maximum of a convex, continuous function

Question

Let $f:\mathbb R^d\rightarrow \mathbb R$ be a convex function and $M\subseteq \mathbb R^d$ compact. Show that $f$ has its maximum over $M$ with $\max\{f(x):x\in conv(M)\}=\max\{f(x):x\in M\}$ where $conv(M)$ is the convex hull. I know that $f$ has to be continuous so it has an maximum over $M$. How can I show that $\max\{f(x):x\in conv(M)\}=\max\{f(x):x\in M\}$?

Answer 1

It's clear that $$B:=\max\{f(x) \backslash x \in Con(M)\} \geq A:=\max\{f(x) \backslash x \in M\}$$, Conversely, let $z\in Conv(M) $, then (by Definition), there's $x_{i} \in M, t_{i}\in \left [0,1\right], \sum_{i=1}^{n}t_{i}=1 $ such that $z=\sum_{i=1}^{n} t_{i} x_{i} $, So $$f(z) \leq \sum_{i=1}^{n}t_{i} f(x_{i}) \leq A\sum_{i=1}^{n}t_{i}\leq A$$ Hence, $B\leq A$, $\cdots$

Answer 2

If $x \in conv (M)$ then $x= \sum\limits_{k=1}^{n} a_kx_k$ with $a_k >0, \sum a_k=1$ and $x_k\in M$ for all $k$. So $f(x)\leq \sum\limits_{k=1}^{n} a_kf(x_k) \leq C \sum\limits_{k=1}^{n} a_k=C$ where $C=\max \{f(y): y \in M\}$. Can you finish?