Holders Inequality

Definition

For sequences and weights , Holders states that

With the equality holding if and only if the sequences are proportional to each other

Proof

1. First note that the inequality trivially holds. Therefore .
2. Define a normalised sequence
4. Thus it suffices to prove
5. By Weighted AM-GM
6. Summing over gives as desired

Variants

Cauchy-Schwarz

Using on Holders Inequality with then transformations )

Note that now we introduced the squares we have expanded the domain so now we get that

Case

Using Holders with two sequences and weights for Let and and , Then

With the equality only holding when the sequences are proportional to each other.

Proof

1. Plugging in conditions to Holders we get
2. Rearranging and raising both sides to the power of gets as desired .

Alternate Proof

1. Since the inequality is homogeneous we can scale both and such that
2. Then by Weighted AM-GM

More concretely the scaling that is used is and . So we yield that and So both sides scale by a positive factor of so we can divide through and now we have

Classical Conjugate-Exponent Form

Using the case and making transformations and renaming exponents so that (This adds the condition ) we get

Titu’s Lemma

Titu’s Lemma states that for and

Note this is just Cauchy-Schwarz applied to the sequences and hence the extra restrictions on . We can extend the domain to as if stays the same and will decrease or remain constant. This is a very useful version for eliminating fractions.