I have a convex function $f(x,y)$, with the equality constraint $x+y=1$. Is this still a convex optimization problem, despite the equality constraint? or is it a nonlinear optimization problem?
Is optimizing a convex function $f(x,y)$ that has a equality constraint $x+y=1$ a convex optimization problem?
2
$\begingroup$
convex-optimization
-
0Why nonlinear? The constraint seems linear to me at least. You can always turn an equality constraint into two inequality constraints. – 2017-01-03
-
0It's possible for an optimization problem to be both nonlinear and convex. Your problem is convex, and depending on $f$ it might also be nonlinear. – 2017-01-03
2 Answers
2
The standard definition of a convex optimization problem is:
Minimize $f(x)$ subject to $x\in S$
where (1) $S$ is a convex set and (2) $f$ is a convex function on set $S$
The set $(x,y)\in R^2$ with $x+y=1$ is clearly convex; hence (1) holds.
And $f$ is convex by your statement; hence (2) holds.
-
0Practically, $S$ needs to be represented by convex functions, and $f$ needs to be convex outside $S$ too: http://math.stackexchange.com/questions/2038759/why-is-it-necessary-to-have-convexity-outside-the-feasible-region – 2017-01-03
-
0@LinAlg But can't we always just redefine $f$ to be infinity outside of $S$? I don't see that it's necessarily important for $f$ to be convex outside of $S$. – 2017-01-03
-
0What if you want to use an infeasible method? – 2017-01-03
0
Linear equality constraints are OK in a convex optimization problem because the set of solutions to a linear system of equality constraints is convex. When you have equality constraints involving a nonlinear function of the variables, then you're out of luck.