1549901369-Elements_of_Real_Analysis__Denlinger_

258 Chapter 5 • Continuous Functions

Solution. Let f ( x) = 3x^2 - 2x - l. This is the same function used in

Example 5.1.2. You a re encouraged to compare this solution with the solution

of that example, to observe the difference between proving continuity at a point

and proving uniform continuity on a set.

(a) Scratchwork: Let c > 0. We want to find 6 > 0 su ch that

't/x, y E [-1, 5], Ix - YI < 6 =? lf(x) - J(y)I < E:. Thus, we want

Ix - YI < 6 =? l(3x^2 - 2x - 1) - (3y^2 - 2y - 1)1 < c

i.e., j3(x^2 -y^2 )-2(x-y)j <t:

i.e., Ix - Yll3(x + y) - 21 < t:.

Note that for -1 :::;; x, y:::;; 5,

-2 :::;; x + y :::;; 10

• 6 < 3(x+y) < 30,
  -8 < 3(x+y)-2 < 28 ,
  l3(x + y) - 21 < 28.

Thus, we want to make sure that Ix - YI < c/28.

c

(b) Proof: Let c > 0. Choose 6 =

28

. Then 't/x,y E [-1,5],
c
Ix - YI < 6 =? -2 :S x + y :S 10 and Ix - YI <
28
c
=? -6 < 3(x + y) < 30 and Ix - YI <
28
c
=? l3(x + y) - 21 < 28 and Ix -yl < -
28 c
=?-8<3(x+y)-2<28 and lx-yl <-
c 28
=? l3(x + y) - 21 < 28 and Ix - YI < -
c 28
=? lx - yll3(x+y)-21 <
28

·28

=? l3(x^2 -y^2 ) - 2(x -y)I < c

=? l(3x^2 - 2x - 8) - (3y^2 - 2y - 8) 1 < E:.

Therefore, the function j(x) = 3x^2 - 2x - 1 is uniformly continuous on the
interval [O, 5]. D

Theorem 5.4.3 If f : V(f)--+ IR is uniformly continuous on a set A~ V(f),
then f : A --+ IR is continuous (on A).

Proof. Exercise 3. •