Darboux functions. Beniamin Bogoşel

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1 Darboux functions Beniamin Bogoşel Abstract The article presents the construction of some real functions with surprising properties, including a discontinuous solution for the Cauchy functional equation having Darboux property. In the end, the notions are generalized in more dimensions. Key words: Darboux property, Cauchy functional equation 2000 MSC: 26A30, 39B22, 26B35, 03E75 In this article we construct some unusual and unintuitive functions which have interesting properties. We will concentrate our attention on the Darboux property. In the first part we prove the existence of functions which map any open real interval onto a certain subset of R. Next, we give a proof for Sierpinski s theorem, which states that every function f : R R can be written as the sum of two functions with the Darboux property, and a theorem related to this one. In the third section we give a very simple example of a function which is a discontinuous solution for the Cauchy functional equation and has the Darboux property. Finally, in the last section, we generalize the notion of Darboux property to the multidimensional case and give a Sierpinski-type theorem in this context. Definition 1. If f : I R is a function, and I R is an interval, f has the Darboux property if for any a, b I, a < b and for any λ (f(a), f(b)) (f(b), f(a)) there is c (a, b) such that f(c) = λ. In applications, the one of the following equivalent definitions is easier to use: f has the Darboux property if and only if f(j) is an interval for any interval J I. f has the Darboux property if and only if f([c, d]) is an interval for any c, d I, c < d. Definition 2. If f : I R is a function, and I R is an interval, f has the Intermediate Value property if f(i) is an interval. Definition 3. If f : I R is a function, and I R is an interval, f has the Weak-Darboux property if f(j) is an interval, for any interval J I. The notions defined above are different from another, the first being the strongest. Darboux property implies Weak-Darboux and Invermediate Value Preprint submitted to Elsevier October 6, 2009

2 properties. We begin by defining an equivalence relation which will be uset throughout the text. Define x y x y Q. This is obviously an equivalence relation and for any x R we will denote [x] = y R : y x} the equivalence class which contains x. It is obvious that R = x R[x] and y / [x] [x] [y] =. In the following, we will denote A = [x] : x R} the set of the equivalence classes, and we will find its cardinal number. The application µ : R A, µ(x) = [x], x R is onto, hence card A card R = ℵ. Lemma The dimension of R as vector space over Q is ℵ, where ℵ = card R. Proof: In [1], page 245 we find the next lemma: If V is a vector space over the field K, and dimv = b is infinite, then card V = b card K. The dimension of R over Q is infinite, otherwise R would be countable, which is not true. Therefore, we can use the lemma above for V = R, K = Q, and b being the cardinal of a basis of R over Q. We have ℵ = ℵ 0 b = b, where ℵ 0 = card N. Last equality holds because b b ℵ 0 b 2 = b, and b is an infinite cardinal number (b 2 = b). Hence, b = ℵ. Theorem 1. card A = card R. Proof: We choose a basis B of R over Q which contains 1 (this is possible because any linearly independent set can be extended to a basis and 1} is such set). If two different numbers, say x and y from B would be in the same equivalence class, then we have x y = q 1, q Q, so 1, x şi y are linear dependent. Contradiction. Therefore, any two different elements of B are in different equivalence classes, and we can define a one to one mapping φ : B A, φ(b) = [b], b B. This yields card B card A. From the proceeding lemma we know that card B = ℵ and also card A ℵ from an anterior remark. Hence, card A = ℵ. Some strange functions with Darboux property Theorem 2.1. There exist non-constant functions f : R R which map any open interval onto a closed one. Proof: We will prove more than this: there exist such functions which map any open interval into the same closed interval. Take a, b R, a < b. Then card [a, b] = ℵ = card A, so we can find a bijective function, g : A [a, b]. We define the function f : R [a, b], f(x) = g([x]). This function is well defined for any x R. We will prove now that the image or any interval (s, t), s, t R, s < t through f is exactly [a, b]. We notice that [x] = x + Q = x + q : q Q}. Because Q is dense R, any of its translations is also dense in R, meaning that any equivalence class from A has at least one element in common with any open interval, (s, t) being such an interval. Therefore f((s, t)) = g(a) = [a, b]. Remark. The function constructed in the prof of Theorem 2 has the Darboux property and it is not continuous at any point in R. 2

