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I'm just learning about convexity and affineness, and I've read over some similar questions asked here, but those were more about general properties. I need some help applying those properties to a few examples that I can't seem to understand. I'd like to see how to show:

  1. $C_{1} = \lbrace x \in \mathbb{R}^{n} \mid h(x) = 0 \rbrace $ is convex iff $h(x)$ is affine in $C_{1}$

  2. $C_{2} = \lbrace x \in \mathbb{R}^{n} \mid g(x) \leq 0 \rbrace $ is convex if $g(x)$ is convex on $C_{2}$

  3. $C_{3} = \lbrace x \in \mathbb{R}^{n} \mid h_{i}(x) = 0,i=1,\ldots,m;\,g_{j}(x) \leq 0,j=1,\ldots,l \rbrace$ is convex if each $h_{i}(x) $ is affine and each $g_{j}(x)$ is convex in $C_{3}$

So far, for (1), I think I can assume $\forall x \in C_{1}, h(x)=a^{T}x+b $ for some $a \in \mathbb{R}^{n}$ and $b \in \mathbb{R}$ but I'm not 100% on where to go from there.

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1.$\rightarrow$. Assume that $C_1$ is convex. Then by definition $$\sum_{i=1}^{n}\lambda_ix_i \in C_1,$$ for all $x_i \in C_1$ and nonnegative numbers $\lambda_1,\ldots,\lambda_n$ such that $\sum \lambda_i=1.$ Then $$h\big(\sum_{i=1}^{n}\lambda_ix_i\big)=0=\sum_{i=1}^{n}\lambda_ih(x_i),$$ where both sides are equal to zero due to the definition of $C_1$ and the fact that $x_i \in C_1$ as well as $\sum_{i=1}^{n}\lambda_ix_i \in C_1$. So $h$ is affine (check in wikipedia the alternative definition of affineness. I use it in the next direction).

2.$\leftarrow$. Assume $h$ is affine. Then by definition $$h\big(\sum_{i=1}^{n}\lambda_ix_i\big)=\sum_{i=1}^{n}\lambda_ih(x_i),$$ for all $x_i \in C_1$ and nonnegative numbers $\lambda_1,\ldots,\lambda_n$ such that $\sum \lambda_i=1.$ Then $$h\big(\sum_{i=1}^{n}\lambda_ix_i\big)=\sum_{i=1}^{n}\lambda_ih(x_i)=\sum_{i=1}^{n}\lambda_1\cdot0=0,$$ Thus,$\sum_{i=1}^{n}\lambda_i(x_i) \in C_1$ and therefore $C_1$ is convex.

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  • $\begingroup$ Analogously proceed with 2 and 3. If you have a problem let me know. $\endgroup$ – Jimmy R. Feb 22 '14 at 17:59
  • $\begingroup$ So, I would think that to start off for (1), you'd have to start with the affineness assumption first? As in prove convexity by using $h(x)$'s affine nature? $\endgroup$ – helposaurus Feb 22 '14 at 21:51
  • $\begingroup$ You have to show two directions. Firstly that convexiry implies affineness and secondly that affineness implies convexity or vise versa. The order in which you will do it is not important. Is that what you asked? $\endgroup$ – Jimmy R. Feb 22 '14 at 22:41
  • $\begingroup$ I guess so, yeah, I just read the problem as: show that the set $C_{1}$ is convex if and only if $h(x)$ is affine in $C_{1}$. Which in my mind just means start with showing convexity, sorry, I'm a newb here hah $\endgroup$ – helposaurus Feb 22 '14 at 23:48
  • $\begingroup$ If and only if means that you have to show two implications. But the order is not important. $\endgroup$ – Jimmy R. Feb 23 '14 at 6:49

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