1.3. Simple estimation techniques 9
Proposition 1.4 (H¨ older inequality). If I is an interval and f and g are
continuous functions on I then
I
|f(x)g(x)| dx
I
|f(x)|p
dx
1/p
I
|g(x)|q
dx
1/q
,
where 1 p, q with 1/p + 1/q = 1.
Proof. The case where
I
|f(x)|p
dx = 0 or
I
|g(x)|q
dx = 0
is trivial. Multiplying f and g by suitable positive real numbers, we may
therefore assume that
I
|f(x)|p
dx =
I
|g(x)|q
dx = 1
by homogeneity. Now
|f(x)g(x)| =
(|f(x)|p)1/p (|g(x)|q)1/q
=
(|f(x)|p)1/p (|g(x)|q)1−1/p,
so
log(|f(x)g(x)|) =
1
p
log
(|f(x)|p)
+ 1
1
p
log
(|g(x)|q).
But the logarithm function is strictly concave, that is to say, all the chords
lie strictly below the graph except for their endpoints. Hence
1
p
log
(|f(x)|p)
+ 1
1
p
log
(|g(x)|q)
log
1
p
log
(|f(x)|p)
+ 1
1
p
log
(|g(x)|q)
,
and so
|f(x)g(x)|
1
p
log
(|f(x)|p)
+ 1
1
p
log
(|g(x)|q).
Integrating over I yields
I
|f(x)g(x)| dx
1
p
·1 + 1
1
p
·1 = 1,
and this proves the inequality.
The case p = 2 is especially important; this is the Cauchy-Schwarz
inequality. The older inequality extends by induction to estimate integrals
of products of more than two functions. We have
I
|f1(x)···fn(x)| dx
I
|f1(x)|p1
dx
1/p1
· · ·
I
|fn(x)|pn
dx
1/pn
if 1 p1,...,pn with 1/p1 + · · · + 1/pn = 1.
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