3 Proof: We can see that f(i) = [a, b] for any interval I R. Hence f has the Darboux property. If x 0 R and ε > 0, we know that f((x 0 ε, x 0 + ε)) = [a, b], so f cannot be continuous at x 0. Another surprising result is given in the following Theorem 2.2. There exist functions f : R R which have the Darboux property and take any real value in any neighborhood of any point in R. Proof: We proceed the same way we did in our previous proof, defining the bijection h : A R, and the function f : R R, f(x) = h([x]). We can see that f((s, t)) = R, for any interval (s, t), using a similar argument to the one above. f(i) = R for any interval I, hence f has the Darboux property. A slight generalization is the following Theorem 2.3. Given T R, there exists a function f : R R such that f maps any open interval onto T. Proof: Because card T card A, there exists a surjection φ : A T. Considering the function f : R R, f(x) = φ([x]) we have the requested function. An even more surprising result is the following, due to Waclaw Sierpinski (see [2]): Theorem 3. (Sierpinski) For any function f : R R there exist functions f 1, f 2 : R R having the Darboux property and being discontinuous at any point in R, such that f = f 1 + f 2. Proof: We take f : R R a function. We consider a bijection g : R A and we denote A 1 = g((, 0)), A 2 = g([0, )). Thus card A 1 = card A 2 = ℵ, and we can define the bijections m : R A 1 and n : R A 2. A 1 m R g R =(, 0) [0, ) g n A 2 R We define the functions: r, t m(r) f 1 (t) = f(t) r, t n(r) f 2 (t) = f(t) r r, t m(r), t n(r) It is clear from their definition that f 1 şi f 2 că f = f 1 + f 2. Let s prove that f 1 and f 2 have the Darboux property. We consider I R an interval. Because any equivalence class in A is dense in R, we find that I m(r), I n(r), r R. Therefore, looking at the definitions of f 1 and f 2 we see that r f 1 (I m(r)) f(i) and r f 2 (I n(r)) f(i), for all r R, 3

4 yielding f 1 (I) = f 2 (I) = R. We conclude that f 1 and f 2 have the Darboux property. We saw that f 1 (I) = f 2 (I) = R for any interval I so the functions f 1 and f 2 are not continuous at any point x 0 in R, because f((x 0 δ, x 0 + δ)) = R (f(x 0 ) ε, f(x 0 ) + ε), ε > 0, δ > 0, x 0 R. Next problem was proposed in a Romanian contest 1 in 2003: We define F = f : [0, 1] [0, 1] : A, B [0, 1], A B =, A B = [0, 1], f(a) B, f(b) A}. Study if F contains continuous functions, functions which have antiderivatives, and functions with Darboux property. The answer to the first two is obvious. If a function f F is continuous or has antiderivatives, then the function g(x) = f(x) x, x [0, 1] has antiderivatives. It is well known that any function which has antiderivatives necessarily has the Darboux property. Since g(0) g(1) 0 and g has the Darboux property we conclude that g has at least one zero in [0, 1]. This yields that f has a fixed point, which contradicts f F. The next theorem gives the answer to the third question. Theorem 4. There exist functions in f F having the Darboux property. Proof: We choose a bijection φ : A R and we denote Y = φ 1 ((-, 0]), Z = φ 1 ((0, )). Take A = x [0, 1] : [x] Y }, B = x [0, 1] : [x] Z}. A and B are disjoint and non void because Y and Z are disjoint and non void. We have Y Z = A, so A B = x [0, 1] : [x] A} = [0, 1]. Because A and B contain all the elements from [0, 1] which are in the same equivalence class, A and B are dense in [0, 1]. From their definitions, card Y = card Z = card A = card B = ℵ. Therefore, we can find the bijections µ : Y B şi ν : Z A. We define the function f : [0, 1] [0, 1] µ([x]), x A f(x) = ν([x]), x B A x [x] µ B B x [x] ν A From the definition of f and sets A, B, Y, Z we find that f(a) B and f(b) A. Let s prove that f has the Darboux property. We take I an interval from [0, 1]. Then I intersects all the classes from A (because any of these is dense in R), which means that I intersects all the classes from Y and Z. Hence f(i) = µ(y ) ν(z) = B A = [0, 1]. Therefore, f F and f has the Darboux property. From the theorem above we have the following 1 Traian Lalescu problem solving contest for university students, a contest held in the memory of the great Romanan mathematician. 4

