Page 172 - DMTH401_REAL ANALYSIS

Real Analysis

Notes but is common to all continuous functions defined on bounded closed intervals of the real line.
We prove it in the following theorem.
Theorem 5: If f is a continuous function on a bounded and closed interval [a, b] then f is uniformly
continuous on [a, b].
Proof: Let f be a continuous function defined on the bounded closed interval [a, b]. Let S be the set
of all real numbers c in the interval [a, b] such that for a given  > 0, there exists positive number
d such that for points x , x belonging to closed interval [a, c],
c 1 2
|f(x ) – f(x )| <  whenever |x – x | < d,.
1 2 2 2
(In other words f is uniformly continuous on the interval [a, c]. Clearly a  S so that S is non-
empty. Also b is an upper bound of S. Prom completeness property of the real line S has least
upper bound which we denote by k. k  b.
f is continuous at k. Hence given E > 0, there exists positive real number d such that
k
|f(x) – f(k)| < /2 whenever |x – k| < d ...(8)
k
1
Since k is the least upper bound of S, k – is not an upper bound of S.
2
Therefore there exists a point c  S such that
k – 1/2 d < c  k. ...(9)
k
Since c S; from the definition of S we see that there exists d such that
c
|f(x ) – f(x )| <  whenever |x – x | < d , x , x  [a, c], ...(10)
1 2 1 2 c 1 2
Let d = min ((1/2) d , d ) and b’ = min. (k + (1/2) d , b).
k c k
Now let x , x [a, b’] and |x – x |. Then if x , x [a, c], |x – x | < d  d bythe choice of d and
1 2 1 2 1 2 1 2 c
d , then |f(x ) – f(x )| <  by (10). If one of x x is not in fa, cl, then both x , x belong to the interval
c 1 2 1 2 1 2
]k – d , k + d [. For x  [a, c], implies b’x > c > k – (1/2)d > k – d by (9) above. This means
k k 1 1 k k
x  b’ implies x  k + (1/2)d < k – d by the choice of b’. i.e.
1 1 k k
k – d < k – (1/2) d < x < k + (1/2)d < k + d ...(11)
k k 1 k k
|x – x | < d implies that x – (1/2) d < x < x + (1/2)d since d  (1/2)d by this choice of d. Thus
1 2 1 k 2 1 k k
we get from (11) above that
æ 1 ö 1
|x – x | < x – (1/2)d < x < x + (1/2)d < k + ç ÷ d + d = k + d ...(12)
1 2 1 k 2 1 k è 2 ø k 2 k k
Then (11) and (12) show that x , x ]k – d , k + d [.
1 2 k k
Thus we get that |x – k| < d and |x – k| < d , which in turn implies, by (8) above, that
1 k 2 y
|f(x ) – (k)| < /2 and |f(x ) – f(k)| < /2 .
1 2
Thus |f(x ) – f(x )| < |f(x ) – f(k)| + |f(k) – f(x )| < /2 + /2 = E. In other words, if |x – x | <
1 2 1 2 1 2
d and x , x are in [a, b’] then |f(x ) – f(x )| < E which proves that b’  S i.e. b’  k. But k  b’ by
1 2 1 2
the choice of b’ since k  k + (1/2) d and k a b. Thus we get that k = b’. This can happen only
k
when k = b. For if k < b. i.e. k = b’ = min (k + (l/2) d , b) < b, then it implies that min (k + (l/2)
k
d , b) = (k + (1/2) d = b’, where b’  S i.e. k + (1/2) d is in S and is greater than k which is a
k k k
contradiction to the fact that k is the l.u.b of S. Thus we have shown that k = b  S. In other
words there exists a positive number d (corresponding to b) such that |x – x | < d , x , x [a,
k 1 2 k 1 2
b] implies |f(x ) – f(x )| < . Therefore f is uniformly continuous in [a, b].
1 2
You may note that uniform continuity always implies continuity but not conversely. Converse
is true when continuity is in the bounded closed interval.

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