Convex Functions and their maximums? Show that the maximum of a convex function on a convex polytope occurs at one of the extreme points of the polytope. I am really stuck here and don't know where to begin.
 A: Hint: For the following reasoning use


*

*The definition of convexity of a function

*The fact that the convex polytope over which the function is maximized contains all convex combinations of points in it. In fact every point in the convex polytope can be expressed as a convex combination of it's extreme points.

*The definition of an exterme point (that it cannot be written as a linear combination of two distinct points).


General idea of proof (you need to write it in more detail): Consider the point $$x_0=λx_1+(1-λ)x_2 \quad \in C$$ where $λ \in [0,1]$ and $x_1, x_2 \in C$. Since $f$ is convex (I named your convex function $f$) you know that $$f(λx_1+(1-λ)x_2)\le λf(x_1)+(1-λ)f(x_2)$$ for every $x_1, x_2 \in C$ (I named the convex polytope $C$), $λ \in [0,1]$. But this implies that the inner point $x_0:=λx_1+(1-λ)x_2$, for $λ \in (0,1)$ (which is in $C$, since $C$ is convex) cannot have a maximum since it's value under $f$ is dominated by the RHS of the above expression (where $x_1$ and $x_2$ can be chosen to be extreme points). 
(Note of course, than when the inequality holds with a equality, then the maximum is attained in the inner point $x_0$ as well as in the extreme points $x_1, x_2$ so the theorem is again valid.) 
A: for example:
Now, note that for any fixed $t\geq0$, the function $g_{t}:R^{r}\rightarrow R$ defined by $g_{t}(x)=-\frac{1}{2}\sum_{k=1}^{r}ln(1+2tx_{k})$ is convex. Hence, its maximum over the simplex $\{x\in R^{r}: \sum_{k=1}^{r}x_{k}=1,x\geq 0\} $ is attained at a vertex of the simplex.
 so $g_{t}(x)=-\frac{1}{2}\sum_{k=1}^{r}ln(1+2tx_{k})\leq-\frac{1}{2}ln(1+2t)$
I want to ask how to understand this?
