Page 60 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 5: Spaces, Hölder

Notes
p
p
Therefore |f g| dx |f g| p |f g| dx .
A 1 A 2
Again, |f + g| (|f| + |g|) p (|g| + |g|) on A and (|f| + |f|) on A
p
p
p
2 1
p
2P |g| on A and 2 |f| on A
p
p
2 1
Integrating, we have
|f g| p 2 p |f| p
A 1 A 1
and |f g| p 2 p |g| p
A 2 A 2

p
Since f, g L [a, b] |f| p and |g| p
A 1 A 2

 |f g| p and |f g| p [by (i) and (ii)]
A 1 A 1

b
p
p
|f g| dx f + g L [a, b]
a
5.1.4 Simple Version of Hölder's Inequality

1 1
Lemma 1: Let p, q > 1 be such that 1 , and let u and v be two non-negative numbers, at
p q
least one being non-zero. Then the function f : [0, 1] R defined by
1
q q
f (t) = ut + v(1 t ) , t [0, 1],
has a unique maximum point at

1
u p q
s = … (1)
u p v p
The maximum value of f is

1
p p
max f(t) = (u p v ) … (2)
t [0, 1]
Proof: If v = 0, then f (t) = tu, t [0, 1] (with u > 0), and in this case, the Lemma is trivial.
Likewise, if u = 0, then
1
q q
f (t) = v(1 t ) , t [0, 1] (with v > 0), … (3)

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