Let f be twice continuously differentiable on (0, +∞) such that

5.3.20. Let f be twice continuously differentiable on (0, +∞) such that

5.3 The Main Theorems 231

Show that

Solution. Since f ′ tends to −∞ and f ′′ tends to +∞ as x tends to 0+, there exists an interval (0, r) such that f ′ (x) < 0 and f ′′ (x) > 0 for all x ∈ (0,r). Hence f is decreasing and f ′ is increasing on (0, r). By the Lagrange mean value theorem, for

every 0 <x<x 0 < r we obtain

f (x) − f (x 0 )=f ′ ( ξ )(x − x 0 )>0, for some ξ ∈ (x,x 0 ). Taking into account that f ′ is increasing, f ′ (x) < f ′ ( ξ ) < 0,

we have

Taking limits as x tends to 0 +, we obtain

f (x)

f (x)

−x 0 ≤ liminf x →0+ f ′ (x) ≤ limsup

≤ 0. Since this happens for all x 0 ∈ (0,r), we deduce that lim x →0+ f (x)/ f ′ (x) exists and

x →0+ f ′ (x)

lim x →0+ f (x)/ f ′ (x) = 0. ⊓ ⊔

5.3.21. Let f :R →R

be twice differentiable such that f (0) = 2 , f ′ (0) = −2 , and f (1) = 1 . Prove that there exists ξ ∈ (0,1) such that f ( ξ )·f ′ ( ξ )+f ′′ ( ξ ) = 0.

International Mathematics Competition for University Students, 1997 Solution. Define the function

g (x) = f 2 (x) + f ′ (x) .

Since g (0) = 0 and f (x)· f ′ (x)+ f ′′ (x) = g ′ (x), it is enough to prove that there exists

a real number 0 < η ≤ 1 such that g( η ) = 0. We first assume that f has no zero. Set

h (x) =

f (x)

Because h (0) = h(1) = −1/2, there exists a real number 0 < η < 1 such that

h ′ ( η ) = 0. But g = f 2 ·h ′ , and we are done. Next, we suppose that f has at least one zero. Since the set of the zeros of a function is closed, we may assume that z 1 is the smallest zero of f and z 2 is the

232 5 Differentiability largest one. Thus, 0 <z 1 ≤z 2 < 1. The function f is positive on the intervals [0, z 1 )

and (z 2 , 1]. Thus, f ′ (z 1 ) ≤ 0 and f ′ (z 2 ) ≥ 0.

Then g (z 1 )=f ′ (z 1 ) ≤ 0 and g(z 2 )=f ′ (z 2 ) ≥ 0, and there exists η ∈ [z 1 ,z 2 ] such that g ( η ) = 0. ⊓ ⊔

be a twice continuously differentiable function such that

5.3.22. Let f : (0, ∞) → R

|f ′′ (x) + 2x f ′ (x) + (x 2 + 1) f (x)| ≤ 1

for all x . Prove that lim x →∞ f (x) = 0 .

Solution. We first point out that l’Hˆopital’s rule is valid if the denominator con- verges to infinity, without any assumption on the numerator. Thus, by l’Hˆopital’s

rule applied twice to the fraction f (x)e x 2 /2 /e x 2 /2 , we deduce that

x (x) + x f (x))e 2 /2 lim f (x) = lim

f 2 (x)e x /2

x (f 2 ′′ (x) + 2x f ′ (x) + (x 2 + 1) f (x))e /2 = lim

The following alternative solution uses just elementary knowledge of integral calculus. Indeed, we first set g (x) = f ′ (x) + x f (x); hence

f ′′ (x) + 2x f ′ (x) + (x 2 + 1) f (x) = g ′ (x) + xg(x) . We claim that if h is a continuously differentiable function such that h ′ (x)+ xh(x)

is bounded, then lim x →∞ h (x) = 0. Applying this lemma for h = g then for h = f , the statement follows. In order to prove our claim, let M be an upper bound of

|h ′ (x) + xh(x)| and let p(x) = h(x)e x 2 /2 . Then |p /2 ′ (x)| = |h ′ x 2 /2 x (x) + xh(x)|e 2 ≤ Me

and

# p 2 (x) p (0) + x x |p(0)| + M t

0 e dt |h(x)| = x 2 /2

0 p ′ (t)dt

e 2 e /2 e x /2 .

Since lim 2

# x →∞

e x /2

= ∞ and, by l’Hˆopital’s rule, lim x →∞ 0 e t 2 /2 dt /e x 2 /2 = 0, we conclude that lim x →∞ h (x) = 0. ⊓ ⊔

5.3.23. Let f (x)

be a continuously differentiable real-valued function on R . Show that if f ′ (x) 2 +f 3 (x)→0 as x →+∞ , then f (x)→0 as x →+∞ .

5.3 The Main Theorems 233 Solution. There are three cases to be considered.

(i) Suppose f (x) changes sign at x n ,x n → + ∞. Then f (x) has a maximum or min- imum at ξ n ,x n < ξ n <x n +1 ,f ′ ( ξ n ) = 0, | f (x)| ≤ | f ( ξ n )| for x n ≤x≤x n +1 , |f( ξ n )|→0, f (x)→0.

(ii) f (x) does not change sign for x ≥ u, say f (x) ≥ 0 for x ≥ u. Then f (x)→0 for x →∞. (iii) f (x) ≤ 0 for x ≥ u. Set g = − f , then g ′2

−g 3 →0, and since (g ′2 −g 3 ) is arbitrarily small for sufficiently large x, g (x) differs arbitrarily little, for suffi-

ciently large x, from h (x) for which h ′2 −h 3 = 0, or (h ′ −h 3 /2 )(h ′ +h 3 /2 ) = 0. If h ′ +h 3 /2

′ +h 3 /2

1 , so h ′ −

h 3 /2 = 0 on I 1 . If I is finite, then there is an abutting interval I 2 on which

h ′ +h 3 /2 = 0. Thus, h(x) = (−1/2x + c 1 ) −2 on I 1 and h (x) = −(1/2x +

c 2 ) −2 on I 2 . Also h ′ (x) = (−1/2x + c 1 ) −3 on I 1 ,h ′ (x) = −(1/2x + c 2 ) −3 on I 2 . At the point x = a where I 1 and I 2 abut, we have (−1/2a + c 1 ) −3 = −(1/2a + c 2 ) −3 ; hence c 1 = −c 2 and h (x) = (1/2x + c) −2 on I 1 ∪I 2 . It fol- lows that h (x) = (1/2x + c) −2 for all x; hence h (x)→0, g(x)→0, f (x)→0 as x → + ∞. ⊓ ⊔