Transitivity of convexity
Dependencies:
Let $x$ be a convex combination of $\{y_1, y_2, \ldots, y_n\}$ and $y_i$ be a convex combination of $\{z_{i,1}, z_{i,2}, \ldots, z_{i,m_i}\}$. Then $x$ is a convex combination of $\{z_{i,j}: 1 \le i \le n, 1 \le j \le m_i\}$.
Proof
Let $x = \sum_{i=1}^n \lambda_i y_i$, where $\sum_{i=1}^n \lambda_i = 1$. Let $y_i = \sum_{j=1}^{m_i} \mu_{i,j} z_{i,j}$, where $\sum_{j=1}^{m_i} \mu_{i,j} = 1$. Let $\eta_{i,j} = \lambda_i\mu_{i,j}$. Then \[ \sum_{i=1}^n \sum_{j=1}^{m_i} \eta_{i,j} = \sum_{i=1}^n \lambda_i \sum_{j=1}^{m_i} \mu_{i,j} = \sum_{i=1}^n \lambda_i = 1, \] \[ \sum_{i=1}^n \sum_{j=1}^{m_i} \eta_{i,j}z_{i,j} = \sum_{i=1}^n \lambda_i \sum_{j=1}^{m_i} \mu_{i,j}z_{i,j} = \sum_{i=1}^n \lambda_iy_i = x. \] Hence, $x$ is a convex combination of $\{z_{i,j}: 1 \le i \le n, 1 \le j \le m_i\}$.
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- Depth: 5
- Number of transitive dependencies: 5