The Real And Complex Number Systems

lim
r n0
fx f lim n
x n
lim n
fx n by continuity of f
lim n
f1 xn by (**)
f1 x
e cx , where log f1 c.
Note: (i) We can prove (2) by the exercise as follows. Note that fx 0 for all x by
the remark (1)-(ii) Consider the composite function gx log fx, then
gx y log fx y log fxfy log fx log fy gx gy. Since log and f
are continuous on R, its composite function g is continuous on R. Use the exercise, we
have gx cx for some c. Therefore, fx e gx e cx .
(ii) We can prove (2) by the remark (1) as follows. It suffices to show that this f is
differentiable at 0 by remark (1) and (1)-(iii). Sincef m n f1 m n
then for every real r,
fr f1 r a
by continuity of f. Note that lim r b r
r0 r exists. Given any sequence r n
with r n 0, and thus consider
fr n f0
1
1
r n
f1rn r n
f1rn rn
r n
exists,
we have f is differentiable at x 0. So,byremark(1),wehavefx e cx .
(3) Give an example such that f is not continuous on R.
Solution: Consider gx y gx gy for all x, y. Then we have gq qg1,
where q Q. ByZorn’s Lemma, we know that every vector space has a basis
v : I. Note that v : I is an uncountable set, so there exists a convergent
sequence s n v : I. Hence, S : v : I s n
n1
sn
n
n1
is a new
basis of R over Q. Givenx, y R, and we can find the same N such that
N
x
k1
N
q k v k and y p k v k ,wherev k S
k1
Define the sume
N
x y : p k q k v k
k1
By uniqueness, we define gx to be the sum of coefficients, i.e.,
N
gx : q k .
k1
Note that
g
s n
1 for all n lim n
g
and
s n 0asn
Hence, g is not continuous at x 0 since if it was, then
s n 1