Proximal Mapping - Derivation of the Proximal Operator from the Resolvent of the Sub Differential

Question

I do not really understand the solution for this question. I do not understand how the zero vector was derived to be an element of the subdifferential and why g(u) is strongly convex. Any help and further elaboration would be appreciated.

Answer 1

Since $x \in z + \lambda \partial f(z)$, and $x, z$ is a vector and $\partial f(z)$ is a set. Thus this means that $x-z \in \lambda \partial f(z)$, or $0 \in \lambda \partial f(z) + (z-x)$. And then according to \begin{equation} \nabla_{z} \|z-x\|^2 = 2(z-x), \end{equation} we have \begin{equation} 0 \in \partial(\lambda f(z) + \|z-x\|^2). \end{equation}

For the $g(u)$, since $g(u) = \lambda f(u) + \frac{1}{2}\|u-x\|^2$ which is the summation of convex function and strictly convex function, and thus it is strictly convex.