Maximization of bilinear objective over convex body

Question

Let $C \in \mathbb{R}^n$ be a convex compact body. Given the following optimization problem:

$$\max_{x \in C\\ \left\| v \right\|_2 \leq 1} \left< x, v \right>$$

where $\left<\cdot,\cdot\right>$ denotes the dot product.

Is this problem considered to be convex or not? And how one would determine it?

The bilinear objective is an indefinite quadratic form and, thus, neither convex nor concave. — Rodrigo de Azevedo, Jun 21 '18 at 15:24

David M. · Accepted Answer · 2018-06-21T14:03:03.880

1

An optimization problem is considered convex if you are minimizing a convex function over a convex feasible region. So you must ask yourself:

Is the function $-\langle x, v\rangle$ convex?
Is the set $C\times\{\|v\|\leqslant1\}$ convex? (See edit below)

A potentially helpful way to attack 1. is to consider the case when $n=1$ (so that $-\langle x,v\rangle=-xv$, where $x,v\in\mathbb{R}$).

Edit

The formal definition of a convex optimization problem is a problem that can be expressed as

$$ \begin{array}{rl} \min\ & f(x) \\ \text{s.t.}\ & g_i(x)\leqslant 0\text{ for all }i=1,\dots,m \end{array} $$

where $f:\mathbb{R}^n\to\mathbb{R}\cup\{\infty\}$ and $g_i:\mathbb{R}^n\to\mathbb{R}\cup\{\infty\}$ are convex functions for $i=1,\dots,m$. However given, any convex set $C$, the indicator function

$$ I_c(x)=\begin{cases}0, & x\in{C}, \\ \infty, &\text{else.}\end{cases} $$

is convex. Hence, if we are minimizing a convex function over a convex set, we can always (mathematically) represent our problem in the standard form above. Note that we are using the extended reals here, which is common in convex optimization (see e.g. Boyd's book).

Caveat: of course, the indicator function is useless for computation--you would need an algebraic description of the set $C$. But for definitions (as we're concerned with here), it will suffice.

edited Jun 21 '18 at 14:03

answered Jun 21 '18 at 12:14

David M.

2,703

You should delete the minus sign in your first point. – vaoy Jun 21 '18 at 12:17
@vaoy No, I meant it to be there. Since I phrased my answer in terms of minimization, the OP’s model would become $\min-\langle x,v\rangle$ (i.e. minimizing the negative of the original objective function). – David M. Jun 21 '18 at 12:20
1

Ou sorry mate, didnt optimisation problem in the question was maximising, my bad. :) – vaoy Jun 21 '18 at 12:22
If you fix one variable and try to solve for the other variable then yes, it's conex! But why are looking at the intersection of a ball and $C$!? (it can be empty!) – user3492773 Jun 21 '18 at 12:22
Oops—see my edited answer in a minute – David M. Jun 21 '18 at 12:24
Keep in mind that the fucntion to be minimised is actually defined over $$\mathbb{R}^n \times \mathbb{R}^n$$ in the two variables $x$ and $y$ – vaoy Jun 21 '18 at 12:25
Then how one can determine convexity of such feasible region? (e.g. how would it look like!?) – user3492773 Jun 21 '18 at 12:27
It is convex beacuse $C$ is convex and ${(x,y): ||y|| \leq 1}$ is convex. And intersection of convex sets are convex. – vaoy Jun 21 '18 at 12:29
The intesection of them, not the cross product! – user3492773 Jun 21 '18 at 12:31
Yes the intersection is convex. – vaoy Jun 21 '18 at 12:31
See this question – David M. Jun 21 '18 at 12:37
Yeah, Thanks. So the problem now is to determine whether the dot product is convex? So should i look at the hessain? (if you fix one variable then yes, it's convex since it's basically linear!) – user3492773 Jun 21 '18 at 12:39
If the hessian is positive semidefinite, then it is convex. – vaoy Jun 21 '18 at 12:41
Your definition of a convex optimization problem is wrong; the constraint functions need to be convex. – LinAlg Jun 21 '18 at 13:21
@LinAlg see my edit – David M. Jun 21 '18 at 14:03

Maximization of bilinear objective over convex body

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