0,C) − c(x0,C) .

Proof. (b) follows directly form the definitions; hence c(x0,C) is a convex cone. Obviously, we have 0 ∈ i(x0,C) ⊂ c(x0,C). Since x0 + c(x0,C) is the set of all closed half-lines from x0 to the points of C (or {x0} if there are no such half-lines), it is contained in aff(C); hence (a) holds. The part (d) follows easily from definitions. Since K ∩ (−K) is a linear set whenever K is a convex cone, we have proved (c), too. Let us show (e). Clearly, x0 + i(x0,C) is an affine set. Let A be an affine set such ←→ that x0 ∈ int(C ∩ A). For every point x ∈ A \ x0 the set C ∩ x0x is a nondegenerate segment containing x0 in its relative interior. Since x0 + i(x0,C) is the union of all possible lines L with this property, we have A ⊂ x0 + i(x0,C). It remains to prove (f). By translation, we can suppose that x0 = 0. Thus, by (b), we have to show that, if 0 ∈ C, then span(C) = cone(C) − cone(C). The inclusion P“⊃” is obvious. To show the other one, consider a (finite) x = i αici with αi 6= 0, ci ∈ C. Then X X x = αici − (−αi)ci ∈ cone(C) − cone(C) .

αi>0 αi<0 ¤

Algebraic relative interior. Let us recall that a relative interior of a convex set C in a normed space is the set ri(C) = intaff(C)(C). Recall also the fact, proved in the chapter on finite-dimensional spaces, that every finite-dimensional convex set has a nonempty relative interior. The algebraic relative interior of C is defined analogously in the following definition. Definition 0.9. Let C be a convex set in a vector space X. The algebraic relative interior of C is the set a-ri(C) = a-intaff(C)(C). Observation 0.10. Let C be a convex set in a vector space X, x ∈ C. Then x ∈ a-ri(C) ⇐⇒ x + i(x, C) = aff(C) ⇐⇒ x + c(x, C) = aff(C) . Proof. Easy exercise. ¤ Observe that Theorem 0.7 immediately gives the following Corollary 0.11. Let C be a convex set in a normed space. Then a-ri(C) = ri(C) whenever at least one of the following conditions is satisfied: (a) ri(C) 6= ∅; (b) dim(C) < ∞; (c) X is a Banach space, aff(C) is closed, and C is an Fσ-set. 4

Inside points. Let C be a convex set in a normed space X. For x ∈ C, let Lx be the family of all lines L ⊂ X, for which C ∩L is a nondegenerate segment containing x in its relative interior. It is easy to see that [ x ∈ a-ri(C) ⇐⇒ (C ∩ L) = C.

L∈Lx Thus the following notion of an “inside point” weakens the notion of an algebraically interior point.

Definition 0.12. Let X, C, x, Lx be asS above. We say that x is an inside point of C (or x is almost surrounded by C) if (C ∩ L) is dense in C. L∈Lx S Since L = x+i(x, C) whenever L 6= ∅, we have the following reformulation L∈Lx x of the above definition. Observation 0.13. A point x is an inside point of C if and only if [x + i(x, C)] ∩ C is dense in C. Proposition 0.14. Let C be a convex set in a normed linear space X. (a) If x is an inside point of C and y ∈ C, then each point of [x, y) is an inside point of C. (b) The set of all inside points of C isS convex. (c) If x is an inside point of C, then L is dense in aff(C). L∈Lx (d) If x is an inside point of C, then every continuous affine function a: C → R that attains its maximum over C at x is constant on C. (e) If C is finite-dimensional, then each inside point of C belongs to ri(C). Proof. (a) Fix z ∈ (0, 1), u ∈ C and ε > 0. Since x is an inside point of C, there exist two distinct points c, d ∈ C such that x ∈ (c, d) and dist(u, [c, d]) < ε. ←→ Consider the line L = cd . If y ∈ L, then obviously z ∈ ri(C ∩ L). If y∈ / L, then ∆ := conv{c, d, y} is a nondegenerate triangle containing z in its relative interior. Thus there exist c0, d0 ∈ ∆ such that z ∈ (c0, d0) and dist(u, [c0, d0]) < ε (indeed, it sufficies take c0 ∈ [c, d] so that kc0 − uk < ε). We have proved that z is an inside point of C. (b) is an immediateS corollary of (a). (c) The set D = (L∩C) is dense in C. Thus aff(D) is dense in aff(C). Observe SL∈Lx that the set A = L = x + i(x, C) is affine (Proposition 0.8(c)), it contains D, L∈Lx and every affine set containing D must contain A. It follows that A = aff(D). So A is dense in aff(C). (d) The functionS a must be constant on each segment C ∩ L (L ∈ Lx); so it is constant on (C ∩ L). By density, a is constant on C. L∈Lx S (e) If C is finite-dimensional, the set affine set L is closed (since it is contained L∈LSx in the finite-dimensional affine set aff(C)). Hence (L∩C) is closed in C. Thus S L∈Lx (L ∩ C) = C which means that x ∈ a-ri(C) = ri(C). (The equality follows L∈Lx from Corollary 0.11). ¤ 5

