While working on the algorithm, I need to solve the following problem
$$ \min_{x \in \mathbb{R}^n} \| x \|_1 + \frac{\alpha}{2}\| x - y \|^2 \\ \mathrm{s.t.} \ \| x - s \|^2 \le r$$ where $y,s \in \mathbb{R}^n, \alpha > 0 $ and $r > 0$.
The Lagrangian function for this problem is given by $$ L(x,\lambda) = \| x \|_1 + \frac{\alpha}{2}\| x -y \|^2 + \lambda(\|x-s\|^2 -r).$$ Let $(x^*,\lambda^*)$ be the optimal solution of this Lagrangian function. If $\lambda^* = 0$, then $x^* = \mathrm{Prox}_{\frac{1}{\alpha}\|\cdot\|_1}(y)$ where $\mathrm{Prox}_h(u) = \arg \min_{v \in \mathbb{R}^n}\{h(v) + \frac{1}{2}\|u - v\|^2 \}$
If $\lambda^* > 0$, I obtain that $$ 0 \in \partial \| x^* \|_1 + \alpha(x^*-y) + 2\lambda^*(x^* -s ) \ \ \mathrm{and} \ \ \|x^* -s \|^2 = r.$$ Using the first relation, I have $$ x^* = \mathrm{Prox}_{\frac{1}{\alpha+2\lambda^*}\|\cdot\|_1}\left(\frac{\alpha}{\alpha + 2\lambda^*}y + \frac{2\lambda^*}{\alpha + 2\lambda^*}s \right)$$
$$ x^*_i =\left\{ \begin{array}{ll} \frac{\alpha}{\alpha + 2\lambda^*}y_i + \frac{2\lambda^*}{\alpha + 2\lambda^*}s_i- \frac{1}{\alpha + 2\lambda^*} & \alpha y_i + 2\lambda^*s_i \ge 1 \\ 0 & -1 \le \alpha y_i + 2\lambda^*s_i \le 1\\ \frac{\alpha}{\alpha + 2\lambda^*}y_i + \frac{2\lambda^*}{\alpha + 2\lambda^*}s_i + \frac{1}{\alpha + 2\lambda^*} & \alpha y_i + 2\lambda^*s_i \le -1 \end{array}\right. $$ and using the second relation, I have $$ \| x^* - s \|^2 = r \Leftrightarrow \sum_{i=1}^n (x^*_i -s_i)^2 = r $$ However, I am not sure how to obtain $\lambda^*$ from here. Could you please tell me how to obtain it?
In case $\lambda^* > 0$, the constraint $\|x^* - s\|^2 \le r$ is active. Thus, you have to choose $\lambda^* > 0$ such that $$ \| x^* - s\|^2 = r, $$ i.e., you have one equation to determine the scalar $\lambda^*$.