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Let $A$ be a set. We define convex hull of $A$ to be $$\text{co}(A) = \bigg\{ \sum_{i=1}^n\lambda_ia_i: 0\leq \lambda_i\leq 1 , \sum_{i=1}^n\lambda_i=1, a_i\in A \text{ for all }1\leq i\leq n \bigg\}.$$

My question is, can we reduce the summation to two terms only? More precisely,

Question: Is it true that $$\text{co}(A) = \{\lambda a + (1-\lambda)b: a,b\in A\}?$$

Clearly the reverse inclusion $\supseteq$ holds. For the forwards inclusion $\subseteq,$ I have some idea.

Let assume that $n=3,$ that is, consider $$\lambda_1 a_1+\lambda_2a_2+\lambda_3a_3.$$ Observe that \begin{align*} \lambda_1 a_1+\lambda_2a_2+\lambda_3a_3 & = (\lambda_1+\lambda_2)\bigg( \frac{\lambda_1}{\lambda_1+\lambda_2}a_1 + \frac{\lambda_2}{\lambda_1+\lambda_2}a_2 \bigg) + \lambda_3a_3 \\ & = (1 - \lambda_3)\bigg( \frac{\lambda_1}{\lambda_1+\lambda_2}a_1 + \frac{\lambda_2}{\lambda_1+\lambda_2}a_2 \bigg) + \lambda_3a_3. \end{align*} However,I do not know that whether the following holds $$\frac{\lambda_1}{\lambda_1+\lambda_2}a_1 + \frac{\lambda_2}{\lambda_1+\lambda_2}a_2\in A.$$

Any hint is appreciated.

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  • $\begingroup$ Maybe I'm missing something, but if A is convex then co(A)=A and all those convex combinations are producing no new elements. $\endgroup$ – Ingix May 11 '18 at 9:27
  • $\begingroup$ My bad. Convexity of $A$ is not assumed. Edited my question. $\endgroup$ – Idonknow May 11 '18 at 13:47
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If $A$ is not convex, then your proposition is not true. Let $A$ be a set consisting of 3 points not on a single line. Then the convex hull $co(A)$ is the triangle spanned by these 3 points (sides and interior), while $\{λa+(1−λ)b:a,b\in A\}$ is just the 3 sides.

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If $A$ is convex then $co(A)=A$ and $\{\lambda a+(1-\lambda)b:0\leq \lambda \leq 1,a \in A,b \in A \} =A$

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  • $\begingroup$ I apologise for my mistake. Convexity of $A$ is not assumed. $\endgroup$ – Idonknow May 11 '18 at 13:50

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