# 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.

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 $f(x_2) \le f(x)$.