The Real And Complex Number Systems

fx
x, ifx 0, 1 4
x 1 ,ifx 1 , 3 .
2 4 4
x 1, if x 3 ,1 4
Exercise Suppose that fx : a, b R is a continuous and non-constant function. Prove
that the function f cannot have any small periods.
Proof: Say f is continuous at q a, b, and by hypothesis that f is
non-constant, there is a point p a, b such that |fq fp| : M 0. Since f is
continuous at q, thengiven M, there is a 0 such that as
x q , q a, b, wehave
|fx fq| M. *
If f has any small periods, then in the set q , q a, b, there is a point
r q , q a, b such that fr fp. It contradicts to (*). Hence, the
function f cannot have any small periods.
Remark 1. There is a function with any small periods.
Solution:The example is Dirichlet function,
0, if x Q
fx
c
1, if x Q .
Since fx q fx, for any rational q, we know that f has any small periods.
2. Prove that there cannot have a non-constant continuous function which has
two period p, andq such that q/p is irrational.
Proof: Since q/p is irrational, there is a sequence qn
p n
Q such that
q n
p n
q p 1 p
p2 |pq n qp n|
n
p n
0asn .
So, f has any small periods, by this exercise, we know that this f cannot a
non-constant continuous function.
Note: The inequality is important; the reader should kepp it in mind. There are
many ways to prove this inequality, we metion two methods without proofs. The
reader can find the proofs in the following references.
(1) An Introduction To The Theory Of Numbers written by G.H. Hardy and
E.M. Wright, charpter X, pp 137-138.
(2) In the text book, exercise 1.15 and 1.16, pp 26.
3. Suppose that fx is differentiable on R prove that if f has any small periods,
then f is constant.
Proof: Givenc R, and consider
fc p n fc
p n
0 for all n.
where p n is a sequence of periods of function such that p n 0. Hence, by
differentiability of f, we know that f c 0. Since c is arbitrary, we know that
f x 0onR. Hence, f is constant.
Continuity in metric spaces