Page 219 - DMTH401_REAL ANALYSIS

Unit 17: Uniform Convergence and Differentiability

Notes
x
t
c
dt
=
f
Part 1: Define ( ): limf c = f n ( ) and ( ): f ( ) + c ò h ( ) .
c
x
n®¥
¢
Note that ( )h t is R-integrable since it is a uniform limit of continuous functions f (Theorem 1).
n
Therefore, according to the definition of ( )f x and by the fundamental theorem of calculus we can
d x d x a d x
c
=
f
see that f ¢ ( ) = ( ( )+ c ò h ( ) ) = ( ( )+ a ò h ( )dt + c ò h ( ) ) = a ò h ( ) h ( ), x [ , ].a bÎ
t
x
dt
t
t
t
x
"
dt
c
f
dx dx dx
b
Part 2: We want to show that f ® f uniformly on [ , ]. By the fundamental theorem of calculus:
a
n
x
f ( ) = f ( )+ f ¢ ( ) .
c
t
dt
x
n n c ò n
c
c
c
¢
c
.
h
Put e > 0. Since f n ( ) ® f ( ) we have f n ( ) - f ( ) < e /2 for all n ³ N 1 Also f ® uniformly,
n
¢
))
-
so we have f - h sup < e /(2(b a for all n ³ N 2 Therefore put N = max{N N 2 Then
,
.
}.
n
1
" n ³ N " Î [ , ] we get
b
a
x
x x
c
t
f n ( )- f ( ) = f n ( )- f ( )+ c ò f n ¢ ( )dt - c ò h ( )dt £
x
x
c
t
x
t
t
c
-
c
£ f n ( ) - f ( ) + c ò f n ¢ ( ) h ( )dt £
x
£ f n ( ) - f ( ) + c ò f n ¢ ( ) h ( ) dt £
c
t
-
t
c
x
¢
c
c
£ f n ( ) - f ( ) + f - h sup ò c 1dt =
n
¢
c
c
-
= f n ( )- f ( ) + f - h sup x c £
n
¢
c
-
£ f ( )- f ( ) + f - h (b a £
)
c
n n sup
e
)
< e /2 + (b a = e
-
2(b a )
-
x
x
Example: Theorem 1 is not true if we replace [ , ]a b by . Look at f n ( ) sin( / ) on .
=
n
1 1
x
Then f n ¢ ( ) = cos( / ) ® 0 uniformly cos( / ) = 1/n ® 0}. Conditions (a) – (d) of
x
n
n
x
n n sup
0
Theorem 1 are satisfied but f ® uniformly.
n
17.2 Series of Functions
¥
}
Definition 1: Let I Ì  and {g n n= 1 be a sequence of real-valued functions on I. Then
¥
å g n
n= 1
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