description/proof of that for real vectors space, convex subset, and finite points on convex subset, convex combination of points is on convex subset
Topics
About: vectors space
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any real vectors space, any convex subset, and any finite points on the convex subset, any convex combination of the points is on the convex subset.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(V\): \(\in \text{ the real vectors spaces }\)
\(S\): \(\in \{\text{ the convex subsets of } V\}\)
\(\{s_0, ..., s_n\} \subseteq S\):
\(v\): \(= t^j s_j\), \(\in \{\text{ the convex combinations of } \{s_1, ..., s_n\}\}\)
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Statements:
\(v \in S\)
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2: Proof
Whole Strategy: see it by the induction principle with respect to \(n\); Step 1: see that it holds when \(n = 0\); Step 2: see that it holds when \(n = 1\); Step 3: suppose that it holds when \(0 \le n \le n' - 1\) for \(2 \le n'\), and see that it holds when \(n = n'\).
Step 1:
Let us suppose that \(n = 0\).
\(v = t^j s_j = t^0 s_0 = 1 s_0 = s_0 \in S\).
Step 2:
Let us suppose that \(n = 1\).
\(v = t^j s_j = t^0 s_0 + t^1 s_1 = (1 - t^1) s_0 + t^1 s_1 = s_0 + t^1 (s_1 - s_0) \in S\).
Step 3:
Let us suppose that \(v \in S\) when \(0 \le n \le n' - 1\) for \(2 \le n'\).
Let us suppose that \(n = n'\).
\(v = t^j s_j = \sum_{j \in \{0, ..., n' - 1\}} t^j s_j + t^{n'} s_{n'}\).
When \(t^{n'} = 1\), \(t^0 = ... = t^{n' - 1} = 0\), and \(v = t^{n'} s_{n'} = s_{n'} \in S\).
When \(t^{n'} \neq 1\), \(v = (1 - t^{n'}) (1 / (1 - t^{n'}) \sum_{j \in \{0, ..., n' - 1\}} t^j s_j) + t^{n'} s_{n'} = (1 - t^{n'}) (\sum_{j \in \{0, ..., n' - 1\}} 1 / (1 - t^{n'}) t^j s_j) + t^{n'} s_{n'}\).
\(0 \le 1 / (1 - t^{n'}) t^j\) and \(\sum_{j \in \{0, ..., n' - 1\}} 1 / (1 - t^{n'}) t^j = 1 / (1 - t^{n'}) \sum_{j \in \{0, ..., n' - 1\}} t^j = 1 / (1 - t^{n'}) (1 - t^{n'}) = 1\), so, \(\sum_{j \in \{0, ..., n' - 1\}} 1 / (1 - t^{n'}) t^j s_j\) is a convex combination of \(\{s_0, ..., s_{n' - 1}\}\), so, by the induction hypothesis, \(\sum_{j \in \{0, ..., n' - 1\}} 1 / (1 - t^{n'}) t^j s_j \in S\).
As \(0 \le 1 - t^{n'}, t^{n'}\) and \(1 - t^{n'} + t^{n'} = 1\), \(v = (1 - t^{n'}) (\sum_{j \in \{0, ..., n' - 1\}} 1 / (1 - t^{n'}) t^j s_j) + t^{n'} s_{n'}\) is a convex combination of \(\{\sum_{j \in \{0, ..., n' - 1\}} 1 / (1 - t^{n'}) t^j s_j, s_{n'}\}\), and so, \(v \in S\).
