# Relationship between proximal mapping and subgradient

Background: I came across this excerpt on Wikipedia

$$0\in \delta(\lambda f(z))+(z-x)$$ $$\Leftrightarrow 0\in \delta(\lambda f(z)+1/2 \lVert z-x\rVert^2)$$

This is probably trivial so apologies in advance.

Your question is a bit confusing as I don't know where the term $$(z-x)$$ in your first expression comes from.
The proximity operator $$\text{prox}_f (x)= \text{argmin}_y \{f(y) + \frac{1}{2}|| x-y||_2^2 \}$$ is used to solve optimization problems involving nonsmooth functions and it computes the "closest" point to $$x$$ that lies in the subdifferential of $$f$$. Designate such a point by $$p$$.
The subdifferential of a function $$f$$ at a point $$x$$, denoted by $$\partial f (x)$$, is the set of all the subgradients $$v$$ of $$f$$ at $$x$$, which have the geometric interpretation of being the supporting hyperplanes of $$f$$ at $$x$$.
We known that if $$x^*$$ minimises $$f$$, then the following holds $$0 \in \partial f(x^*),$$ which indicates that $$x^*$$ is a stationary point of $$f$$.
Now, if $$x^*$$ is optimal, we want $$\text{prox}_f$$ to give us $$x^*$$, that means $$x^*=p=\text{prox}_f (x^*),$$ which means that the optimal solution $$x^*$$ is a fixed point for $$\text{prox}_f$$. The equivalence with the stationarity condition involving the subdifferential follows automatically.