# Kazdan inequality

Let $V$ be a real $N$-dimensional scalar inner product space (cf. also Inner product; Pre-Hilbert space), let $\mathcal{L} ( V )$ be the space of linear operators of $V$, and let $\mathcal{R} ( t ) \in \mathcal{L} ( V )$ be a given family of symmetric linear operators depending continuously on $t \in \mathbf{R}$. For $s \in \mathbf{R}$, denote by $C _ { S } : \mathbf{R} \rightarrow \mathcal{L} ( V )$ the solution of the initial value problem

\begin{equation*} X ^ { \prime \prime } ( t ) + \mathcal{R} ( t ) \circ X ( t ) = 0 \end{equation*}

\begin{equation*} X ( s ) = 0 , X ^ { \prime } ( s ) = I. \end{equation*}

Suppose that $C _ { 0 } ( t )$ is invertible for all $t \in ( 0 , \pi )$. Then for every positive $C ^ { 2 }$-function $f$ on $( 0 , \pi )$ satisfying $f ( \pi - t ) = f ( t )$ on $( 0 , \pi )$, one has

\begin{equation*} \int _ { 0 } ^ { \pi } d s \int _ { s } ^ { \pi } f ( t - s ) \operatorname { det } C _ { s } ( t ) d t \geq \end{equation*}

\begin{equation*} \geq \int _ { 0 } ^ { \pi } d s \int _ { s } ^ { \pi } f ( t - s ) \operatorname { sin } ^ { N } ( t - s ) d t, \end{equation*}

with equality if and only if $\mathcal{R} ( t ) = I$ for all $t \in ( 0 , \pi )$.

How to Cite This Entry:
Kazdan inequality. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Kazdan_inequality&oldid=49998
This article was adapted from an original article by H. Kaul (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article