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P. 191
Unit 14: Sequences and Series of Functions
m
If the same m serves for all x for a given > 0 then x < for all x, 0 < x < 1. This implies that Notes
log
m > (since log x is negative). This is not possible since log x decreases to zero as x tends to
log x
1 and so log /log x is unbounded.
Thus we have shown that the sequence (f ) does not converge to the function f uniformly in [0, 1]
n
even though it converges pointwise.
x
Example: Show that the sequence (g,) where g (x) = , x [0,¥[ is uniformly
n 1 nx
+
convergent in [0, ¥[.
Solution: lim g (x) = 0 for all x in the interval [0, ¥[. Thus (g ) is pointwise converge of where
n®¥ n n
f(x) = 0 " x [0, ¥[.
x 1
Now |g (x) – f(x)| = < for all x in [0, ¥[.
+
n 1 nx n
1 1
Since lim = 0, therefore given > 0, there exists a positive integer m such that < for n m
n®¥ n n
Thus m depends only on . Therefore,
|g (x) – f(x)| < for n m and " x [0, ¥[.
n
Therefore (g,) ® f uniformly in [0, ¥[.
Just as you have studied Cauchy’s Criterion for convergence of sequence of real numbers, we
have Cauchy’s Criterion for uniform convergence of sequence of functions which we now state
and prove.
Theorem 1: Cauchy’s Principle of Uniform Convergence
The necessary and sufficient condition for a sequence of functions (f ) defined on A to converge
n
uniformly on A is that for every > 0, there exists a positive integer m such that
|f (x) –f (x)| < for n > k m and " x A
n k
Proof: Condition is necessary. It is given that (f ) is uniformly convergent on A.
n
Let f – f uniformly on A. Then given > 0, there exists a positive integer m such that
n
|f (x) – f(x)| < /2 for n m and " x A.
n
|f (x) – f (x)| = |f (x) – f(x)| + |f(x) – f (x)|
n k n k
< |f (x) – f(x)| + |f(x) – f (x)| (By triangular inequality)
n k
< + + for n > k m and " x A
2 2
This proves the necessary part. Now we prove the sufficient part.
Condition is sufficient: It is given that for every > 0, there exists a positive integer m such that
|f (x) – f (x)| < for n > k m and for all x in A. But by Cauchy’s principle of convergence of
n k
sequence of real numbers, for each fixed point x of A, the sequence of numbers (f (x)) converges.
n
In other words, (f ) is pointwise convergent say to f on A. Now for each > 0, there exists a
n
positive integer m such that
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