512 Real Analysis
B A
C D
xyFigure 67Remark.This result is known as the “pancake theorem.’’
402.Assume thatfis not continuous at some pointa. Then there exists>0 and a
sequencexn→asuch that|f(xn)−f(a)|>for alln≥1. Without loss of generality,
we may assume that there is a subsequence(xnk)ksuch thatf(xnk)<f(a), for allk,
in which casef(xnk)≤f(a)−. Chooseγ in the interval(f (a)−, f (a)). Sincef
has the intermediate value property, andf(xnk)<γ <f(a), for eachkthere existsyk
betweenxnkandasuch thatf(yk)=γ. The setf−^1 (γ )contains the sequence(yk)k,
but does not contain its limita, which contradicts the fact that the set is closed. This
contradiction proves that the initial assumption was false; hencefis continuous on the
intervalI.
(A.M. Gleason)
403.The function is continuous off 0, so it maps any interval that does not contain 0 onto
an interval. Any interval containing 0 is mapped onto[− 1 , 1 ], which proves thatfhas
the intermediate value property for anya∈[− 1 , 1 ].
For the second part of the problem, we introduce the function
F(x)={
x^2 sin^1 x forx = 0 ,
0 forx= 0.One can verify easily that
F′(x)={
2 xsin^1 x forx = 0 ,
0 forx= 0 ,+
{
cos^1 x forx = 0 ,
0 forx= 0.The only place where this computation might pose some difficulty isx =0, which
can be done using L’Hôpital’s theorem. The first function is continuous; hence it is the
derivative of a function. Because the differentiation operator is linear we find that the
second function, which isf 0 (x), is a derivative. And because whena =0,