Advanced book on Mathematics Olympiad

586 Real Analysis

g(x)=

1

3

g

(

x− 1

2

)

, for allx∈R.

Forx=−1 we haveg(− 1 )=^13 g(− 1 ); henceg(− 1 )=0. In general, for an arbitrary
x, define the recursive sequencex 0 =x,xn+ 1 = xn− 21 forn≥ 0. It is not hard to
see that this sequence is Cauchy, for example, because|xm+n−xm|≤ 2 m^1 − 2 max( 1 ,|x|).
This sequence is therefore convergent, and its limitLsatisfies the equationL=L− 21 .It
follows thatL=−1. Using the functional equation, we obtain

g(x)=

1

3

g(x 1 )=

1

9

g(x 2 )= ··· =

1

3 n

g(xn).

Passing to the limit, we obtaing(x)=0. This shows thatf(x)=x−^32 is the unique
solution to the functional equation.
(B.J. Venkatachala,Functional Equations: A Problem Solving Approach, Prism
Books PVT Ltd., 2002)

543.We will first show thatf(x)≥xfor allx. From (i) we deduce thatf( 3 x)≥ 2 x,
sof(x)≥^23 x. Also, note that if there existsksuch thatf(x)≥ kxfor allx, then

f(x)≥k

(^3) + 2
3 xfor allxas well. We can iterate and obtainf(x)≥knx, whereknare the
terms of the recursive sequence defined byk 1 =^23 , andkn+ 1 =k
n^3 +^2
3 fork≥1. Let us
examine this sequence.
By the AM–GM inequality,
kn+ 1 =
kn^3 + 13 + 13
3
≥kn,
so the sequence is increasing. Inductively we prove thatkn<1. Weierstrass’ criterion
implies that(kn)nis convergent. Its limitLshould satisfy the equation

L=

L^3 + 2

3

,

which shows thatLis a root of the polynomial equationL^3 − 3 L+ 2 =0. This equation
has only one root in[ 0 , 1 ], namelyL=1. Hence limn→∞kn=1, and sof(x)≥xfor
allx.
It follows immediately thatf( 3 x)≥ 2 x+f(x)for allx. Iterating, we obtain that
for alln≥1,

f( 3 nx)−f(x)≥( 3 n− 1 )x.

Therefore,f(x)−x ≤ f( 3 nx)− 3 nx. Ifweletn →∞and use (ii), we obtain
f(x)−x≤0, that is,f(x)≤x. We conclude thatf(x)=xfor allx>0. Thus the
identity function is the unique solution to the functional equation.
(G. Dospinescu)