Consider the following primal problem
$$ f=\max_{x}c^{\top}x -{\varepsilon ||x||_2} \quad \text{s.t.} \quad Ax\leq b\\ $$ with $\epsilon>0$. Could you help me to write down the dual of this problem? My suggestion is reported below, but the professor told me that this is the wrong answer. I don't understand what I'm doing wrong.
$$ f=\min_{y\geq 0} y^\top b\quad \text{s.t.} \quad ||A^\top y -c||_2 \leq \epsilon$$
Time to tell your 'professor' to find a new job.
The Lagrangian is $$L(x,y)=c^{\top}x -{\varepsilon ||x||_2} + y^T(b-Ax)$$
The primal is $\max_x \min_{y \geq 0} L(x,y)$ and the dual is $\min_{y \geq 0} \max_x L(x,y)$. Now: \begin{align} \max_x L(x,y) &= \max_x \left\{ b^Ty - (A^Ty-c)^Tx-{\varepsilon ||x||_2} \right\} \\ &= \begin{cases} b^Ty & \text{if } ||A^Ty-c||_2 \leq \varepsilon \\ \infty & \text{otherwise.}\end{cases} \end{align} Therefore, the dual is $\min_{y \geq 0} \left\{ b^Ty : ||A^Ty-c||_2 \leq \varepsilon\right\}$.