Why are close points in different regions close to the boundary?

Question

I have a problem and it is bugging the heck out of me. It seems very obvious but now has become a major and frustrating stumbling block to a more productive line of thought.

I have used a regular $n$-simplex $\sigma$ to split $\mathbb{R}^n$ into $n+1$ regions, where the regions are the cones over the large faces of the simplex. I label them arbitrarily from $1$ to $n+1$ and I select a (proper, nonempty) set $S\subset\{1,2,\dots,n+1\}$ of labels.

(A bit of notation: I've found it useful to write $[T]$ to talk about the intersection of the regions with labels in $T$. Then region $k$ is $[\{k\}]$ and larger sets are cones over smaller faces.)

For $s\in S$, let $a_s$ be some element in $[\{s\}]$. Suppose that we know $||a_s-a_t||<\delta$ for all $s,t\in S$. Here is what should be true: that there is some point $a\in[S]$ such that $||a_s-a||<\delta$ for all $s$. I don't need something this strong; it would be fine if $d(a_s,[S])<\delta$ for all $s$, but the first one is almost certainly also correct.

Can anyone help me out, or at least provide a possible direction?

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Edit 1: I'd like to put out a partial result I have found for barycentric coordinates, in case it helps anyone out. Let $\sigma$ have edge length $L$, vertices $x_1, x_2 \dots x_{n+1}$, and the origin as its barycenter. Some facts which can be proved without too much trouble include...

Characterization of $[S]$:

Denote $\{1,2,\dots,n+1\}-S$ by $S^C$. Then $ [S] = \{x_s=x_t\leq x_p : s,t\in S ,~ p\in S^C\}.$ In other words, the $S$-coordinates are equal and minimal.

Inner Products:

Suppose that $a$ and $b$ are in $\mathbb{R}^n$ and notate the barycentric coordinates (with respect to $\sigma$) of $a$ and $b$ by $(a_1,\dots,a_{n+1})$ and $(b_1,\dots,b_{n+1})$, respectively. Then $$\frac{2}{L^2}\langle a,b\rangle = \sum_{i=1}^{n+1}{\left(a_i-\frac1{n+1}\right)\left(b_i-\frac1{n+1}\right)} = \left[\sum_{i=1}^{n+1}a_ib_i\right]-\frac{1}{n+1}.$$

Closeness:

If $||a-b||<\varepsilon$ for all $a$ and $b$ in a bounded set, there is a $k$ such that $|a_i-b_i|<k\varepsilon$. }

As far as I can tell, the bounded set restriction is necessary because those $\frac{1}{n+1}$ terms from the inner product have the wrong sign to be simply removed by the inequality. But they are being added, which needs to end up being something multiplied if you want to keep the arbitrary smallness of $\varepsilon$. In particular, if we could drop $k$ down to one, that would be fantastic.

Now, this is what we can do with all of these. If instead of $||a_s-a_t||<\varepsilon$ we let $||a_s-a_t||<\frac12\varepsilon$, then we can create the following inequality chain, which holds for any $s$ and $t$ in $[S]$: $$\textstyle a_{tt} ~\leq~ a_{ts} ~<~ a_{ss}+\frac12k\varepsilon ~\leq~ a_{st}+\frac12k\varepsilon ~<~ a_{tt}+k\varepsilon.$$ Since the outer terms differ by $k\varepsilon$, and the inequalities that relate them to the third and fourth terms are strict, this implies that $a_{ss}-a_{tt}<k\varepsilon$. This must apply for the reverse direction as well, which means that $|a_{ss}-a_{st}|<k\varepsilon$.

What I cannot do from here is the last step. This seems to be moving in the right direction since now we have a relationship between the $s^\text{th}$ and $t^\text{th}$ coordinates, which is required to be near $[S]$. But the last step has evaded me for a good while now.

