How to compute proximal mapping of function The proximal mapping is $$\text{prox}_{\eta, g}(w) = \arg\min_z \Big[\frac{1}{2\eta} ||z-w||_2^2 + g(z) \Big]$$
Now we want to consider the function $$g(w) = \sum_{i=1}^d \frac{\lambda}{\alpha} \ln(1 + \alpha|w_j|)$$
What is the proximal mapping $\text{prox}_{\eta, g}(w)?$
I tried to plug in $g$, then get the derivative and equate it to zero, but eventually it involved something like $\frac{z_j-w_j}{\eta} + \frac{\lambda \cdot sign(z_j)}{1 + \alpha|z_j|} = 0$, which I'm not sure is correct and I'm not sure also how to proceed.
EDIT:
Full task is we want to solve $\min_w \frac{1}{n}\sum_{i=1}^n f_i(w) + g(w)$ using proximal stochastic gradient, whose algorithm update step is $w^{(t)} = \text{prox}_{\eta, g}(w^{(t-1)} -\eta\nabla f_i(w))$ for a randomly chosen $i$.
 A: Since $g$ is a separable sum, we can just focus on the 1D case. The general proximal mapping is then just the 1D proximal mapping applied on each component. In 1D we have
$$g(w) = \frac{\lambda}{\alpha}\ln(1+\alpha|w|)$$
Then the derivative of the function to be minimized is
$$\frac{d}{dz} \left(\frac{1}{2\eta}(z-w)^2+\frac{\lambda}{\alpha}\ln(1+\alpha|z|)\right) = \frac{z-w}{\eta}+\frac{\lambda\operatorname{sign}(z)}{1+\alpha|z|}$$
similar to what you computed. So we need to solve
$$\frac{z-w}{\eta}+\frac{\lambda\operatorname{sign}(z)}{1+\alpha|z|} = \frac{z-w}{\eta}+\frac{\lambda}{\operatorname{sign}(z)+\alpha z} = 0$$
$$\iff \alpha z^2+(\operatorname{sign}(z)-\alpha w)z+(\lambda\eta-\operatorname{sign}(z)w) = 0$$
So using the quadratic formula
$$
\begin{aligned}
z &= \frac{1}{2\alpha}\left((\alpha w-\operatorname{sign}(z))\pm\sqrt{(\operatorname{sign}(z)-\alpha w)^2-4\alpha(\lambda\eta-\operatorname{sign}(z)w)}\right)\\
&= \frac{1}{2\alpha}\left((\alpha w-\operatorname{sign}(z))\pm\sqrt{1+2\operatorname{sign}(z)\alpha w+\alpha^2w^2-4\alpha\lambda\eta}\right)
\end{aligned}
$$
where
$$\operatorname{sign}(z)=\pm 1$$
From a computational point of view, this is a good solution. It gives 4 possible values for $z$, some of them might be non-real. By simply discarding the non-real solutions and filling in the rest in the function we are minimizing, we can see where it is the lowest.
You can verify the solutions with wolfram alpha