5 Corollary. If a, b R, a < b then we can find functions f : [a, b] [a, b] which have the Darboux property and have no fixed points. Proof: We do the same construction from previous theorem for [a, b]. If this function has a fixed point then we have a contradiction with the definitions of the sets A and B. Cauchy functional equation and Darboux property We will prove in the following that we can find nontrivial solutions to the Cauchy functional equation which have the Darboux property. For doing this, we need some preliminary results: 0) A function f : R R satisfies the Cauchy functional equation if f(x+y) = f(x) + f(y), x, y R (C). i) A function which satisfies (C) also satisfies f(qx) = qf(x), q Q, x R ii) Any function of the form f(x) = ax, a R, satisfies the equation (C). We will call these the trivial solutions of equation (C). It is well known that any continuous solution of the equation (C) is trivial. iii) If we consider R as vector space over Q, then, according to 0) and i), f is a linear map. This implies that f is well and uniquely defined if we know its values on a basis of R over Q. For proofs see [3], page 193. The next theorem allows us to speak of nontrivial solution of (C). Theorem 5. There exist nontrivial solution of the equation (C). Proof: Using iii), we take a basis B which contains 1 and consider f(1) = 1, f(b) = 0, b B\1}. Thus f(x) will be the coefficient of 1 in the representation of x in the basis B. Because f(r) = Q, f is not a trivial solution for (C). Once we established the existence of nontrivial solutions for (C), we can give the following theorem: Theorem 6. If f is a nontrivial solution for (C), then it s graph, G f = (x, y) R 2 : y = f(x)}, is everywhere dense in R 2. For a proof, see [3], page 200. Remark. We can see from the results discussed above that the solution of the equation (C) have a common property: If f is a solution for (C) then f has the Weak-Darboux property. Proof: If f is a trivial solution, it is continuous and has the Darboux property, so it has the Weak-Darboux property, also. Let s assume that f is a nontrivial solution for (C). We proved that G f is everywhere dense in R 2. Let I R be an interval. Then for all a, b R, a < b, G f I (a, b). Therefore exists t I such that f(t) (a, b), hence f(i) = R, which is an interval. Consequently, f has the Weak-Darboux property. We ask ourselves if a solution of (C) with Darboux property is necessary trivial. We will answer this question in the following theorem. First, we can see that there exist nontrivial solutions for (C) without Darboux property, because we have such an example in the proof of Theorem 5. Theorem 7. There exist nontrivial solutions for equation (C) which have the Darboux property. 5

6 Proof: We use iii) and choose a basis B of R over Q which contains 1. We take f(1) = 0 and we establish a bijection φ : B \ 1} R. We define f(b) = φ(b), b B \ 1}. It is clear that f(q) = 0, q Q. Considering the equivalence relation defined we have x [y] f(x) = f(y) + f(x y) = f(y). Let y R be a real number. Then we can find b B such that f(b) = φ(b) = y. We take b 0 [b] I which is non void. We have f(b 0 ) = f(b) = y, so y f(i), and because y was chosen randomly, it follows that f(i) = R. Therefore, f has the Darboux property. We see that f(q) = 0} and f(r 0}, which proves that f is a nontrivial solution for (C). The multi dimensional case In the one dimensional case we saw that a function has Darboux property if and only if f(j) is an interval whenever J is an interval. In R the intervals are the only connected or closed sets. We say that A R p, p > 1 is a region if it is open and connected. If A is also convex, then we say that A is a convex region. We wish to extend the Darboux property in the multi dimensional case in the following way: Definition. 4 If f : R p R q, p, q > 1, we say that f has the Darboux property if f maps any region of R p onto a region of R q. Furthermore, we say that f has the Convex-Darboux property if f maps any convex region in R p onto a convex region of R q. In the following, we wish to prove an analogue to the Sierpinski theorem in the one dimensional case. Theorem 8. If f : R p R q p, q > 1 is an arbitrary function, then we can write f as the sum of two functions with the Darboux property, and also, we can write f as the sum of two functions with Convex-Darboux property. Proof: We define the relation p on R p by x p y x y Q p. This turns out to be an equivalence relation, so we denote [x] the equivalence class containing x and A p the set of all equivalence classes. If x p y then any two of their corresponding coordinates are in the same equivalence class in A. This yields A p = A p, so card A p = ℵ p = ℵ. This holds for any p. We can establish a bijection g : R p A p. Set A 1 = g((, 0) R p 1 ) and A 2 = g([0, ) R p 1 ). We can see that A 1 A 2 = A p, A 1 A 2 = and card A 1 = card A 2 = ℵ. Therefore we can find the bijections m : R q A 1 and n : R q A 2. Given the function f, we define f 1 (t) = r f(t) r, t m(r), t n(r) f 2 (t) = f(t) r r, t m(r), t n(r), functions which satisfy f 1 + f 2 = f and have the desired properties. To prove this, we take a region (convex region) A R p. Then A contains an open disk D. Since every equivalence class in A p is a translation of Q p, every one of them is dense in R p. Now, taking y R p we see that m(y) D and n(y) D. 6

7 Therefore y f 1 (D) f 2 (D) f 1 (A) f 2 (A). This yields f 1 (A) = f 2 (A) = R q, which is a connected and convex region of R q. Thus f 1 and f 2 have the Darboux and Convex-Darboux properties. Moreover, f 1 and f 2 are not continuous at any point in R p. 7

8 References [1] Nathan Jacobson, Lectures in Abstract Algebra, vol II, Linear Algebra, D. Van Nostrand Company, Inc., 1953 [2] Waclaw Sierpinski, Sur la représentation d une function arbitraire par des fonctions jouissant de la propriété de Darboux, Transactions of the American Mathematical Society, 1960, 95, 3, [3] A. B. Kharazishvili, Strange Functions in Real Analysis, Chapman & Hall/CRC 2006 student Faculty of Mathematics and Informatics West University, Timişoara, Romania beni22sof@yahoo.com 8