Let f :R → [0,∞) be a continuously differentiable function. Prove that

10.2.6. Let f :R → [0,∞) be a continuously differentiable function. Prove that

f 3 2 1 (x) dx − f 1 (0) f (x) dx ≤ max ′ .

0 |f (x)|

f (x) dx

0 0 ≤x≤1

Solution. Let M = max 0 ≤x≤1 |f ′ (x)|. We have −M f (x) ≤ f (x) f ′ (x) ≤ M f (x), for all x ∈ [0,1].

Therefore " x

f 2 2 (t) dt ≤ x f (x) − f (0) ≤ M f (t) dt, for all x ∈ [0,1].

−M

We deduce that " x

1 1 " x −M f (x)

f (t) dt ≤ f 3 (x) − f 2 (0) f (x) ≤ M f (x)

f (t) dt .

10.2 Integral Inequalities 379 Integrating the last inequality on [0, 1], it follows that " 1 2 " 1 " 1 " 1 2

which concludes the proof. ⊓ ⊔

be a function of class C 1 with f (0) = 0 . Show that

10.2.7. (Poincar´e’s inequality, the smooth case) . Let f : [0, 1]→R

" 1 1 /2 sup | f (x)| ≤

(f ′ (x)) 2 dx

0 ≤x≤1

Solution. Let x ∈ [0,1]. Using f (0) = 0 and applying the Cauchy–Schwarz inequality, we obtain

| f (x)| =

f ′ (t)dt

1 /2 " x

2 ≤ 2 |f (t)| dt 1 dt

|f ′

(t)| 2 dt

which concludes the proof. ⊓ ⊔

10.2.8. (Poincar´e’s inequality, the L p variant) . Let p ≥1 be a real number, a , b real numbers, and let f : (a, b)→R be an arbitrary function such that f (0) = 0 and

a | f (x)| p dx <∞ . Prove that there exists a positive constant C depending only on

b −a such that

| f (x)| dx ≤C

|f ′ (x)| dx ,

0 (a, b) . Solution. By the Cauchy–Schwarz inequality we obtain

for all ,p f ∈W 1

" b | f (x)| = | f (x) − f (a)| =

f ′ (t)dt ≤

|f ′ (t)|dt,

for any x ∈ (a,b). So, by the H¨older inequality,

p dx | f (x)| ′ ≤ |f (t)|dt

(x)| p dx

380 10 Applications of the Integral Calculus

= (b − a) p · |f (x)| dx .

p −1

The conclusion follows with C = (b − a) p −1 . ⊓ ⊔

10.2.9. Let f : [0, 1]→R

be a function of class C 1 such that f (1) = 0 . Prove that

f 2 (x)dx ≤ 2

x 2 f 2 (x)dx ·

f ′2 (x)dx .

Solution. Integrating by parts and using f (1) = 0, we obtain

f 2 (x)dx =

x ′ f 2 (x)dx = −

2x f (x) f ′ (x)dx.

Applying the Cauchy–Schwarz inequality, we have

xf (x) f ′ (x)dx ≤

x 2 f 2 (x)dx

f ′2 (x)dx . ⊓ ⊔

be a continuous function and consider K >0 such that

10.2.10. Let f : [0, a]→[0,∞)

f (t) ≤ K

f (s)ds, for all t ∈ [0,a].

Prove that f ≡0 . Solution. The most direct argument is based on iteration. By hypothesis and a

recurrence argument we obtain K n " t

f (t) ≤ (t − s) n −1 f (s)ds, ∀t ∈ [0,a], ∀n ≥ 1. (n − 1)! 0

Let M = max t ∈[0,a] f (t). Hence MK n " t

Mt n K n

M (aK) n

f (t) ≤ n (t − s) −1 ds =

−→ 0, ∀t ∈ [0,a],

(n − 1)! 0 n !

which shows that f (t) = 0, for all t ∈ [0,a]. The following alternative proof can be used in similar situations. Let

F (t) = K

f (s)ds .

By hypothesis, we have

F ′ (t) = K f (t) ≤ KF(t).

Hence

−Kt

F ′ (t) ≤ 0.

