Uniform Continuity

Definition. The function f is uniformly continuous on an interval A if for every e > 0 there is some d > 0 such that, for all x and y in A,
if | x-y| < d, then | f(x)-f(y)| < e. 
 
Lemma. Let a < b < c and let f be continuous on the interval [a,c]. Let e > 0, and suppose that statements (i) and (ii) hold:

(i) if x and y are in [a,b]  and | x-y| < d1, then | f(x)-f(y)| < e,

(ii) if x and y are in [b,c]  and |x-y| < d2, then |f(x)-f(y)| < e.

Then there is a d > 0 such that

if x,y are in [a,c] and |x-y| < d, then |f(x)-f(y)| < e. 
 
Proof. Since f is continuous at b, there is a d3 > 0 such that
if |x-b| < d3, then |f(x)-f(b)| <  e 
2		 . 
 
It follows that

(iii) if |x-b| < d3 and |y-b| < d3 then |f(x)-f(y)| < e.

Choose d to be the minimum of d1,d2 and d3. We claim that this d works. In fact, suppose that x and y are any two points in [a,c] with |x-y| < d. If x and y are both in [a,b], then |f(x)-f(y)| < e  by (i); and if x and y are both in [b,c], then |f(x)-f(y)| < e  by (ii). The only other possibility is that

x < b < y or y < b < x.

In either case, since | x-y| < d , we also have |x-b| < d and |y-b| < d. So |f(x)-f(y)| < e  by (iii).

THEOREM If f is continuous on [a,b], then f is uniformly continuous on [a,b].

PROOF: For e > 0 let's say that f  is e-good on [a,b] if there is some d > 0 such that, for all y and z in [a,b],if |y-z| < d, then |f(x)-f(y)| < e. Then we're trying to prove that f is e-good on [a,b] for all e > 0.

Consider any particular e > 0. Let

A = {x Î [a,b]: f  is e-good on [a,x]}. 
 
Then A ¹ f  (since a is in A), and A is bounded above (by b), so A has a least upper bound a. We really should write ae , since A and a might depend on e. But we won't since we intend to prove that a = b, no matter what e  is.

Suppose that we had a < b. Since f is continuous at a, there is some d0 > 0 such that, if |y-a| < d0, then |f(y)-f(a)| < e/2. Consequently, if |y-a| < d0 and |z-a| < d0, then |f(y)-f(z)| < e. So f is surely e-good on the interval [a-d0,a+d0]. On the other hand, since a is the least upper bound of A, it is also clear that f is e-good on [a,a-d0]. Then the above Lemma implies that f is e-good on [a,a+d0], so a+d0 is in A, contradicting the fact that a is an upper bound.

To complete the proof we just have to show that a = b is actually in A . The argument for this is practically the same: Since f is continuous at b, there is some d0 > 0 such that, if |y-b| < d0, then |f(y)-f(b)| < e/2.  So f is e-good on [b-d0,b]. But f is also e-good on [a,b-d0], so the Lemma implies that f is e-good on [a,b].