Let g : (0, +∞)→(0,+∞) be a continuous function such that

5.5.5. Let g : (0, +∞)→(0,+∞) be a continuous function such that

for some α >0 . Let f :R →(0,+∞)

be a twice differentiable function. Assume that there exist a >0 and x 0 ∈R such that

f ′′ (x) + f ′ (x) > ag( f (x)), for all x ≥x 0 . (5.15) Prove that lim x → +∞ f (x) exists, is finite and compute its value.

Vicent¸iu R˘adulescu, Amer. Math. Monthly, Problem 11024 Solution. If x 1 >x 0 is a critical point of f , then by (5.15), f ′′ (x 1 ) > 0, so x 1 is

a relative minimum point of f . This implies that f ′ (x) does not change sign if x is sufficiently large. Consequently, we can assume that f is monotone on (x 0 , +∞); hence ℓ := lim x → +∞ f (x) exists. The difficult part of the proof is to show that ℓ is finite. This will be deduced after applying in a decisive manner our superlinear growth assumption (5.14). Arguing by contradiction, let us assume that ℓ = +∞. In particular, it follows that f is monotone

increasing on (x 0 , +∞). Define the function

u (x) = e x /2 f (x), x ≥x 0 . Then u is increasing, and for any x ≥x 0 ,

u (x) + ae x /2 g ( f (x)). (5.16)

Our hypothesis (5.14) and the assumption ℓ = +∞ yield some x 1 >x 0 such that

g 1 ( f (x)) ≥ f +α (x), for all x ≥x 1 . (5.17) So, by (5.16) and (5.17),

1 u ′′ (x) > u (x) + Cu(x) f α (x) > Cu(x) f α (x), for all x ≥x 1 , (5.18)

5.5 Differential Equations and Inequalities 243 for some C > 0. In particular, since ℓ = +∞, there exists x 2 >x 1 such that

(5.19) We claim a little more, namely that there exists C 0 > 0 such that

u ′′ (x) > u(x), for all x ≥x 2 .

u ′′ (x) > C 0 u 1 +α/2 (x), for all x ≥x 2 . (5.20) Indeed, let us first choose 0 < δ < min{e −x 2 u (x 2 ), e −x 2 u ′ (x 2 )}. We prove that

u (x) > δ e x , for all x ≥x 2 .

(5.21) For this purpose, consider the function v (x) = u(x) − δ e x . Arguing by contradiction

and using v (x 2 ) > 0 and v ′ (x 2 ) > 0, we deduce the existence of a relative maximum point x 3 >x 2 of v. So, v (x 3 ) > 0, v ′ (x 3 ) = 0, and v ′′ (x 3 ) ≤ 0. Hence δ e x 3 =u ′ (x 3 ) < u(x 3 ). But by (5.19), u ′′ (x 3 ) > u(x 3 ), which yields v ′′ (x 3 ) > 0, a contradiction. This concludes the proof of (5.21). Returning to (5.18) and using (5.21), we obtain

u ′′ (x) > Cu 1 +α/2 (x)u α/2 (x)e −αx/2 >C u 1 0 +α/2 (x), for all x >x 2 , where C 0 =C δ α/2 . This proves our claim (5.20). So u ′ (x)u ′′ (x) > C u 1 0 +α/2 (x)u ′ (x), for all x >x 2 .

Hence

u ′2 (x) −C 1 u 2 +β (x) > 0, for all x >x 2 ,

2 where C 1 = 2C 0 /(4 + α ) and β = α /2 > 0. Therefore

u ′2 (x) ≥ C 2 +C 3 u 2 +β (x), for all x >x 2 , for some positive constants C 2 and C 3 . So, since u is unbounded, there exists x 3 >x 2

and C 4 > 0 such that u ′

4 u 1 (x), for all x >x 3 , where γ = β /2 > 0. Applying the mean value theorem, we obtain

(x) ≥ C +γ

u −γ (x 3 )−u −γ (x) = γ (x − x 3 )u −γ−1 ( ξ x )u ′ ( ξ x )≥C 4 γ (x − x 3 ), for all x >x 3 , where ξ x ∈ (x 3 , x). Taking x→+∞ in the above inequality, we obtain a contradiction,

since the left-hand side converges to u −γ (x 3 ) (because ℓ = +∞), while the right-hand side diverges to +∞. This contradiction shows that ℓ = lim x → +∞ f (x) must be finite. We prove in what follows that ℓ = 0. Arguing by contradiction, let us assume that ℓ > 0. We first observe that relation (5.15) yields, by integration,

f ′ (x) − f ′ (x 0 ) + f (x) − f (x 0 )≥a

g ( f (t))dt. (5.22)

244 5 Differentiability Since ℓ is finite, it follows by (5.22) that lim x → +∞ f ′ (x) = +∞. But this contradicts

the fact that lim x → +∞ f (x) is finite. ⊓ ⊔ Remark. The result stated in our problem does not remain true if g has linear

growth at +∞, so if (5.14) fails. Indeed, it is enough to choose f (x) = e x and g the identity map. We also remark that “ ℓ is finite” does not follow if the growth

hypothesis (5.14) is replaced by the weaker one lim x → ∞ g (x)/x = +∞. Indeed, if

g (x) = x ln(1 + x) and f (x) = e x 2 , then ℓ = +∞.

The next problem is inspired by the following classical framework. Consider the linear differential equation g ′′ + g = 0 on [0, ∞). All solutions of this equation are of

the form g (x) = C 1 cos x +C 2 sin x, where C 1 and C 2 are real constants. In particular, this implies that there are no solutions that are positive on the whole semiaxis [0, ∞). The purpose of this problem is to find a class of functions f such that f for all x > 0 and the nonlinear differential equation g ′′ + f ◦ g = 0 has no positive solution on the positive semiaxis.