Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose we have a measure $\mu$ on $\mathbb R_+$ such that $\forall s>-1$ $t^s\in L^1(\mathrm d\mu(t))$.

I'd like to impose some conditions on $\mu$ so the function

$$f:p\to \frac{\int_0^\infty t^p \mathrm d\mu(t)}{\int_0^\infty t^{p-1}\mathrm d\mu(t) },\quad p>0 $$ is monotone on $(0,\infty)$. For example, if $\mathrm d\mu(t)=e^{-t}\mathrm d t $, then $f(p)=p$. Another example is $\mathrm d\mu(t)= \mathbf{1}_{t\in (0,1)}\mathrm d t $, $f(p)=p/(p+1)$ (monotone, too). If $\mathrm d\mu(t) = e^{-t^2}$, then for $p\in [0,5]$ (I checked numerically) $f(p)$ is monotone. In fact, I can't find a counterexample to the desired result.

Somehow this reminds me the Riesz–Thorin theorem, but I don't see how I can apply it here.

Are there any results on this (or similar) subject? What are possible ways to approach this problem?

share|improve this question
2  
This is a simple consequence of log-convexity of the function $p \mapsto \int_{0}^{\infty} t^{p} \mu(dt)$, which can be estabilished by Holder's inequality. –  Mateusz Wasilewski Jun 23 '13 at 8:39
    
@MateuszWasilewski Thanks a lot! If you put your comment in an answer, I'll accept it. –  TZakrevskiy Jun 30 '13 at 13:32
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.