请求求解L₁范数与L₂,₁范数之和的近端算子

Got it, let's work through the proximal operator for this composite regularization function step by step. First, recall the definition of the proximal operator for a function $f$:
$$\text{prox}_f(v) = \arg\min_x \frac{1}{2}|x - v|_2^2 + f(x)$$

For your specific function, this translates to solving:
$$\text{prox}_f(v) = \arg\min_x \frac{1}{2}|x - v|_2^2 + \lambda_1|x|_1 + \lambda_2\sum_g|x_g|_2$$

Since the objective splits into independent subproblems for each group $g$, we can solve each group separately and concatenate the results. Here's the breakdown for a single group $g$:

Step 1: Element-wise L1 Soft Thresholding #

First, handle the $\lambda_1|x|_1$ term, which enforces element-wise sparsity. For each element $v_i$ in group $g$, compute:
$$x^{(1)}_i = \text{sign}(v_i) \cdot \max\left(|v_i| - \lambda_1, 0\right)$$
This zeros out any element where $|v_i| \leq \lambda_1$, and shrinks elements with $|v_i| > \lambda_1$ by $\lambda_1$ while keeping their sign intact. Let $x^{(1)}_g$ be the vector of these processed elements for the group.

Step 2: Group-wise L2 Soft Thresholding #

Next, address the $\lambda_2|x|_{2,1}$ term, which enforces group-level sparsity. Calculate the L2 norm of the intermediate vector $x^{(1)}_g$:

• If $|x^{(1)}_g|_2 > \lambda_2$: Shrink the entire group proportionally to preserve the relative magnitudes of elements within the group:
  $$x_g = x^{(1)}_g \cdot \frac{|x^{(1)}_g|_2 - \lambda_2}{|x^{(1)}_g|_2}$$
• If $|x^{(1)}_g|_2 \leq \lambda_2$: The group's total "strength" is too small to justify retaining it, so set $x_g = 0$ (the all-zero vector for the group).

Final Result #

Concatenate the processed vectors $x_g$ across all groups to get the full proximal operator output $\text{prox}_f(v)$.

Verification & Edge Cases #

• When $\lambda_1 = 0$, this reduces to the standard proximal operator for the group L2 ($|\cdot|_{2,1}$) norm, which aligns with expected behavior.
• When $\lambda_2 = 0$, this simplifies to element-wise soft thresholding for the L1 norm, which is correct.
• This solution satisfies the KKT optimality conditions: for non-zero elements in non-zero groups, the gradient of the squared loss plus the subgradients of the two regularization terms sum to zero, confirming its validity.

内容的提问来源于stack exchange，提问作者NicNic8