Now, we are going to prove a theorem about existence of inside points. Before proving it, we shall need an important lemma about infinite convex combinations. We say that a point x ∈ X is an infinite of elements of a set A ⊂ X if X+∞ x = λnyn n=1 P+∞ with yn ∈ A, λn ≥ 0, 1 λj = 1. Lemma 0.15. Let C be a convex set in a normed space X. Suppose that C is either closed or open. Then every infinite convex combination of elements of C belongs to C.

Proof. We can (and do) suppose that λn > 0 for each n. Let C be closed. Denoting PN σN = 1 λj (N ∈ N), we have XN XN XN λn λn x = lim λncn = lim σN cn = lim cn ∈ C = C. N N σ N σ n=1 n=1 N n=1 N Now, let C be open. By translation, we can suppose that 0 ∈ C. Suppose that x 6= 0 (otherwise we are done). There exists δ > 0 such that ec1 := c1 + δx ∈ C. Then, by the first part of the proof, P+∞ xe := (1 + λ1δ)x = λ1ec1 + n=2 λncn ∈ C. ¡ ¢ Since 0 ∈ int(C), xe ∈ C and x ∈ 0, xe . Applying Corollary 0.2 and Theorem 0.3(a), we get x ∈ int(C) = int(C) = C. ¤ The following proposition is both an immediate corollary and a generalization of Lemma 0.15. Proposition 0.16. Let C be a convex set in a normed space X. If C is the inter- section of a family of open convex sets, then every infinite convex combination of elements of C belongs to C. Remark 0.17. The sets that are intersections of families of open convex sets are called evenly convex sets. An easy application of the Hahn-Banach theorem gives that a convex set is evenly convex if and only if it is the intersection of a family of open halfspaces. There exist convex sets C that are not evenly convex but still they are closed under making infinite convex combinations of their elements: consider, in R2, the union of an open halfplane and one of its points. The following theorem is essentially due to V. Klee. Theorem 0.18 (Klee, 1955). Let X be a Banach space. If C ⊂ X is a separable non-singleton set which is the intersection of a family of open convex sets (e.g., a separable closed convex set), then C has an inside point.

Proof. There exist countably many pairwise distinct points cn ∈ C (n ∈ N) such that P+∞ the set {cn}n∈N is dense in C. Fix a sequence {λn} ⊂ (0, 1) such that 1 λn = 1 P+∞ P+∞ and 1 λnkcnk < +∞. Then the series 1 λncn converges in X (since it 6

P+∞ converges absolutely and X is complete). Let us prove that the point x = 1 λncn is an inside point of C. P For each n ∈ N, x = λ c +(1−λ )x where x = λk c ∈ C (Lemma 0.15). n n n n n k6=n 1−λn k If cn = xn for some n, then cn = x, and hence such n is unique; put n0 = n + 1 in this case. Otherwise define n0 = 1. The segments [cn, xn](n ≥ n0) are nondegener- ate and contained in C, and their union is dense in C. Since x ∈ (cn, xn) for each n ≥ n0, the proof is complete. ¤ Corollary 0.19. Let X be a normed space, and C ⊂ X a separable complete convex set. Then C has an inside point. Proof. Consider C as a closed convex set in the completion X of X, and apply the previous theorem. ¤

