Advanced book on Mathematics Olympiad

Real Analysis 587

544.We should keep in mind thatf(x)=sinxandg(x)=cosxsatisfy the condition.
As we proceed with the solution to the problem, we try to recover some properties of sinx
and cosx. First, note that the conditionf(t)=1 andg(t)=0 for somet =0 implies
g( 0 )=1; hencegis nonconstant. Also, 0=g(t)=g( 0 )g(t)+f( 0 )f (t)=f( 0 );
hencefis nonconstant. Substitutingx=0 in the relation yieldsg(−y)=g(y),sog
is even.
Substitutingy=t, we obtaing(x−t)=f(x), with its shifted versionf(x+t)=
g(x). Sincegis even, it follows thatf(−x)=g(x+t). Now let us combine these facts
to obtain

f(x−y)=g(x−y−t)=g(x)g(y+t)+f(x)f(y+t)

=g(x)f (−y)+f (x)g(y).

Changeyto−yto obtainf(x+y)=f (x)g(y)+g(x)f (y)(the addition formula

for sine).

The remaining two identities are consequences of this and the fact thatfis odd. Let

us prove this fact. Fromg(x−(−y))=g(x+y)=g(−x−y), we obtain

f(x)f(−y)=f(y)f(−x)

for allxandyinR. Settingy=tandx=−tyieldsf(−t)^2 =1, sof(−t)=±1. The

choicef(−t)=1 givesf(x)=f(x)f(−t)=f(−x)f (t)=f(−x); hencefis even.

But then

f(x−y)=f(x)g(−y)+g(x)f (−y)=f (x)g(y)+g(x)f (y)=f(x+y),

for allxandy. Forx=z+ 2 w,y=z− 2 w, we havef(z)=f(w), and sofis constant, a

contradiction. Forf(−t)=−1, we obtainf(−x)=−f(−x)f (−t)=−f(x)f(t)=

−f(x); hencefis odd. It is now straightforward that

f(x−y)=f (x)g(y)+g(x)f (−y)=f (x)g(y)−g(x)f (y)

and

g(x+y)=g(x−(−y))=g(x)g(−y)+f(x)f(−y)=g(x)g(y)−f(x)f(y),

where in the last equality we also used the fact, proved above, thatgis even.

(American Mathematical Monthly, proposed by V.L. Klee, solution by P.L. Kannap-

pan)

545.Becausef(x)=f^2 (x/ 2 )>0, the functiong(x)=lnf(x)is well defined. It

satisfies Cauchy’s equation and is continuous; therefore,g(x)=αxfor some constant

α. We obtainf(x)=cx, withc=eα.