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Let $z \in \operatorname{conv}(\{x_{1}, x_{2},x_{3}\})$. Find $y \in \operatorname{conv}(\{x_{1}, x_{2}\})$, so that $z \in \operatorname{conv}(\{y, x_{3}\})$

My idea so far: since $z \in \operatorname{conv}(\{x_{1}, x_{2},x_{3}\})\implies z=\lambda_{1}x_{1}+\lambda_{2}x_{2}+\lambda_{3}x_{3}\implies z-\lambda_{3}x_{3}=\lambda_{2}x_{2}+\lambda_{1}x_{1}\implies \frac{z}{\lambda_{1}+\lambda_{2}}-\frac{\lambda_{3}x_{3}}{\lambda_{1}+\lambda_{2}}=\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}}x_{2}+\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}x_{1} (*)$

But $\lambda_{1}+\lambda_{2}=1-\lambda_{3}$ and hence:

$\frac{1}{\lambda_{1}+\lambda_{2}}z-\frac{\lambda_{3}}{\lambda_{1}+\lambda_{2}}x_{3}=\frac{1}{1-\lambda_{3}}z-\frac{\lambda_{3}}{1-\lambda_{3}}x_{3}=:y$, thus defining it as $y$ and $y \in \operatorname{conv}(\{z,x_{3}\})$ and by $(*)$ it is clear that $y \in \operatorname{conv}(\{x_{1},x_{2}\})$

but I cannot say that $y \in \operatorname{conv}(\{z,x_{3}\})$ implies that $z \in \operatorname{conv}(\{y,x_{3}\})$, so I am lost.

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  • $\begingroup$ You can see directly that $z=(\lambda_1+\lambda_2)y+\lambda_3 x_3$. $\endgroup$ – Stinking Bishop Nov 13 at 21:46
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Yes, $$y = \sum_{i=1}^2 \frac{\lambda_i}{\lambda_1+\lambda_2} x_i$$ is the right candidate. Now solve $$y = \frac{z-\lambda_3 x_3}{\lambda_1+\lambda_2}$$ for $z$, obtaining $$z=(\lambda_1+\lambda_2)y+\lambda_3 x_3 = \mu_1 y+\mu_2 x_3,$$ where $\mu_1 = \lambda_1+\lambda_2$ and $\mu_2 = \lambda_3$. Now $z$ is a convex combination of $y$ and $x_3$ because $\mu_1+\mu_2=\sum_{i=1}^3 \lambda_i = 1$ and $\lambda_i \ge 0$ for all $i$ so $\mu_j \ge 0$ for all $j$.

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