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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 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 concave function. If either of the inequalities are strict, then we say that $f$ is a strictly convex function, or a strictly concave function, respectively.
- 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.
- On $\sR$ , a continuous function is convex if and only if for all $x,y\in \sR$ , we have $$f\left(\frac{x+y}{2}\right)\le\frac{f(x)+f(y)}{2}.$$
- On $\sR$ , a once differentiable function is convex if and only if $f'$ is monotone increasing.
- 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.
- A local minimum of a convex function is a global minimum. See this page.
- $e^x$ ,$e^{-x}$ , and $x^2$ are convex functions on $\sR$ . Also, $x^4$ is strictly convex, but $12x^2$ vanishes at $x=0$ .
- A norm is a convex function.
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 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 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 convex if its 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 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 $z\in \operatorname{eff.dom}(f)$ .
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