10.2 Integral Inequalities 381 Thus, the mapping t

−Kt F (t) is decreasing in [0, a]. Since F(0) = 0, it follows that F (t) ≤ 0. On the other hand, f (t) ≥ 0, so F(t) ≥ 0. Therefore F(t) ≡ 0, that is,

f ≡ 0. ⊓ ⊔

be a continuous function. Prove that for any x ∈R and λ >0 there exists a unique real y such that for all z ∈R ,

10.2.11. Fix a real number p >1 and let f :R →R

" x +λ

|y − f (t)| p dt ≤ |z − f (t)| dt .

x −λ

x −λ

Ecole Normale Sup´erieure, Paris, 2003 ´

Solution. For any fixed (x, λ ) ∈ R × (0,+∞) we define the continuous function

g # (z) = x +λ x −λ |z− f (t)| p dt . Let us observe that for any z ∈ R with |z| ≥ max{| f (t)|; x− λ ≤t≤x+ λ } =: M we have g(z) ≥ 2 λ (|z| − M) p , which implies g (z)→ + ∞ as

|z|→ + ∞. Since g is continuous, it follows that g achieves its minimum. On the other hand, by the strict convexity of the mapping

we deduce that g is strictly convex, too. In particular, this implies that the minimum point of g is unique.

The important inequality we prove in what follows does not belong to the the- ory of linear differential equations per se, but is very useful to establish the global existence of solutions of homogeneous differential linear systems. ⊓ ⊔

10.2.12. (Gronwall’s inequality, the differential form) (a) Consider the differ- entiable function f : [0, T ]→[0,∞) satisfying

f ′ (x) ≤ a(x) f (x) + b(x), ∀x ∈ [0,T ],

where a , b : [0, T ]→[0,∞) are continuous functions. Prove that

(x) ≤ e

f 0 f (t)dt

f (0) +

f (t)dt , ∀x ∈ [0,T ].

(b) In particular, if f ′ (x) ≤ a(x) f (x) pe [0, T ] and f (0) = 0 , prove that f ≡0 in [0, T ] .

Solution. (a) By hypothesis it follows that

f # (x)e 0 0 x a =e (t)dt (f ′ − (x) − a(x) f (x)) ≤ e 0 x a (t)dt b (x), for any x ∈ [0,T ]. Hence

− # x a (t)dt ′

f (x)e

a (t)dt

e − # 0 s a ≤ f (0) + (t)dt b (s)ds ≤ f (0) +

b (s)ds,

which concludes the proof. ⊓ ⊔ (b) Follows from (a).

382 10 Applications of the Integral Calculus

10.2.13. (Gronwall’s inequality, the integral form) (a) Consider the continuous function f : [0, T ]→[0,∞) satisfying

f (x) ≤ C 1 f (t)dt + C 2 , ∀x ∈ [0,T ],

where C 1 , C 2 ≥0 are constants. Prove that

f C 1 (x) ≤ C x 2 1 x e ∀0 ≤ x ≤ T. (b) In particular, if

f (x) ≤ C 1 f (t)dt, ∀x ∈ [0,T ],

prove that f ≡0 ˆın [0, T ] .

0 ≤C 1 g +C 2 ˆın [0, T ]. Applying the Gronwall inequality in differential form, we obtain

Solution. (a) Let g # x (x) = f (t)dt. Then g ′

g (x) ≤ e C 1 x (g(0) + C 2 x )=C 2 x e C 1 x . Using now the hypothesis, we obtain

f (x) ≤ C 1 g (x) + C 2 ≤C 2 (1 + C 1 x e C 1 x ). ⊓ ⊔ Independent Study. (i) What can we say about a continuous function f :

R →[0,∞) if

f (x) ≤

f (t)dt for all x ∈ [a,b]?

(ii) Assume that for all t ∈ [a,b],

f (x) ≤ C 1 (x − a) +C 2 f (t)dt + C 3 ,

where f is a nonnegative continuous function on [a, b], and C 1 ≥ 0, C 2 > 0, and

C 3 ≥ 0 are constants. Prove that for all x ∈ [a,b],

f 1 C (x) ≤ 1 +C C 2

C 3 e (x−a) −

The following inequality was proved by G.H. Hardy in 1920. He was looking for a nice proof of Hilbert’s double series inequality. Some six years later, Landau was able to supply the best constant in this inequality. The interest illustrated for this result is reflected by the fact that twenty papers giving new proofs or modify- ing slightly the inequality appeared before 1934. One of these modified inequalities is the Carleman inequality, which in some sense is a limiting version of Hardy’s

10.2 Integral Inequalities 383 inequality. We also point out that special cases of the following result become

Poincar´e’s inequality, Friedrich’s inequality, or the weighted Sobolev’s inequalities. All these inequalities have many applications in embedding theories and in differen- tial equations theory, both ordinary and partial. We state below the Hardy inequality in its simplest and original form.

