(a) f (x) = L and f (x) = M.
(b) a is not an accumulation point of D fl (a, oo), a E D and lim., tea- f (x) = M. In this case, L will denote the value of f (a).
(c) a is not an accumulation point of D n (- oo, a), a E D and lim, a+ f (x) = L. In this case, M will denote the value of f(a).
Suppose further that L M. Then f has a jump discontinuity at x = a. The value Ja (f) = L - M is called the jump of f at x = a.
Note that if D represents an interval, then parts (b) and (c) of Definition 4.2.5 occur only at the endpoints at which points both removable and jump discontinuities may exist. For example, the function
0 if x = 0
f (x) -
1 if x >0
has both removable and jump discontinuities at the point x = 0, the endpoint of the interval [0, oo}. Note also that the function
0 if x < 0 r (x) = -sin-1 1 -
if x > 0
x x
has a discontinuity at x = 0, which is neither removable nor jump. Why?
If at x = a there exists a removable and/or jump discontinuity of the function f, then x = a is referred to as a point of simple discontinuity of f, or a discontinuity of the first kind. All other discontinuities are discontinuities of the second kind.
Consider the function
ifx 0
k(x} = x
0 if x = 0
which is discontinuous at the point x = 0 and fits neither Definition 4.2.3 nor Definition 4.2.5
since k(0) = -oo and k(0) = +oo. Suspecting an infinite jump at x = 0 can lead to
the conclusion that this discontinuity is an "infinite" discontinuity. This is precisely true. We generalize the idea of this discontinuity of the second kind next.
Definition 4.2.6. Suppose that for a function f D -* fit, one of the following three conditions is satisfied.
(a) f (x) and limx.a- f(x) are both infinite.
(x) _ 1 1
:
(b) a is not an accumulation point of D fl (a, c) and f (x) is infinite.
(c) a is not an accumulation point of D fl (-x, a) and limx+a+ f (x) is infinite.
Then f has an infinite discontinuity at x = a.
For now, we will group together all nonremovable discontinuities that are neither jump nor infinite, and call them oscillating discontinuities. This group will be broken up into several classes at some point in the future. Before closing the section, we just wish to present one more definition.
Definition 4.2.7. The function f is piecewise continuous on D c.JR if and only if there exists finitely many points x1, x2,... , x, suchthat
(a) f is continuous on D except at x 1, x2, ... , x, and (b) f has simple discontinuities at x1, x2, ... , x,.
As commented earlier, the simple discontinuities, that is, those of the first kind, are consid- ered as removable or jump discontinuities. Note that if D is an open interval (a, b), then for x1, x2, ... , x, as given in Definition 4.2.7, both
1im f (x) and lim f (x)
are finite for each i = 1, 2, ... , n. In this definition, f is not required to be defined at any of the points x 1,x2, ... ,x,..
Exercises 4.2
1. Determine if x = 0 is a point of removable discontinuity for the given functions. If it is, what should f(0) be defined to be to make f continuous at x = o?
(a) f (x) = sgn x (See Exercise 3(h) of Section 4.1.)
(b) f() = sin x (See Exercise 3(c) of Section 4.1.) x
sin(sin x) (c) .f (x) =
x
(d) .f(x) _ sin(sin x2)
(e) f(x) = Lxi - fxlx
x + li
(f) f (x) = 2 (See Exercise 8(g) of Section 1.2.) (g) .f (x) x
1 - x otherwise x ifx=? n E Z
(h) f (x) = n ' (See Exercise 4 of Section 4.1.) 1 - x otherwise
i x = sin 1 See Example 3.2.8.
.
2
x
Sec. 4.2 * Discontinuity of a Function 161
x = x sin 1 See Exercise 3(d) of Section 4.1 and Example 4.2.2.
x
(k) f (x) =
(1) f (x) = eXP(1 )
x
ifx =±-n EN1
n otherwise
2. For each given function, locate and classify all the points of discontinuity. Then graph the functions.
(a) f (x) = 2x --- L2x , x E [-1, 1]. (Functions of this type are called sawtooth func- tions.)
b x = x I x = ±1 and n E N See the discussion Example 4.2.2.
()fC) x n ( preceding )
Lxi+ x 3
(c) f (x) = 2 , x E - 2 ,1
(d) f (x) =(-1)n, x E [n, n+ 1), and n EN
-,1 if x = in lowest terms with p, q E N
(e) f (x) = q q (See Example 3.2.10.)
0, if x is irrational
(f) f : [--1, 1] - defined by f (x ) Section 3.2.)
1 ifx =±1'nEZ
(g) f (x) = x n
0 otherwise
-1 if x >0
(h) f (x) _ x1 if x 0, x rational
0 ifx=±1 neN
n' (See Exercise 8 from
1 otherwise
1-1 if x <0, x irrational.
1
i x = exp -
x ifx 0
See Exercise 4(c) of Section 3.3.
0 if x < 0
-exp ( 1 /x) - ifx 0
f (x) = 1 + exP(1 Ix) (See Exercise 9(c) of Section 3.3.)
1 ifx = 0
2
k x = ex 1 sin1 x > 0 See Exercise 4(g) of Section 3.3.
()f{) P x + x ( Cg) )
x = ex (-)p sin1 x <0 See Exercise 4(h) of Section 3.3.
