Let f : [0, 1] → R be a continuous function such that for any x , y ∈ [0,1] ,

10.2.3. Let f : [0, 1] → R be a continuous function such that for any x , y ∈ [0,1] ,

xf (y) + y f (x) ≤ 1.

(i) Show that

f (x)dx ≤ .

(ii) Find a function satisfying the condition for which there is equality.

376 10 Applications of the Integral Calculus Solution. (i) We observe that

f (x)dx =

f (sin θ ) cos θ d θ =

f (cos θ ) sin θ d θ .

So, by hypothesis,

2 f (x)dx ≤

1d θ = .

√ (ii) Let f (x) = 1 −x 2 . If x = sin θ and y = sin Φ , then

xf (y) + y f (x) = sin θ cos Φ + sin Φ cos θ = sin( θ + Φ ) ≤ 1. # √

On the other hand, 1 1 −x 0 2 dx = π /4. ⊓ ⊔

10.2.4. Let # f : [a, b]→R be a function of class C 1 such that b

a f (t)dt = 0 . (i) For any integer n ≥1 define the function f n (t) = cos(2 π nt ) . Prove that f n

satisfies the above hypotheses and compute

sup 0 |f n (t)| # dt

0 |f n ′ (t)| dt

n ≥1 1 2

(ii) Show that

f 2 2 (t)dt ≤ (b − a) b (f ′ (t)) 2 dt .

be a function of class C 1 such that g (a) = g(b) = 0 (in this case, # b

(iii) Let g : [a, b]→R

a g (t)dt is not necessarily zero). Prove that

" b (b − a) 2 "

(t)dt ≤

(g ′ (t)) 2 dt .

Solution. (i) The required value is 4 π −1 . (ii) Since f is continuous and # b

a f (t)dt =0, there exists c ∈ (a,b) such that f (c)=0. Hence

f (t) = f (c) +

f ′ (u)du =

f ′ (u)du, ∀t ∈ [a,b].

By the Cauchy–Schwarz inequality we obtain for all t ∈ [a,b], " t

1 /2 | f (t)| ≤ |

f ′ (u)du| ≤ |t − c|

(f ′ (u)) 2 du .

Therefore

| f (t)| 2 dt ≤

(f ′ (u)) 2 du

|t − c|dt.

10.2 Integral Inequalities 377 To evaluate the second term, we observe that # b |t −c| ≤ b−a, so

a |t −c|dt ≤ (b−a) 2 , and the proof is concluded. An alternative proof is based on the remark that

2 +b

|t − c|dt = c − c(a + b) +

This is a quadratic function in c, whose minimum is achieved at c = (a + b)/2. The maximum is attained either at a or at b, and this value equals (b − a) 2 /2. We conclude that

2 dt

(b − a)

| f (t)|

(f ′ (u)) 2 du .

(iii) We intend to apply the above result for f =g ′ . In this case, we would find # b

a f (t)dt = g(b) − g(a) = 0. However, this in not a good idea, since it does not yield our conclusion. We remark that the points a and b play the same role

as c. This implies that it is better to restart the proof, but considering the differ- ent cases t ≤ (b + a)/2 and t ≥ (b + a)/2.

If t ≤ (b + a)/2 we have

f (t) =

f ′ (u)du,

so

| f (t)| 2 ≤ (t − a) (f ′ (u)) 2 du ≤ (t − a)

" (b+a)/2

(f ′ (u)) 2 du .

By integration we obtain " (b+a)/2

2 (b − a) " (b+a)/2

| f (t)| 2 dt ≤

(f ′ (t)) 2 dt .

If t ≥ (b + a)/2, we obtain

| f (t)| 2 dt

(b − a)

(f ′ (t)) 2 dt .

(b+a)/2

8 (b+a)/2

Adding the above two inequalities, we obtain

" b 2 (b − a) " b

| f (t)| 2 dt ≤

(f ′ (t)) 2 dt ,

which concludes the proof. ⊓ ⊔ We give in what follows an elementary proof of the Young inequality in a partic-

ular case.

10.2.5. Let f : [0, +∞)→R

be a strictly increasing function with a continuous derivative such that f (0) = 0 . Let f −1 denote the inverse of f . Prove that for all

positive numbers a and b , with b < f (a) ,

ab <

f (x)dx +

f −1 (y)dy .

378 10 Applications of the Integral Calculus Solution. We first observe that if f is a strictly increasing function with a

continuous derivative, then integration by parts yields

f (x) dx = b f (b) − a f (a) − xf ′ (x) dx.

Let y = f (x), x = f −1 (y). We obtain

" b " f (b)

f (x) dx = b f (b) − a f (a) −

f −1 (y) dy. (10.2)

a f (a)

Now if u = f (a), v = f (b), relation (10.2) becomes " f −1 (v)

Thus, we can always express # f −1 (y) dy in terms of f (x) dx. Returning to our problem, assume that 0 < r < a. Then

f (x)dx < r f (r) −

f (x) dx .

Applying relation (10.2) to # r

0 f (x)dx, we obtain

" a " f (r)

af (r) −

f (x)dx <

f −1 (y) dy.

Since 0 < b < f (a), we can take r = f −1 (b), and our conclusion follows. ⊓ ⊔