Suppose that the differentiable functions a , b, f , g : R → R satisfy

5.2.34. Suppose that the differentiable functions a , b, f , g : R → R satisfy

f (x) ≥ 0, f ′ (x) ≥ 0, g(x) > 0, g ′ (x) > 0 for all x ∈ R,

+ a(x)

= b(x).

g (x)

g (x)

Prove that

Solution. Let 0 < ε < A be an arbitrary real number. If x is sufficiently large, then

f (x) > 0, g(x) > 0, |a(x) − A| < ε , |b(x) − B| < ε , and

B − ε < b(x) = ′

+ a(x)

+ (A + ε )

g (x) (A + ε )(A + 1) f ′ (x)(g(x)) A + A · f (x) · (g(x)) A −1 ·g ′ (x)

A (A + 1) · (g(x)) A ·g ′ (x) (A + ′ ε )(A + 1) A

A · ((g(x)) A +1 ) ′ .

Hence

A (B − ε )

((g(x)) A +1 ) ′ >

. (A + ε )(A + 1)

214 5 Differentiability Similarly, for sufficiently large x,

A (B + ε )

((g(x)) A +1 ) ′ <

. (A − ε )(A + 1)

As ε → 0, we have

lim

((g(x)) A +1 ) ′ =

x →∞

A +1

By l’Hˆopital’s rule we deduce that

f (x)

(x) · (g(x)) A

→∞ g (x) x →∞ (g(x)) A +1

5.2.35. Prove that there is no differentiable f : (0, 1)→R for which sup x ∈E |f ′ (x)| is finite, where E is a dense subset of the domain, and |f| is a nowhere differentiable function.

P. Perfetti Solution. Assume that a nonempty interval

0 for all x ∈ (a,b). Since f is differentiable on (a,b), thus continuous, f does not change sign on (a, b). Thus, either | f | = f on all of (a,b) or | f | = − f on (a,b). This shows that | f | is differentiable on (a,b), which is a contradiction. Therefore, in every nonempty subinterval of (0, 1), f vanishes at least once. This shows that

f is identically zero in (0, 1), hence | f | ≡ 0, a contradiction. There is therefore no differentiable function f : (0, 1)→R such that | f | is nowhere differentiable. ⊓ ⊔

5.2.36. Let f : (0, 1) → [0,∞)

be a function that is zero except at the distinct points a n , n ≥1 . Let b n = f (a n ) .

(i) Prove that if ∑ ∞ n =1 b n <∞ , then f is differentiable at least once in (0, 1) . (ii) Prove that for any sequence of nonnegative real numbers (b n ) n ≥1 , with ∑ ∞ n =1 b n =∞ , there exists a sequence (a n ) n ≥1 such that the function f defined as above is nowhere differentiable.

Solution. (i) We first construct a sequence (c n ) n ≥1 of positive numbers such that

n → ∞ as n→∞ and ∑ n =1 c n b n < 1/2. Let B = ∑ n =1 b n , and for each k = 0, 1, . . . denote by N k the first positive integer for which

n =N k

Now set c n =2 k /5B for each n, N k ≤n<N k +1 . Thus, c n → ∞ and

n =1 k =0 N k ≤n<N k +1

k =0 5B n =N k

k =0 5B 4 5

5.2 Introductory Problems 215 Consider the intervals I n = (a n −c n b n ,a n +c n b n ). The sum of their lengths is

2 ∑ n ≥1 c n b n < 1; thus there exists a point x 0 ∈ (0,1) that is not contained in any

I n . We show that f is differentiable at x 0 and f ′ (x 0 ) = 0. Since x 0 is outside of the intervals I n , then x 0 n for any n and f (x 0 ) = 0. For arbitrary x ∈ (0,1) \ {x 0 }, if x =a n for some n, then

f (x) − f (x 0 )

f (a n )−0

c n Otherwise, ( f (x) − f (x 0 )) /(x−x 0 ) = 0. Since c n → ∞, this implies that for arbitrary

x −x 0 |a n −x 0 |

ε > 0 there are only finitely many x ∈ (0,1) \ {x 0 } for which

f (x) − f (x 0 ) < ε x −x 0

does not hold. This shows that f is differentiable at x 0 and f ′ (x 0 ) = 0. (ii) We remove the zero elements from the sequence (b n ) n ≥1 . Since f (x) = 0 exce-

pting a countable subset of (0, 1), we deduce that if f is differentiable at some

point x 0 , then f (x 0 ) and f ′ (x 0 ) must be 0.

be a sequence satisfying 0 < β n ≤b n ,b n → 0 as n→∞, and ∑ n =1 β n = ∞. Choose the numbers a n (n ≥ 1) such that the intervals I n = (a n − β n ,a n + β n ) cover each point of (0,1) infinitely many times (this is possible, since the sum of lengths is 2 ∞ ∑ n =1 b n = ∞). Fix x 0 ∈ (0,1) such that f (x 0 ) = 0. For any ε > 0, there

∞ n ≥1

Let ( β n )

exists n ≥ 1 such that β n < ε and x 0 ∈I n . Therefore

hence f is not differentiable at x 0 . ⊓ ⊔