Interchange Argmin and Monotone Function

Question

Suppose that $f:X\rightarrow (0,\infty)$ is lower semi-continuous, and coercive (so it admits a minimizer) and suppose that $g:(0,\infty)\rightarrow (0,\infty)$ is monotone increasing and smooth. Then is it true that $$ \operatorname{argmin}_{x \in X} f(x) = \operatorname{argmin}_{x \in X}g \circ f(x)? $$

Answer 1

$\implies$ Suppose $x^* \in \underset{x \in X}{\mathrm{arg\min}} \: f(x)$. Then $f(x^*) \leq f(x)$, $\forall x \in X$. Since $g$ is monotone increasing, this implies $g(f(x^*)) \leq g(f(x))$, $\forall x \in X$. Therefore $x^* \in \underset{x \in X}{\mathrm{arg\min}} \: g(f(x))$.

$\impliedby$ Suppose $x^* \in \underset{x \in X}{\mathrm{arg\min}} \: g(f(x))$. Then $g(f(x^*)) \leq g(f(x))$, $\forall x \in X$. Since $g$ is monotone increasing, it has an inverse that is also monotone increasing. Taking the inverse on both sides yields $f(x^*) \leq f(x)$, $\forall x \in X$. Therefore, $x^* \in \underset{x \in X}{\mathrm{arg\min}} \: f(x)$.