Prove that there does not exist a positive continuously differentiable func-
5.5.1. Prove that there does not exist a positive continuously differentiable func-
tion f on [0, ∞) such that f ′ (x) ≥ f ( f (x)) , for all x ≥0 . Solution. Assume that f is such a function. Since f ′ (x) ≥ f ( f (x)) > 0, f
is increasing, so that f ′ (x) ≥ f ( f (x)) ≥ f ( f (0)) > 0. This means that f ′ (x) is bounded away from 0, and so lim x → ∞ f (x) = ∞. Hence lim x → ∞ f ( f (x)) = ∞, and so lim x → ∞ f ′ (x) = ∞. Set g(x) := f (x) −x−1. Then g ′ (x) also approaches ∞, which implies that g (x) approaches ∞. Thus there exists some x 0 such that f (x 0 )>x 0 + 1. Next, applying the mean value theorem to the interval [x 0 , f (x 0 )], we obtain a point ξ ∈ (x 0 , f (x 0 )) for which
f ( f (x 0 )) = f (x 0 )+f ′ ( ξ ) ( f (x 0 )−x 0 )>f ′ ( ξ ) ( f (x 0 )−x 0 ) ≥f(f( ξ )) ( f (x 0 )−x 0 ) > f ( f (x 0 )) ( f (x 0 )−x 0 ) > f ( f (x 0 )),
which is a contradiction. ⊓ ⊔ The following inequality asserts that if a twice differentiable function f : R →R is
bounded together with its second derivative, then the derivative of f is bounded, too. In 1932, the British mathematicians Godfrey Harold Hardy (1877–1947) and John Edensor Littlewood (1885–1977) [42] extended this inequality to larger classes of functions.
be a function of class C 2 . Ass- ume that both f and f ′′ are bounded and set
5.5.2. (Landau’s Inequality, [63]). Let f :R →R
M 0 = sup | f (x)|, M 2 = sup |f ′′ (x)|.
x ∈R
x ∈R
Prove that f ′ is bounded and, moreover,
5.5 Differential Equations and Inequalities 239
sup |f ′ (x)| ≤ 2 M
x ∈R
Solution. We first observe that if M 2 = 0, then the only functions satisfying our hypotheses are the constant mappings. So, we can assume without loss of generality that M 2 > 0. Let x ∈ R and fix arbitrarily h > 0. By Taylor’s formula, there is some t ∈ (x,x + 2h ) such that
f (x + 2h) = f (x) + 2h f ′ (x) + 2h 2 f ′′ (t), that is,
f (x + 2h) − f (x)
f (x) =
−hf ′′ (t).
2h
Taking the modulus, and applying our hypothesis we obtain |f 0 ′ M
(x)| ≤
+M 2 h for all x ∈ R.
h Choosing now h = (M 0 /M 2 ) 1 /2 , we obtain our conclusion. ⊓ ⊔
Remark. Applying Taylor’s formula twice, namely between x and both x ±h, we can obtain (exercise!) the better estimate
sup |f ′ (x)| ≤ 2M
x ∈R
We also point out that the above Landau inequality establishes that #f $ ′ #
L ∞ (R) ≤2 #f# L ∞ (R) ·#f ′′ # L ∞ (R) , where
denotes the norm in the Banach space L #·# ∞ L (R) (R) and is defined by
#u# L ∞ (R) : = sup |u(x)|.
x ∈R
The above L ∞ -inequality was established in the so-called L 2 framework by Hardy and Littlewood [42] as follows:
|f ′ (x)| 2 dx ≤2
| f (x)| 2 dx
|f ′′ (x)| 2 dx .
Hardy and Littlewood’s inequality has been given considerable space and even several proofs in the famous book [43], and has subsequently attracted further atten- tion.
A variant of the above inequality for higher-order derivatives is stated in what follows. This result is due to the Russian mathematician Andrey Nikolaevich Kolmogorov (1903–1987), who is known for his major contributions in probability
240 5 Differentiability theory and topology. The Landau and Kolmogorov inequalities are extended in what
follows.
be a function of class C 3 . Ass- ume that both f and f ′′′ are bounded and set
5.5.3. (Kolmogorov’s Inequality). Let f :R →R
M 0 = sup | f (x)|, M 3 = sup |f ′′′ (x)|.
x ∈R
x ∈R
(a) Prove that f ′ is bounded and, moreover,
(b) Is f ′′ bounded, too? Solution. (a) Fix x
x + h, we obtain, successively,
2 ≤M 3 6 . Therefore
|f (x)| ≤
=: ψ (h).
But ψ ′ M 0 M 3 (h) = − h
h 2 + 3 , which shows that ψ achieves its minimum for
1 h /3
0 M 3 , our conclusion follows. (b) By hypothesis and (a), the functions f ′ and f ′′′ =(f ′ ) ′′ are bounded. Applying
0 M 3 . Since ψ (h) = 2 −1
Landau’s inequality to these functions, we deduce that f ′′ is bounded. ⊓ ⊔
5.5 Differential Equations and Inequalities 241
be a nonconstant function of class C n such that both f and f (n) are bounded.
5.5.4. (Landau–Kolmogorov Generalized Inequalities). Let f :R →R
(a) Prove that f (n−1) is bounded. (b) Deduce that all the derivatives f (k) are bounded, 1 ≤k≤n−1 . For any integer 0 ≤k≤n , set M
k = sup x ∈R |f (x)| .
(k)
(c) Show that M k >0 , for any integer 0 ≤k≤n .
(d) Using the functions u k =2 k −1 M k M −1 k −1 , 0 ≤k≤n , as needed, prove that
k (n−k)/2 1 k M /n −k/n
k ≤2
Solution. (a) The Taylor formula implies
n −1 k j
∑ f ( j)
M n k n n −k−1 k
∑ (−1) C n −1 ∑ f (x) ≤ k ∑ C n −1 2M 0 +
n ! which implies
n −1−k C k k j
≤ ∑ n −1
C ∑ k 2M 0 +
n ! Using the elementary formula
if 1 ∑ ≤ j ≤ m − 1,
we observe that all the terms on the left-hand side of the above inequality are zero, excepting that corresponding to j = n − 1. It follows that for all x ∈ R we have
n |f (n−1)
n −1
(x)| ≤ ∑ C k n −1 2M 0 + n k .
k =1
(b) We use a standard argument by induction. (c) If M k = 0 then f is a polynomial of degree at most k − 1. Since f is bounded,
it follows that f must be constant, which is excluded by our hypotheses. So, M k > 0 for all 0 ≤ k ≤ n.
(d) Applying Landau’s inequality, we obtain M k ≤ 2M k −1 M k +1 for all 1 ≤k≤ n − 1. This inequality shows that u 1 ≤u 2 ≤ ··· ≤ u n , which yields
(u 1 u 2 ··· u k ) n ≤ (u 1 u 2 ···u n ) k .
242 5 Differentiability Applying this inequality in our case, we obtain
2 nk (k−1)/2 k
n ≤2 k M ,
kn (n−1)/2
which concludes the proof. ⊓ ⊔ The function g in the next example behaves like at least like a superlinear power
at +∞. So, the next problem involves a second-order differential inequality with superlinear growth in the nonlinear term.