Let $\Omega$ be an open bounded subset of $\mathbb{R}^2$ and $f\in L^2(\Omega)$ be a given function. Consider the optimization problem $$\mathrm{min} \int_\Omega u(x) f(x) \,dx\,,$$ where a minimum is considered over all functions $u\in X(\Omega): = \{ v \in L^2(\Omega) : v(x)\in\{0,1\} \ \forall x\in \Omega\}$. The set $X(\Omega)$ is obviously non-convex. Is it true that this non-convex optimization problem has a unique minimum which is $u(x)=0$ if $f(x)\geq 0$ and $u(x)=1$ if $f(x)<0$?
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$\begingroup$ I would say, assume that there is another minimizer and show that you improve it by modifying it towards the one you stated. $\endgroup$Hannes– Hannes2023-07-19 14:11:31 +00:00Commented Jul 19, 2023 at 14:11
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$\begingroup$ Thank you for your answer below. I just think if some additional assumption about $f$ could be helpful, e.g., $f\in BV(\Omega)$ and $|f = 0|=\emptyset$. $\endgroup$mlogm– mlogm2023-07-24 20:38:45 +00:00Commented Jul 24, 2023 at 20:38
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$\begingroup$ @mlogm : I don't understand your comment. Are you satisfied with the answer? If not, why not? $\endgroup$Iosif Pinelis– Iosif Pinelis2023-07-24 21:59:07 +00:00Commented Jul 24, 2023 at 21:59
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$\begingroup$ I think one should also take into account the fact that if $u_\ast$ is a mnimizer then $u_{\ast\ast}=u_\ast$ a.e. on the set $|f\neq 0|$ and $u_{\ast\ast}\neq u_\ast$ on the set $|f=0|$ is also a minimizer. $\endgroup$mlogm– mlogm2023-07-25 07:27:45 +00:00Commented Jul 25, 2023 at 7:27
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$\begingroup$ Your latter comment is incomprehensible to me as well: (i) What is $u_{**}$ is your comment? (ii) What do you mean by "take into account" in this case? -- Concerning (i), if you mean $u_{**}$ is as in my answer, it is stated there: "If $u_{**}\in X(\Omega)$ and $u_{**}=u_*$ almost everywhere (a.e.) on the set $[f\ne0]:=\{x\colon f(x)\ne0\}$, then $\int u_{**}f=\int u_*f$, so that $u_{**}$ is also a minimizer." $\endgroup$Iosif Pinelis– Iosif Pinelis2023-07-25 13:18:53 +00:00Commented Jul 25, 2023 at 13:18
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1 Answer
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Yes (mainly). That $u_*:=1_{f<0}$ is a minimizer (of $\int uf$ over all $u\in X(\Omega)$) follows because, if $0\le u\le1$, then $$uf-u_*f=(u-u_*)f\ge0, \tag{1}\label{1}$$ whence $$\int u_*f\le\int uf.$$
The minimizer is not unique, though. If $u_{**}\in X(\Omega)$ and $u_{**}=u_*$ almost everywhere (a.e.) on the set $[f\ne0]:=\{x\colon f(x)\ne0\}$, then $\int u_{**}f=\int u_*f$, so that $u_{**}$ is also a minimizer. Vice versa, if $u_{**}\in X(\Omega)$ is a minimizer, then, by \eqref{1}, $$\int|u_{**}-u_*|\,|f|=\int(u_{**}-u_*)\,f=\int u_{**}\,f-\int u_*\,f=0,$$ so that $u_{**}=u_*$ a.e. on the set $[f\ne0]$. Thus, $u_{**}\in X(\Omega)$ is a minimizer if and only if $u_{**}=u_*$ a.e. on the set $[f\ne0]$.
The same conclusion holds with $\{v\in L^2(\Omega)\colon v(x)\in[0,1] \ \forall x\in\Omega\}$ in place of $X(\Omega)=\{v\in L^2(\Omega)\colon v(x)\in\{0,1\} \ \forall x\in\Omega\}$.
