Page 63 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes We now apply Lemma (1) stated above, which immediately gives us (7).
Let us examine when equality holds.
If a = a = 0, the equality obviously holds, and in this case (a , a ) is clearly proportional to (b ,
1 2 1 2 1
b ). Assume (a , a ) (0, 0).
2 1 2
Again by Lemma (1), we know that equality holds in (7), exactly when

1
a p q
r = 1
a p 1 a p 2
1
b a p q
that is 1 = 1 ,
1 a p a p
b q 1 b q q 1 2
2
or equivalently

b q 1 a p 1
b q b q = a p a p .
1 2 1 2
Obviously this forces

b q a q
1 q = 1 p ,
b q 1 b 2 a p 1 a 2

q
p
q
so indeed a , a p and b , b are proportional.
1 2 1 2
Having proven the case n = 2, we now proceed with the proof of:
The implication: Case n = k case n = k + 1, start with two sequences (a , a , …, a , a ) and
1 2 k k+1
(b , b , …, a , b ).
1 2 k k+1
Define the numbers
1 1
k p k q
a = a p and b = b q .
j j
j 1 j 1
Using the assumption that the case n = k holds, we have
1 1
k 1 k p k q
a b j a p j b q j + a b k+1
j
k+1
j 1 j 1 j 1
= ab + a b … (8)
k+1 k+1
Using the case n = 2, we also have
1 1
ab + a b a p a p p b q b q q
k+1 k+1 k 1 k 1
1 1
k 1 p k 1 q
= a p b q , … (9)
j j
j 1 j 1
so combining with (8) we see that the desired inequality (5) holds for n = k + 1.

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