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I am trying to prove the following problem:

Given a set of points $S = \{(x_i,t_i)_{i = 1}^K \}$ where $x_i \in R^n, t_i >0 ,\forall i = 1,...,K$ and $Y = \{y \in R^n: y = \frac{x}{t},(x,t) \in conv(S) \}$.

I have proved that $Y \subseteq conv(x_1/t_1,...,x_K/t_K)$. But I am stuck on proving that $conv(x_1/t_1,...,x_K/t_K) \subseteq Y$. Can anyone help me and give me some hint? Great thanks!

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To give a motivation as to my answer, let me first write out the proof that $Y\subseteq conv(x_1/t_1,\ldots,x_K/t_K)$. In this case, there exist $x\in\mathbb R^n,t>0$ and non-negative numbers $\lambda_1,\ldots,\lambda_K$ such that $y=\frac xt$ and $$ \sum\limits_{i=1}^K\lambda_ix_i=x,\qquad\sum\limits_{i=1}^K\lambda_it_i=t,\qquad\sum\limits_{i=1}^K\lambda_i=1, $$ by definition. Observe that $$ y=\frac xt=\sum\limits_{i=1}^K\frac{\lambda_ix_i}t=\sum\limits_{i=1}^K\left(\frac{t_i}{t_i\sum\limits_{j=1}^K\lambda_jt_j}\right)\lambda_ix_i=\sum\limits_{i=1}^K\left(\frac{\lambda_it_i}{\sum\limits_{j=1}^K\lambda_jt_j}\right)\frac{x_i}{t_i}, $$ and since $$ \sum\limits_{i=1}^K\frac{\lambda_it_i}{\sum\limits_{j=1}^K\lambda_jt_j}=1, $$ it follows $y\in conv(x_1/t_1,\ldots,x_K/t_K)$.

Conversely, if $z\in conv(x_1/t_1,\ldots,x_K/t_K)$, then there exist non-negative numbers $\lambda_1,\ldots,\lambda_K$ such that $z=\sum\limits_{i=1}^K\lambda_i\frac{x_i}{t_i}$, and $\sum\limits_{i=1}^K\lambda_i=1$. The problem is resolved as soon as we have a list of non-negative numbers $\delta_1,\ldots,\delta_K$ such that $\sum\limits_{i=1}^K\delta_i=1$ and $$ \lambda_i=\frac{\delta_i t_i}{\sum\limits_{j=1}^K\delta_jt_j},\quad\text{for each }i=1,2,\ldots,K. $$ This is because, in this case, we can write $$ z=\sum\limits_{i=1}^K\left(\frac{\delta_i t_i}{\sum\limits_{j=1}^K\delta_jt_j}\right)\frac{x_i}{t_i}=\frac{\sum\limits_{i=1}^K\delta_ix_i}{\sum\limits_{j=1}^K\delta_jt_j}=:\frac{x}t, $$ and, by construction, $(x,t)\in conv(S)$. So the problem is to find the list of numbers $\delta_i$'s. However, this is a linear algebra problem ($K$ variables with $K+1$ equations, one of which is non-homogeneous), and one can construct the $\delta_i$'s directly from the $\lambda_i$'s. For instance, for any $c>0$, if we let $$ \delta_i':=c\frac{\lambda_i}{t_i},\quad\text{for each }i=1,2,\ldots,K, $$ then by multiplying through $t_i$ in each of the above definitions and then adding up the equations, we get: $$ c=\sum\limits_{j=1}^K\delta_j't_j, $$ and upon letting $\delta_i=\frac{\delta_i'}{\sum\limits_{j=1}^K\delta_j'}$, it follows the list $\delta_1,\ldots,\delta_K$ is of the desired form.

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  • $\begingroup$ Why can we construct $\delta_i's$ via c? Seems like we are recurrently using the $\delta'$ $\endgroup$
    – Chloe
    Sep 5, 2017 at 3:06
  • $\begingroup$ No; we first fix $c>0$, then define each $\delta_i'$ through the mentioned formula. Might as well take $c=1$, for instance. The thing is, if done in this way, the sum of the $\delta_i'$'s need not be $1$. This is the reason why at the end we have to divide through by the sum of the $\delta_i'$'s to define the $\delta_i$'s. $\endgroup$
    – Lentes
    Sep 5, 2017 at 4:23
  • $\begingroup$ Notice that $c=\sum_j\delta_j't_j$ is a consequence of the definition of the $\delta_i'$'s. $\endgroup$
    – Lentes
    Sep 5, 2017 at 4:30

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