Question 15389
 Let f: R to R be a function. Show that if f is continuous on 
[0,infinitive) and uniformly continuous on [a,infinitive) for some 
positive real number a, then f is uniformly continuous on [0,infinitive)
(f is uniformly continuous on [0,b] for any b>a).
 
 First of all, I wonder why you ask this of calculus or real analysis here.

 It seems you should use the theorem that if a function is
 continuous on a compact set C(closed and bounded set in R),then
 f is continuous on C.

 Proof: Now f is continuous on[0,+oo), so as a subset, 
 f is continuous on [0,a] and thus unif. conti. on [0,a]. 
 Also, we know that f is unif. conti. on [a,+oo).
 But, [0,+oo) = [0,a] U [a,+oo) , we conclude that f is unif. conti. on [0,+oo).

 More precisely, since f is unif. conti. on [a,+oo) 
for any {{{epsilon}}} > 0, there exists {{{delta[1]}}} > 0 such that
 |x-y|< {{{delta[1]}}} implies {f(x) -f(y)| < {{{epsilon/2}}}
  for all x, y in [a,+oo)
 And, f is unif. conti. on [0,a]. for the same {{{epsilon}}},
 there exists {{{delta[2]}}} > 0 such that
 |x-y|< {{{delta[2]}}} implies |f(x) -f(y)| < {{{epsilon/2}}}
  for all x, y in [0,a].

 Choose {{{delta}}} =min ({{{delta [1]}}}, {{{delta [2]}}}) the we have
  |x-y| < {{{delta}}} implies |f(x) -f(y)| < {{{epsilon/2}}} 
  for all x,y in [0,a] (or [a,+oo))
 if x is in [0,a] and y is in [a,+oo) we have (Note:a btwn x & y)
  |x-y| < {{{delta}}} implies |x-a| < {{{delta}}} and |y-a| < {{{delta}}}
  and hence |f(x) - f(y)| <=   |f(x) - f(a)| + |f(a) - f(y)|
                          <  {{{epsilon/2}}} +  {{{epsilon/2}}} = {{{epsilon}}}
 
 This shows f is unif. conti.on [0,+oo)

 Try to read carefully and draw diagram to understand the details.


 Kenny
 [Note: you don't have to worry about the interval, [0, b] for any b.
 Since [0,b] is compact subset of R. However, f is unif. conti. on [0,b]
 for any b > 0 does not imply f is unif. conti. on [0,+oo) as
 the example f(x) = {{{x^2}}} on [0,+oo) shows.