1
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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
2
$\begingroup$

$\implies$ 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, $x^* \in \underset{x \in X}{\mathrm{arg\min}} \: f(x)$.

$\endgroup$
3
  • $\begingroup$ Is it necessary that $f$ is lower semi-continuous? $\endgroup$ – Mikal Apr 2 '20 at 15:13
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.