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I recently asked this question.

I think, if the following were true, then I would solve my problem.

Let $E\subset\{(x_1,\dots,x_n)\in\mathbb R^n\;|\;x_i\geq 0\, \&\, \sum_ix_i=1\}$ be a convex set. Let $\bar E$ denote the closure of $E$ in $\{(x_1,\dots,x_n)\in\mathbb R^n\;|\;x_i\geq 0 \,\&\, \sum_ix_i=1\}$.

$\bar E$ is a submanifold of $\mathbb R^n$ (($\bar E$ will only be a $C^0$ manifold, but I hope that this is enough for my purposes. If $\bar E$ needs to be a $C^\infty$-manifold, then that wouldn't be too big of a problem.)). For all small enough $\epsilon>0$ I can thus pick a tubular neighbourhood around it.

Is there any way to show that I can pick such a neighbourhood with a boundary which is nowhere piece-wise linear and such that the neighbourhood (together with $\bar E$) is convex? By nowhere piece-wise linear I mean: ``it doesn't contain any line segment''.

All help greatly appreciated.

Dear all:

thanks for your comments. I have updated my question.

@ Fedja, by ``tubular neighbourhood'' I mean Wiki on tubes

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  • $\begingroup$ What exactly do you mean by "tubular"? $\endgroup$
    – fedja
    Commented Mar 22, 2014 at 11:27
  • $\begingroup$ Maybe you need some more hypotheses: convexity of the set E does not imply that E is a manifold. For example take E to be the n-1 simplex you described, minus an arbitrary subset of the boundary of the n-simplex. I might also suggest that you clarify what you mean by "nowhere piecewise linear". Does that mean that it doesn't contain any line segment? $\endgroup$
    – Danny Ruberman
    Commented Mar 22, 2014 at 18:21

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