Page 65 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes Or A B 1– – A (1 – ) B or
A B 1– A + B (1 – ) … (2)
Obviously equality holds good only for x = 1, i.e. only when A = B.

Proof of Theorem

Note that when p = 1, q = , the proof of theorem is obvious. Let us assume that 1 < p < and
1 < q < .
1
Now set = ; p > 1 < 1
p

1 1 1
Therefore = 1 –  1
q p q
Putting these values of and 1 – in (2), we get

1 1 A B
p
A B q … (3)
p q
If one of the functions f (x) and g (x) is zero a.e. then the theorem is trivial. Thus, we assume that
f 0, g 0 a.e. and hence the integrals
b b
p
q
|f| dx and |g| dx
a a
are strictly positive and hence f > 0, g > 0.
p q
f(x) g(x)
Set f (x) = , g(x)
f g
p q
1 1
and A p = |f(x)|, B q g(x) .

Then (3) gives

p q
f(x) g(x)
|f(x) g(x)|
p q
Integrating, we get

b b b
1 1
p
p
|f(x) g(x)|dx |f(x)| dx |g(x)| dx
p q
a a a
b b
1 |f(x)| p 1 |g(x)| q
= b dx b dx
p q
|f| dx |g| dx
a p a q
a a

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