Let $(X,||\cdot||)$ be a normed space where $||\cdot||$ is the sup-norm and let $E$ be a convex and compact subset. Let $f:E\to [0,1]$ be continuous and affine, i.e. for all $x,y\in E$ and all $\lambda \in [0,1]$, $f(\lambda x+(1-\lambda)y)=\lambda f(x)+(1-\lambda)f(y)$. 


Is $f$ Lipschitz? If not, can you give a simple counter-example?


Technically, I can always modify the metric on $E$ such that $d(x,y)=||x-y||+|f(x)-f(y)|$. Then $f$ is Lipschitz. But this metric is not well-defined over the entire space $X$. I was hoping to show (or not show) that $f$ is Lipschitz under the original sup-norm metric.



affine vs lipschitz
For instance, consider the convex subset $E:=\{x\in\ell\_\infty: 0\le x\_k\le 2^{-k} \text{ for all } k\ge0 \}$ of $\ell\_\infty$. By dominated convergence, $E$ is compact and its relative topology coincides with the topology induced by $\ell\_1$. 


In particular, the affine function $f:E\to [0,1]$ defined by $f(x):=\sum\_{k=1}^\infty x\_k$ is continuous on $E$.


One has $0\in E$ and $f(0)=0$, while for $u=2^{-n}\sum\_{k=1}^n e\_k\in E$ one has $\|u\|\_\infty=2^{-n} $ and $f(u)=2^{-n}n=n\|u\|\_\infty $, so $f$ is not Lipschitz. 


