Page 220 - DMTH401_REAL ANALYSIS

Real Analysis

Notes n ¥
x
is a series of functions. Partial sums: f ( ) = å g ( ). We say that å g converges pointwise/
x
n i= 1 i n= 1 n
:
uniformly to f I ®  if f ® f pointwise/unformly.
n
¥ n
Example: Consider a series å x on [–1/2, 1/2] and on (–1, 1). We want to investigate
n= 0
convergence of this series, whether it is pointwise of n uniform and eventually, what is the limit.
n i 2 n x n+ 1 - 1
+
+
Partial sums are in this case f n ( ) = å i= 0 x = 1 x x + ... x = x - 1 . If we fix x Î ( 1, 1) we
x
+
-
x n+ 1 - 1 1
see that f ( ) = ® pointwise on (–1, 1) and therefore also on [–1/2, 1/2]. Does not
x
n
x - 1 1 x
-
1
series converge to the also uniformly on the same intervals? Look at the modulus
1 x
-
x n+ 1 - 1 1 x n+ 1
f n ( )- f ( ) = - =
x
x
x - 1 1 x 1 x
-
-
On [–1/2, 1/2] our series converges uniformly to f since partial sums do
x n+ 1 1 1
f - f = sup £ n+ 1 2 = n ® 0.
n
sup
[ 1/2, 1/2] 1 x-
x Î - 2 2
On (–1, 1) the series does not converge to f since partial sums do not.
x n+ 1
f - f = sup = ¥ if x ® 1.
n
sup
( 1, 1) 1 x-
x Î -
¥
Let us fix a sequence of real-valued functions {g n n= 1 on [ , ].a b
}
Theorem 2: Continuity for Series
If (a) all g are continuous and (b) å ¥ g converges uniformly then å ¥ g is continuous.
n n= 1 n n= 1 n
.
+
Proof: Consider partial sum f = g + g + ... g We see that f is continuous, since it is a sum of
n
n
1
n
2
continuous functions. Also, by (b), f ® å ¥ g uniformly. Therefore, the continuity of the
n n= 1 n
¥
uniform limit å g is Riemann-integrable and we can integrate the series term by term
n= 1 n
¥ ¥
b b
a ò å g dx = å a ò g dx .
n
n
n= 1 n= 1
Proof: Since sum of R-integrable functions is R-integrable, we see that the partial sum
n n ¥
å i= 1 g = g + g + ... g is R-integrable and by (b) å i= 1 g ® å i= 1 g i
+
1
2
i
n
i
¥ b n b b ¥
å a ò g dx = limåò g dx = lim g dx = å g dx
a ò
i
i
i
i
i= 1 n®¥ i= 1 a n®¥ i= 1
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