260 Chapter 5 • Continuous FunctionsLemma 5.4.5 (Negation of Uniform Continuity) A function f : D(f) ---+
JR is not uniformly continuous on a set A ~ D(f) iff 3 c > 0 3 VS > 0,
3 x,y EA 3 Ix -yl < S but lf(x) - f(y)I 2 E.
Proof. Exercise 4. •We now give two proofs that the function f(x)continuous on (0, 1).
-^1 is. not un11·c orm^1 y
x
First proof that f(x) = ~ is not uniformly continuous on (0, 1):
Choose E = l. Let 8 > 0 and choose >. = min{2, 8}, x = ~ and y = i.
>. >. >. 8
Ix -YI = 3 - 6 = 6 :::; 6 < 8, andlf(x) - f(y)I = - - - = -- = - > - > E.
13 6I I 3I 3 3
>. >. >. >.-2Thus, 3 c > 0 3 \:/8 > 0, 3 x,y E (0, 1) 3 Ix -yl < 8 but lf(x) - f(y)l 2 E.
That is , by Lemma 5.4.5, f(x) = ~ is not uniformly continuous on (0, 1). •
Despite the fact that the "First Proof" is technically correct, it doesn't
seem very satisfying. It seems to be more technical than necessary; there must
be a n easier way to see that f is not uniformly continuous on (0, 1). The next
theorem will a llow us to give an (easier) Second Proof.
Theorem 5.4.6 If f is uniformly continuous on a bounded se t A , then f is
bounded^16 on A.
Proof. Suppose f is uniformly continuous on a bounded set A. Since A is
a bounded set, 3 a, b E JR 3 A~ [a, b].
Letting E = 1 in Definit ion 5.4.1,
38 > 0 3 \:/x,y EA, lx-yl < S ~ lf(x) -f(y)I < l. (8)
Keep this S fixed in what follows. By the Archimedean property, 3 n E N 3
--:;;:-b- a < u. s: F or i · = , , ,^0 1 2 · · · , n,^1 e t Xi = a + i· b- a -:;;:-.a bFigure 5.11- But, recall Exercise 5.3.6.