$l_1$ and $l_2$ norm minimization with a constraint 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 $\lambda^*$.
