Minimizing quadratic objective function on the unit $\ell_1$ sphere I would like to solve the following optimization problem using a quadratic programming solver
$$\begin{array}{ll} \text{minimize} & \dfrac{1}{2} x^T Q x + f^T x\\ \text{subject to} & \displaystyle\sum_{i=1}^{n} |x| = 1\end{array}$$
How can I re-write the problem using linear constraints?
Note: I have read other similar questions. However, they define the absolute value differently. 
 A: In one of the other answers (unfortunately now deleted) it is incorrectly assumed that we can apply a standard variable splitting technique, without worrying that both the positive and negative parts can become nonzero. I posted a comment already that this probably needs some additional binary variables to fix this.  Let me try to show how I would implement this. I assume we have some upper bounds and lower bounds $U_i, L_i$ on $x_i$, i.e. $x_i \in [L_i,U_i]$ where $U_i>0$ and $L_i<0$. Then:
$$\begin{align}
   &x_i = x^{plus}_i - x_i^{min} \\
   &x_i^{abs} = x^{plus}_i + x_i^{min} \\
   &x_i^{plus} \le U_i \delta_i \\
   &x_i^{min} \le -L_i (1-\delta_i) \\
   &\sum x_i^{abs} =1 \\
   &\delta_i \in \{0,1\}\\
   &x^{plus}_i \ge 0 \\
   &x^{min}_i \ge 0 \\
   &x^{abs}_i \ge 0 \\    
\end{align}$$
You need a MIQP solver to handle this (Cplex and Gurobi can do this).
A: This is not possible. The set
$$\{x \in \mathbb R^n \mid \sum_{i=1}^n |x_i| = 1\}$$
is not convex, but the feasible set of a quadratic program is convex.
