Advanced book on Mathematics Olympiad

156 3 Real Analysis

3.2.10 Inequalities for Integrals...................................

A very simple inequality states that iff :[a, b]→Ris a nonnegative continuous
function, then
∫b

a

f(x)dx≥ 0 ,

with equality if and only iffis identically equal to zero. Easy as this inequality looks,

its applications are often tricky. This is the case with a problem from the 1982 Romanian

Mathematical Olympiad, proposed by the second author of the book.

Example.Find all continuous functionsf:[ 0 , 1 ]→Rsatisfying

∫ 1

0

f(x)dx=

1

3

+

∫ 1

0

f^2 (x^2 )dx.

Solution.First, we would like the functions in both integrals to have the same variable.
A substitution in the first integral changes it to

∫ 1

0 f(x

(^2) ) 2 xdx. Next, we would like to
express the number^13 as an integral, and it is natural to choose

∫ 1

0 x

(^2) dx. The condition
from the statement becomes
∫ 1
0
2 xf (x^2 )dx=

∫ 1

0

x^2 +

∫ 1

0

f^2 (x^2 )dx.

This is the same as

∫ 1

0

[f^2 (x^2 )− 2 xf (x^2 )+x^2 ]dx= 0.

Note that the function under the integral,f^2 (x^2 )− 2 xf (x^2 )+x^2 =(f (x^2 )−x)^2 ,isa

perfect square, so it is nonnegative. Therefore, its integral on[ 0 , 1 ]is nonnegative, and

it can equal zero only if the function itself is identically zero. We find thatf(x^2 )=x.

Sof(x)=

√

xis the unique function satisfying the condition from the statement. 

473.Determine the continuous functionsf:[ 0 , 1 ]→Rthat satisfy

∫ 1

0

f (x)(x−f (x))dx=

1

12

.

474.Letnbe an odd integer greater than 1. Determine all continuous functionsf :

[ 0 , 1 ]→Rsuch that

∫ 1

0

(

f

(

x

(^1) k))n−k
dx=
k
n
,k= 1 , 2 ,...,n− 1.