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I'm doing convex analysis studies and I have the following problem to prove:

Show that, if $C$ is a convex cone, then

$\lambda_1\textbf{x}_1+\lambda_2\textbf{x}_2\in C$, with $\textbf{x}_1, \textbf{x}_2\in C$ and $\lambda_1, \lambda_2 \geq0$

I have tried solving this myself and I just want verification for my solution :) I mean could I have done it in a smarter/cooler way etc ;D (assuming it's correct)

Here is my attempt:

Because $\textbf{x}_1, \textbf{x}_2\in C$ and $C$ is a cone, I know that:

$\lambda_1\textbf{x}_1\in C$ and $\lambda_2\textbf{x}_2\in C, $ for all $\lambda_1, \lambda_2 \geq0$. Now because $C$ is also convex set, I know that:

$$\lambda\lambda_1\textbf{x}_1+(1-\lambda)\lambda_2\textbf{x}_2\in C, $$ for some $\lambda\in[0,1]$. I can assign $a :=\lambda\lambda_1$ and $b:=(1-\lambda)\lambda_2$,

where $:=$ is the assignment operator.

So I get

$$a\textbf{x}_1+b\textbf{x}_2\in C,\;\;\;\;\text{with} \;\;a,b\geq0 \;\;\;\;\;\blacksquare$$

Is my answer valid or not? =) Thank you for any help!

jjepsuomi
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    Seems alright to me. – SPK.z Sep 11 '14 at 12:21
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    $\lambda_1:=\lambda\lambda_1$? Then $\lambda=1$ or $\lambda_1=0$. – drhab Sep 11 '14 at 12:24
  • B.t.w., I think you mean "for all $\lambda_1, \lambda_2$" and "for all $\lambda$" instead of " for some $\lambda_1, \lambda_2$" and "for some $\lambda$". And as @drhab suggests, you should give your final $\lambda_1$ and $\lambda_2$ a different name. – SPK.z Sep 11 '14 at 12:25
  • +1 @drhad I assign the new value $\lambda\lambda_1$ in place of $\lambda_1$, like if I would assign a variable in C++ etc. =) I'm not sure but doesn't $:=$ mean assignment and $=$ means equality =) – jjepsuomi Sep 11 '14 at 12:27

1 Answers1

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Trivial if $\lambda_{1}=\lambda_{2}=0$.

If $\lambda_{1}+\lambda_{2}>0$ then $\lambda_{1}\mathbf{x}_{1}+\lambda_{2}\mathbf{x}_{2}=\left(\lambda_{1}+\lambda_{2}\right)\left[\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}\mathbf{x}_{1}+\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}}\mathbf{x}_{2}\right]$.

If $\mathbf{x}_{1},\mathbf{x}_{2}\in C$ then $\mathbf{x:=}\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}\mathbf{x}_{1}+\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}}\mathbf{x}_{2}$ since $C$ is convex.

Secondly $\left(\lambda_{1}+\lambda_{2}\right)\mathbf{x}\in C$ because $C$ is a cone.

drhab
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