Page 62 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 5: Spaces, Hölder

and f (t) < 0 t (s, 1). Notes
This means that, f is increasing on [0, s] and decreasing on [s, 1], and we are done.
The maximum value of f is then given by

max f(t) = f (s),
t [0, 1]
and the fact that f (s) equals the value in (2) follows from an easy computation.

5.1.5 Hölder's Inequality

Statement: Let a , a , …, a , b , b , …, b be non-negative numbers. Let p, q > 1 be real number
1 2 n 1 2 n
1 1
with the property 1 . Then
p q

1 1
n
n p n q
a b p q … (5)
j j a j b j
j 1
j 1 j 1
q
p
Moreover, one has equality only when the sequences a , , a p and b , , b q are proportional.
1 n n
Proof: The proof will be carried on by induction on n. The case n = 1 is trivial.
Case n = 2.
Assume (b , b ) (0, 0). (otherwise everything is trivial).
1 2
Define the number

b
r = 1 1 .
b q 1 b q q
2
Notice that r [0, 1] and we have

b 2 1
1 = 1 r q q
b q b q q
1 2

1
Notice also that, upon dividing by b q b q q , the desired inequality
1 2
1 1
a b + a b a q a q p b q b q q … (6)
1 1 2 2 1 2 1 2
reads

1 1
a r + a 1 r q q a q a q p … (7)
1 2 1 2
It is obvious that this is an equality when a = a = 0. Assume (a , a ) (0, 0), and set up the
1 2 1 2
function.
1
f (t) = a t a 1 t q q , t [0, 1].
2
1

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