Finding the limit of an integral over a finite measure set I have the following problem and I'm not sure how to prove the statements.

Let $(E,\mathcal{A},\mu)$ be a measure space such that $0<\mu(E)<\infty$. Given $\varepsilon\in(0,1)$ and $f:E\to[\varepsilon,\infty)$ integrable over $E$.

*

*For $\alpha\in[0,1]$, show that $f^\alpha$ and $\log{f}$ are integrable over $E$.


*For each $\alpha\in[0,1]$, define $F(\alpha)=\int_{E}f^\alpha\,d\mu$. Show that $F$ is differentiable in $\left[0,\dfrac{1}{2}\right)$ and find its derivative.


*Find the value of $$\lim_{\alpha\to 0}\left(\dfrac{1}{\mu(E)}\int_{E}f^\alpha\,d\mu\right)^{\frac{1}{\alpha}}.$$

My idea for the first one is just bound for $\alpha>0$ and for 0, use the finiteness of $E$.
For the second one, use the theorem that says the derivative is the integral of the partial derivative on $\alpha$ under the assumption of some conditions.
For the third one, just try taking logarithm and L'Hôpital.
Can someone help me? Are my ideas correct?
 A: *

*As was mentioned in the comments, notice that $f/\varepsilon:E \to [1,\infty)$. Since $\alpha \in (0,1)$, then $(f/\varepsilon)^{\alpha} \leq f/\varepsilon$. Thus $f^{\alpha} \leq \varepsilon^{\alpha-1}f \in L^{1}$. For the logarithm,
$$ \int_{E} |\log f| d\mu=\int_{\{f \geq 1\}} \log f d\mu-\int_{\{\varepsilon \leq f<1\}} \log f d\mu \leq \|f\|_{L^{1}}-\log(\varepsilon) \mu(E)<\infty.$$


*Let $\alpha \in [0,1/2)$. First observe that $\partial_{\alpha}f^{\alpha}=\log(f)f^{\alpha}$. By Bunyakovsky--Cauchy--Schwarz you have that
$$\| \partial_{\alpha}f^{\alpha}\|_{L^{1}} \leq \|f^{2\alpha}\|_{L^{1}}^{1/2} \| \log(f) \|_{L^{2}}.$$
Note that $\|f^{2\alpha}\|_{L^{1}}^{1/2}<\infty$ since $2\alpha \in [0,1)$. If we prove that $\| \log(f) \|_{L^{2}}<\infty$, by a well-known corollary of the DCT we obtain that $F^{\alpha}$ is differentiable with
$$ \frac{d}{d\alpha} F(\alpha)=\int_{E} \log (f) f^{\alpha} d\mu. $$
To see why the $L^{2}$ norm is bounded, we proceed as before:
\begin{align*}
    \int_{E} |\log f|^{2} d\mu &=\int_{\{f \geq 1\}} |\log f|^{2} d\mu+\int_{\{\varepsilon \leq f<1\}} |\log f|^{2} d\mu \\
    & \leq \frac{1}{\alpha^{2}} \int_{\{f \geq 1\}} |f|^{2\alpha} d\mu+ |\log \varepsilon|^{2}\mu(E) \\
    &<\infty,
\end{align*}
where I used that $\log f^{\alpha} \leq f^{\alpha}$ since we are working in $\{f \geq 1\}$.


*As you mentioned, you can compute the limit using L'Hôpital rule. Let $L$ be the limit. Then
$$\log(L)=\lim_{\alpha \to 0} \frac{1}{\alpha} \log \left(\frac{1}{\mu(E)} \int_{E}f^{\alpha} d\mu \right).$$
Since this is a limit of the form $\frac{0}{0}$ (this is justified by the DCT), you can apply L'Hôpital rule (and DCT) to obtain
\begin{align*}
    \log(L)=\lim_{\alpha \to 0} \frac{1}{1} \log \left(\frac{1}{\mu(E)} \int_{E}  \log(f)f^{\alpha} d\mu \right).
\end{align*}
Finally
$$L=\frac{1}{\mu(E)}\int_{E}\log(f)d\mu.$$
As always, your intuition is very good :).
