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The above is taken from John Lee's Introduction to Smooth Manifolds (p. 121). In Proposition 5.47, one supposes that $M$ is a smooth manifold. Does Proposition 5.47 also hold if $M$ is changed to a smooth manifold with boundary?

Proposition 5.47. Suppose $M$ is a smooth manifold and $f \in C^\infty(M)$.

  1. For each regular value $b$ of $f$, the sublevel set $f^{-1}((-\infty,b])$ is a regular domain in $M$.

  2. If $a$ and $b$ are regular values of $f$ with $a < b$, then $f^{-1}([a,b])$ is a regular domain in $M$.


1 Answer 1


Definitely not.

First recall the definition of a regular domain: it's a properly embedded (hence closed) codimension-$0$ smooth submanifold with boundary in $M$.

The basic problem is that wherever the boundary of a sublevel set intersects the boundary of $M$, you're likely to get a corner or worse. A simple counterexample is to take $M$ to be the closed upper half-plane $\mathbb R \times [0,\infty)$, and take $f(x,y) = x$. Then $f^{-1}((-\infty,0])$ is the quadrant $\{(x,y): x \le 0,\ y\ge 0\}$, which is a smooth manifold with corners but not a regular domain.

But it can be much worse than that -- for example, the "corner points" can have an accumulation point, which prevents the sublevel set from even being a smooth manifold with corners. For example, with $M$ as above, define $f\colon M\to\mathbb R$ by $f(x,y) = u(x)-y$, where $u\colon\mathbb R\to \mathbb R$ is given by \begin{equation*} u(x) = \begin{cases} e^{-1/x^2} \sin \frac{1}{x}, & x\ne 0,\\ 0, & x=0. \end{cases} \end{equation*} Then there are infinitely many points on the $x$-axis where the boundary of the sublevel set $f^{-1}((-\infty,0])$ is not smooth, and they accumulate at the origin.


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