Page 223 - DMTH401_REAL ANALYSIS

Unit 17: Uniform Convergence and Differentiability

Notes
Example: Look at the uniform convergence of series å ¥ n= 1 x n both on [ r- , ], 0 r < 1
r
<
and (–1, 1). In the first case we see that the series converges uniformly by Weierstrass M-test
Since
¥
n
n
x £ r n and å r < ¥
n= 1
In the second one we will try to show that there is no uniform convergence. Look at the partial
sums f If we can prove that { }f n is not a uniform Cauchy sequence then { }f n is not uniformly
.
n
convergent and therefore the series will not converge uniformly. Often it suffices to look at
f - f and show that it does not converge to 0.
n+ 1 n sup
n+ 1 n
i
i å
f - f = sup å x - x = sup x n+ 1 = 1.
n+ 1 n sup
x Î - 1 i= 1 xÎ ( 1,1)
-
( 1,1) i=
Therefore, take e = 1/2, and N" e  put n = N + 1 and m = N + 1. We see that f - f m sup =
n
>
1 1/2 = e .
In conclusion, use M-test to prove uniform convergence.
17.4 Power Series and Uniform Convergence

Recall, from Analysis 2, that a power series is the series of functions of the form å ¥ a x n ,
n= 1 n
where a is sequence of real numbers. We define a radius of convergence R of the series such
n
that å ¥ a x converges absolutely on ( R- , ) and diverges for x > . R
n
R
n= 1 n
Example: Consider å ¥ x n . The series converges pointwise on (–1, 1), but this
n= 0
convergence is not uniform, whereas on [ r- , ] converges uniformly.
r
¥
n
r
Theorem 6: Let å a x be a power series with a radius of convergence R. Then for any 0 £ < R
n= 0 n
r
the series converges uniformly on [ r- , ].
-
Proof: Fix r Î ( R , ) and define a sequence M = a r n . å ¥ M converges absolutely by our
R
n n n= 0 n
choice of r and we get
¥
n
n
r
-
" x Î [ r , ] : a x £ a r = M n and å M < . ¥
n
n
n
n= 0
¥
n
a x converges uniformly on [–r, r].
Therefore by Weierstrass M-test, the power series å n
n = 0

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