Impact of constraints on convex optimization Let me start by saying I know almost nothing about optimization so please bear with me. Basically, I am wondering whether it is possible to solve a problem with two constraints by solving the problem with each constraint individually and somehow combining the solutions. However, I'm having trouble visualizing the scenario. More specifically:
Suppose $V$ is some Banach space and $f:V \to \mathbb{R}$ is convex and continuous. Furthermore, let $K_1, K_2 \subset V$ be convex and consider the solutions $x_i$ of the optimization problems
$$\quad \min_{x \in K_i}  f(x).$$
For $K = K_1 \cap K_2$ let $x$ solve
$$\min_{x \in K} f(x).$$


*

*Is $x$ the projection of $x_1$ onto $K_2$?

*Is $x$ a linear combination of $x_1$ and $x_2$?

*Are there any relations between $x$, $x_1$ and $x_2$ at all?

*Does the situation change if we further restrict $f$? For example to be quadratic?


Thank you very much in advance.
 A: For general convex sets, the sorts of simple relationships you're suggesting don't exist. I find it helps to draw two dimensional pictures to see why. (Even if the space is dimension higher than 2, we could restrict the problems to the plane spanned by $x_1,x_2,x$ and none of the answers to your questions would change.) Here's an example of what can happen even in the simple case where $f$ is linear:

The red arrow is the direction of the negative gradient of $f$. Minimizing $f$ means moving as far as possible in that direction. To address your specific questions:


*

*No. In this case, the projection of $x_2$ onto $K_1$ is not in $K_2$, so it can't equal $x$. The projection of $x_1$ onto $K_2$ is in $K$, but still doesn't equal $x$

*The origin is not depicted in the diagram, but if $x_1$ and $x_2$ are collinear with the origin, then $x$ is not a linear combination of $x_1$, $x_2$.

*The only easy, simple relationship is that $f(x_1) \le f(x)$ and $f(x_2) \le f(x)$.

*In this example, the function is linear. That's about as simple as you can get without making the problem trivial.


That said, there are methods (like ADMM) that can use solvers for subproblems to converge to a solution of a larger problem.
