lqm.tex 3. marca 2005

Hölder inequality and averages

Let us denote

A_{p} = (\frac{∑_{k=1}^{n} a_{k}^{p}}{n} )^{\frac{1}{p}} .

I.e. A_{p} is a generalization of arithmetic average. Proposition 1. 1 ≤ p ≤ q ⇒ A_{p}(\overline{a}) ≤ A_{q}(\overline{a}) Proof. The inequality

(\frac{∑_{k=1}^{n} a_{k}^{p}}{n} )^{\frac{1}{p}} ≤ (\frac{∑_{k=1}^{n} a_{k}^{q}}{n} )^{\frac{1}{q}}

is equivalent to

\frac{∑_{k=1}^{n} a_{k}^{p}}{n} ≤ (\frac{∑_{k=1}^{n} a_{k}^{q}}{n} )^{\frac{p}{q}} .

Let b_{k} := a_{k}^{p} and u = \frac{q}{p}. Then u ≥ 1 and we can rewrite this inequality as

\frac{∑_{k=1}^{n} b_{k}}{n} ≤ (\frac{∑_{k=1}^{n} b_{k}^{u}}{n} )^{\frac{1}{u}}

∑_{k=1}^{n}b_{k} ≤ n^{(1− \frac{1}{u})} (∑_{k=1}^{n}b_{k}^{u})^{\frac{1}{u}}

The last inequality follows from Hölder inequality (for vectors (1, 1, …, 1) and (b_{1}, …, b_{n})).