Proximal Mapping of Least Squares with $ {L}_{1} $ and $ {L}_{2} $ Norm Terms Regularization (Similar to Elastic Net)

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I was trying to solve $$\min_x \frac{1}{2} \|x - b\|^2_2 + \lambda_1\|x\|_1 + \lambda_2\|x\|_2,$$

where $ b \in \mathbb{R}^n$ is a fixed vector, and $\lambda_1,\lambda_2$ are fixed scalars. Let $f = \lambda_1\|x\|_1 + \lambda_2\|x\|_2$, that is to say my question is how to find out the proximal mapping of $f$. It formulates as

$$\begin{equation} prox_f(b)=arg\min_x\{ \frac{1}{2}\|x - b\|_2 + \lambda_1\| x \|_1 + \lambda_2\| x \|_2 \}. \label{eq1} \end{equation}$$

There are two ways to get proximal mapping of $l_2$-norm and $l_1$-norm respectively.

For $l_1$-norm, soft threshold operator was given in Derivation of soft threshold operator. For $l_2$-norm, block soft threshold was given in deriving block soft threshold from l2 norm.

EDIT: I got stuck to find the subgradient of the object function. I followed above-mention methods to solve my problem. The subgradient of original target shows as, $$\begin{equation} 0 \in x - b + \lambda_1 \partial \|x\|_2 + \lambda_2 \partial \|x\|_1. \label{eq2} \end{equation}$$

I guess that it should be discussed for different conditions:

  • If $x = 0$, then $\partial \|x\|_1 = \{g: g_i \in [-1,1]\}$ and $\partial \|x\|_2 = \{g: \|g\|_2 \leq 1\}$, where $g_i$ denotes $i$th element of $g$. Thus, I got $$ 0\in \lambda_1 \{g: g_i \in [-1,1]\} + \lambda_2 \{g:\|g\|_2 \leq 1 \} - b \\ \Leftrightarrow b \in \lambda_1 \{g: g_i \in [-1,1]\} + \lambda_2 \{g:\|g\|_2 \leq 1 \}. $$ It implies that for $\|b\|_2 < \lambda_2$ or for all $|b_i|_1 < \lambda_1$, the optimal condition is $x = 0$.
  • If $x \neq 0$, then $\partial \|x\|_2 = x/\|x\|_2$, and the optimal is $$ b \in \lambda_1 \partial \|x\|_1 + \lambda_2 \frac {x}{\|x\|_2} + x. $$ If $x_i = 0$, then $\partial |x_i|= sgn(x_i)$, where $sgn(x_i)$ takes the sign of $x_i$. I guess that it also should be discussed the conditions that whether or not $x_i = 0$ by each component. But the question is that I don't know how to discuss. The reason is that $x_i$ is restricted by $\|x\|_2$, and each dimension cannot be separated.

I would really appreciate help on solving my problem. Thanks very much.