0

Suppose $X$ is a salient convex cone. That is, if $x,y$$\in$$X$ and $\alpha,\beta$$\geq$$0$ are scalars, then $\alpha$$x$+$\beta$$y$$\in$$X$ and if $0\neq x$$\in$$X$, then $-x$ is not.

Then for any $\bar{x}$$\in$$X$, $$\bigcup_{\sum_{i=1}^{n}{x_i=\bar{x}},x_i\in X}\bigcap_{i=1}^{n}{X+\bar{x}-x_i}=X+(n-1)\bar{x}/n$$ where I use the notation $X+y=${${x+y:x\in X}$}.

Is this claim true? It holds when $X=\mathbb{R}_+$, which I show below.

For $X=\mathbb{R}_+$, take an arbitrary $\bar{x} \in X$ and $x_i \in X$ such that $x_1+...+x_n=\bar{x}$. Then $$\bigcap_{i=1}^{n}{X+\bar{x}-x_i}=[\bar{x}-x_k,\infty)$$ for some $k$ such that $x_k\leq \bar{x}/n$. This implies that $$\bigcap_{i=1}^{n}{X+\bar{x}-x_i} \subseteq X+(n-1)\bar{x}/n$$ and therefore the union is also contained in $X+(n-1)\bar{x}/n$. The reverse inclusion also holds because $\bar{x}/n + ... + \bar{x}/n = \bar{x}$.

Mike
  • 498
  • Thanks. The condition for a convex cone to be salient is that for all non-zero vectors $x$ in $X$, $-x$ is not in $X$. – Mike Dec 02 '14 at 23:19
  • 1
    I think there's a technical glitch in your example with $X = \mathbb R_+.$ Using the notation from there, we have $\cap_{i=1}^nX+\overline x - x_i = \cap_{i=1}^n[\overline x - x_i,\infty) = [\max_{1\leq i\leq n}\overline x - x_i,\infty) = [\overline x - \min_{1\leq i\leq n}x_i,\infty).$ Now, $\min_{1\leq i\leq n}x_i \leq \overline x / n,$ since not every summand $x_i$ can be strictly larger than the average $\overline x / n.$ – jflipp Dec 03 '14 at 13:20
  • 1
    [continued] So, $\overline x - \min_{1\leq i\leq n}x_i \geq \overline x - \overline x / n$ and $[\overline x - \min_{1\leq i\leq n}x_i,\infty) \subseteq [\overline x - \overline x / n, \infty).$ By setting identically $x_i = \overline x / n,$ we see that $[\overline x - \overline x / n, \infty)$ can be reached and conclude that the union over all decompositions of $\overline x$ is $X + \overline x - \overline x / n,$ which is different from your claim. You might want to adjust your claim accordingly. – jflipp Dec 03 '14 at 13:25
  • Thanks. The original claim has been modified. (I hastily generalized from n=2 to n>2 in my statement of the claim without looking back at my notes.) – Mike Dec 03 '14 at 14:17

1 Answers1

0

Lemma 1: Suppose $X_1,…,X_N$ are convex cones. Then $$\bigcap_{i=1}^{N}{X_i}\subset \sum_{i=1}^{N}{\frac{1}{N}X_i}$$

Proof of Lemma 1: Take $x\in\cap_{i=1}^{N}X_i$. Then $x\in X_i$. Since $X_i$ is a convex cone, $\frac{1}{N}x\in X_i$ for each $i$. Therefore, $$x=\sum_{i=1}^{N}\frac{1}{N}x\in \sum_{i=1}^{N}\frac{1}{N}X_i.\blacksquare $$

Lemma 2: Let $X$ be a convex cone. For any $\bar{x}\in X$,$$\bigcup_{x_1+…+x_N=\bar{x}}_{x_i\in X}\bigcap_{i=1}^{N}X+x_i=X+\frac{1}{N}\bar{x}.$$

Proof of Lemma 2: To prove $(\subset)$, take $x_1,…,x_N$ such that $\sum_{i=1}^{N}{x_i=\bar{x}}$. From Lemma 1, we know that $$\bigcap_{i=1}^{N}{X+x_i}\subset \sum_{i=1}^{N}\frac{1}{N}{(X+x_i)}.$$

Since $X$ is a convex cone, $\frac{1}{N}X=X$. We therefore have that $$\sum_{i=1}^{N}{\frac{1}{N}{(X+x_i)}=X+\sum_{i=1}^{N}{\frac{1}{N}}x_i=X+\frac{1}{N}\bar{x}},$$

since $x_i$ is just a scalar. Since $x_1,…,x_N$ was arbitrary, this holds for all $x_1,…,x_N$ with $x_i\in X$ and $\sum_{i=1}^{N}{x_i}=\bar{x}$. This establishes $(\subset).$

To prove $(\supset)$, suppose $x\in X+\frac{1}{N}\bar{x}$. Let $x_i=\frac{1}{N}x$. Then $\sum_{i=1}^{N}{\frac{1}{N}x_i}=x$, and $\frac{1}{N}x\in X+\frac{1}{N}x$, establishing the claim.$\blacksquare$

Lemma 2 applies to all convex cones, including salient convex cones (which is the non-trivial case). The claim in the question is then a straightforward corollary (involving minor relabeling).

Mike
  • 498