510 Real Analysis
395.LetLbe the length of the trail andT the total duration of the climb, which is the
same as the total duration of the descent. Counting the time from the beginning of the
voyage, denote byf(t)andg(t)the distances from the monk to the temple at timeton
the first and second day, respectively. The functionsfandgare continuous; hence so is
φ:[ 0 ,T]→R,φ(t)=f(t)−g(t). It follows thatφhas the intermediate value property.
Becauseφ( 0 )=f( 0 )−g( 0 )=L− 0 =L>0 andφ(T )=f(T)−g(T )= 0 −L<0,
there is a timet 0 withφ(t 0 )=0. Att=t 0 the monk reached the same spot on both days.
396.The fact thatfis decreasing implies immediately that
x→−∞lim(f (x)−x)=∞ and xlim→∞(f (x)−x)=−∞.By the intermediate value property, there isx 0 such thatf(x 0 )−x 0 =0, that is,f(x 0 )=
x 0. The function cannot have another fixed point because ifxandyare fixed points, with
x<y, thenx=f(x)≥f(y)=y, impossible.
The triple(x 0 ,x 0 ,x 0 )is a solution to the system. And if(x,y,z)is a solution then
f(f(f(x)))=x. The functionf◦f◦fis also continuous and decreasing, so it has
a unique fixed point. And this fixed point can only bex 0. Therefore,x=y=z=x 0 ,
proving that the solution is unique.
397.The inequality from the statement implies right away thatfis injective, and also
thatftransforms unbounded intervals into unbounded intervals. The setsf((−∞, 0 ])
andf([ 0 ,∞))are unbounded intervals that intersect at one point. They must be two
intervals that cover the entire real axis.
(P.N. de Souza, J.N. Silva,Berkeley Problems in Mathematics, Springer, 2004)
398.Letxdenote the distance along the course, measured in miles from the starting line.
For eachx∈[ 0 , 5 ], letf(x)denote the time that elapses for the mile from the pointxto
the pointx+1. Note thatfdepends continuously onx. We are given that
f( 0 )+f( 1 )+f( 2 )+f( 3 )+f( 4 )+f( 5 )= 30.It follows that not all off( 0 ), f ( 1 ),...,f( 5 )are smaller than 5, and not all of them
are larger than 5. Choosea, b∈{ 0 , 1 ,..., 5 }such thatf(a)≤ 5 ≤f(b). By the
intermediatevalue property, there existscbetweenaandbsuch thatf(c)=5. The mile
betweencandc+1 was run in exactly 5 minutes.
(L.C. Larson,Problem-Solving Through Problems, Springer-Verlag, 1990)
399.Without loss of generality, we may assume that the cars traveled on one day from
AtoBkeeping a distance of at most one mile between them, and on the next day they
traveled in opposite directions in the same time interval, which we assume to be of length
one unit of time.
Since the first car travels in both days on the same road and in the same direction, it
defines two parametrizations of that road. Composing the motions of both cars during the