# Hölder inequality

The Hölder inequality for sums. Let $\{a_s\}$ and $\{b_s\}$ be certain sets of complex numbers, $s\in S$, where $S$ is a finite or an infinite set of indices. The following inequality of Hölder is valid: \begin{equation}\label{eq:1} \Bigl|\sum\limits_{s\in S}a_sb_s\Bigr| \leq \Bigl(\sum\limits_{s\in S}|a_s|^p\Bigr)^{\frac1p}\Bigl(\sum\limits_{s\in S}|b_s|^q\Bigr)^{\frac1q}, \end{equation} where $p>1$ and $\frac1p + \frac1q =1$; this inequality becomes an equality if and only if $|a_s|^p = C|b_s|^q$, and $\arg(a_sb_s)$ and $C$ are independent of $s\in S$. In the limit case, when $p=1$, $q=+\infty$, Hölder's inequality has the form \begin{equation*} \Bigl|\sum\limits_{s\in S}a_sb_s\Bigr| \leq \Bigl(\sum\limits_{s\in S}|a_s|\Bigr)\sup\limits_{s\in S}|b_s|. \end{equation*} If $0<p<1$, Hölder's inequality is reversed. The converse proposition of Hölder's inequality for sums is also true (M. Riesz): If \begin{equation*} \Bigl|\sum\limits_{s\in S}a_sb_s\Bigr| \leq AB \end{equation*} for all $\{a_s\}$ such that \begin{equation*} \sum\limits_{s\in S}|a_s|^p \leq A^p, \end{equation*} then \begin{equation*} \sum\limits_{s\in S}|b_s|^q \leq B^q. \end{equation*}

For sums of a more general form, Hölder's inequality takes the form \begin{equation}\label{eq:2} \Bigl|\sum\limits_{s\in S}\rho_sa_{1s}\cdot\dots \cdot a_{ms}\Bigr|\leq \prod\limits_{k=1}^m\Bigl(\sum\limits_{s\in S}\rho_s|a_{ks}|^{p_k}\Bigr)^{\frac{1}{p_k}}\quad \rho_s \geq 0, \end{equation} if $\frac{1}{p_1} +\dots + \frac{1}{p_m}=1,\quad p_k>1$, and $1\leq k\leq m$.

The Hölder inequality for integrals. Let $S$ be a Lebesgue-measurable set in an $n$-dimensional Euclidean space $\mathbb R^n$ and let the functions \begin{equation*} a_k(s) = a_k(s^1,\dots,s^n),\quad 1\leq k\leq m, \end{equation*} belong to $L_{p_k}(S)$. The following inequality of Hölder is then valid: \begin{equation*} \Bigl|\int\limits_{S}a_1(s)\cdot\dots \cdot a_m(s)\, ds\Bigr|\leq \prod\limits_{k=1}^m\Bigl(\int\limits_{S}|a_k(s)|^{p_k}\Bigr)^{\frac{1}{p_k}}. \end{equation*}

If $m=p=q=2$, one obtains the Bunyakovskii inequality. Analogous remarks (concerning the sign and the limit case) as were made for the Hölder inequality

are also valid for the Hölder inequality for integrals.

In the Hölder inequality the set $S$ may be any set with an additive function $\mu$ (e.g. a measure) specified on some algebra of its subsets, while the functions $a_k(s)$, $1\leq k\leq m$, are $\mu$-measurable and $\mu$-integrable to degree $p_k$.

The generalized Hölder inequality. Let $S$ be an arbitrary set, let a (finite or infinite) functional $\phi:a\to\phi(a)$ be defined on the totality of all positive functions $a:S\to\mathbb R$ and let this functional satisfy the following conditions: a) $\phi(0)=0$; b) $\phi(\lambda a)=\lambda\phi(a)$ for all numbers $\lambda>0$; c) if $0<a(s)\leq b(s)$, then the inequality $\phi(a)\leq \phi(b)$ is valid; and d) $\phi(a+b) \leq \phi(a) + \phi(b)$. If the conditions are also met, the generalized Hölder inequality is valid for the functional: \begin{equation*} \phi(|a_1\cdot\dots \cdot a_m|)\leq \prod\limits_{k=1}^m[\phi(|a_k|^{p_k})]^{\frac{1}{p_k}}. \end{equation*}

How to Cite This Entry:
Hölder inequality. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=H%C3%B6lder_inequality&oldid=28956
This article was adapted from an original article by L.P. Kuptsov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article