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

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.

• You can see directly that $z=(\lambda_1+\lambda_2)y+\lambda_3 x_3$. – Stinking Bishop Nov 13 '19 at 21:46

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 $$j$$.