If a set defined as $-xy+\alpha <0$, $\forall x,y > 0$ and $\alpha$ given as an arbitrary positive constant is a convex set, is the same true for a set given by $-xy+z<0$ $\forall x,y,z > 0$ ?
Asked
Active
Viewed 50 times
0
-
1It is not clear where the set is. Your formulation is ambiguous: one may understand the set to be either two or three dimensional. Please clarify your question. – uniquesolution Apr 06 '17 at 19:33
-
the set is three-dimensional, and x,y and z variables are strictly positive – M992 Apr 06 '17 at 19:52
-
No, the set ${(x,y,z) \in \mathbb{R}^3: x,y,z>0, z<xy}$ is not a convex set, even though it is convex in $(x,y)$ for every fixed $z>0$. You can play around to get an example of two points in the set such that the midpoint is not in the set. – Michael Apr 06 '17 at 19:54
-
I will try to be more clear. So, I have a bilinear function which I want to relax , i.e. $ z = xy $. I also know the bounds for x and y, hence I can get the bounds for z from there, i.e. $x\in [{0,x_1} ], y\in [{0,y_1}] $and$ z\in [{0,x_1y_1}] $. I am trying to prove that $-x*y+z<0$ is not convex using the midpoint convex set formula and I am just not getting it. Can the given set be convex in such bounded case? – M992 Apr 06 '17 at 22:32
-
You want to find $(x,y,z)$ and $(a,b,c)$ in the set $\mathcal{A}$, but the midpoint $\frac{1}{2}(x+a,y+b,z+c)$ is not in $\mathcal{A}$. What if you try $x=y=high$ and $a=b=low$? – Michael Apr 06 '17 at 22:48
-
1I got it. Thanks a lot – M992 Apr 06 '17 at 23:05
-
@M992 : If you like you can answer your own question, which then completes it as a question (this is standard practice on this website). You can also notify particular users about comments with the @[username] command. – Michael Apr 06 '17 at 23:16