Let f be differentiable with f ′ continuous on [a, b] . Show that if there is

5.3.24. Let f be differentiable with f ′ continuous on [a, b] . Show that if there is

a number c in (a, b] such that f ′ (c) = 0 , then we can find a number ξ in (a, b) such that

f ′ ( ξ )=

f ( ξ ) − f (a)

b −a

S. Penner, Math. Magazine, Problem 987 Solution. For x in [a, b], define

g (x) = f (x) − f (a) (x) − .

b −a

Assume first that f (c) > f (a). Choose d in (a, c] such that f (d) is a maximum for f in [a, c], and let e be a point in (a, d) such that f ′ (e) = ( f (d) − f (a)) /(d − a). Then g(e) > 0 > g(d), so, since g is continuous, there is a point ξ in (e, d) such that g ( ξ ) = 0. A similar argument takes care of the cases f (c) < f (a) and

f (c) = f (a). ⊓ ⊔

n =1 (−1) −1 sin (log n)/n α converges if and only if α >0 . Solution. Set f (t) = sin(logt)/t α . We have

5.3.25. Show that ∞ ∑

cos (logt) (t) = − α+1 sin (logt) +

. It follows that |f ′ (t)| ≤ (1 + α )/t α+1 for any α > 0. Thus, by the mean value theo-

t α+1

rem, there exists θ ∈ (0,1) such that

1 + α | f (n + 1) − f (n)| = | f ′ (n + θ )| ≤ n α+1

234 5 Differentiability

Since ∑ ∞ n =1 (1 + α )/n α+1 < +∞ for α > 0 and f (n) → 0 as n→∞, we deduce that

∑ (−1) n −1 f (n) = ∑ ( f (2n − 1) − f (2n))

n =1

n =1

converges. Next, we prove that sin (log n)/n α does not converge to 0 for α ≤ 0. It suffices to consider α = 0. We show that a n = sin(log n) does not tend to zero. Assume the contrary. Then there exist a positive integer k n and λ n ∈ [−1/2,1/2] for n > e 2 such that log n / π =k n + λ n . Then |a n | = sin π | λ n |. Since a n → 0, we obtain λ n → 0. We have

log 1 1 k n +1 −k n =

(n + 1) − logn −( λ n +1 − λ n )= log 1 +

−( λ n +1 − λ n ). π

n Then |k n +1 −k n | < 1 for all n big enough. Hence there exists n 0 such that k n =k n 0 for

n >n 0 . So log n / π =k n 0 + λ n for all n >n 0 . Since λ n → 0, we obtain a contradiction to log n → ∞ as n→∞. ⊓ ⊔

The geometric interpretation of the following straddle lemma is that if the points u and v “straddle” z, then the slope of the chord between the points (u, f (u)) and (v, f (v)) on the graph of f : [a, b]→R is close to the slope of the tangent line at (z, f (z)).

be differentiable at z ∈ [a,b] . Prove that for each ε >0 , there exists δ >0 such that

5.3.26. Let f : [a, b]→R

| f (v) − f (u) − f ′ (z)(v − u)| ≤ ε (v − u)

whenever u ≤z≤v and [u, v] ⊆ [a,b] ∩ (z − δ ,z+ δ ) . Solution. Since f is differentiable at z, there exists δ > 0 such that

f (x) − f (z)

−f ′ (z) < x ε

−z

for 0 < |x − z| < δ ,x ∈ [a,b]. If z = u or z = v, the conclusion is immediate, so suppose u < z < v. Then

| f (v) − f (u) − f ′ (z)(v − u)| ≤ | f (v) − f (z) − f ′ (z)(v − z)| + | f (z) − f (u) − f ′ (z)(z − u)|

< ε (v − z) + ε (z − u) = ε (v − u).

This concludes the proof. ⊓ ⊔

5.4 The Maximum Principle 235