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 H¨ 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.