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 June 28th, 2012, 03:06 AM #1 Senior Member   Joined: Apr 2012 Posts: 106 Thanks: 0 Convex set, convex function Let's have a convex function $f:\mathbb{R}^n \rightarrow \mathbb{R}$. We need to prove that $A= \left\{(x_1,...,x_n, x_{n+1}) \in \mathbb{R}^{n+1} | f(x_1, ..., x_n) \leq x_{n+1}\right\}$ is a convex set. Any help would be appreciated.
 June 30th, 2012, 02:57 PM #2 Member   Joined: Jun 2012 From: Uzbekistan Posts: 59 Thanks: 0 Re: Convex set, convex function Let $a,b\in A$ and $0\leq\alpha\leq 1$. Then $a=(a_1,a_2,\ldots a_n,a_{n+1}),\: b=(b_1,b_2,\ldots b_n, b_{n+1})$ and $f(a_1,a_2,\ldots a_n)\leq a_{n+1},\: f(b_1,b_2,\ldots b_n)\leq b_{n+1}$. We have: $c=\alpha a+(1-\alpha)b=(\alpha a_1+(1-\alpha)b_1,\ldots \alpha a_n+(1-\alpha)b_n, \alpha a_{n+1}+(1-\alpha)b_{n+1})$. As $f$ is convex so $f(\alpha a_1+(1-\alpha)b_1,\ldots \alpha a_n+(1-\alpha)b_n)=f(\alpha(a_1,\ldots a_n)+(1-\alpha)(b_1,\ldots b_n))\leq \alpha f(a_1,\ldots a_n)+(1-\alpha)f(b_1,\ldots b_n)\leq \alpha a_{n+1}+(1-\alpha)b_{n+1}$. Thereby $c\in A$, i.e. $A-$convex set.

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