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

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

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.

$E$ is a submanifold of $\mathbb R^n$. 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 $E$) is convex?

All help greatly apreciated.

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