Page 76 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 6: Minkowski Inequalities

Adding these two, Notes
1 1 1
|f g| p/q (|f| |g|) |f| p p |g| p p |f g| p q … (iii)

1 1 p
Also 1 1 p
p q q

p p
q
p
|f + g| = |f g||f g| q |f| |g| |f g| , as f 0, g 0
1
p 1 1 q
|f g| = |f| p p |g| p p |f g| p

1
q
Dividing by |f g| p , we get
1 (1/q)
|f g| p f g
p p

1/q
|f g| p f + g
p p
f + g f + g
p p p
6.1.2 Minkowski Inequality in Integral Form

Statement: Suppose f :  ×   is Lebesgue measurable and 1 p < . Then

1/p 1/p
p
h(x, y) dy dx h(x, y) dx dy

Proof: By an approximation argument we need only consider h of the form
N
h(x,y) f (x)1 f (y), (x,y)   ,
j j
j 1
where N is a positive integer, f is Lebesgue measurable, and F L , j = 1, … N and F F = if
j j n i i j
1 i < j N. We use Minkowski’s inequality to estimate

1/p
p
1/p N N 1/p
p
h(x,y) dy dx = F f (x) F j f (x) dx
j
i
j
j 1 j 1
dx
But
p 1/p N 1/p N 1/p
p
p
h(x,y) dx dy = |h(x,y)| dx |f (x)| dx
j
F i F j
j 1 j 1

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