BOUNDED DERIVATIVE AND UNIFORM CONTINUITY
MOLLY REEDY
Abstract. A theorem concerning uniform continuity will be presented. I will show that a function
f whose derivative is bounded on the open interval (a, b) is uniformly continuous on this interval.
A function f is uniformly continuous if for every ✏ > 0 there will exist a > 0 such that for any
two elements x and c, in the interval, if |x c| < then |f (x) f (c)| < ✏. Using the Mean Value
Theorem and other topics covered in Real Analysis at Bu↵alo State College, a formal proof will be
given. After the proof, I will show that the converse fails to hold; namely, I will identify a function
f which is di↵erentiable and uniformly continuous on the interval ( 1, 1) and yet the derivative f 0
is unbounded on this interval.
Theorem (The Mean Value Theorem). If a function f is continuous on the interval [a, b] and
di↵erentiable on (a, b), then for any distinct x 2 (a, b) and c 2 (a, b) there is there is a point z
between x and c such that
f (x) f (c)
f 0 (z) =
.
x c
Theorem 1. Let f : (a, b) ! R be di↵erentiable on the open interval (a, b). Suppose for some
M > 0 we have |f 0 (x)|  M for all x 2 (a, b). Then f is uniformly continuous on (a, b).
Proof. Let f : (a, b) ! R be di↵erentiable on the open interval. We know,
|f 0 (x)|  M
(1)
for all x 2 (a, b).
This tells us the derivative of our function is bounded. Thus the graph is contained by M. The
Mean Value Theorem, together with (1) implies that for any distinct x 2 (a, b) and c 2 (a, b), there
is a point z between x and c such that
f (x)
x
(2)
f (c)
= |f 0 (z)|  M.
c
Thus, for any x 2 (a, b) and c 2 (a, b), we have
|f (x)
(3)
f (c)|  M |x
c|.
To prove that f is uniformly continuous on (a, b), let ✏ > 0 be arbitrary. We then choose = M✏ .
Next let x 2 (a, b) and c 2 (a, b) be arbitrary. Assume that |x c| < . We will show that
|f (x) f (c)| < ✏ as follows:
|f (x)
Hrnce, |f (b)
f (c)|  M |x c|
<M
✏
=M
M
=✏
by (3)
because |x
c| < .
✏
because =
.
M
by arithmetic.
f (a)| < ✏. Therefore we conclude that f is uniformly continuous.
Date: May 3, 2014.
1
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2
MOLLY REEDY
I just proved above that since a di↵erentiable function that has a bounded first derivative is is
uniformly continuous. This however does not mean that a function that is uniformly continuous
will have a bounded first derivative. The function f defined by,
(
x2 sin x12 , when x 6= 0;
f (x) =
0,
when x = 0.
This function is di↵erentiable and uniformly continuous on [ 1, 1] but its derivative is unbounded
on [ 1, 1].
Department of Mathematics, Buffalo State College
URL: http://math.buffalostate.edu/~cunnindw/491