The norm in the n-simplex

Question

İ have this question and my attempt to solve it as the following :

Consider an n-simplex $[p_0, p_1,...,p_n]$. If $a,b \in [p_0, > p_1,...,p_n]$, then show that $||a − b|| \leq \sup_i ||a − p_i||$.

My attempt : since $b \in [p_0,p_1,...,p_n]$ then $b = \sum_{i=0}^{n} t_i p_i$

where $t_i \geq 0$ for all $i$ and $\sum_{i=0}^{n} t_i = 1$.

Then using the equation \begin{align} ||a − b| &= ||(\sum_{i=0}^{n} t_i)a − $\sum_{i=0}^{n} t_i p_i|| \\ &= ||\sum_{i=0}^{n} t_i (a-p_i)|| \\ &\leq \sum_{i=0}^{n} t_i ||a-p_i|| \leq \sum_{i=0}^{n} t_i \sum_{i=0}^{n} ||a-p_i|| \end{align}

but $\sum_{i=0}^{n} t_i = 1$ so we have

$$||a − b| \leq \sum_{i=0}^{n} ||a-p_i||$$

but I don't know how I can complete , any help please!

Answer 1

You got very close! You showed the first inequality in this chain. \begin{align} ||a-b|| &\leq \sum_{i=0}^n t_i ||a-p_i|| \\ & \leq \sum_{i=0}^n t_i \left (\sup_j ||a-p_j|| \right ) \\ &= \sup_j ||a-p_j||= \sup_i ||a-p_i|| \end{align} In the second line we have used the bound, valid for all $i=0,1,...,n$ $$||a-p_i|| \leq \sup_j ||a - p_j||$$ to upper bound each term in the second sum. The third line is just $\sum_{i=0}^n t_i = 1$.