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Edit 2: As another useful tool, there is a direct translation between the barycentric coordinates and the coordinates suggested by scaling over faces of $\sigma$:

Let $p$ have barycentric coordinates $(p_1, p_2,\dots, p_n)$. If $p\in [S]$, then we can write $p=\alpha\displaystyle\sum_{j\in S^C} \lambda_jx_j$. These variables are related by the following equations: $$\alpha = 1 + (n+1)p_s \qquad\qquad \lambda_j = \frac{1}{\alpha}(p_j-p_s).$$ (where $s\in S$ can be chosen arbitrarily because all $S$-coordinates of $p$ are equal)

Answer 1

I figured it out on my own (earlier today!), so I'm just putting this answer up to help anyone else who might get hung up on this silly problem. Ultimately, the inner product result was misleading because it seemed to suggest $||a-b||=\sum_{i=1}^{n+1}\left(a_i-b_i-\frac{1}{n+1}\right)^2$. But this is not true because $a_i-b_i$ are not the barycentric coordinates of any point; they sum to zero, not one. This allowed for a closeness result that was exactly what you would expect in Euclidean space.

We wish to find a suitable expression for $||a-b||$ in terms of barycentric coordinates. To begin, we apply definitions to deduce $$||a-b||^2 = \left|\left|\sum_{i=1}^{n+1} (a_i-b_i)x_i \right|\right|^2 = \left\langle\sum_{i=1}^{n+1} (a_i-b_i)x_i ,~ \sum_{i=1}^{n+1} (a_j-b_j)x_j\right\rangle.$$ which simplifies to $$||a-b||^2=\mathop{\sum\sum}_{1\leq\, i,j\, \leq n+1}(a_i-b_i)(a_j-b_j)\langle x_i,x_j\rangle.$$ Let $K=\langle x_i,x_j\rangle$ and $C=\langle x_i,x_i\rangle=||x_i||^2$; these are independent of the choice of $i$ and $j$, so we can split the summation into two parts which give $C$ and $K$ as coefficients for each part: $$||a-b||^2 = \sum_{i=1}^{n+1} C(a_i-b_i)^2 + \mathop{\sum\sum}_{1\leq i\neq j \leq n+1} K(a_i-b_i)(a_j-b_j).$$

It can be shown without a great deal of trouble that $C=\frac{L^2n}{2(n+1)}$ and $K=-\frac {L^2}{2(n+1)}$ by considering the standard simplex.

Now, we return to the expression for the norm, plugging the values of $C$ and $K$: $$||a-b||^2 = \left(\frac{L^2n}{2(n+1)}\right)\sum_{i=1}^{n+1} (a_i-b_i)^2 + \left(-\frac{L^2}{2(n+1)}\right)\mathop{\sum\sum}_{1\leq i\neq j \leq n+1} (a_i-b_i)(a_j-b_j).$$ Adding $\frac{L^2}{2(n+1)}\sum (a_i-b_i)^2$ to the first summation and subtracting it from the second yields $$||a-b||^2 = \frac{L^2}{2}\sum_{i=1}^{n+1} (a_i-b_i)^2 - \frac{L^2}{2(n+1)}\sum_{i=1}^{n+1}{\left( (a_i-b_i) \sum_{i=1}^{n+1} a_j-b_j \right)}.$$

Recall that $\sum a_j=\sum b_j = 1$, so that $\sum(a_j-b_j)=0$. This means the second term in the above expression vanishes, which leaves only the scaled sum of squares: $$||a-b||^2=\frac{L^2}{2}\sum_{i=1}^{n+1} (a_i-b_i)^2.$$

From here, the problem becomes familiar. Because $||a-b||<\varepsilon$, it follows that $(a_i-b_i)^2<\frac{2}{L^2}\varepsilon^2$. Taking the root of both sides gives $|a_i-b_i|<\frac{\sqrt2}{L}\varepsilon$. Conversely, if all of $|a_i-b_i|$ are less than $\delta$, then $||a-b||<\sqrt{n+1}\frac{L}{\sqrt2}\delta$.

Answer 2

I want to post more later, but for now, take the smaller simplex spanned by the points given. It has diameter $<\delta$, and intersects the cone over the cell $[S]$ of the original simplex in a single point. That point lies in th small simplex, so is less than $\delta$ from each of the points.