Minimizing $\left \| X \right \|_{\infty}$ and $\left \| X \right \|_{2}$ under linear constraints

Question

I'm trying to solve these 2 problems:

$$\min_{X \in R^{n}} \left \| X \right \|_{\infty}$$ $$\min_{X \in R^{n}} \left \| X \right \|_{2}$$ $$s.t$$ $$p\geq 1$$ $$x_{1} + ... + x_{n} = S$$ $$B_L\leq X\leq B_{U}$$ $$B_{L},B_{U} \in R^{n}$$

It is obvious that different results are produced depending on the p-value.

(p = 2 or infinity)

But what about the vector X?

Here, My assumption is,

"Even though the optimization results are different, the vector X will be the same regardless of p."

(p = 2 or infinity)

and followings are brief reasons for the assumption.

1. given constraints are linear and X exist on the hyperplane.

2. So their arg min value will be determined only by the given boundary condition.

3. Arg min of the problem given above will be identical because they have the same boundary conditions.

Is there any error in my assumption?

Answer 1

The $p=\infty$ problem is not uniquely solvable in general:

Take $n=3$, $S=1$, $$ B_L=\pmatrix{1\\-\epsilon\\-\epsilon}, B_R=\pmatrix{1\\ \epsilon\\ \epsilon} $$ for some $\epsilon\in(0,1)$. Then $$ X_1=\pmatrix{1\\-\epsilon\\+\epsilon}, X_2=\pmatrix{1\\ +\epsilon\\ -\epsilon} $$ are solutions, as well as their convex combinations.

The $p=2$ problem is uniquely solvable. So there cannot be equivalence.

I would not be surprised if the solution for $p=2$ also solves the $p=\infty$ problem due to the special form of the linear equation constraint.