Advanced book on Mathematics Olympiad

Real Analysis 601

∑n

i= 1

ln|P(x)−xi|=n^2 lnC|x|,

whereCis some positive constant. After adding the logarithms on the left we have

ln

∏n

i= 1

|P(x)−xi|=lnCn

2

|x|n

2

,

and so ∣

∣∣

∣∣

∏n

i= 1

(P (x)−xi)

∣

∣∣

∣∣=k|x|n

2

,

withk=Cn^2. Eliminating the absolute values, we obtain

P(P(x))=λxn

2

,λ∈R.

We end up with an algebraic equation. An easy induction can prove that the coefficient

of the term ofkth degree is 0 fork<n. HenceP(x)=axn, withasome constant, are

the only polynomials that satisfy the relation from the statement.

(Revista Matematica din Timi ̧soara ̆ (Timi ̧soara Mathematics Gazette), proposed by

T. Andreescu)

572.The idea is to use an “integrating factor’’ that transforms the quantity under the inte-

gral into the derivative of a function. We already encountered this situation in a previous

problem, and should recognize that the integrating factor ise−x. We can therefore write

∫ 1

0

|f′(x)−f(x)|dx=

∫ 1

0

|f′(x)e−x−f(x)e−x|exdx=

∫ 1

0

|(f (x)e−x)′|exdx

≥

∫ 1

0

(f (x)e−x)′|dx=f( 1 )e−^1 −f( 0 )e−^0 =

1

e

.

We have found a lower bound. We will prove that it is the greatest lower bound. Define

fa:[ 0 , 1 ]→R,

fa(x)=

{

ea−^1

a x forx∈[^0 ,a],

ex−^1 forx∈[a, 1 ].

The functionsfaare continuous but not differentiable ata, but we can smooth this

“corner’’ without altering too much the function or its derivative. Ignoring this problem,

we can write

∫ 1

0

|fa′(x)−fa(x)|dx=

∫a

0

∣∣

∣∣e

a− 1

a

−

ea−^1

a

x

∣∣

∣∣dx=e

a− 1

a

(

a−

a^2

2

)

=ea−^1

(

1 −

a

2

)

.

Asa→0, this expression approaches^1 e. This proves that^1 eis the desired greatest lower
bound.
(41st W.L. Putnam Mathematical Competition, 1980)