From Bazaraa, Sherali, and Shetty's Nonlinear Programming. Probem 2.16:
Let $S$ be a non-empty set in $\mathbb{R}^n$ and let $\bar{x} \in S$. Consider the set $C=\{y:y=\lambda(x− \bar{x}), \lambda \geq 0, x \in S\}$
(a) Show that $C$ is a cone and interpret it geometrically.
(b) Show that $C$ is convex if $S$ is convex.
(c) Suppose that $S$ is closed. Is it necessarily true that $C$ is closed? If not, under what conditions would $C$ be closed?
I have (a) fine.
I'm getting caught with (b). I've tried directly showing that if: $y_1, y_2 \in C \implies \lambda_1(x_1 - \bar{x}_1), \lambda_2(x_2 - \bar{x}_2) \in C$, then: $\lambda_3[\lambda_1(x_1 - \bar{x}_1)]+(1-\lambda_3)[\lambda_2(x_2 - \bar{x}_2)]$ directly rearranges to $\lambda[\bar{\lambda}x_1+(1-\bar{\lambda})\bar{x}_1-\hat{\lambda}x_2-(1-\hat{\lambda})\bar{x}_2]$ with arbitrary lambdas.
Can't seem to make that work. I also tried saying that $x,\bar{x} \in S \implies (x-\bar{x}) \in S \implies \lambda(x-\bar{x}) +(1-\lambda)(x-\bar{x}) \in S$ and just multiplying by $\lambda$. That worked, but I feel like it relies on $S$ being closed under addition, and I don't know if that holds.
Let $y_1=\lambda_1(x_1-\bar x)\in C$ and $y_2=\lambda_2(x_2-\bar x)\in C$ where $\lambda _1,\lambda _2>0$ and $x_1,x_2\in S$. Therefore for any $0<\lambda <1$ $$ \lambda y_1+(1-\lambda)y_2=\lambda \lambda_1x_1+(1-\lambda)\lambda_2x_2-[\lambda \lambda_1+(1-\lambda)\lambda_2]\bar x $$ We know that since $\bar x\in S$ and $S$ is convex, then ${\lambda \lambda_1x_1+(1-\lambda)\lambda_2x_2\over \lambda \lambda_1+(1-\lambda)\lambda_2}\in S$. Hence ${\lambda \lambda_1x_1+(1-\lambda)\lambda_2x_2\over \lambda \lambda_1+(1-\lambda)\lambda_2}-\bar x\in C$ and since $C$ is a cone, ${\lambda \lambda_1x_1+(1-\lambda)\lambda_2x_2}-[{\lambda \lambda_1+(1-\lambda)\lambda_2}]\bar x\in C$.