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Let $C_1$ and $C_2$ be convex cones in $\mathbb{R}^n$. Show that $C_1\bigoplus C_2$ is also a convex cone and that $C_1 \bigoplus C_2 = $ conv$(C_1 \cup C_2)$.

My attempt:

Need to show that $\boldsymbol{x}\in C_1$, $\boldsymbol{y}\in C_2$ implies that $\lambda\boldsymbol{z}\in C_1\bigoplus C_2$ for all $\lambda\ge 0$.

Let $\boldsymbol{x}\in C_1$ and $\boldsymbol{y}\in C_2$.

$C_1$ and $C_2$ are convex cones, so $\lambda \boldsymbol{x}\in C_1$ and $\lambda \boldsymbol{y}\in C_2$.

Now, let $\boldsymbol{z}=\boldsymbol{x}\bigoplus\boldsymbol{y}$. Assume that $\boldsymbol{z}\in$

I am stuck at this point. I have seen a similar definition somewhere that begins by assuming that $\boldsymbol{x}\in C_1\bigoplus C_2$, but I think it is better (more correct?) to begin with $\boldsymbol{x}, \boldsymbol{y}$ only in $C_1, C_2$ and deduce the conclusion from there. Unless that somehow isn't possible.

I am sure this is, in fact, simple, but I would appreciate polite feedback and advice. Thank you for your time

user391838
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  • There are three claims: (1) $C_1\oplus C_2$ is a convex cone, (2) $C_1\oplus C_2\subset\mathrm{conv}(C_1\cup C_2)$, and (3) $C_1\oplus C_2\supset\mathrm{conv}(C_1\cup C_2)$. Which claim are you addressing? – Chris Culter Oct 24 '17 at 17:46
  • The claim that knowing $C_1$, $C_2$ are convex cones implies that $C_1\bigoplus C_2$ is a convex cone. – user391838 Oct 24 '17 at 18:25
  • Okay. And are you verifying some defining characteristic of convex cones, or are you verifying the two separate sub-claims that it is a convex set and that it is a cone? – Chris Culter Oct 24 '17 at 18:28
  • I'm asking if you can explain the proof, or tell me how to finish my proof attempt if it is correct so far. It would be necessary to show that the direct sum is a convex cone. – user391838 Oct 24 '17 at 18:56
  • Sure, and I'm asking what the goal of your proof is. Whether you choose $x\in C_1$ or $x\in C_1\oplus C_2$ depends entirely on what definition or theorem you're trying to take advantage of. Put it this way: What is your working definition of a convex cone, and of $C_1\oplus C_2$? – Chris Culter Oct 24 '17 at 19:45
  • A nonempty, convex set $C$ in $\mathbb{R}^n$ is called a convex cone with vertex zero if $\boldsymbol{x}\in C$ implies that $\lambda\boldsymbol{x}\in C$ for all $\lambda \ge 0$. – user391838 Oct 28 '17 at 23:21
  • Convex Hull: Let $S$ be an arbitrary set in $\mathbb{R}^n$. The convex hull of $S$, denoted conv$(S)$, is the collection of all convex combinations of $S$. In other words, \linebreak $\boldsymbol{x} \in$ conv$(x)$ if and only if $\boldsymbol{x}$ can be represented as

    $\boldsymbol{x}=\sum_{j=1}^k \lambda_j\boldsymbol{x}_j$, where

    $\sum_{j=1}^k \lambda_j = 1$,

    $\lambda_j \ge 0 \quad$ for $j=1,...,k$, where $k$ is a positive integer and $\boldsymbol{x}_1$,...,$\boldsymbol{x}_k \in S$

     – user391838 Oct 28 '17 at 23:22
  • @ChrisCulter I erroneously assumed you were belittling me with your earlier questions, but on closer inspection I see that you were just inviting me to understand that the version of the definition I need to satisfy was important (which I did understand). I was anticipating rude comments because I have received them before. I apologize for doubting your intentions. Fact is, I have had the working definitions inside my own proof from the beginning but still do not see how to get there. I suppose I can start by letting my convex set equal the direct sum (or union) in the working def, rather than – user391838 Oct 28 '17 at 23:27
  • ...go piece-by-piece – user391838 Oct 28 '17 at 23:30

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