1

Consider a function $f: \mathbb{R}^4 \to \mathbb{R}$ such that

$$f(x_1, x_2, y_1, y_2) = p x_1 (x_1 - y_1) + (1 - p) x_2 (x_2 - y_2)$$

for some $p \in (0, 1)$. I'd like to check whether the set

$$C = \{ z \in \text{(feasible area)} \subseteq \mathbb{R}^4 \mid f(z) \leq 0 \}$$

is convex or not in the following feasible area: for both $i = 1, 2$, \begin{align} x_i + y_i &= 1 \land x_i \geq 0 \land y_i \geq 0. \end{align}

If I first substitute two equality constraints into the original objective function $f$, then $f$ is convex w.r.t. $(x_1, x_2)$, which implies $C$ is convex.

Then is there a way to check the convexity of $C$, without reducing the number of variables first?

Rodrigo de Azevedo
  • 1
myu
  • 31
  • You can write $$f = \begin{bmatrix} x_1\ x_2\ y_1\ y_2 \end{bmatrix}^\top \begin{bmatrix} p & 0 & -\frac{p}{2} & 0\ 0 & 1-p & 0 & -\frac{1-p}{2}\ -\frac{p}{2} & 0 & 0 & 0\ 0 & -\frac{1-p}{2} & 0 & 0\end{bmatrix} \begin{bmatrix} x_1\ x_2\ y_1\ y_2 \end{bmatrix}$$ and then use Sylvester's criterion to determine for which values of $p$ the matrix is positive semidefinite. – Rodrigo de Azevedo Dec 25 '19 at 08:47
  • I know that $f$ is not convex (therefore the hessian of $f$ is not positive semi-definite) in general. I'd like to find how to check the convexity of $f$ on $x_i + y_i = 1$ line. – myu Dec 25 '19 at 15:25
  • It's not a line. It's a line segment. – Rodrigo de Azevedo Dec 25 '19 at 15:56

0 Answers0