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For a subset $A\subset V$ of a vector space over $\mathbb{R}$, let

$\mbox{conv}(A) := \left\{ \sum_{i=1}^n a_i x_i\, \middle| \,x_i\in A, a_i \ge 0\text{ with } \sum_{i=1}^n a_i = 1 \right\}$.

I want to show that if for all $x,y\in A$ and $\alpha\in[0,1]$ the vector $(1-\alpha)x + \alpha y $ is an element of $A$, then $A = \mbox{conv}(A)$.

Is there an elegant solution? Thank you for your help!

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2 Answers 2

up vote 4 down vote accepted

The inclusion $A \subset conv(A)$ is obvious.

Now, if A has that property, you can prove that $conv(A) \subset A$ by induction by $n$. Note that

$$\sum_{i=1}^{n+1} a_i x_i = (\sum_{i=1}^n a_i x_i ) + a_{n+1}x_{n+1}=(1-a_{n+1})(\sum_{i=1}^n \frac{a_i}{1-a_{n+1}} x_i ) + a_{n+1}x_{n+1}$$

and

$$\sum_{i=1}^n \frac{a_i}{1-a_{n+1}}=1 \,.$$

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Elegant is subjective.

Since $x= 1\cdot x$, $A\subset \operatorname{conv}(A)$. Let $$C_n:=\{\sum_{i=1}^na_ix_i,x_i\in A,a_i\geq 0, \sum_{i=1}^na_i=1\}$$ for a fixed integer $n$. We have $\operatorname{conv}(A)=\bigcup_{n\geq 1}C_n$, so we have to show that for each $n$, $C_n\subset A$. For $n=1$ and $n=2$, it's assumed. Now, we assume it's true for $n$. Let $x=\sum_{i=1}^{n+1}a_ix_i,a_i\geq 0, \sum_{i=1}^{n+1}a_i=1,x_i\in A$. Let, for $i\in\{1,\dots,n\}$, $b_i:=\frac{a_i}{\sum_{j=1}^na_j}$ (assuming the denominator is not $0$, otherwise the result is true). We have that $\sum_{i=1}^nb_ix_i\in A$ by the induction hypothesis and $x_{n+1}\in A$. Hence $$\left(\sum_{j=1}^nb_j\right)\sum_{i=1}^nb_ix_i+a_{n+1}x_{n+1}\in A.$$ But this term is $x$.

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