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 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 


*

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

*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
$$\operatorname{prox}_{\alpha, f}(z) = \arg\min_{x\in \mathbb{R}_+^n}\bigg(\frac{1}{2\alpha}\|x-z\|^2 + \frac{1}{2}\|x\|_2^2 \bigg)$$ 

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 $I$.
