I encounter the following problem today. It seems a simple question. 


Let $U$ be a real function from $R^+\rightarrow \bar{R}$ satisfying the following conditions:


(1) $U$ is concave, continuous, and strictly increasing,


(2) $\limsup\_{x\rightarrow +\infty}\frac{xU^{'}(x)}{U(x)} <1.$


(3) $ U^{'}(0+) = +\infty, \mbox{ and }U^{'}(+\infty) = 0.$


Is the following statement true?


$\bf Claim:$ For any non-negative random variable $\xi,$ if $E[U(\xi)] < +\infty,$ then we have 
$$E[\xi U^{'}(\xi)] < +\infty.$$


$\bf Remark:$ If $U(0) >-\infty,$ it is trivial if one notices that $$0\leq \xi U^{'}(\xi) \leq U(\xi) - U(0).$$ But in general, the property $U(0) >-\infty$ does NOT hold. For example $\ln(x).$


Any comment and suggestion are welcome. Thanks. 



Is the following statement true? $E[\xi U^{'}(\xi)] < +\infty$?
Take $U(x) = -e^{1/x^2}$ and the probability density function $f(x) = C e^{-1/x^2}$, $x \in [0;1]$, where $C$ is normalization constant.


Then, $U'(x) = \frac{2 e^{1/x^2}}{x^3}$, $E(U(\xi)) = \int\_0^1 U(x) f(x) dx = -C$,
$E(\xi U'(\xi)) = \int\_0^1 x U'(x) f(x) dx = 2C \int\_0^1 \frac{dx}{x^2} = +\infty$


