Let $H$ be some Hilbert space, which we can take to be the usual finite-dimensional Euclidean space if needed. For a function $f : H \to \mathbb{R}$, let $f^* : H \to \mathbb{R}$ be its conjugate dual, defined as \begin{align*} f^*(u) = \sup_{x \in H}\{ \langle u,x \rangle - f(x)\}. \end{align*} Let $f$ be proper, convex and l.s.c. so that $f^{**} = f$. Let $\text{Prox}_f: H \to H$ be its proximal map, defined as the unique solution of the following optimization problem \begin{align*} \text{Prox}_f(x) := \arg \min_{y \in H} f(y) + \frac12 \|x-y\|^2. \end{align*} Then, Moreau decomposition states that $\text{Id} = \text{Prox}_f + \text{Prox}_{f^*}$, where $\text{Id}$ is the identity map. That is, $x = \text{Prox}_f(x) + \text{Prox}_{f^*}(x)$. Is this decomposition unique? By uniqueness, I mean something like this:
Let $\text{Fix} (\text{Prox}_f)$ be the set of fixed points of $\text{Prox}_f$ (which is known to be equal to the set of minimizers of $f$). Assume that $x = u +v$ where $u \in \text{Fix} (\text{Prox}_f)$ and $v \in \text{Fix} (\text{Prox}_{f^*})$. Is it true that $u = \text{Prox}_f(x)$ and $v = \text{Prox}_{f^*}(x)$?
