Let $\Omega$ be a bounded domain in $\mathbb R^n$, $n\geq 2$, with a smooth boundary and let $f$ be a smooth function on $\bar\Omega$. Is there a natural condition that one can impose on $f$ ( say in terms of for example saying that the function is absolutely continuous or something else like a bounded norm in a certain functional space) so that as a result one may be able to conclude that given any $\lambda \in \mathbb R$, the level set $$\{f(x)=\lambda\,:\, x \in \bar{\Omega}\},$$ is either the empty set or is the union of a finite number of disjoint closed sets $U_1,\ldots,U_{k_\lambda}$ and such that the boundaries of each of these closed sets is smooth?
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$\begingroup$ For instance, a smooth immersion of $\mathbb S^1$ like a figure eight, is not ok for your request as a level set, right? But there always be such level sets if there are critical points besides local minimum or maximum points $\endgroup$– Pietro MajerNov 3, 2021 at 19:25
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You cannot say anything good about the level sets for all $\lambda$, no matter what smoothness conditions you impose on your function, except that this set is closed. However Sard's Lemma tells you something about ALMOST all level sets (for almost all lambda).
If you really need information about ALL level sets, you need to impose the condition that your function is analytic in some region containing $\overline{\Omega}$.
answered Nov 3, 2021 at 16:57