See discussions, stats, and author profiles for this publication at: https://www.researchgate.net/publication/257821297 Problem 11402: Squares on Graphs Article in The American Mathematical Monthly * October 2008 CITATIONS 0 READS 21 1 author: Some of the authors of this publication are also working on these related projects: Metatheoretical analysis and structural alternatives of the resolving of the philosophical problem of the applicability of mathematics in natural sciences and Wigner's puzzle View project Editing/Proofreading/Commenting/Research Designing in philosophy of science/philosophy of mathematics Collaboration and co-authorship View project Catalin Barboianu University of Bucharest 27 PUBLICATIONS 5 CITATIONS SEE PROFILE All content following this page was uploaded by Catalin Barboianu on 09 May 2014. The user has requested enhancement of the downloaded file. By the uniqueness of the minimum, t ′ = φ(λ). Since 〈tn〉 is bounded we conclude that 〈tn〉 converges to φ(λ). This shows that φ is continuous. Lemma 2. limλ→+∞ φ(λ) = 0 and limλ→0+ |φ(λ)| = +∞. Proof. Let 〈λn〉 be a sequence such that limn→∞ λn = +∞, and let tn = φ(λn). For t ∈ R, we have f (tn)/λn + |tn| ≤ f (t)/λn + |t |, and in particular f (tn)/λn + |tn| ≤ f (0)/λn. Let λ0 be a fixed positive value, and let m = infR[ f (t) + λ0|t |]. Now f (tn) ≥ m − λ0|tn|, so (1 − λ0/λn)|tn| ≤ ( f (0) − m)/λn. Therefore limn→∞ tn = 0. For the other claim of the lemma, let 〈λn〉 be a positive sequence that tends to zero, let tn = φ(λn), and let t ′ be a limit point of 〈tn〉 (if one exists). The argument of Lemma 1 proves that for any real t , f (t) ≥ f (t ′). That makes f (t ′) a global minimum for f , contrary to the hypothesis. Since 〈tn〉 has no limit point, limn→∞ |tn| = +∞. From these two lemmas, we see that the range of φ contains (0, ∞) or (−∞, 0) (but not both). We will show that in the first case conclusion (a) holds. Similarly, the second case leads to (b). Assume the range contains (0, ∞), and let m(λ) = infR ( f (t) + λ|t |). Now f (t) ≥ supλ ( m(λ) − λ|t |). If t = φ(λ), then f (φ(λ)) = m(λ) − λ|φ(λ)|. Thus f is the pointwise supremum of a family of affine functions on (0, ∞), so f is convex there. We claim that f is actually strictly convex. Indeed, if f is affine on some interval [a, b] with 0 < a < b, then we can choose λ such that the function fλ given by fλ(t) = f (t) + λ|t | reaches its infimum at a point of (a, b). Since fλ is is affine on this interval, it is minimized at an interior point only if it is constant on that interval, which contradicts the uniqueness of the minimum point. Let s, t be given with t > 0 and −t ≤ s < t . There exists λ such that t = φ(λ). Thus f (s) + λ|s| > f (t) + λ|t | ≥ f (t) + λ|s|. We obtain f (s) > f (t). (If −t ≤ s ≤ t , we obtain f (s) ≥ f (t).) For the integral inequality, we have −|u(x)| ≤ u(x) ≤ |u(x)|. So f (u(x)) ≥ f (|u(x)|). Since f is convex, Jensen's inequality yields∫  f (u) ≥ ∫  f (|u|) ≥ f (∫  |u| ) . It is a strict inequality since u is not essentially constant and f is strictly convex. Also solved by R. Stong. Squares On Graphs 11402 [2008, 949]. Proposed by Doru Catalin Barboianu, Infarom Publishing, Craiova, Romania Let f : [0, 1] → [0, ∞) be a continuous function such that f (0) = f (1) = 0 and f (x) > 0 for 0 < x < 1. Show that there exists a square with two vertices in the interval (0,1) on the x-axis and the other two vertices on the graph of f . Solution by Byron Schmuland and Peter Hooper, University of Alberta, Edmonton, AB, Canada. Extend f by letting f (x) = 0 for x ≥ 1. Define g(x) = f (x + f (x)) − f (x) for x ≥ 0. If there exists x ∈ (0, 1) with g(x) = 0, then a square as required exists with vertices (x, 0), (x + f (x), 0), (x, f (x)), (x + f (x), f (x)). June–July 2010] PROBLEMS AND SOLUTIONS 563 Now g is continuous, so to show that such x exists we will show that y, z ∈ (0, 1) exist with g(y) ≥ 0 and g(z) ≤ 0. Let z be a value where f takes its maximum. Then f (z) ≥ f (z + f (z)), so that g(z) ≤ 0. Since 0 + f (0) = 0 < z < z + f (z), by continuity there is a value y ∈ (0, z) so that y + f (y) = z. Hence g(y) = f (y + f (y)) − f (y) = f (z) − f (y) ≥ 0. Editorial comment. Pál Péter Dályay (Hungary) noted a generalization: Given any p > 0, there exists a rectangle with base-to-height ratio p having two vertices on the x-axis and the other two vertices on the graph of f . Also solved by B. M. Ábrego & S. Fernández-Merchant, F. D. Ancel, K. F. Andersen (Canada), R. Bagby, N. Caro (Brazil), D. Chakerian, R. Chapman (U.K.), B. Cipra, P. Corn, C. Curtis, P. P. Dályay (Hungary), C. Diminnie & R. Zarnowski, P. J. Fitzsimmons, D. Fleischman, T. Forgács, O. Geupel (Germany), D. Grinberg, J. Grivaux (France), J. M. Groah, E. A. Herman, S. J. Herschkorn, E. J. Ionascu, A. Kumar & C. Gibbard (U.S.A. & Canada), S. C. Locke, O. P. Lossers (Netherlands), R. Martin (Germany), K. McInturff, M. McMullen, M. D. Meyerson R. Mortini M. J. Nielsen, M. Nyenhuis (Canada), Á. Plaza & S. Falcón (Spain), K. A. Ross, T. Rucker, J. Schaer (Canada), K. Schilling, E. Shrader, A. Stadler (Switzerland), R. Stong, B. Taber, M. Tetiva (Romania), T. Thomas (U.K.), J. B. Zacharias & K. Greeson, BSI Problems Group (Germany), GCHQ Problem Solving Group (U.K.), Lafayette College Problem Group, Microsoft Research Problems Group, Missouri State University Problem Solving Group, Northwestern University Math Problem Solving Group, NSA Problems Group, and the proposer. A Trig Series Rate 11410 [2009, 83]. Proposed by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. For 0 < φ < π/2, find lim x→0 x−2 ( 1 2 log cos φ + ∞∑ n=1 (−1)n−1 n sin2(nx) (nx)2 sin2(nφ) ) . Solution by Otto B. Ruehr, Michigan Technological University, Houghton, MI. We begin with three elementary identities. The first is ∞∑ n=1 r n sin2 nφ = r(r + 1) sin 2 φ (1 − r)[(1 − r)2 + 4r sin2 φ] . (i) This is derived by writing sin2 nφ in terms of exponentials and summing the resulting geometric series. Now divide (i) by r and integrate with respect to r to get ∞∑ n=1 r n n sin2 nφ = 1 4 log [ (1 − r)2 + 4r sin2 φ (1 − r)2 ] . (ii) Differentiate (i) with respect to r to obtain ∞∑ n=1 nrn−1 sin2 nφ = 1 2(1 − r)2 − 1 2 [ (r − 1)2 − 2(r 2 + 1) sin2 φ [(1 − r)2 + 4r sin2 φ]2 ] . (iii) The limit at r = −1 in (ii) gives us ∞∑ n=1 (−1)n−1 n sin2 nφ = −1 2 log cos φ. Now we can write the requested limit as lim x→0 x−2 lim r→−1+ ∞∑ n=1 r n n [ 1 − sin 2 nx n2x2 ] sin2 nφ. 564 c© THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 117 View publication stats