The Real And Complex Number Systems

f
x 1 ...x n
n
f
x 1 ...x n1
n x n
fx n
fx 1 ...fx n
n by induction hypothesis.
So, we have
f
x 1 ...x n1
fx 1 ...fx n1
.
n 1
n 1
Hence, we have proved (**). Given a rational number m/n 0, 1, where
g. c. d. m, n 1; we choose x : x 1 ... x m ,andy : x m1 ... x n , thenby(**),we
finally have
f mx n my
n n mfx n mfy
n n m n fx 1 m n
fy. ***
Given 0, 1, then there is a sequence q n Q such that q n as n . Then
by continuity and (***), we get
fx 1 y fx 1 fy.
Remark: The Reverse Induction is that let S N and S has two properties:(1) For
every k 0, 2 k S and (2) k S and k 1 N, then k 1 S. Then S N.
(Lemma) Letf be a convex function on a, b, then f is bounded.
Proof: LetM maxfa, fb, then every point z I, write z a 1 b, we
have
fz fa 1 b fa 1 fb M.
In addition, we may write z ab t, wheret is chosen so that z runs through a, b. So,
2
we have
which implies that
2f
f
a b
2
a b
2
1 2 f
f
a b
2
a b
2
t 1 2 f a b
2
t f a b
2
t
t
fz
which implies that
2f a b M : m fz.
2
Hence, we have proved that f is bounded above by M and bounded below by m.
(Theorem) Iff : I R is convex, then f satisfies a Lipschitz condition on any closed
interval a, b intI. In addition, f is absolutely continuous on a, b and continuous on
intI.
Proof: We choose 0 so that a , b intI. By preceding lemma, we
know that f is bounded, say m fx M on a , b . Given any two points x, andy,
with a x y b We consider an auxiliary point z y , and a suitable
then y z 1 x. So,
fy fz 1 x fz 1 fx fz fx fx
which implies that
fy fx M m y x
Change roles of x and y, we finally have
M m.
yx
yx ,