Comparison of several properties. We can define another two properties of a point x of a convex set C in a normed linear space X. Let L be as above. We shall say that: x S • x is a quasi-inside point (“quasi inside point”) of C if L is dense in aff(C); L∈Lx • x is a non-support point of C if every x∗ ∈ X∗ that attains its supremum over C at x is constant on C. Theorem 0.20. Let X, C, x be as above. Let us consider the following properties: (a) x ∈ ri(C); (b) x ∈ a-ri(C); (c) x is an inside point of C; (d) x is a q-inside point of C; (e) x is a non-support point of C. Then the following implications hold (a) ⇒ (b) ⇒ (c) ⇒ (d) ⇒ (e) , and none of these implications can be reversed. However, for finite-dimensional C, the properties (a)-(e) are all equivalent. Proof. The implications (a) ⇒ (b) ⇒ (c) are easy, while (c) ⇒ (d) is contained in Proposi- tion 0.14(c). ∗ ∗ ∗ ∗ ∗ (d)S⇒ (e). Let x ∈ X be such that x (x) = sup x (C). Then necessarily x is constant on L, and hence also on aff(C) by continuity. L∈Lx Let C be finite-dimensional and (e) holds. Then ri(C) 6= ∅ (by the “relative interior theo- rem”). We can suppose that 0 ∈ C (thus aff(C) = span(C)). If x∈ / ri(C), the Hahn-Banach separation theorem implies existence of a nonzero functional ϕ ∈ [span(C)]∗ that attains its supremum over C at x. Obviously, ϕ is not constant on C By the Hahn-Banach extension theorem, ϕ can be extended to an alement of X∗. But this contradicts (e). (b) 6⇒ (a). Let X be an infinite dimensional normed space and f : X → R a discontinuous linear functional. Consider C = {y ∈ X : f(y) > 0} and any x ∈ C. (c) 6⇒ (b). Let f : X → R be a discontinuousS linear functional. Consider the set C = {y ∈ X : f(y) ≥ 0} and x = 0. Then L = Ker(f) and aff(C) = X. L∈Lx (d) 6⇒ (c). Let ei (i ∈ N) be the canonical unit elements of c0 (i.e., ei = (0,..., 0, 1, 0, 0,...) where 1 is on the i-th position). Let X = span{ei}i∈N (the space of all finitely supported 7 sequences), equipped with the sup-. Define a (dsicontinuous) linear functional f : X → R by f(ei) = i (i ∈ N) and consider C = {y ∈ X : 0 ≤ f(y) ≤ 1, |y(i)| ≤ 2−i ∀i ∈ N} , x = 0 . S It is easy to see that 0 ∈ C, aff(C) = span(C) = X and L = Ker(f). Thus (d) holds. L∈L0 To showS that (c) does not hold, it suffices to show that f is continuous on C; indeed, in this case, (L ∩ C) = Ker(f) ∩ C is closed in C and hence not dense in C. Let us proceed L∈L0 by contradiction. If f|C is not continuous, there exist % > 0 and a sequence {cn} ⊂ C converging to some c ∈ C, such that |f(cn) − f(c)| > % for each n. By definition of C, the sequences {cn(i)}n∈N (i ∈ N) and {f(cn)}n∈N are bounded (in R) and hence contain convergent subsequences. By a standard diagonal argument and passing to a subsequence, we can suppose that

lim cn(i) = z(i) ∈ R (i ∈ N), lim f(cn) = λ ∈ R . n n −i P Since |zi| ≤ 2 (i ∈ N), the series i∈N z(i)ei converges absolutely, and hence its sum is an element z of c0. We have P P kcn − zk = k i∈N(cn(i) − z(i))eik ≤ i∈N |cn(i) − z(i)| . By the Lebesgue Dominated Convergence Theorem (applied to the counting measure on N), the last expression tends to 0 as n → ∞. It follows that z = c ∈ X; in particular, z(i) = c(i) for each i.. For each n ∈ N, denote N(n) = max supp(cn − c). Then we have ¯ ¯ ¯ ¯ PN(n) P f(cn − c) = | i=1 (cn(i) − c(i))f(en)| ≤ i∈N i|cn(i) − c(i)| . −i P∞ −i Since i|cn(i)| ≤ i2 and icn(i) → ic(i)(i ∈ N), and 1 i2 < +∞P, we can apply again the Lebesgue Dominated Convergence Theorem to conclude that limn i∈N i|cn(i) − c(i)| = 0. Hence we get f(cn) → f(c), a contradiction. (e) 6⇒ (d). Let f : X → R be a discontinuous linear functional. Let C = {y ∈ X : f(y) > 0} ∪ {0} , x = 0. S Then L = ∅, hence (d) is false. If some x∗ ∈ X∗ satisfies x∗| ≤ 0, then it must be L∈Lx C constant on the level sets of the type f −1(λ) with λ > 0. It follows that Ker(x∗) contains the Ker(f); hence x∗ = 0. This proves (e). ¤