10.2.14. (Hardy’s inequality) . Let p >1 and let f : [0, ∞)→R be a differentiable

increasing function such that # f (0) = 0 and ∞

0 (f ′ (x)) p dx is finite. Show that

x −p | f (x)| p dx ≤ (f ′ (x)) p dx ,

0 p −1

and equality holds if and only if f vanishes identically. Moreover, the best constant is given by p p (p − 1) −p .

Solution. If f vanishes identically, then equality holds. So, we will assume that

0 g n (t)dt. Let A be sufficiently large that f ′

f # n = min{ f ′ , n}, G n (x) = x

is satisfied by g n and G n . After integration by parts, we obtain " X n p (x)

dx

=− 1 −p ′

0 n (x)(x

g (x)dx

p " X n p (x) −1

g n (x)dx .

p −1 0 x

We have used above the fact that G n (x) = o(x) for x→0. So, by H¨older’s inequality, " X p

g p (x)dx . (10.3)

0 x p −1 n

Since the left-hand side is positive and finite, it follows from (10.3) that

n (x)

dx

≤ p g (x)dx.

p −1

Passing to the limit as n →∞, it follows that

Next, taking X →∞, we obtain " ∞

dx p (x) dx . (10.4)

p −1

384 10 Applications of the Integral Calculus This is exactly the Hardy inequality in the weak form, that is, for < replaced

by Indeed, taking n →∞ and X→∞ in (10.3), we obtain

1 /p dx

1 /p ′ " ∞

(f ′ (x)) ≤ p dx . (10.5)

dx

0 x p −1

Using now (10.4) in the right-hand side of (10.5), we deduce that " ∞

dx p < (x) dx ,

p −1 with equality if and only if x −p f p (x) and ( f ′ (x)) p are proportional. This situation

is impossible, since it would imply that f # ′ is a certain power of x, so ∞

0 (f ′ (x)) p dx diverges, contradiction. The fact that p p (p − 1) −p is the best constant in Hardy’s inequality follows by

choosing the family of functions if 0 ≤ x ≤ 1,

f ε (x) =

p −1−pε x (p−1−pε)/p −1 , if x > 1, for ε > 0 sufficiently small. The following inequality due to Hardy is closely related to the uncertainty prin-

ciple. Let f : [a, b]→R be a continuously differentiable function such that f (a) =

f (b) = 0. Prove that

where d (x) = min{|x − a|,|x − b|}. It is sufficient to prove that

where 2c = a + b, and a similar inequality for the other half-interval. Without loss of generality, we consider only the case a = 0. We then have

′ . 2 c 1 /2 −1/2 ′ f (x) 2

f (x) dx =

x (x

f (x)) +

x −1/2 f (x)(x −1/2 f (x)) ′

2 0 dx 4x

10.2 Integral Inequalities 385

(x −1/2 f (x)) 2 + (x)

as required. ⊓ ⊔ Comments. The N-dimensional version of the Hardy’s inequality can be stated

as follows. For this purpose, if Ω is a region in R and u ∈R N such that |u| = 1, we define

This definition implies that 0 ≤d u (x) ≤ +∞ and d u Ω . Defining the vectors

e 1 = (1, 0, . . . , 0), e 2 = (0, 1, . . . , 0), . . . , e 2 = (0, 0, . . . , 1), we put d i (x) = d e i (x), for any 1 ≤ i ≤ n.

be an arbitrary open set in R N and assume that f is a continuously differ- entiable real-valued function with compact support in Ω . Then

Let Ω

f 2 (x)

2 dx ≤

|∇ f (x)| 2 dx , (10.7)

i =1 Ω 4d i (x)

where ∇ f (x) = ( f x 1 (x), . . . , f x N (x)).

Indeed, applying Hardy’s inequality (10.6), we deduce that for any 1 ≤ i ≤ n,

f 2 (x)

2 dx ≤

|f x i (x)| dx .

Ω 4d i (x)

By summation from i = 1 to N we deduce (10.7). Remark. Opic and Kufner [87] were looking at higher-dimensional versional of

Hardy’s inequality. If Ω is a bounded set in R N , they established the inequality

1 /p q

1 /q

| f (x)| v (x)dx

≤C ∑

(x) w (x)dx .

i =1 Ω ∂ x i

Here the constant C is proved to exist if and only if (sample result for N = 1 and Ω = (a, b))

1 /p ′ sup

" b 1 /q " x

v (t)dt

w (t) 1 −p ′ dt

x ∈(a,b)

where p ′ denotes the conjugate exponent of p.

386 10 Applications of the Integral Calculus