Page 166 - DMTH401_REAL ANALYSIS

Real Analysis

Notes This property called the completeness property of R states as follows:
Any non-empty subset of the Seal Hue R which is bounded above has the least upper bound or
equivalently, any non-empty subset of R which is bounded below has the greatest lower bound.

In the following theorems we prove the properties of functions continuous on bounded closed
intervals. In the first two theorems we show that a continuous function on a bounded closed
interval is bounded and attains its bounds in the interval. Recall that f is bounded on a set S, if
there exists a constant M > 0 such that |f(x)|  M for all x  S. Note also that a real function f
defined on a domain D (whether bounded or not) is bounded if and only if its range f(D) is a
bounded subset of R.
Theorem 1: A function f continuous on a bounded and closed interval [a, b] is necessarily a
bounded function.
Proof: Let S be the collection of all real numbers c in the interval [a, b] such that f is bounded on
the interval [a, c]. That is, a real number c in [a, b] belongs to S if and only if there exists a constant
M such that |f(x)| M for all x in [a, c]. Clearly, S  since a  S and b is an upper bound for S.
c c
Hence, by completeness property of R, there exists a least upper bound for S. Let it be k (say).
Clearly, k  b. We prove that k  S and k = b which will complete the proof of the theorem.
Corresponding to  = 1, by the continuity of f at k( b) there exists a d > 0 such that

|f(x) – f(k)| <  = 1 whenever |x – k| < d, x [a, b].
By the triangle inequality we have
|f(x)| – |f (k)|  |f(x) – f(k)| < 1
Hence, for all x in [a. b] for which |x – k| < d, we have that

|f(x)| < |f(k)| + 1 ...(1)
Since k is the least upper bound of S, k – d is not an upper bound of S. Therefore, there is a number
c  S such that

k – d < c  k
Consider any t such that k  t < k + d. If x belongs to the interval [c, t] then |x – k| < d. For,
x  [c, t] =  c  x  t  k – d < c  x  t < k + d ...(2)
Now c  S implies that there exists M > 0 such that for all
c
x  [a, c], |f(x)|  M
c
x  [a, t] = [a, c] U [c, t]  either x [a, c] or x [c, t].
If x  [a, c], by (3) we have
|f(x)|  M < M + |f(k)| + 1.
c c
If, however, x [c, t] then by (1) and (2) we have
|f(x)| < |f(k)| + 1 < M + |f(k)| + 1
c
In any case we get that x  [a, t] implies that
|f(x)| < M + |f(k)| + 1
c
This shows that f is bounded in the interval [a, t] thus proving that t  S whenever k  t < k + d.
In particular k  S. In such a case k = b. For otherwise we can choose a ‘t’ such that k < t < k + d and
t  S which will contradict the fact that k is an upper bound. This completes the proof of the
theorem.

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