The Real And Complex Number Systems

Proof: Given a partition P on a, x, then we have
n
k1
|f k | |f k | |f k |
kAP
kAP
kBP
f k |f k |,
kBP
which implies that (taking supermum)
Vx px nx. *
Remark: The existence of px and qx is clear, so we know that (*) holds by
Theorem 1.15.
(b) 0 px Vx and 0 nx Vx.
Proof: Consider a, x, and since
n
Vx |f k | |f k |,
k1 kAP
we know that 0 px Vx. Similarly for 0 nx Vx.
(c) p and n are increasing on a, b.
Proof: Letx, y in a, b with x y, and consider py px as follows. Since
py f k f k ,
kAP, a,y kAP, a,x
we know that
py px.
That is, p is increasing on a, b. Similarly for n.
(d) fx fa px nx. Part (d) gives an alternative proof of Theorem 6.13.
Proof: Consider a, x, and since
which implies that
fx fa
k1
n
f k f k f k
kAP kBP
fx fa |f k | f k
kBP kAP
which implies that fx fa px nx.
(e) 2px Vx fx fa, 2nx Vx fx fa.
Proof: By (d) and (a), the statement is obvious.
(f) Every point of continuity of f is also a point of continuity of p and of n.
Proof: By (e) and Theorem 6.14, the statement is obvious.
Curves
6.8 Let f and g be complex-valued functions defined as follows:
ft e 2it if t 0, 1, gt e 2it if t 0, 2.
(a) Prove that f and g have the same graph but are not equivalent according to defintion