Real Analysis 591the functional equation, we can findx 0 such thatf(x 0 ) =x 0. Rewrite the functional
equation asf(f(x))−f(x)=f(x)−x.Iff(x 0 )<x 0 , thenf(f(x 0 ))<f(x 0 ), and if
f(x 0 )>x 0 , thenf(f(x 0 ))>f(x 0 ), which both contradict the fact thatfis decreasing.
Thus any functionfthat satisfies the given condition is increasing.
Pick somea>b, and set f (a)=f(a)−aand f (b)=f(b)−b. By adding a
constant tof(which yields again a solution to the functional equation), we may assume
that f (a)and f (b)are positive. Composingfwith itselfntimes, we obtainf(n)(a)=
a+n f (a)andf(n)(b)=b+n f (b). Recall thatfis an increasing function, so
f(n)is increasing, and hencef(n)(a)>f(n)(b), for alln. This can happen only if
f (a)≥ f (b).
On the other hand, there existsmsuch thatb+m f (b)=f(m)(b)>a, and the
same argument shows that f ( f(m−^1 )(b)) > f (a). But f ( f(m−^1 )(b))= f (b),so
f (b)≥ f (a). We conclude that f (a)= f (b), and hence f (a)=f(a)−ais
independent ofa. Therefore,f(x)=x+c, withc∈R, and clearly any function of this
type satisfies the equation from the statement.
553.The answer is yes! We have to prove that forf(x)=ex
2
, the equationf′g+fg′=
f′g′has nontrivial solutions on some interval(a, b). Explicitly, this is the first-order
linear equation ing,
( 1 − 2 x)ex2
g′+ 2 xex2
g= 0.Separating the variables, we obtain
g′
g=
2 x
2 x− 1= 1 +
1
2 x− 1,
which yields by integration lng(x)=x+^12 ln| 2 x− 1 |+C. We obtain the one-parameter
family of solutions
g(x)=aex√
| 2 x− 1 |,a∈R,on any interval that does not contain^12.
(49th W.L. Putnam Mathematical Competition, 1988)
554.Rewrite the equationf^2 +g^2 =f′^2 +g′^2 as
(f+g)^2 +(f−g)^2 =(f′+g′)^2 +(g′−f′)^2.This, combined withf+g=g′−f′, implies that(f−g)^2 =(f′+g′)^2.
Letx 0 be the second root of the equationf(x)= g(x). On the intervalsI 1 =
(−∞, 0 ),I 2 =( 0 ,x 0 ), andI 3 = (x 0 ,∞)the functionf−gis nonzero; hence so
isf′+g′. These two functions maintain constant sign on the three intervals; hence
f−g=j(f′+g′)onIj, for somej∈{− 1 , 1 },j= 1 , 2 ,3.