Page 64 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 5: Spaces, Hölder

Assume now we have equality. Then we must have equality in both (8) and in (9). Notes
q
q
p
p
p
On one hand, the equality in (8) forces a , a , , a and b , b , , b q k to be proportional (since
k
1
2
2
1
q
p
we assume the case n = k). On the other hand, the equality in (9) forces a , a p and b , b q k 1 to
k 1
be proportional (by the case m = 2). Since
k k
p
a p a and b q b q
j j ,
j 1 j 1
p
p
q
q
q
p
it is clear that a , a , , a , a p and b , b , , b , b q are proportional.
1 2 k k 1 1 2 k k 1
5.1.6 Riesz-Hölder's Inequality
p
q
Statement: Let p and q be conjugate indices or exponents (numbers) and f L [a, b], g L [a, b];
then show that
(i) f g L[a, b]
(ii) fg f g i.e.
p q
1 1
p q
|fg| |f| p |g| q
with equality only when |f| = |g| a.e. for some non-zero constants and .
q
p
Lemma: If A and B are any two non-negative real numbers and 0 < < 1, then
–
A B1 A + (1 – ) B, with equality when A = B.
Proof: If either A = 0 or B = 0, then the result is trivial.
Let A > 0, B > 0
Consider the function
(x) = x – x, where 0 x < and 0 < < 1

d d 2 1
x 1 and ( 1)x .
dx dx 2
d
Now solving = 0, we get x = 1.
dx

d 2
Also at x = 1, < 0 as 0 < < 1.
dx 2
By calculus, (x) is maximum at x = 1, so
(x) (1) i.e. x – x l – . … (1)
A
Now, putting x = , we get
B
A A A
1 or A B 1
B B B

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