(1) fC) P ( )
x x
1 1
(m) f (x) = exp - + sin -, x E R
x x
x ifx = ±-,n1 E N
(n) f (x) = n
x2 otherwise
\ /
2
_
-
- -
,
+
(o) f (x) _
x-2 n
(p) f (x) = x (See Exercise 15 of Section 3.3 and Exercise 1(i) of Sec- tion8.1.)
4.3 Properties of Continuous Functions
Continuous functions, when defined on a particular set, are guaranteed to posses some wonder- ful properties. It is a goal of this section to study some of these properties. First, however, we wish to prepare the stage by defining a closed and open set.
Definition 4.3.1. A set E c Ill is said to be closed if and only if every accumulation point of E is in E.
Examples of closed sets are intervals [-1, 2], [3, oo), (-x, 5], iJI, a set S where S = {x l x E [1,2] U (3}}, etc.
Definition 4.3.2. A set E C Jl is said to be open if and only if for each x E E there exists a neighborhood I of x such that I is entirely contained in E.
Examples of open sets are intervals (-1,2), (3, oo), (-oo, 5), iR, etc. Also, the interval [-1, 2) is not closed and not open. However, the interval (0, oo) is open but not closed. The
sets 111 and are both open and closed. Are isolated points closed or open?
THEOREM 4.3.3. A set E c Ill is closed if and only if IJI \ E is open.
A proof of Theorem 4.3.3 is Exercise 20. Since our first property for continuous functions deals with boundedness, a review of Definition 1.2.14 and/or Definition 2.1.10 is recommended.
As observed from Exercise 6(b) of Section 4.1, a continuous function need not be bounded.
However, if the domain is a closed and bounded interval, then continuity implies boundedness.
We record this as follows.
THEOREM 4.3.4. If a function f is continuous on a closed and bounded interval [a, b], then f is bounded on [a, b].
Theorem 4.3.4 tells us that if f : [a, b] -- s)I is continuous function, then the range of f forms a set that is bounded. Thus, the graph of f lies between some two horizontal lines. It is important to realize that the assumptions in Theorem 4.3.4 are sufficient but not necessary. See Exercise 6(a) from Section 4.1. In addition, the result is not necessary true if any one of the conditions in Theorem 4.3.4 is not satisfied. For further details, see Exercise 1.
Proof We will prove this theorem by assuming to the contrary that f is not bounded on [a, b].
If f is not bounded on [a, b], there exists a bounded sequence (xn} in [a, b] such that I f (xn)I >
n for all n. However, by the Bolzano-Weierstrass theorem for sequences, {xn } has a convergent subsequence. So let {x1zk } be this subsequence, which converges to, say, c. Now, since [a, b] is a closed set, and thus contains all of its accumulation points, c E [a, b] (see Definition 4.3.1). But
is continuous at c. Thus, by Exercise 6(k) of Section 4.1, we have limk f (xnk) = f (c).
This, however, contradicts the fact that I f (xnk) E > nk for all k E N. Why? Hence, the proof is
complete. Q
2
Sec. 4.3 Properties of Continuous Functions 163 The next theorem is among the most powerful results in mathematics. It says that if f [a, b] - ill is continuous function, then f must have a maximum and a minimum value.
(Review part (b) of Definition 1.2.15.) Thus, the graph of f must have the very highest and very lowest points, which means that there is the largest value in the range of f for some input out of the domain [a, b].
THEOREM 4.3.5. (Extreme Value Theorem) If f is a continuous function on an interval [a, b], then f attains its maximum and minimum values on [a,b].
Proof A popular method that mathematicians use to prove this result is by contradiction. (See Exercise 2 for an alternative proof.) First we will prove the existence of the maximum value of f. In view of Theorem 4.3.4, f is bounded. Hence, f has the least upper bound, call it M.
Thus,
M = sup f (x) x E [a, b] } = least upper bound off on [a, b].
We assume that there is no value c E [a, b] for which f (c) = M. Therefore, f (x) C M for all x E [a, b] . Now define a new function g by
g (x) = M _1 f(x).
(Why this choice?) Observe that g(x) > 0 for every x E [a, b] and that, by Theorem 4.1.8, g is continuous on [a, b]. By Theorem 4.3.4, g is bounded on [a, b]. Thus, there exists K > 0
such that x K for eve x E a b. Since for each x E a b x= 1 K
Mrf(x)
is equivalent to f (x) M - 1
K, we have contradicted the fact that M was assumed to be the least upper bound of f on [a, b]. Hence, there must be a value c E [a, b] such that f (c) = M.
In proving the second part of this theorem, the case for attaining the minimum value, we either rewrite the preceding proof, or we apply the completed argument to the function h (x) =
- f (x) . Details are left to the reader. D
THEOREM 4.3.6. (Bolzano's Intermediate Value Theorem) If a function f is continuous on [a, b] and if k is a real number between f (a) and f (b), then there exists a rear number c E (a, b) such that f (c) = k.
The geometric meaning of Bolzano's intermediate value theorem is quite simple. If we pick any value k between f (a) and f (b) and draw a horizontal line through the point (0, k), then this line will intersect the function f at a point whose x-coordinate is between a and b. There may very well be more than one such value. This idea is stated formally in Definition 4.3.7 and demonstrated in Figure 4.3.1 where f (a) was chosen to be smaller than f (b). Note that in Theorem 4.3.6, f (a) need not be the minimum value of f nor f (b) the maximum. But they could. See Corollary 4.3.9.