Let $g: \mathbb{R}\rightarrow \mathbb{R}$. Let $\mathcal{X}\equiv \{(x,y)\in \mathbb{R}^2: x=y\}$. Take the function $$ (x,x)\in \mathcal{X} \mapsto f(g(x), x)\in \mathbb{R}^+_0, $$ where $\mathbb{R}^+_0$ is the set of positive reals including zero.
Assume that:
(1) $f(g(3),3)=0$. This means that $f(g(x),x)$ achieves its minimum when $x=3$.
(2) $3\underbrace{=}_{\text{unique solution!}} \text{argmin}_x f(g(3), x)$.
Question: does this imply that $$ 3\underbrace{=}_{\text{unique solution!}} \text{argmin}_x f(g(x), x) $$ ? If not, under which conditions we can claim so (like continuity, etc.)?
If we set $f(a,b)=|b-3|-a$ and $g(a)=a$, then certainly $3$ is the unique minimizer of $\min_{b\in\mathbb{R}}\;f(g(3),b)$. However, when we consider minimizing the general function $f(g(a),a)$, we can see graphically that while $3$ is indeed a minimizer, it is no longer unique.
I am not aware of an equivalent condition, but a sufficient condition would be that $f(g(\cdot),\cdot)$ is strongly convex (uniformly convex, strictly convex, or convex+supercoercive would also be fine here, since they also guarantee uniqueness).