Let $f:[0,1]\to\mathbb{R}$ be a bounded (Lebesgue) measurable function.


Consider the function $$w(p)=\int\_0^1|f|^p\,d\mu$$.


Is $w(p)$ differentiable at any $0<p<\infty$? I.e. does $w'(p)$ exist for **all** (not just almost all) $0<p<\infty$?


I hope this is not too easy a question. I have asked on Math.SE (few days ago with bounty), classmates, none of them know how to prove or disprove it. It is kind of a research question since it is not known whether this result is true or false.


Thanks.




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My attempt: I can prove it with an additional assumption that $|f|\geq\epsilon$ for some $\epsilon>0$. But this technique does not work for the general case.


First let $E=\{x\in [0,1]: f(x)>0\}$. Then $w(p)=\int\_E |f|^p\,d\mu$.


Basically I hope to apply "differentiation under the integral" Theorem 2 in <http://planetmath.org/differentiationundertheintegralsign>.


We check the conditions:


1. $|f|^p$ is measurable since $f$ is, it is clearly integrable since it is bounded, and $|E|<\infty$.
2. $\frac{\partial}{\partial p}|f|^p=|f|^p\ln|f|$ exists since on $E$, $|f|>0$.
3. $|f|^p\ln|f|$ is also bounded (since $f$ is bounded, and $|f|\geq\epsilon$) so it is dominated by its upper bound, which is integrable over $|E|<\infty$.


So we apply "Differentiation under the integral", $w'(p)=\int\_E |f|^p\ln|f|\,d\mu$, which exists again since $|f|^p\ln|f|$ is measurable (composition of $f$ with continuous $\phi=x^p\ln x$), and also bounded.



Differentiate an integral (Lebesgue integral)
Yes, we need the boundedness of $|f|^p \ln|f|$ when both $x$ and $p$ vary, but important note is that $p$ may vary on a given segment $[p\_1,p\_2]$, $1<p\_1<p\_2$ (this would imply that the derivative $d/dp(\int)$ exists at all points $p\in (p\_1,p\_2)$, and since the segment is arbitrary, it exists for all $p>1$.) Now this boundedness is pretty clear. 


