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

  • $\begingroup$ maybe im missing something, but doesn't $g(x)=x+1 $ is a counter example? (take $f=1$ ) $\endgroup$ – infinity Feb 20 '20 at 20:34
  • $\begingroup$ 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 $\operatorname{arginf}$). $\endgroup$ – 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, $x^* \in \underset{x \in X}{\mathrm{arg\min}} \: f(x)$.

  • $\begingroup$ Is it necessary that $f$ is lower semi-continuous? $\endgroup$ – Mikal Apr 2 '20 at 15:13
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    $\begingroup$ @Mikal I don't think so - we only require the optimization problems to have minimizers. $\endgroup$ – madnessweasley Apr 2 '20 at 15:20
  • $\begingroup$ +1 Cool stuff thanks $\endgroup$ – Mikal Apr 2 '20 at 21:51

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