An elementary proof of a converse mean value theorem Ricardo Almeida
[email protected] Department of Mathematics, University of Aveiro, 3810-193 Aveiro, Portugal Abstract We present a new converse mean value theorem, with a rather elementary proof.
Keywords: mean value theorem, converse, differentiability.
1
Introduction
The goal of this note is to present a simple proof for a converse mean value theorem. Given a differentiable function f : [a, b] → R and a point c ∈]a, b[, are there reals a1 , b1 ∈]a, b[ such that f (b1 ) − f (a1 ) = f 0 (c)(b1 − a1 )? Sufficient conditions for the above converse to hold are established in [1]. However, there appear to be a gap in the proof. They claim that, since f is differentiable at x = ci then f 0 (ci ) = limx,y→ci (f (x) − f (y))/(x − y). The following example shows that this can be false. Let 2 x sin x1 if x 6= 0 f (x) = 0 if x = 0 Then f is differentiable at x = 0 and f 0 (0) = 0. However, given N ∈ N, let k = 2N + 1. Then 1 1 1 4(kπ)2 + 4kπ + 1 2(kπ)2 + f − f sin −0 kπ 2(kπ)2 kπ 4(kπ)4 2kπ + 1 = 1 1 2(kπ)2 2(kπ)2 kπ 4(kπ)2 + 4kπ + 1 = sin kπ − 2(kπ)2 2kπ + 1 2 4(kπ) + 4kπ + 1 kπ 4(kπ)2 + 4kπ + 1 kπ k+1 = (−1) sin = sin 2 2 2(kπ) 2kπ + 1 2(kπ) 2kπ + 1 (since k = 2N + 1) and so the limit is 2 sin
2
1 if N → ∞. 2
The theorem
Theorem 1. Let f : [a, b] → R be a continuous function and differentiable in ]a, b[. Given c ∈]a, b[, let k0 > 0 be such that ]c − k0 , c + k0 [⊆]a, b[. If, for all k ∈]0, k0 [, 1. f 0 (c − k) < f 0 (c) < f 0 (c + k) then there exist a1 , b1 ∈]a, b[ with c ∈]a1 , b1 [ and f (b1 ) − f (a1 ) = f 0 (c)(b1 − a1 ). 2. f 0 (c−k) ≤ f 0 (c) ≤ f 0 (c+k) then there exist a1 , b1 ∈]a, b[ with c ∈ [a1 , b1 ] and f (b1 )−f (a1 ) = f 0 (c)(b1 − a1 ).
1
Proof. First we prove 1. Define the function g(x) = f (x)−f 0 (c)x, x ∈ [a, b]. Then g is differentiable and g 0 (x) = f 0 (x) − f 0 (c). By the hypothesis, for all k ∈]0, k0 [, g 0 (c) = 0 , g 0 (c + k) > 0 and g 0 (c − k) < 0 and so g has a strict local minimum at c. Therefore we have two possible cases: (a) g(c) < g(c −
k0 2 )
≤ g(c +
k0 2 )
(b) g(c) < g(c +
k0 2 )
≤ g(c −
k0 2 )
If the first one occurs, then by the intermediate value theorem, there exists b1 ∈]c, c + k20 ] with g(b1 ) = g(c − k20 ). If we take a1 = c − k20 then c ∈]a1 , b1 [ and g(a1 ) = g(b1 ), which is equivalent to f (b1 ) − f (a1 ) = f 0 (c)(b1 − a1 ). The second case is analogous. To proof the second part of the theorem, observe that in this case g 0 (c) = 0 , g 0 (c + k) ≥ 0 and g 0 (c − k) ≤ 0 and so c is a local minimum of g. Therefore one of the two following cases holds (a) g(c) ≤ g(c −
k0 2 )
≤ g(c +
k0 2 )
(b) g(c) ≤ g(c +
k0 2 )
≤ g(c −
k0 2 )
If (a) is true, there exists b1 ∈ [c, c + Analogous for (b).
k0 2 ]
with g(a1 ) = g(b1 ), where a1 = c −
k0 2
and c ∈ [a1 , b1 ].
References [1]
J. Tong and P.A. Braza, A converse of the mean value theorem, Amer. Math. Monthly 106 (1997), pp. 939–942.
2
