The Real And Complex Number Systems

2n
h P 2 hx i hx i1
i1
n
2n
hx i hx i1 hx i hx i1
i1
in1
n
i1
n
i1
x i x i1 2 fx i fx i1 2
2n
1/2
in1
x i x i1 2 gx i gx i1 2 1/2
x i x i1 2 fx i fx i1 2 1/2
xi x i1 2 gx i gx i1 2 1/2
From (*) and (**), we know that
2 H a, b 2 H P 1 h P 2 ***
which implies that
H a,2b a 2 H a, b h a,2b a.
So, we know that the length of 0 does not exceed the length of .
Remark: Definex i x i1 a i , fx i fx i1 b i ,andgx i gx i1 c i , then we
have
4a i2 b i c i 2 1/2
ai2 b i2 1/2 a i2 c i2 1/2 .
Hence we have the result (***).
Proof: It suffices to square both side. We leave it to the reader.
Absolutely continuous functions
A real-valued function f defined on a, b is said to be absolutely continuous on a, b
if for every 0, there is a 0 such that
n
k1
|fb k fa k |
for every n disjoint open subintervals a k , b k of a, b, n 1, 2, . . . , the sum of whose
n
lengths k1
b k a k is less than .
Absolutely continuous functions occur in the Lebesgue theory of integration and
differentiation. The following exercises give some of their elementary properties.
6.11 Prove that every absolutely continuous function on a, b is continuous and of
bounded variation on a, b.
Proof: Letf be absolutely continuous on a, b. Then 0, there is a 0 such that
n
k1
|fb k fa k |
for every n disjoint open subintervals a k , b k of a, b, n 1, 2, . . . , the sum of whose
n
lengths k1
b k a k is less than . So,as|x y| , wherex, y a, b, wehave
|fx fy| .
That is, f is uniformly continuous on a, b. So,f is continuous on a, b.
n
In addition, given any 1, there exists a 0 such that as k1
b k a k ,
where a k , b k s are disjoint open intervals in a, b, wehave
**