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 $C_0 = 100$.


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.

  • $\begingroup$ Do you mean $0 \le B_r \le \alpha(B_s + B_d)$, or is it correct as you have written it? $\endgroup$ Dec 30 '17 at 6:15
  • 1
    $\begingroup$ 2) does not look convex at first glance. All the rest are. $\endgroup$
    – copper.hat
    Dec 30 '17 at 7:51
  • $\begingroup$ @MaithreyaSitaraman, yes $0 \leq B_r \leq \alpha (B_s + B_d)$, my mistake $\endgroup$
    – kou
    Dec 30 '17 at 18:32
  • $\begingroup$ My first thoughts are to try and define the domain in relation to the epigraph of the function $g(\alpha, B_s, B_d) = \alpha (B_s + B_d)$ and the interval given in 5) in the hopes of using the second derivative test. I'm tired but I'll try and make this more precise in the morning if it still makes sense to me then. $\endgroup$ Jan 1 '18 at 9:31

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 $$\frac{1}{2} \nleq \frac{1}{2}\left(\frac{1}{2}+0\right).$$

  • 1
    $\begingroup$ An easier cache of 100 bounty points one is unlikely to find! ;-) $\endgroup$ Jan 2 '18 at 22:14

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