Advanced book on Mathematics Olympiad

Real Analysis 543

We are left to compute the limit of this expression asngoes to infinity. Ifa≤1, this
limit is equal to 0. Ifa>1, the limit is

lim

n→∞

πlnn

√

a^2 n− 1

a^2 − 1

= 2 πlna.

(S.D. Poisson)

472.The conditionf(x)f( 2 x)···f (nx)≤ankcan be written equivalently as

∑n

j= 1

lnf(jx)≤lna+klnn, for allx∈R,n≥ 1.

Takingα>0 andx=αn, we obtain

∑n

j= 1

lnf

(

jα

n

)

≤lna+klnn,

or

∑n

j= 1

α

n

lnf

(

jα

n

)

≤

αlna+kαlnn

n

.

The left-hand side is a Riemann sum for the function lnfon the interval[ 0 ,α]. Because
fis continuous, so is lnf, and thus lnfis integrable. Lettingntend to infinity, we
obtain
∫ 1

0

lnf(x)dx≤nlim→∞

αlna+kαlnn

n

= 0.

The fact thatf(x)≥1 implies that lnf(x)≥0 for allx. Hence lnf(x)=0 for all
x∈[ 0 ,α]. Sinceαis an arbitrary positive number,f(x)=1 for allx≥0. A similar
argument yieldsf(x)=1 forx<0. So there is only one such function, the constant
function equal to 1.
(Romanian Mathematical Olympiad, 1999, proposed by R. Gologan)

473.The relation from the statement can be rewritten as
∫ 1

0

(xf (x)−f^2 (x))dx=

∫ 1

0

x^2

4

dx.

Moving everything to one side, we obtain
∫ 1

0

(

f^2 (x)−xf (x)+

x^2

4

)

dx= 0.