Page 162 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 12: General Convergence Theorems

Notes
nx sinx
1
1 (nx)
If (x) = 1, x, then | fn (x) | (x), x.
Hence by dominated convergence theorem, we get

1 1 1
nx sinx nx sinx
Lim dx Lim dx (0) dx 0
n 1 (nx) n 1 (nx)
0 0 0
1
x sinx
Limn dx = 0
n 1 (nx)
0
1
x sinx
dx = 0 (n ).
–1
1 (nx)
0
2 2
2
n xe n x
Example: Show that Lim dx = 0, if a > 0, but not for a = 0.
n 1 x 2
a
putting nx u
Solution: If a > 0, , we get
and du ndx
2
n xe n 2 x 2 ue u 2 du ue u 2
dx du
2
2
1 x 2 1 u /n 2 (na, ) 1 u /n 2
a na o
2
u
u.e 2
Also (na, ) u.e u L [0, ]
2
2
1 (u /n )
u.e u 2
and Lim (ne, ) 0 as ( , ) 0 .
2
n 1 u /n 2
Hence by Lebesgue dominated convergence theorem, we obtain
2 2 u 2
n x
2
n xe u.e
Lim dx Lim du
2
n 1 x 2 n (na, ) 1 u /n 2
a a
u.e u 2
= Lim (na, ) du 0 dx 0 .
2
n 1 u /n 2
a o
Now when a = 0,
2 2
2 2
2
2
n xe n x 1 n xe n x
dx dx
1 x 2 1 x 2
o 0

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