# Interchange Argmin and Monotone Function

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 $$\operatorname{argmin}_{x \in X} f(x) = \operatorname{argmin}_{x \in X}g \circ f(x)?$$

• maybe im missing something, but doesn't $g(x)=x+1$ is a counter example? (take $f=1$ ) – infinity Feb 20 '20 at 20:34
• No, in both those cases $X=argmin_{x \in X} 1 = argmin_{x \in X} f(x) \& argmin_{x \in X} g\circ f(x) = argmin_{x \in X} 2=X$. (It's not true for $\inf$ but I'm considering $\operatorname{arginf}$). – AIM_BLB Feb 20 '20 at 20:40

$$\implies$$ 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, $$x^* \in \underset{x \in X}{\mathrm{arg\min}} \: f(x)$$.
• Is it necessary that $f$ is lower semi-continuous? – Mikal Apr 2 '20 at 15:13