[Math] Prove that the intersection of convex sets is convex using the following three points…

convex optimizationconvex-analysisoptimization

I want to prove each point, then, use points (1) and (2) to prove (3).

  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 $C_{3}$

Once I have (3), I have an actual application to try, where I need to show the following set is convex:

$C = \lbrace x \in \mathbb{R}^{3} \mid 3x_{1} – 2x_{2} + x_{3}max(3,x_{1}^2 -2x_1x_2 -6)=3,x_1+x_2+x_3 \leq 5,(x_3-3)^2-(x_1+2x_2-13)^3 \leq10,(x_1-3)^2+(x_2-3)^2 \leq 1.5 \rbrace$

I don't really understand the notation of the last part, it seems like a huge mess, so making sense of that would be nice too.

I have part (1), I know part (2) is trivial almost, but I can't show it. I know (3) is the intersection proof, which requires (1) and (2), and itself would be convex if $h$ was affine and $g$ was convex, but again I'm lost.

Best Answer

Here is the gist of part 2. Let me know if you have any questions.

Let $x_1, x_2 \in C_2$. Then as $g$ convex over $C_2$ this implies that $$g(tx_1 +(1-t)x_2) \le tg(x_1) + (1-t)g(x_2) \le 0$$ as $x_1,x_2\in C_2$ where $t\in [0,1]$ implying that $tx_1 + (1-t)x_2 \in C_2$ and thus $C_2$ is convex.

Part 3 is very similar to part 1 and 2.

Let $x_1, x_2 \in C_3$ Then for any $g_i$ as $g_i$ convex in $C_3$, $$g_i(tx_1 +(1-t)x_2) \le tg_i(x_1) + (1-t)g_i(x_2) \le 0$$ as $x_1,x_2\in C_2$ where $t\in [0,1]$

Now you need to show the affine part for $x_1,x_2$ ;)

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