I came across this question while studying duality
$\min$ $\alpha$
s.t $Aq=\alpha f$
$||q||^2\leq\epsilon$
where $\alpha\in\mathbb{R},q\in\mathbb{R}^n$ are the variables and $A\in\mathbb{R}^{m\times n},f\in\mathbb{R}^m,\epsilon\in\mathbb{R}_{++}$ and the rows of A are linearly independent. Find the dual problem to and don't assign a lagrange multiplier to the quadratic constraint.
The lagragian is :
$$\underset{||q||^2\leq\epsilon}{L(\alpha,q,\lambda)}=\alpha(1-\lambda^Tf)+\lambda^TAq$$
Solving first for $\alpha$ I get that $\underset{\alpha}{\min} \alpha(1-\lambda^Tf)=\left\{\begin{array}{rcl}0&1-\lambda^Tf=0\\-\infty&else\end{array}\right.$
And for $q$ i get the following $\underset{||q||^2\leq\epsilon}{\min}\lambda^TAq$ and I'm not sure how to continue from here. I think that I need to demand that $\lambda^TA=0$ or else I $-\infty$ but not sure.
any hint?
First, rewrite for simplicity
$$ L'(q, \lambda) := \min_{\alpha} L(\alpha, q, \lambda) = \delta_{C}(\lambda) + \langle A^* \lambda, q \rangle, $$
where $\delta_{C}(\lambda)$ is $0$ if $1 - \lambda^* f = 0$ and $+\infty$ otherwise. From there, you can minimize over $q$ since
$$ \min_{q \in \sqrt{\varepsilon} B} \left\{ \delta_{C}(\lambda) + \langle A^* \lambda, q \rangle \right\} = \delta_{C}(\lambda) + \min_{q \in \sqrt{\varepsilon} B} \langle A^* \lambda, q \rangle \\ = \delta_{C}(\lambda) + \frac{1}{\sqrt{\varepsilon}} \min_{z \in B} \langle A^* \lambda, q \rangle \overset{(*)}{=} \delta_{C}(\lambda) - \frac{\| A^* \lambda \|_2}{\sqrt{\varepsilon}}, $$
where in $(*)$ we made use of the fact that the minimum of $x \mapsto \langle y, x \rangle$ over the unit ball is $- \| x \|_2$.