Hölder's Inequality

Hölder's Inequality is the "big brother" of Cauchy-Schwarz. It provides a bound for the sum of products involving different powers ( $p$ and $q$ ).

Statement

Theorem (Hölder's Inequality)

Let

a_{1}, \dots, a_{n}

and

b_{1}, \dots, b_{n}

be non-negative real numbers. Let

p, q > 1

be "conjugate exponents" such that

\frac{1}{p} + \frac{1}{q} = 1

. Then:

i = 1 \sum n a_{i} b_{i} \leq (i = 1 \sum n a_{i}^{p})^{1/ p} (i = 1 \sum n b_{i}^{q})^{1/ q}

If we set

p = 2

and

q = 2

(since

1/2 + 1/2 = 1

), Hölder becomes:

\sum a_{i} b_{i} \leq (\sum a_{i}^{2})^{1/2} (\sum b_{i}^{2})^{1/2}

Squaring both sides gives the standard Cauchy-Schwarz inequality.

Hölder can prove bounds involving sums of cubes, fourth powers, etc., where Cauchy-Schwarz (squares) might not fit.

Example

Prove that for

a, b, c > 0

(a^{3} + b^{3} + c^{3})^{1/3} (1 + 1 + 1)^{2/3} \geq a + b + c

Proof. Apply Hölder with sequences $(a, b, c)$ , $(1, 1, 1)$ , $(1, 1, 1)$ ? Wait, simpler form: Using $p = 3, q = 3/2$ :

\sum a \cdot 1 \leq (\sum a^{3})^{1/3} (\sum 1^{3/2})^{2/3}

a + b + c \leq (a^{3} + b^{3} + c^{3})^{1/3} (3)^{2/3}

Cubing both sides:

(a + b + c)^{3} \leq 9 (a^{3} + b^{3} + c^{3})

. ∎

Exercise (Problem 1)

Prove that if

x, y, z > 0

and

x + y + z = 1

, then

x^{4} + y^{4} + z^{4} \geq \frac{1}{27}

Exercise (Problem 2)

State Hölder's inequality for integrals.

Exercise (Problem 3)

Use Hölder's to find the minimum of

x^{5} + y^{5}

subject to

x + y = 2