The Real And Complex Number Systems

hx gfx if x S. Iff is uniformly continuous on S and if g is uniformly continuous
on fS, show that h is uniformly continuous on S.
Proof: Given 0, we want to find a 0 such that as x y R
n , x, y S, we
have
hx hy gfx gfy .
For the same , sinceg is uniformly continuous on fS, then there exists a 0 such
that as fx fy R
m ,wehave
gfx gfy .
For this ,sincef is uniformly continuous on S, then there exists a 0 such that as
x y R
n , x, y S, wehave
fx fy R
m .
So, given 0, there is a 0 such that as x y R
n , x, y S, wehave
hx hy .
That is, h is uniformly continuous on S.
Remark: It should be noted that (Assume that all functions written are continuous)
(1) uniform continuity uniform continuity uniform continuity.
(2) uniform continuity NOT uniform continuity
(3) NOT uniform continuity uniform continuity
(a) NOT uniform continuity, or
(b) uniform continuity.
(a) NOT uniform continuity, or
(b) uniform continuity.
(4) NOT uniform continuity NOT uniform continuity
(a) NOT uniform continuity, or
(b) uniform continuity.
For (1), it is from the exercise.
For (2), (a) let fx x, andgx x 2 , x R fgx fx 2 x 2 .
(b) let fx x ,andgx x 2 , x 0, fgx fx 2 x.
For (3), (a) let fx x 2 ,andgx x, x R fgx fx x 2 .
(b) let fx x 2 ,andgx x , x 0, fgx f x x.
For (4), (a) let fx x 2 ,andgx x 3 , x R fgx fx 3 x 6 .
(b) let fx 1/x, andgx 1 , x 0, 1 fgx f 1
x .
x x
Note. In (4), we have x r is not uniformly continuous on 0, 1, forr 0. Here is a
proof.
Proof: Let r 0, and assume that x r is not uniformly continuous on 0, 1. Given
1, there is a 0 such that as |x y| , wehave
|x r y r | 1. *
Let x n 2/n, andy n 1/n. Then x n y n 1/n. Choose n large enough so that 1/n .
So, we have