Let K be a convex set. Suppose $x_1,x_2,...,x_n \in$ K. Prove that: $$ x=\sum_1^n a_jx_j \in K, \space\space\space\space\space\space\space where \space\space\space\space\space\space\space\sum_1^na_j=1 $$ I tried to prove this by induction but failed because I couldn't handle the condition $\sum_1^na_j=1$ for the $(n+1)^{th}$ case.
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3Maybe this answer to a related question helps. – dxiv Feb 01 '18 at 07:13
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Don't you also have to assume that $a_j\ge0$? – bof Feb 01 '18 at 08:21
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$a_j \geq 0$ must hold aswell. If not, you have an affine combination. – Douglas Molin Feb 01 '18 at 08:24
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$\sum_1 ^{n+1} a_j x_j =\alpha \sum_1 ^{n} b_j x_j +(1-\alpha) x_{n+1}$ where $\alpha =\sum_1 ^{n} a_j$, and $b_j = \frac {a_j} { \sum_1 ^{n} a_j}$. First conclude that $\sum_1 ^{n} b_j x_j$ is in K and then use definition of convexity again to finish the proof. ( In writing above identity I have used the fact that $1-\alpha =1-\sum_1 ^{n} a_j =a_{n+1}$.
Kavi Rama Murthy
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