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.

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:

$|f|^p$ is measurable since $f$ is, it is clearly integrable since it is bounded, and $|E|<\infty$.

$\frac{\partial}{\partial p}|f|^p=|f|^p\ln|f|$ exists since on $E$, $|f|>0$.

$|f|^p\ln|f|$ 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.