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!
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.