Advanced book on Mathematics Olympiad

Real Analysis 589

It now becomes natural to letg(x)=f(x)x , which satisfies the equation

g(xy)=g(x)+g(y).

The particular casex =yyieldsg(x)=^12 g(x^2 ), and henceg(−x)=^12 g((−x)^2 )=
1
2 g(x

(^2) )=g(x). Thus we only need to consider the casex>0.
Note thatgis continuous on( 0 ,∞). If we composegwith the continuous function
h:R→( 0 ,∞),h(x)=ex, we obtain a continuous function onRthat satisfies Cauchy’s
equation. Henceg◦his linear, which then impliesg(x)=logaxfor some positive base
a. It follows thatf(x)=xlogaxforx>0 andf(x)=xloga|x|ifx<0.
All that is missing is the value off at 0. This can be computed directly setting
x=y=0, and it is seen to be 0. We conclude thatf(x)=xloga|x|ifx =0, and
f( 0 )=0, whereais some positive number. The fact that any such function is continuous
at zero follows from
lim
x→ 0 +
xlogax= 0 ,
which can be proved by applying the L’Hôpital’s theorem to the functions logaxand^1 x.
This concludes the solution.
549.Settingy =z=0 yieldsφ(x)=f(x)+g( 0 )+h( 0 ), and similarlyφ(y)=
g(y)+f( 0 )+h( 0 ). Substituting these three relations in the original equation and letting
z=0 gives rise to a functional equation forφ, namely
φ(x+y)=φ(x)+φ(y)−(f ( 0 )+g( 0 )+h( 0 )).
This should remind us of the Cauchy equation, which it becomes after changing the
functionφtoψ(x)=φ(x)−(f ( 0 )+g( 0 )+h( 0 )). The relationψ(x+y)=ψ(x)+ψ(y)
together with the continuity ofψshows thatψ(x)=cxfor some constantc. We obtain
the solution to the original equation
φ(x)=cx+α+β+ γ, f(x)=cx+α, g(x)=cx+β, h(x)=cx+γ,
whereα, β, γare arbitrary real numbers.
(Gazeta Matematic ̆a(Mathematics Gazette, Bucharest), proposed by M. Vlada)
550.This is a generalization of Cauchy’s equation. Trying small values ofn, one can
guess that the answer consists of all polynomial functions of degree at mostn−1 with
no constant term (i.e., withf( 0 )=0). We prove by induction onnthat this is the case.
The casen=2 is Cauchy’s equation. Assume that the claim is true forn− 1
and let us prove it forn. Fixxnand consider the functiongxn :R→R,gxn(x)=
f(x+xn)−f(x)−f(xn). It is continuous. More importantly, it satisfies the functional
equation forn−1. Hencegxn(x)is a polynomial of degreen−2. And this is true for
allxn.