Page 222 - DMTH401_REAL ANALYSIS

Real Analysis

Notes
x
" n ³ N " Î : I f ( )- f ( ) £ e /2 < e
x
x
n
what by definition means that f ® f uniformly.
n
Theorem 5: Weierstrass M-test
.
Let {g } ¥ be a sequence of real-valued functions on I Ì  Let {M } ¥ be a sequence of
n n= 1 n n= 1
x
x
number s.t. å ¥ M < . ¥ If g n ( ) £ M n , " Î " Î  then å ¥ g converges uniformly.
n
I
n= 1 n n= 1 n
¥ n } ¥
Proof: å M < ¥ means {å M i converges and therefore is a Cauchy sequence i.e. given
i= 1 i i= 1 n= 1
e > 0 we can find N Î s.t. without loss of generality
n m n n
i å
" , n m Î , n > m ³ N : å M - M = å M = å M < e /2.
i
i
i
i= 1 i= 1 i m+ 1 i m+ 1
=
=
n ¥
Let’s prove that {å g i } is a uniform Cauchy sequence. Given e > 0, pick N as above. Then
1
i=
n= 1
" n > m ³ N Î " x Î I
n m n
å g i ( )- å g i ( ) = å g i ( ) £
x
x
x
i= 1 i= 1 i m+ 1
=
n n
£ å g i ( ) £ å M < e /2.
x
i
=
i m+ 1 i m+ 1
=
We get
n m
i å
å g - g i £ e /2 < e .
i= 1 i= 1 sup
Therefore {å n g } ¥ is a uniform Cauchy sequence and converges uniformly and so does
i= 1 i n= 1
¥
å i= 1 g i .
Example: Consider series å ¥ sin(nx ) on  By Weierstrass M-test, we see that this
.
i= 1 2 n
series converges uniformly on  since
sin(nx ) 1 ¥ 1
£ and å < ¥
2 n 2 n n= 1 2 n
¥ 1
Example: Consider series å n= 1 n + x on [0, ¥ By Weierstrass M-test, we can obtain
).
2
uniform convergence of this series on [0, )¥ since
¥
1 £ 1 and å 1 < ¥
2
n + x n 2 n= 1 n 2

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