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Reffering to this question Clarke's tangent cone, Bouligand's tangent cone, and set regularity I'm asking myself if may exist a closed bounded set $S\in\mathbb{R}^2$ and a point $x\in\partial S$ such that the two cones (Bouligand and Clarke cones) reduces both to $\{0\}$, namely only the origin. Is this possible if the boundary is piecewise the graph of a continuous function?

I think the answer has to be no, at least for the second part of the question but I'm not able to find a proof.

Any reference to books or articles will be greatly appreciated.

Thank you in advance.

Mathland
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1 Answers1

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It might depend on your notion of "piecewise the graph of a continuous function".

Define $r \colon [0,1] \to [0,1/e]$ via $r(\theta) = \exp(-1/\theta)$ for $\theta > 0$ and $r(0) = 0$. Then $r$ is continuous and $$ C := \{ r(\theta) (\cos(\theta),\sin(\theta)) \mid \theta \in [0,1]\} \subset \mathbb R^2 $$ is the image of a continuous function. Then, a straightforward but tedious computation yields that both cones at $0$ reduce to $\{0\}$.

gerw
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  • The first example that come in mind to me was a spiral but it fails to be a graph (both locally or piecewise) at the origin. For me the definition of piecewise graph of a continuous function is the one in https://math.stackexchange.com/questions/4511335/on-a-property-of-curved-polygons – Mathland Apr 25 '23 at 11:08
  • Spiral is the first idea but it fails to be a graph (both locally or piecewise) at (0,0). Mine definition of piecewise graph of a continuous function is the one in https://math.stackexchange.com/questions/4511335/on-a-property-of-curved-polygons. @gerw The property of being locally graph is the same except for the fact that in this case, using the notation in the linked page, for every $t\in[0,1]$ there exist an interval [a,b] containing t such that on each interval [a,b], restricted to [a,b] is the graph of a continuous function g() and the interior of S "stays below" the graph of f. – Mathland Apr 25 '23 at 11:16
  • But if $S$ is "below" $\partial S$, we should always have that $(0,-1)$ belongs to the tangent cones. – gerw Apr 26 '23 at 06:46
  • as I suspect...Thank you – Mathland Apr 27 '23 at 05:34