extreme points and choquet theorem

**Prometheus** · July 15th 2010, 11:58 PM

define the set A to be the functions $f:[0,1]\rightarrow[-1,1]$ such that f is measurable and its integral is $\intop _0 ^1 fdt = 0$ .
let $\varphi(f)=\intop_0 ^1 e^{\alpha f(t)} dt$ .
I need to show that $\varphi(f) \leq e^{\alpha^2/2}$ for all $f \in A$ , using the inequality $\frac{e^t+e^{-t}}{2} \leq e^{t^2/2}$ , and using the extreme points of A.
I know how to prove that the extreme points are $\chi_B -\chi _{B^c}$ where B is a set of measure 1/2, and it is easy to see that for these type of functions the inequality holds. so now I want to show that the maximum of $\varphi$ is attained on these extreme points.

One way I thought of proving this, is by showing that $\varphi$ is continuous, and so has a maximum, because A is compact in the weak* topology. Then I can use the fact that the function is convex to show that the maximum must be on the extreme points.
The other way is somehow use choquet theorem that connects between the functions in A to its extreme points.

don't really know how to continue from here, so any help would be appreciated

**Ackbeet** · July 16th 2010, 02:45 AM

Use Jensen's inequality on $\frac{e^t+e^{-t}}{2} \leq e^{t^2/2}$ . I think that would get you where you need to go.

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