Is this function differentiable? (Potential Differentiate under integral question)

Question

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 $0<p<\infty$?

---

I am not entirely sure whether this statement is true or false. Here is my attempt to prove it, but I may have made some mistakes:

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 so it is dominated by its upper bound, which is integrable over $|E|<\infty$. (Update: This is wrong! Thanks to @Martin Argerami)

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.

Is the above sketch of proof valid?

Thanks for any comment!

**Update: Now I have some doubts on whether the statement is true. Hence, looking for a counter-example.

Theorem 3 (http://planetmath.org/differentiationundertheintegralsign) looks promising however one small caveat is the derivative only exists almost everywhere.

I'm waiting for more answers before accepting. Any answer/comment will receive an upvote from me.**

16 Sep 2016: I just realized a problem: We need $|f(x)|^p\ln|f(x)|\leq g(x)$ for all $p$ in order for the "dominating" to work. I think my previous argument on boundedness of $|f(x)|^p\ln|f(x)|$ does not apply, we need boundedness as $p$ varies not as $x$ varies.

Answer 1

There is a problem with 3., as $\ln|f|$ is not necessarily bounded (it fails when $f(t)=t$, for instance).

So the derivative will exist for any $p$ where $|f|^p\log|f|$ is integrable. As far as I can tell, this is a sufficient condition, but it is not clear that it is necessary.