The Real And Complex Number Systems

|fy fx| K|y x|, whereK M m .
That is, f satisfies a Lipschitz condition on any closed interval a, b.
We call that f is absolutely continuous on a, b if given any 0, there is a 0
n
such that for any collection of a i , b i i1
of disjoint open intervals of a, b with
n
b i a i , wehave
k1
n
|fb i fa i | .
k1
Clearly, the choice /K meets this requirement. Finally, the continuity of f on intI is
obvious.
(Theorem) Letf be a differentiable real function defined on a, b. Prove that f is
convex if and only if f is monotonically increasing.
Proof: () Suppose f is convex, and given x y, we want to show that f x f y.
Choose s and t such that x u s y, then it is clear that we have
fu fx
u x
fs fu
s u
Let s y , wehaveby(*)
fu fx
u x f y
fy fs
y s . *
which implies that, let u x
f x f y.
() Suppose that f is monotonically increasing, it suffices to consider 1/2, if
x y, then
fx fy
2
f
x y
2
fx f
xy
2
fy f
xy
2
2
x y
4 f 1 f 2 , where 1 2
0.
Similarly for x y, and there is nothing to prove x y. Hence, we know that f is convex.
(Corollary 1) Assume next that f x exists for every x a, b, and prove that f is
convex if and only if f x 0 for all x a, b.
Proof: () Suppose that f is convex, we have shown that f is monotonically increasing.
So, we know that f x 0 for all x a, b.
() Suppose that f x 0 for all x a, b, it implies that f is monotonically
increasing. So, we know that f is convex.
(Corollary 2) Let0 , then we have
|y 1 | ...|y n|
n
1/
|y 1 | ...|y n|
n
Proof: Letp 1, and since x p pp 1x p2 0 for all x 0, we know that
fx x p is convex. So, we have (let p )
by
x 1 ...x n
n
1/
/
x 1 / ...x n
/
n *