Page 169 - DMTH401_REAL ANALYSIS

Unit 12: Properties of Continuous Functions

Let S denote the collection of all real numbers x in [a, b] such that f(x) < K. Clearly S contains a, so Notes
S  and b is an upper bound for S. Hence, by completeness property of R, S has least upper bound
and let us denote this least upper bound by c. Then a  c  b. We want to show that f(c) = K.
Since f is continuous on [a, b], f is continuous at c. Therefore, given  > 0, there exists a 6 > 0 such
that whenever x is in [a, b] and |x – c| < 6, |f(x) – f(c) ( < G,

i.e., f(c) –  < f(x) < f(c) + . .. . (4)
If c  b, we can clearly assume that c + 6 < b. Now c is the least upper bound of S. So c –  is not
‘an upper bound’ of S. Hence, there exists a y in S such that c – 6 < y  c. Clearly |y – c| <  and
so by (4) above, we have
f(c) –  < f(y) < f(c) + .
Since y is in S, therefore f(y) < K. Thus, we get

f(c) – S < K
If now c = b then K –  < K < f(b) = f(c), i.e., K < f(c) + E. If c  b, then c < b; then there exists an x
such that c < x < c + 6, 6, x [a, b] and for this x, f(x) < f(c) +  by (4) above. Since x > c, K  f(x), for
otherwise x would be in S which will imply that c is not an upper bound of S. Thus, again we
have K  f(x) < f(c) + E.

In any case,
K < f(c) +  ...(6)
Combining (5) and (6), we get for every  > 0
f (c) –  < K < f(c) + 

which proves that K = f(c), since  is arbitrary while K, f(c) are fixed. In fact, when f(a) < K < f(b)
and f(c) = K, then a < c < b.
Corollary 1: If f is a continuous function on the closed interval [a, b] and If f(a) and f(b) have
opposite signs (i.e., f(a) f(b) < 0), then there is a point x in ]a, b[ at which f vanishes. (i.e., f(x) = 0).
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Corollary follows by taking K = 0 in the theorem.
Corollary 2: Let f be a continuous function defined on a bounded closed interval [a, b] with
values in [a, b]. Then there exists a point c in [a, b] such that f(c) = c. (i.e., there exists a fixed point
c for the function f on [a, b]).

Proof: If f(a) = a or f(b) = b then there is nothing to prove. Hence, we assume that f(a)  a and f(b)
 b.
Consider the function g defined by g(x) = f(x) – x, x [a, b]. The function being the difference of
two continuous functions, is continuous on [a, b]. Further, since f(a), f(b) are in [a, b], f(a) > a
(since f(a)  a, f(a) [a, b]) and f(b) < b. (Since f(b)  b, f(b) [a, b]). So, g(a) > 0 and g(b) < 0. Hence,
by Corollary 1, there exists a c in ]a, b[ such that g(c) = 0, i.e., f(c) = c. Hence, there exists ac in [a,
b] such that f(c) = c.

The above Corollary 1 helps us sometimes to locate some of the roots of polynomials. We
illustrate this with the following example.

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Example: The equation x + 2x – 11 = 0 has a real root lying between 1 and 2.
Solution: The function f(x) = x + 2x – 11 is a continuous function on the closed interval [1, 2],
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f(1) = –8 and f(2) = 9. Hence, by Corollary 1, there exists an x ]1, 2[ such that f(x ) = 0, i.e., x is
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a real root of the equation x + 2x – 11 = 0 lying in the interval ]1, 2[.
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