Martin Flashman's Courses
MATH 316 Real Analysis I Spring, 2016
Compactness and Uniform Continuity

Definition : Suppose f: R -> R and S is an interval. We say that f is uniformly continuous on S if
for any epsilon > 0, there is a real number delta >0 so that
for any a and x in S, if |x-a| < delta then |f(x)-f(a)| < epsilon.

Demonstration on wolfram.com of Uniform Continuity

Proposition: If f is uniformly continuous on and interval,S, then f is continuous on S.
Proof: Check the logic of the quantifiers for uniform continuity and continuity.

Example: f(x) = 1/x for x \ne 0, f(0) = 0; S = (0, 1). f is continuous on S, but f is not uniformly continuous on S.
Example: f(x) = mx + b is uniformly continuous for (-oo,oo).
Discussion: Suppose epsilon > 0 is given. If m=0 then let delta = epsilon will do. If m \ne 0, let delta = epsilon/{|m|}. The verification that for any x and a if  |x-a| < delta then |f(x)-f(a)| < epsilon is left as a routine exercise for the reader.

Theorem: A continuous function on a compact set of real numbers, K is uniformly continuous.
Proof:
Plan- Given epsilon >0, find an open cover of K and use a finite subcover of K to find adelta >0
where if |x-a| < delta then |f(x)-f(a)| < epsilon.

Start:
For each a in K,
there is a delta_a where if  |x-a| < delta_a, then |f(x) -f(a)| < epsilon/3. [1]
Consider N(a, delta_a/2) and open set for each a in K. Then O = {N(a,delta_a/2), a in K} is an open cover of K.
Using K is compact, there is an open subcover determined by {a_1, a_2, ..., a_n} with related delta_j= delta_{a_j}/2.
Then K sub N(a_1, delta_1) uu N(a_2,delta_2) uu ... uu N(a_n,delta_n).
Let
delta = min{delta_1, delta_2,...,delta_n} = 1/2 min{delta_{x_1},delta_{x_2},...,delta_{x_n}} .

Now suppose  x,a, in K and |x-a| < delta.
Then for some j with 1  le j le n we have s in N(a_j,delta_j) or |a-a_j|<delta_j.
Then |x-a_j| le |x-a| + |a-a_j| < delta  + delta_j < delta_j  + delta_j =2 cdot delta_{a_j}/2 = delta _{a_j}
Thus by [1]  |f(x) - f(a_j) < epsilon/3 and |f(a) -f(a_j)| <epsilon/3.
So |f(x) -f(a)| le |f(a)-f(a_j)| + |f(a_j) - f(x)| < 2 epsilon/3 < epsilon.
Thus f is uniformly continuous on K.
EOP

For a well done proof with figures see http://pirate.shu.edu/~wachsmut/ira/cont/proofs/ctunifct.html

Theorem: Any continuous function on [a,b] is Darboux (or Riemann) integrable.
Proof:
Suppose f:[a,b] -> R is continuous and epsilon > 0.
To show f is Darboux integrable we must find a partition P where U(P,f) -L(P,f) < epsilon.
Since f is continuous on [a,b] which is compact, f is uniformly continous on [a,b].
Choose delta>0 so when |x-t| < delta,  |f(x)-f(t)|< epsilon/{b-a}.
Then by the extreme value theorem, for any interval, I, with length l(I)<delta, lub{f(x): x in I} - glb{f(x):x in I} = f(c_M)-f(c_m) < epsilon/{b-a}.
Now choose N so that  Delta x = {b-a)/N <delta, x_k = a + k Delta x, and P = {x_kP: k = 0,1, ..., n}.

Then
U(P,f) -L(P,f) < \Sigma_1^N [epsilon/{b-a}] \ Delta x = [epsilon/{b-a}] \Sigma_1^N \ Delta x = [ epsilon/{b-a}] (b-a) = epsilon
.
EOP