I want to run an optimization function of following variables. How do I know that the following multivariable domain is convex?
$f(\alpha, B_s, B_r, B_d)$, and the domain is consisted of $\alpha, B_r, B_d, B_s$, where:
1) $0\leq \alpha \leq1$
2) $0\leq B_r \leq \alpha(B_s + B_d)$
3) $0\leq B_d \leq C_0$
4) $0\leq B_s \leq C_0$
5) $0\leq B_r \leq C_0$
and $C_0$ is a positive constant. Say $C_0 = 100$.
Attempt:
The definition of a convex set is that given two points $x,y$ in the set, if $\lambda x + (1-\lambda)y$ is also in the set, then this domian is a convex set, for $\lambda $ between 0 and 1.
Let $x = [\alpha, B_s, B_r, B_d]^T, y = [\alpha', B_s', B_r', B_d']^T$.
Then, $\lambda x + (1-\lambda)y$ =
$[\lambda \alpha + (1-\lambda) \alpha', \lambda B_s + (1-\lambda) B_s', \lambda B_r + (1-\lambda) B_r', \lambda B_d + (1-\lambda) B_d']^T$.
Then I got stuck. It is very obvious that $\alpha, B_d, Bs$ they are convex. But $B_r$ is a little bit tricky.
This domain is not convex. Suppose $\lambda=\frac{1}{2}$ with $$x=(\alpha,B_s,B_r,B_d)=(1,1,1,0)$$ $$y=(\alpha',B_s',B_r',B_d')=(0,0,0,0)$$ Then $x$ and $y$ both satisfy all the inequalities but $\lambda x + (1-\lambda y)=(\frac{1}{2},\frac{1}{2},\frac{1}{2},0)$ does not satisfy 2) because $$\frac{1}{2} \nleq \frac{1}{2}\left(\frac{1}{2}+0\right).$$