Let $f:B_\delta(y)\rightarrow \mathbb{R}^n$ be a smooth function for some $\delta>0$, $y\in\mathbb{R}^m$. Let us define the set $G:= f(B_\delta(y)) + \mathbb{R}^n_\geq$ (using the Minkowski sum). I want to investigate whether a point $\tilde{z}:=f(\tilde{x})$ for some $\tilde{x}\in\text{int}(B_\delta(y))$ (with int the interior of the set) is a smooth boundary point of $G$, i.e. whether there exists an open neighbourhood $U=U(z)$ and a smooth function $\rho:U\rightarrow \mathbb{R}$ that fulfils $$ U\cap G = \{z\in U: \rho(z)<0\} $$ and $$ \frac{d}{dz} \rho(z) \neq 0\ \forall z\in U. $$ Does this hold in general or do I need additional requirements? Could you point me to a book that is treating a similar problem? I am not very familiar with manifolds and therefore struggling with looking into the relevant literature.

Could you give me a hint on how to construct a function $\rho$?

Thank you for your help!

  • $\begingroup$ I would believe that you need further assumptions. Consider a function $f:(-2;2)\rightarrow\mathbb{R}^2$ such that $$f(t)=\begin{cases} (0,-1/2-t),& t\in (-2;-1),\\ (3/2-t,0),&t\in (1;2).\end{cases}$$ Extend $f$ to a smooth function such that $f([-1;1])$ is contained in the first quadrant. Then $G$ looks like a staircase (with a single step). So the boundary of $G$ will not be smooth in this case. $\endgroup$ Apr 13, 2023 at 16:08
  • $\begingroup$ Sorry, should have been $(-3/2+t,0)$ for $t\in (1;2)$ and $f([-1;1])$ should be contained in the upper left quadrant. Also it would be enough to just require $f([-1;2))$ to be in the upper left quadrant and for some $t_0\in (-1;2)$ we have $f(t_0)=(-1,0)$, i.e. no self-intersection needed. $\endgroup$ Apr 13, 2023 at 16:19
  • $\begingroup$ Even making $f$ real-analytic seems not to save us. We can construct the staircase also with $$f: I \rightarrow \mathbb{R}^2, f(t)= (a,b) + (\cos(t), \sin(t))$$ for suitable $a,b$ and a suitable interval $I\subset [-\pi/2; \pi]$. Translate the circle such that it intersects the $y$-axis once at the origin and once below the origin. Pick the starting point of the interval $I$ in such a way that it corresponds to the intersection point below the origin. Pick the endpoint of $I$ in such a way that it corresponds to intersection point of the circle and $x$-axis to the left of the origin. $\endgroup$ Apr 13, 2023 at 16:35
  • $\begingroup$ Thank you for your help, @SeverinSchraven, and for finding a counterexample embarrassingly fast. Do you think that convexity would help? If I see that correctly, in your examples there is always a point in the image of $f$ that is not at the boundary of $G$. $\endgroup$
    – Ina
    Apr 13, 2023 at 17:28
  • $\begingroup$ I have not quite understood your last comment. You want to assume that the image of $f$ is convex? Or that $G$ is convex? None of this will help us. Consider $$f:(0;1)\rightarrow\mathbb{R}^2, f(t)=(0,0).$$ Then $G=\mathbb{R}_{\geq 0}^2$ which does not have a smooth boundary, despite $G$ being convex (and the image of $f$ is contained in the boundary of $G$). Also the entire image of $f$ is at the point where boundary is not smooth. $\endgroup$ Apr 14, 2023 at 0:17


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