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It is known that for any $\hat{x}$ in a smooth manifold $\mathcal{M}$, there is a neighbourhood $\mathcal{U}(\hat{x})$ such that for all $x\in \mathcal{U}(\hat{x})$ there is a unique projection $\bar{x} \in \mathcal{M}$ which is the global minimum of the distance function $d(x,\cdot)$ belonging to the manifold. Is it true, that the neighbourhood can be chosen so small that there is no other local minimum of $d(x,\cdot)$ within $\mathcal{U}(\hat{x})$?

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  • $\begingroup$ Isn't the closest point to $x$ just $x$, so you can take $U(\hat{x})=M$ for the first statement and similarly for the second? Is the metric Riemannian, Finsler, continuous, or arbitrary? I think you are looking the the phrase "geodesically convex". $\endgroup$
    – Ben McKay
    Sep 10, 2016 at 20:05
  • $\begingroup$ Perhaps your manifold is embedded in some metric space? Or immersed? $\endgroup$
    – Ben McKay
    Sep 10, 2016 at 20:06
  • $\begingroup$ $\mathcal{M}$ is a submanifold of some $\mathbb{R}^n$. So $U(\hat{x})$ is an open subset of $\mathbb{R}^n$. $\endgroup$ Sep 12, 2016 at 8:11
  • $\begingroup$ Careful: your statement should include the hypothesis that $M$ is a submanifold of $\mathbb{R}^n$, and more importantly your statement is not true unless $M$ is an embedded submanifold of $\mathbb{R}^n$: en.wikipedia.org/wiki/Submanifold#/media/… $\endgroup$
    – Ben McKay
    Sep 12, 2016 at 8:14
  • $\begingroup$ The rough idea of a proof: The local minima of distance to $x$ are the points where a normal ray points to $x$. You want to prevent existence of sequences of arbitrarily small triangles, two sides of which are normal to $M$. In the limit, the third side becomes very small, and $M$ is nearly flat, when you zoom in, so such triangles are not possible. Of course, this is only a rough idea of a proof. $\endgroup$
    – Ben McKay
    Sep 12, 2016 at 8:28

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