Let $X \subset {\Bbb R}^n$ be a compact and convex set, and let $f : X \times X \to {\Bbb R}$ be a continuous and differentiable scalar field defined by $$ f(x,x^*)=g(x) + \sum_{i=1}^n x_ih_i(x^*). $$ Consider the mapping $T : X \to {\Bbb R}$ defined by
$$ T(x^*) := \arg\min_{x \in X} f(x,x^*).$$
Suppose that $g$ is strictly convex so that the minimization problem is always well-defined and has a unique solution.
I am interested in the set of fixed points of $T$, that is the set of $x^*$ such that $T(x^*)=x^*$. Are there sufficient conditions that we can impose on $f$ such that $T$ has a unique fixed point? Is there a good reference for these types of problems?
(Partial Answer):
Consider the setting when $f$ is a separable sum, i.e., $$f(x,x^*)=g_1(x)+g_2(x^*),$$ where $g_1$ and $g_2$ are strictly convex functions. A sufficient condition is $$\textrm{argmin}g_1=\textrm{argmin}g_2.$$ In fact, the fixed-point iteration $x_{n+1}=T(x_n)$ converges in one iteration. Let $x_0\in X$. Then $$x_1=T(x_0)=\textrm{Argmin}_{x\in X}\;g_1(x)+g_2(x_0)=\textrm{Argmin}_{x\in X}\;g_1(x)=\textrm{Argmin}\;g_2.$$ Therefore, $$x_2=T(x_1)\in\textrm{Argmin}_{x\in X}g_1(x)+g_2(x_1)=\textrm{Argmin} g_1=x_1$$ hence $x_1$ is a fixed point. Then, uniqueness of the fixed point follows from the fact that $g_1$ and $g_2$ are strictly convex.