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I have a doubt regarding a constrained optimisation problem.

Suppose my original constrained minimisation problem is $$ \min_x f(g(x), x) \quad \text{ s.t. }\quad g(x)=3 $$ I would like to know if this equivalent to solving the unconstrained minimisation problem $$ \min_x f(3, x) $$ If not, when are these two problems equivalent?

Star
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  • I would immediately say that these are equivalent, but the devil is in the details: where does $x$ take values? Is it closed set? Are the functions continuous? Is the set $g^{-1}({3})$ non-empty? – topolosaurus Mar 28 '22 at 10:00
  • (1) $x\in \mathcal{X}\subseteq \mathbb{R}^4$. $g$ has domain $\mathcal{X}$. I can assume $\mathcal{X}$ closed if needed. (2) $\mathcal{X}_g\equiv {x\in \mathcal{X}: g(x)=3}\neq \emptyset$. (3) $g(x)$ can be assumed continuous with respect to $x$ if needed. Regarding $f(g(x),x)$, do we need continuity with respect to $x$, or with respect to both $g$ and $x$? – Star Mar 28 '22 at 10:19
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    They are almost never equivalent. They would only be equivalent if you knew that $g$ is equal to $3$ on the set of optimal solutions to the second problem, which seems like too much of a coincidence to ask for. – Michal Adamaszek Mar 28 '22 at 10:41

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The problems are certainly not equivalent in general. Take $\ f:\mathbb{R}\times\mathbb{R}^4\rightarrow\mathbb{R}\ $ to be given by $$ f(y,x)=y^2+\|x\|^2\ , $$ and $\ g:\mathbb{R}^4\rightarrow\mathbb{R}\ $ to be given by $$ g(x)=\mathbb{1}^\top x\ . $$ for instance. Then the minimum value of $\ f(3,x)\ $ is $\ 9\ $, and occurs when $\ x=0\ $, whereas $$ \min_x f(g(x),x)\ \ \text{ s.t. }\ \ \ g(x)=3 $$ is $\ 11\frac{1}{4}\ $, and occurs when $\ x=\frac{3}{4}\mathbb{1}\ $. I agree with Michal Adamszek's comment that the two problems will almost never be equivalent.

lonza leggiera
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