convex function ConvexFunction 2001-10-15 22:34:11 2007-08-01 10:20:55 Definition concave function strictly convex function strictly concave function strictly convex strictly concave epigraph effective domain concave % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newcommand{\sR}[0]{\mathbb{R}} \newcommand{\sC}[0]{\mathbb{C}} \newcommand{\sN}[0]{\mathbb{N}} \newcommand{\sZ}[0]{\mathbb{Z}} {\bf Definition} Suppose $\Omega$ is a convex set in a vector space over $\sR$ (or $\sC$), and suppose $f$ is a function $f:\Omega\to \sR$. If for any $x,y\in \Omega$, $x\neq y$ and any $\lambda \in (0,1)$, we have $$f\Big( \lambda x + (1-\lambda)y\Big)\leq \lambda f(x)+(1-\lambda)f(y),$$ we say that $f$ is a \emph{convex function}. If for any $x,y\in \Omega$ and any $\lambda \in (0,1)$, we have $$f\Big( \lambda x + (1-\lambda)y\Big)\geq \lambda f(x)+(1-\lambda)f(y),$$ we say that $f$ is a \emph{concave function}. If either of the inequalities are strict, then we say that $f$ is a \emph{strictly convex function}, or a \emph{strictly concave function}, respectively. \subsubsection*{Properties} \begin{itemize} \item A function $f$ is a (strictly) convex function if and only if $-f$ is a (strictly) concave function. For this reason, most of the below discussion only focuses on convex functions. Analogous result holds for concave functions. \item On $\sR$, a continuous function is convex if and only if for all $x,y\in \sR$, we have $$f\left(\frac{x+y}{2}\right)\le\frac{f(x)+f(y)}{2}.$$ \item On $\sR$, a once differentiable function is convex if and only if $f'$ is monotone increasing. \item Suppose $f$ is twice continuously differentiable function on $\sR$. Then $f$ is convex if and only if $f'' \ge 0$. If $f''>0$, then $f$ is strictly convex. \item A local minimum of a convex function is a global minimum. See \PMlinkname{this page}{ExtremalValueOfConvexconcaveFunctions}. \end{itemize} \subsubsection*{Examples} \begin{itemize} \item $e^x$,$e^{-x}$, and $x^2$ are convex functions on $\sR$. Also, $x^4$ is strictly convex, but $12x^2$ vanishes at $x=0$. \item A \PMlinkname{norm}{NormedVectorSpace} is a convex function. \end{itemize} \subsubsection*{Remark.} We may generalize the above definition of a convex function to an that of an extended real-valued function whose domain is not necessarily a convex set. First, we define what an \emph{epigraph} of a function is. Let $\Omega$ be a subset of a vector space over the reals, and $f$ an extended real-valued function defined on $\Omega$. The \emph{epigraph} of $f$, denoted by $\operatorname{epi}(f)$, is the set $$\lbrace (x,r)\mid x\in \Omega,\mbox{ } r\ge f(x)\rbrace.$$ An extended real-valued function $f$ defined on a subset $\Omega$ of a vector space $V$ over the reals is said to be \emph{convex} if its \emph{epigraph} is a convex subset of $V\times\mathbb{R}$. With this definition, the domain $\Omega$ of $f$ need not be convex. However, its subset $\lbrace x\in \Omega\mid f(x) < \infty\rbrace$, called the \emph{effective domain} and denoted by $\operatorname{eff.dom}(f)$, is convex. To see this, suppose $x,y\in \operatorname{eff.dom}(f)$ and $z=\lambda x+(1-\lambda) y$ with $0\le \lambda \le 0$. Then $ (z,\overline{z})=\lambda(x,f(x))+(1-\lambda)(y,f(y))\in \operatorname{epi}(f)$, where $\overline{z}=\lambda f(x)+(1-\lambda)f(y)$, since $\operatorname{epi}(f)$ is convex by definition. Therefore, $z\in \operatorname{dom}(f)$. In fact, $f(z)\le \overline{z}=\lambda f(x)+(1-\lambda)f(y)<\infty$, which implies that $z\in \operatorname{eff.dom}(f)$.