264 Chapter 5 • Continuous Functions*Definition 5.4.11 Suppose f : V(f) . JR and V(f) ~ A. Then a function
g : A. JR is said to be an extension of f to A if Vx E V(f), g(x) = f(x).
That is , the two functions agree on V(f); i.e., glv(f) = f.
*Theorem 5.4.12 Suppose a < b. A function f : (a, b) ___. JR is uniformly
continuous on (a, b) ¢:;> f has a continuous extension g :. [a, b] _.JR.Proof. Suppose f : (a, b) . JR where a < b.
Part 1 ( ~): This direction is trivial. Suppose f has a continuous extension
g: [a, b] . R Then, by Theorem 5.4.7, g is uniformly continuous on [a, b], and
hence on (a, b). But 'ix E (a, b), f(x) = g(x). Thus, f is uniformly continuous
on (a, b).
Part 2 ( =?): Suppose f is uniformly continuous on (a, b). Then f is contin-
uous at every point of (a, b), by Theorem 5.4.3. We claim that both lim f(x)
x --+a+
and lim f(x) exist.
X->b-
Claim #1: lim f(x) exists.
X--+a+
Proof: Suppose that {xn} is any sequence in (a, b) 3 Xn .a. Then {xn} is
a Cauchy sequence in (a, b), so by Theorem 5.4.8, {f(xn)} is a Cauchy sequence.
Hence, {f(xn)} converges. Let L = lim f(xn)·
n->oo
Now, suppose {yn} is any other sequence in (a, b) 3 Yn .a. Consider the
sequence {Zn} defined by
{zn} = {x1, Y1, X2, Y2, · · ·, Xn, Yn · · · }.
Then Zn ___. a (see Exercises 2.6.6). Thus, {zn} is a Cauchy sequence, so
by Theorem 5.4.8, {f(zn)} is a Ca uchy sequence, hence converges. But every
subsequence of a convergent sequence must have the same limit. Thus,
lim f (xn) = lim f (Yn) = L.
n--+ ex:> n--+ cx:>
Therefore, V sequences {xn } in (a,b) converging to a, {f(xn)} converges
to the same limit, L. By the sequential criterion for one-sided limits (Theorem
4.3.5) lim f(x) = L.
X--+a+
Claim #2: lim f(x) exists. (The proof is just like that of Claim #1.)
X->b-
Now, we are ready to define the extension g: [a, b] ___.JR off by
{X--+a+ lim f(x) if x =a; }
g(x)= f(x) ifa<x<b;.
lim f ( x) if x = b
x--+a-By Exercise 5.3.2, g is continuous on [a, b]. •