Here are standard definitions of what it means for a function to be concave up/down, withoutassuming differentiability (and not even continuity!). Let f be a function defined on the interval I R.| We say that f is concave up on I if and only if for every x, y 2 I and every t 2 [0, 1] we havef (tx + (1 t)y) t f (x) + (1 t) f (y).~ We say that f is strictly concave up on I if and only if for every x, y 2 I , x 6 = y and every t 2 [0, 1]we have f (tx + (1 t)y) < t f (x) + (1 t) f (y).

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We say that f is concave down on I if and only if the function f is concave up on I .} We say that f is strictly concave down on I if and only if the function f is strictly concave up on IThe goal problems 1 and 2 is to explore what kind of continuity properties a concave up/down functionmay or may not enjoy.1)(a) Let A and B be two real numbers such that A < B. Prove that for every x 2 [ A, B] there exists aunique t 2 [0, 1] such that x = tA + (1 t)B. Use this fact to sketch a picture of what what thedefinitions above mean geometrically.(b) Observe that a function can be concave up (or down) on an interval without being continuous onthat interval.

To see a counterexample, prove that the functionf (x) =(1 if x = 0px if 0 < x < 1is strictly concave up on the interval [0, 1), but it is not continuous on [0, 1).2) Let us restrict our attention to the case where I is an open interval of the form I = (a, b), wherea < b are real numbers. Our goal is to prove thatif f is concave up on (a, b), then f is continuous on (a, b).In order to prove this statement, we will prove its contrapositive. Namely:if f is not continuous on (a, b), then f is not concave up on (a, b).Let us break down the proof of the latter into a few steps.(a) Note that the function f is not continuous on (a, b) if and only if there exist a point x02 (a, b)such that f is not continuous at x0. Recall how continuity is defined using limits.

Write down thestatement “ f is not continuous at x0” using the (negation of the) e d definition of limit.(b) Consider the picture belowDeduce from (2a) that for any fixed d > 0 there must exist infinitely many points in the graph off belonging to one of the regions I, II, III, IV, and whose x-coordinate is arbitrarily close to thevertical line y = x0. We now continue the proof differently, depending on what this region is.(c) Suppose that the region from (2b) is region I or II. Assume for simplicity that this is regionII. Show that you can pick two numbers x1and x2as in the picture below, with the propertythat (x2, f (x2)) lies above the segment joining (x0, f (x0)) and (x1, f (x1)). Explain how this factimplies that f is not concave up on (a, b). (The case when the region in (2b) is region I is similarand you don’t need to discuss it.)(d) Suppose that the region from (2b) is region III or IV.

Assume for simplicity that this is regionIII. Show that you can pick three numbers x1, x2and x0(the latter could also be = ¥) as inthe picture below, with the property that x0< x2 < x0 < x1 such that the point (x0, f (x0)) liesabove the segment joining (x2, f (x2)) and (x1, f (x1)). Explain how this fact implies that f is notconcave up on (a, b). (The case when the region in (2b) is region IV is similar and you don’t needto discuss it.)Congratulations! You have proved that if a function is concave up on the interval (a, b), then it isautomatically continuous on (a, b). Of course the same holds for concave down functions! This isbecause f is concave down if and only if f is concave up, and f is continuous if and only f iscontinuous.

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