Page 193 - DMTH401_REAL ANALYSIS

Unit 14: Sequences and Series of Functions

Notes
1
2
So (1 – x | > for |x| < which is a contradiction.
m
2
Consequently (f,,) is not uniformly convergent in [–2, 2].
Example: Discuss, for continuity, the convergence of the sequence ((f ) where
n
x
f (x) = x [0, ¥[.
n 1 nx
+
Solution: As you have seen that (f,,) ® f uniformly in [0, ¥[ where f(x) = 0, x [0,¥[.
Here each f,, is continuous in [0, ¥[ and also the uniform limit is continuous in [0, ¥[.

Example: Discuss for differentiability the sequence (f ) where
n
sin nx
f (x) = , " x [0, ¥[.
n
n
Solution: Here (f ) ® f uniformly where f(x) = 0 " x R. You can see that each f and fare
n n
differentiable in R and

f ’(x) = n cos nx and f’(x) = 0 " x R.
n
f ’(0) = n ®¥ whereas f’(0) = 0
n
lim f’ (0)  f’(0)
n®¥ n
i.e. limit of the derivatives is not equal to the derivative of the limit.

As you will see in the theorem for the differentiability of f and the equality of the limit of the
derivatives and the derivative of the limit, we require the uniform convergence of the sequence
(f ).
n

Example: Discuss for integrability the sequence (f,,) where

f(x) = n x e - nx 2 , x [0, 1],

Solution: If x = 0, then f (0) = 0
n
nx ¥
and lim f (0) = 0. If x  0, lim f (x) = lim which is of the form .
n n nx 2
n®¥ n®¥ e ¥
Applying L’ Hopital’s Rule, we have

nx x
lim = lim 2 = 0
e
n ® ¥ nx 2 n®¥ 2nxe nx
So (f ) ® f, pointwise, where f(x) = 0, " x [0, 1]
n
1 1 1
n
-
n ò
-
You may find that f (x) dx = (1 e ) and f(x) dx = 0
ò
0 2 0
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