The Real And Complex Number Systems

1 lim n
g
s n
g lim
s n n by continuity of g at 0
g0
0
which is absurb. Hence, g is not continuous on R by the exercise. To find such f, it suffices
to consider fx e gx .
Note: Such g (or f) isnot measurable by Lusin Theorem.
4.19 Let f be continuous on a, b and define g as follows: ga fa and, for
a x b, letgx be the maximum value of f in the subinterval a, x. Show that g is
continuous on a, b.
Proof: Define gx maxft : t a, x, and choose any point c a, b, wewant
to show that g is continuous at c. Given 0, we want to find a 0 such that as
x c , c a, b, wehave
|gx gc| .
Since f is continuous at x c, then there exists a 0 such that as
x c , c a, b, wehave
fc /2 fx fc /2. *
Consider two cases as follows.
(1) maxft : t a, c a, b fp 1 ,wherep 1 c .
As x c , c a, b, wehavegx fp 1 and gc fp 1 .
Hence, |gx gc| 0.
(2) maxft : t a, c a, b fp 1 ,wherep 1 c .
As x c , c a, b, wehaveby(*)fc /2 gx fc /2.
Hence, |gx gc| .
So, if we choose ,thenforx c , c a, b,
|gx gc| by (1) and (2).
Hence, gx is continuous at c. And since c is arbitrary, we have gx is continuous on
a, b.
Remark: It is the same result for minft : t a, x by the preceding method.
4.20 Let f 1 ,...,f m be m real-valued functions defined on R n . Assume that each f k is
continuous at the point a of S. Define a new function f as follows: For each x in S, fx is
the largest of the m numbers f 1 x,...,f m x. Discuss the continuity of f at a.
Proof: Assume that each f k is continuous at the point a of S, then we have f i f j and
|f i f j | are continuous at a, where 1 i, j m. Sincemaxa, b ab|ab| , then
2
maxf 1 , f 2 is continuous at a since both f 1 f 2 and |f 1 f 2 | are continuous at a. Define
fx maxf 1 ,...f m ,useMathematical Induction to show that fx is continuous at
x a as follows. As m 2, we have proved it. Suppose m k holds, i.e., maxf 1 ,...f k is
continuous at x a. Then as m k 1, we have
maxf 1 ,...f k1 maxmaxf 1 ,...f k , f k1
is continuous at x a by induction hypothesis. Hence, by Mathematical Induction, we
have prove that f is continuos at x a.
It is possible that f and g is not continuous on R whihc implies that maxf, g is
continuous on R. For example, let fx 0ifx Q, andfx 1ifx Q c and gx 1