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Let $\mu$ a probability measure over $\mathbb R$. Show that if for some strictly convex function $f: \mathbb R \rightarrow \mathbb R$ occurs

$$ \int_{\mathbb R} f(x)d\mu (x) = f(x_0). $$

for some $x_0 \in \mathbb R$, then $\mu = \delta_{x_0}.$

I use the definition about strictly convex function, but I cannot get $\mu$

I try to use a function $$h(x) = \mu(x) - \delta_{x_0}(x)$$ nothing happened.

I get a inequality

$$ \int_{\mathbb R}f((1-\lambda)x + \lambda y)d\mu(x) \leq (1- \lambda)f(x_0) + \lambda f(x_0).$$

Could you give me a hand?

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The problem is false as stated. Let $\mu$ be Lebesgue measure on $[0,1]$ and zero elsewhere, and let $f(x) = e^x$. Then $\int f d \mu = \int_0^1 e^x dx = e - 1$, and as $e-1 > 0$, certainly $e-1 = e^{x_0}$ for some $x_0 \in \mathbb{R}$. In fact, we have $x_0 \in (0,1)$ by the intermediate value theorem! – A Blumenthal Jan 31 '13 at 4:25
I think you must need some additional information about $x_0$. Now if $\int x d \mu(x) = x_0 \in \mathbb{R}$, in addition to the circumstances you already have, then we'd be in business: just look carefully at the proof of the Jensen inequality. – A Blumenthal Jan 31 '13 at 4:27
@ A Blumenthal Thanks. – Malaq Jan 31 '13 at 12:09

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