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Let $K_1$ and $K_2$ be two convex cones, including the origin, in a real vector space. Show that $K_1 + K_2 = \text{conv}(K_1 \cup K_2)$.

It is straight forward to show that $K_1 + K_2$ is a convex cone. Tow show the statement we need to proof $K_1 + K_2 \subseteq \text{conv}(K_1 \cup K_2)$ and $\text{conv}(K_1 \cup K_2) \subseteq K_1 + K_2$.

To show $K_1 + K_2 \subseteq \text{conv}(K_1 \cup K_2)$ we can write

Let $x \in K_1+ K_2$ and $y \in K_1+ K_2$, therefore

$\lambda x+(1-\lambda)y \in K_1+ K_2$ because $K_1 + K_2$ is a convex cone.

To show $\text{conv}(K_1 \cup K_2) \subseteq K_1 + K_2$

Let $z \in \text{conv}(K_1 \cup K_2)$.

How can we proceed?

Saeed
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  • do the cones have the vertex at $0$ ? – Hayk Oct 25 '18 at 18:28
  • I do not see how the proof for the first case is complete – LinAlg Oct 26 '18 at 00:39
  • Proof for the first case is has not been completely written because when $K_1$ and $K_2$ are convex cone, we are done for the first case. – Saeed Oct 26 '18 at 00:46
  • if you only need to show $\mathrm{conv} (K_1 \cup K_2) \subset K_1 + K_2$ then it's a trivial consequence of the definition of $\mathrm{conv}$ and the fact that both $K_1$ and $K_2$ contain the origin. Indeed, $K_1 = K_1 + 0\subset K_1 + K_2$ and $K_2 = K_2 + 0 \subset K_1 + K_2$ as $0 \in K_1 \cap K_2$. Since $K_1 + K_2$ is convex, and $\mathrm{conv}(K\cup K_2)$ is the smallest convex set containing the union, the desired inclusion follows. – Hayk Oct 26 '18 at 18:05
  • To prove that we need to take an element in the first set and show that it is in the second set. What you are doing does not make sense to me in this sense. Can you write it rigorously? – Saeed Oct 26 '18 at 18:21
  • @Sepide, please write my name with @ in the comment, so that I get a notification. Now the proof, what I wrote is a rigorous proof of the last inclusion in your question. Just check the definition of $\mathrm{conv}$ - the convex hull (on Wikipedia e.g.) and then the argument should become clear to you. Taking an element from one set and showing that it's in the second one, is one way of doing things here. That is not the only approach, however. Again, if $K_1 \cup K_2 \subset X$ for some convex set $X$ then necessarily $\mathrm{conv}(K_1\cup K_2)$ must be a subset of $X$ by definition of conv. – Hayk Oct 26 '18 at 18:45
  • @ Hayk: I want to show in the way that I wrote not using other implications. Can you help me to write it in that way? – Saeed Oct 26 '18 at 18:49
  • @Sepide, what is the definition of $\mathrm{conv}$ you can use? also, please do not leave space between @ and the name. – Hayk Oct 26 '18 at 19:11

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