Page 85 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes
f f .
2
n k
Choosing a number k, as large that k > n and n > n , we get
k
f n f f n f f f ,
2 2
n k
n k
n > n , we obtain f – f < , where is an arbitrary quantity.
n
f f X and hence X is a complete space.
n
p
p
Theorem: Let {f } be a sequence in L , 1 p < , such that f f a.e. and that f L .
n n
If lim f = f , then
n n p p
lim f n f = 0.
n p
Proof: Without any loss of generality, we may assume that each f 0 a.e. so that f is also 0 a.e.
n
–
+
since the result in general case follows by considering f = f – f .
Now, let a and b be any pair of non-negative real numbers, we have
|a – b| p 2 (|a| + |b| ),
q
p
p
1 p <
So, we get
p
p
2 (|f | + (|f| ) – |f – f| p 0 a.e.
p
n n
Thus, by Fatou’s Lemma and by the given hypothesis,
We get
p
2 p 1 |f| = lim 2 p f p f p f f p
n n n
lim inf 2 p f p f p f f p
n n n

= 2 p 1 lim f n p 2 p f p liminf f n f p
n n

p
= 2 p 1 f p limsup f n f .
n
p
Since f it follows that

p
limsup f n f 0.
n
p p
Therefore limsup f n f liminf f n f 0 ,
n n
p
So that lim f n f = 0
n

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