Proximal Operator / Mapping of $\frac{1}{2} {\|x\|}^2 + \delta_{\mathbb{R}_+^n}\left(x\right)$: Sum of $L_2$ Norm Squared and Indicator Function

Question

Let $$f(x) = \frac{1}{2}\|x\|^2 + \delta_{\mathbb{R}_+^n}(x)$$

(componentwise nonnegtive). How to find $$\operatorname{prox}_{\alpha, f}(z)$$

I know

1. $\operatorname{prox}_{\alpha, \frac{1}{2}\|\cdot\|^2}(z) = \frac{1}{1+\alpha}z$

2. The definition: $$\operatorname{prox}_{\alpha, f}(z) = \arg\min_x\bigg(\frac{1}{2\alpha}\|x-z\|^2 + f(x)\bigg)$$

So we have $$\operatorname{prox}_{\alpha, f}(z) = \arg\min_x\bigg(\frac{1}{2\alpha}\|x-z\|^2 + \frac{1}{2}\|x\|_2^2 + \delta_{\mathbb{R}_+^n}(x) \bigg)$$ and then $$\operatorname{prox}_{\alpha, f}(z) = \arg\min_{x\in \mathbb{R}_+^n}\bigg(\frac{1}{2\alpha}\|x-z\|^2 + \frac{1}{2}\|x\|_2^2 \bigg)$$

Answer 1

For simplicity I'll assume that $\alpha = 1$. You want to evaluate $$ x^\star = \arg \min_x \quad I(x) + \frac12 \|x\|^2 + \frac12 \|x - z \|^2 $$ where $I$ is the indicator function of the nonnegative orthant. We'll combine the two quadratic terms into a single quadratic term by completing the square. Notice that \begin{align} &\frac12 \|x \|^2 +\frac{1}{2} \|x - z \|^2 \\ &= \frac12 \|x\|^2 + \frac12 \|x\|^2 - \langle x,z \rangle + \frac12\|z\|^2 \\ &= \|x\|^2 - \langle x,z \rangle + \frac12 \| z\|^2 \\ &= \underbrace{\|x\|^2 - 2 \langle x,z/2\rangle + \|z/2\|^2}_{\text{perfect square}} - \|z/2\|^2 + \frac12 \|z\|^2\\ &= \left\|x - \frac{z}{2} \right\|^2 + \text{terms that do not depend on $x$}. \end{align} Therefore, $$ x^\star = \arg \min_x \quad I(x) + \|x - \frac{z}{2} \|^2. $$ Computing $x^\star$ has now been reduced to evaluating the prox-operator of $I$.