The Real And Complex Number Systems

dx, y , x, y A, wehave
dfx, fy .
Let x, y clA, and thus we have x n A with x n x, andy n A with y n y.
Choose /3, then we have
dx n , x /3 and dy n , y /3 as n N 1
So, as dx, y /3, wehave(n N 1 )
dx n , y n dx n , x dx, y dy, y n /3 /3 /3 .
Hence, we have as dx, y , (n N 1 )
dgx, gy dgx, fx n dfx n , fy n dfy n , fy
*
dgx, fx n dfy n , gy
And since lim n fx n gx, and lim n fy n gy, we can choose N N 1 such that
dgx, fx n and
dfy n , gy .
So, as dx, y , (n N) wehave
dgx, gy 3 by (*).
That is, g is uniformly on clA.
It remains to show that g is a unique extension of f to clA which is uniformly
continuous on clA. If there is another extension h of f to clA which is uniformly
continuous on clA, thengivenx A A, we have, by continuity, (Say x n x)
hx h lim n
x n lim n
hx n lim n
fx n lim n
gx n g lim n
x n gx
which implies that hx gx for all x A A. Hence, we have hx gx for all
x clA. Thatis,g is a unique extension of f to clA which is uniformly continuous on
clA.
Remark: 1. We do not require that A is bounded, in fact, A is any non-empty set in a
metric space.
2. The exercise is a criterion for us to check that a given function is NOT uniformly
continuous. For example, let f : 0, 1 R by fx 1/x. Sincef0 does not exist, we
know that f is not uniformly continuous. The reader should feel that a uniformly continuous
is sometimes regarded as a smooth function. So, it is not surprising for us to know the
exercise. Similarly to check fx x 2 , x R, and so on.
3. Here is an exercise to make us know that a uniformly continuous is a smooth
function. Let f : R R be uniformly continuous, then there exist , 0 such that
|fx| |x| .
Proof: Since f is uniformly continuous on R, given 1, there is a 0 such that as
|x y| , wehave
|fx fy| 1. *
Given any x R, then there is the positive integer N such that N |x| N 1. If
x 0, we consider
y 0 0, y 1 /2, y 2 ,...,y 2N1 N 2 , y 2N x.
Then we have