The Real And Complex Number Systems

Note: We have the following interesting results:. Prove that, for x 0, y 0,
|x p y p |
|x y| p if 0 p 1,
p|x y|x p1 y p1 if 1 p .
Proof: (As 0 p 1) Without loss of generality, let x y, consider
fx x y p x p y p , then
f x p x y p1 x p1 0, note that p 1 0.
So, we have f is an increasing function defined on 0, for all given y 0. Hence, we
have fx f0 0. So,
x p y p x y p if x y 0
which implies that
|x p y p | |x y| p
for x 0, y 0.
Ps: The inequality, we can prove the case p 1/2 directly. Thus the inequality is not
surprising for us.
(As 1 p Without loss of generality, let x y, consider
x p y p pz p1 x y, wherez y, x, byMean Value Theorem.
which implies
for x 0, y 0.
px p1 x y, note that p 1 0,
px p1 y p1 x y
|x p y p | p|x y|x p1 y p1
5. In general, we have
x r is uniformly continuous on 0, , ifr 0, 1,
is NOT uniformly continuous on 0, , ifr 1,
and
sinx r
is uniformly continuous on 0, , ifr 0, 1,
is NOT uniformly continuous on 0, , ifr 1.
Proof: (x r )Asr 0, it means that x r is a constant function. So, it is obviuos. As
r 0, 1, thengiven 0, there is a 1/r 0 such that as |x y| , x, y 0, ,
we have
|x r y r | |x y| r by note in the exercise
r
.
So, x r is uniformly continuous on 0, , ifr 0, 1.
As r 1, assume that x r is uniformly continuous on 0, , thengiven 1 0,
there exists a 0 such that as |x y| , x, y 0, , wehave
|x r y r | 1. *
By Mean Value Theorem, we have (let x y /2, y 0)