Show that : $ \sup_{x\in \overline{\text{co}} (A)}\|x\|=\sup_{x\in A} \|x\| $

Question

Let $(X,\|.\|)$ be a Banach space and $A\subset X$.

Show that :

$$ \sup_{x\in \overline{\text{co}} (A)}\|x\|=\sup_{x\in A} \|x\| $$

with : $$ \overline{\text{co}}(A)=\overline{\{y~:~y=\sum_{i=1}^{n}{\lambda_ix_i~,~\text{with}~\lambda_i\in[0,1]~, ~\text{such that}~\sum_{i=1}^{n}{\lambda_i}=1,~\text{and }x_i\in A,~n\in\mathbb{N} }\}} $$ An idea please.

Answer 1

How about

$$\left\|\sum_{i=1}^{n}\lambda_ix_i\right\|\leq \sum_{i=1}^{n}\lambda_i\|x_i\|\leq \sum_{i=1}^{n}\lambda_i\sup_{x\in A}\|x\| = \sup_{x\in A}\|x\|$$

The reverse inequality is even simpler