Page 167 - DMTH401_REAL ANALYSIS

Unit 12: Properties of Continuous Functions

Having proved the boundedness of the : function continuous on a bounded closed interval, we Notes
now prove that the function attains its bounds, that is, it has the greatest and the smallest values.

Theorem 2: If f is a continuous function on the bounded closed interval [a, b] then there exists
points x and x in [a, b] such that f(x ) S f(x)  f(x ) for all x [a, b] (i.e. f attains its bounds).
1 2 1 2
Proof: From Theorem 1, we know that f is bounded on [a, b].

Therefore there exists M such that |f(x)|  M " x [a, b].

Hence, the collection {f(x) : a a x  b] has an upper bound, since f(x)  |f(x)|  M " x [a, b].
So by the completeness property of R, the set (f(x) : a  x  b) has a least upper bound.

Let us denote by K the least upper bound of {f(x) : a  x b].
Then f(x)  K for all x such that a  x  b. We claim that there exists x in [a, b] such that f(x ) = K.
2 2
If there is no such x , then K – f(x) > 0 for all a  x  b. Hence, the function g given by,
2
1
g(x) =
K f(x)
-
is defined for all x in [a, b] and g is continuous since f is continuous. Therefore by Theorem 1,
there exists a constant M’ > 0 such that
|g(x)|  M’ " x [a, b]

Thus, we get
1 1
|g(x)| = =  M’
-
-
|K f(x)| K f(x)
1
i.e., f(x)  K – " x [a, b].
M
But this contradicts the choice of K as the least upper bound of the set (f(x) : a | x  b). This
contradiction, therefore, proves the existence of an x in [a, b] such that f(x ) = K  f(x) for a  x  b.
2 2
The existence of x in [a, b] such that f(x )  f(x) for a  x | b can be proved on exactly similar lines
1 1
by taking the g.l.b. of {f(x) : a  x  b} instead of the l.u.b. or else by considering –f instead of f.
Theorems 1 and 2 are usually proved using what is called the Heine-Borel property on the real
line or other equivalent properties. The proofs given in this unit straightaway appeal to the
completeness property of the red line (Unit 2) namely that any subset of the real line bounded
above has least upper bound. These proofs may be slightly longer than the conventional ones
but it does not make use of any other theorem except the property of the real line stated above.
As remarked earlier, the properties of continuous functions fail if the intervals are not bounded
or closed, that is, the intervals of the type
]a, b[, 1a, b], [a, b[, [a,’[, 1a, [, ] –, a], ] –, a [ or ] –, [.

Example: Show that the function f defined by f(x) = 3 V x  [0, [ is continuous but not
bounded.

Solution: The function f being a polynomial function is continuous in [0, [. The domain of the
function is an unbounded closed interval. The function is not bounded since the set of values of
the function that is the range of the function is {x : x  [0, [ } = [0, [ which is not bounded.
2

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