3
$\begingroup$

I have a signed measure $\mu$ on a convex subset $C\subset \mathbb{R}^n$, and I want to prove that $\mu$ is a probability measure, most importantly that it is positive everywhere.

I do know that $\int f(x)d\mu(x)\geq 0$ for any positive CONVEX function $f$. So if I could get this inequality for indicator functions I'd be done.

Do you know if this suffices to get that the measure is positive, or maybe have a counterexample?

$\endgroup$
1
  • 1
    $\begingroup$ If $C=(a,b)$ is an interval, your assumption is that $\langle \mu,f\rangle\ge0$ for every $f$ such that $f''\ge0$ in distributional sense. Therefore an equivalent statement is that $\mu=N''$ where $N\ge0$ and $N(a)=N(b)=0$. If $n\ge2$, a characterization must be more involved. $\endgroup$ Oct 13, 2011 at 6:40

1 Answer 1

11
$\begingroup$

A counterexample is a signed measure on the interval $I:=[-1,1]$ concentrated in the points $\{-1\}$, $\{0\}$, $\{1\}$ with weights respectively $1/2$, $-1$, $1/2$. (Thus $\int_If d\mu= f(1)/2+f(-1)/2 - f(0)\ge0$ is just the convexity inequality).

$\endgroup$
1
  • 1
    $\begingroup$ Thanks, that's a very simple example. I actually had the problem stated otherwise before, and I thought that the formulation I put here was simpler, but equivalent, but I see that was not really the case. Maybe I'll put out another question with the alternative formulation. $\endgroup$ Oct 12, 2011 at 18:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.