Points at Which Continuous Functions Have the Same Height

Transcription

1 GENERAL ARTICLE Points at Which Continuous Functions Have the Same Height Parth Prashant Karnawat As an easy application of the intermediate value theorem, one can show that for any continuous function f : [0, 1] R with f (0) = f (1), there are points a, a + 1/2 both in [0, 1] such that f (a) = f (a + 1/2). In this note, we show that this property holds with 1/2 replaced by any number of the form 1/n for a positive integer n. More interestingly, we show that this is false for every number not of the form 1/n. Continuous Functions The author is a first year student of the B.Math. Hons. Programme at the Indian Statistical Institute, Bangalore. Consider a real-valued continuous function f on [0, 1] that has the same height at 0 and 1; that is, f (0) = f (1). Then there must be a point a [0, 1/2] so that f (a) = f (a + 1/2). The proof is an application of the intermediate value theorem to the continuous function g(x) = f (x) f (x + 1/2) defined on [0, 1/2] (see [1 Exercise 18.10]). Indeed, the values g(0) = f (0) f (1/2) and g(1/2) = f (1/2) f (1) = f (1/2) f (0) have opposite signs (or are both zero), which implies by the intermediate value theorem that g takes the value zero at some point a between 0 and 1/2; this is the same as saying that f (a) = f (a + 1/2). This raises the following natural problem. Find all t (0, 1] such that for every continuous function f : [0, 1] R satisfying f (0) = f (1), there exists a point a [0, 1 t] (depending on f ) such that f (a) = f (a + t). In this note, we discuss and solve this problem and the answer turns out to be rather interesting. We prove the following proposition. Proposition. For every integer n > 1, and for every continuous function f : [0, 1] R satisfying f (0) = f (1), there exists a RESONANCE May 2018 Keywords Intermediate value theorem. 591

2 [0, 1 1/n] (depending on f ) such that f (a) = f (a+1/n). Further, if 0 < t < 1 is such that t is not of the form 1/k for any positive integer k, then there exists a continuous function f :[0, 1] [ n, n] such that f (0) = f (1) but f (a) 1 = f (a + t) for all a [0, 1 t]. Here, n = 1/t, the greatest integer not exceeding 1/t. Proof. We first show the assertion for any n > 1. Given any continuous function f :[0, 1] R, consider the function g(x) = f (x) f (x + 1/n) defined for x [0, 1 1/n]. Note that g is continuous and the sum n 1 i=0 g(i/n) = f (0) f (1) = 0. Therefore, either g(i/n) = 0 for all i (in which case clearly f (0) = f (1/n)) or there exist i < j among 0, 1,, n 1sothatg(i/n)g( j/n) < 0. Then, the intermediate value theorem applied to g (see [1, 18.2]) implies the existence of a point a (i/n, j/n) such that g(a) = 0. In other words, f (a) = f (a + 1/n). Now, to prove the necessity assertion of the proposition, we assume that 0 < t < 1 is not of the form 1/k for any positive integer k. This assumption is equivalent to the assertion 1/t 1/t. Writing n = 1/t 1/t, wehave nt < 1 < (n + 1)t. We will construct a continuous function f on [0, 1] whose range is [ n, n] for which f (0) = f (1) and f (a + t) = f (a) 1 for all a [0, 1 t]. Note that we may decompose the interval [0, 1] as a disjoint union [0, 1] = {0} A B where A = n n 1 (kt, 1 (n k)t], B = (1 (n k)t, (k + 1)t]. Note that each interval in A is a translate by t of the previous one, and the same thing holds for the intervals in B. Consider the 592 RESONANCE May 2018

3 Figure 1. A function with different heights at width t = 3/4. function f defined on [0, 1] by 0, x = 0 n(x kt) f (x) = 1 nt k, x (kt, 1 (n k)t] (k n) (n+1)(x (k+1)t) 1 (n+1)t (k + 1), x (1 (n k)t, (k + 1)t] (k n 1). Note that f (A) = n ( k, n k], f (B) = n 1 [ (k + 1), n k). Figures 1 and 2 display the two special cases t = 0.75 and t = 0.4 and Figure 3 depicts the general case. It is easy to check that f is continuous on [0, 1]. Indeed, clearly, f is continuous at all internal points of A and B; it is continuous at other points as it is left continuous and right continuous at these points with the same limiting values. Consider any two points of the form a, a+t in [0, 1]. We will show f (a) f (a + t). We may assume a 0as f (0) = 0, f (t) = 1. RESONANCE May

4 Figure 2. A function with different heights at width t = 2/5. First, suppose a A (0, 1 t]; say, a (kt, 1 (n k)t]forsome k n 1; then a + t ((k + 1)t, 1 (n k 1)t]. Thus, n(a kt) f (a) = k, 1 nt n(a + t (k + 1)t) f (a + t) = (k + 1), 1 nt so that f (a + t) = f (a) 1. Now, suppose a B (0, 1 t]; say, a (1 (n k)t, (k + 1)t]for some k n 2; then a + t (1 (n k 1)t, (k + 2)t]. So, (n + 1)(a (k + 1)t) f (a) = (k + 1), 1 (n + 1)t (n + 1)(a + t (k + 2)t) f (a + t) = (k + 2), 1 (n + 1)t so that f (a + t) = f (a) 1. Therefore, we have proved that f is a continuous function from [0, 1] to [ n, n] and, for any a [0, 1 t], that f (a + t) = f (a) RESONANCE May 2018

5 This completes the proof. Editor s Comment This very interesting result was accepted to be published in the American Mathematical Monthly. However, when the proof sheets were to be sent to the author, simultaneously another paper appeared whose motivation and methods though quite different have intersection with the results of this piece. Even though the editors of the Monthly suggested they could publish this in adifferent undergraduate research journal, the author decided to publish it in Resonance. For the interested reader, the reference to the other paper is given below as [2]. Figure 3. A function with different heights at width t 1/n. RESONANCE May

6 Suggested Reading Address for Correspondence Parth Prashant Karnawat B Math Hons. First year Indian Statistical Institute 8th Mile Mysore Road Bangalore , India bmat1715@ms.isibang.ac.in [1] K A Ross, Elementary Analysis: The Theory of Calculus, Springer International Edition, 4th Indian Reprint. New Delhi, India: Springer-Verlag, [2] K Burns, O Davidovich and D Davis, Average Pace and Horizontal Chords, arxiv: v2[math.ho], 21 February RESONANCE May 2018