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Background: I came across this excerpt on Wikipedia

enter image description here Can anybody please help in computing exactly how

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

This is probably trivial so apologies in advance.

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Your question is a bit confusing as I don't know where the term $(z-x)$ in your first expression comes from.

But I'll try to do my best to explain the relationship between the proximity operator and the subdifferential for a convex function.

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.

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