Is the map $p\mapsto\int_X |f|^p d\mu$ differentiable when $\log|f|\notin L^1$? Let $(X,\mathscr{M},\mu)$ be a measure space and $f$ be a measurable function. Suppose $J$ is a nonempty open subinterval of $(0,\infty)$ such that $f\in L^p$ for all $p\in J$, and let $\varphi:J\rightarrow\mathbb{R}$ be the map given by $\varphi(p)=\int_X |f|^p d\mu$. My goal is to show that $\varphi$ is differentiable  on $J$, with the obvious derivative $\varphi'(p)=\int_X |f|^p\log|f| d\mu$. I have succeeded in proving this under the assumption that $\log|f|\in L^1$. My argument goes as follows:
Let $p\in J$ and $(p_n)_{n\geq 1}$ be a sequence in $J$ converging to $p$ with $p_n\ne p$ for all $n\geq 1$. Choose a point $q\in J$ such that $p_n<q$ for all $n\geq 1$, a small $\epsilon>0$ such that $q+\epsilon\in J$, and a constant $C_{\epsilon}$ satisfying the inequality $\log t\leq C_{\epsilon}t^{\epsilon}$ for all $t\geq 1$.
We have $$\frac{\varphi(p_n)-\varphi(p)}{p_n-p}=\int_{|f|>1}\frac{|f(x)|^{p_n}-|f(x)|^{p}}{p_n-p} d\mu+\int_{|f|\leq 1}\frac{|f(x)|^{p_n}-|f(x)|^{p}}{p_n-p} d\mu$$ and the mean value theorem gives (for each $x$) points $p'_n$ lying between $p_n$ and $p$ such that $\frac{|f(x)|^{p_n}-|f(x)|^{p}}{p_n-p}=|f(x)|^{p'_n} \log|f(x)|$. By our choice of $q$, we have $p'_n<q$ regardless of $x$ and $$\left|\frac{|f(x)|^{p_n}-|f(x)|^{p}}{p_n-p}\right|=|f(x)|^{p'_n}|\log|f(x)||\leq|f(x)|^q\log|f(x)|\leq C_{\epsilon}|f(x)|^{q+\epsilon}$$ whenever $|f(x)|>1$, and $$|f(x)|^{p'_n}|\log|f(x)||\leq|\log|f(x)||$$ whenever $|f(x)|\leq1$.
Since $q+\epsilon\in J$, both the functions $|f|^{q+\epsilon}$ and $|\log|f||$ are in $L^1$ and the dominated convergence theorem applies. Hence we have $$\lim_{n\rightarrow\infty}\frac{\varphi(p_n)-\varphi(p)}{p_n-p}=\int_{|f|>1} |f|^p\log|f| d\mu+\int_{|f|\leq 1} |f|^p\log|f| d\mu=\int_X |f|^p\log|f| d\mu$$ and we conclude that $\varphi$ is differentiable at $p$ with the derivative $\int_X |f|^p\log|f| d\mu$.
As above, I have used the additional assumption $\log|f|\in L^1$ in the argument. My questions are:


*

*Can this assumption be dropped? Does the expression $\int_X |f|^p\log|f| d\mu$ still make sense when $\log|f|\notin L^1$ ?

*When $\log|f|\notin L^1$, the integral $\int_X \log|f| d\mu$ exists but equals $-\infty$. In this case, if we further assume that $\mu(X)=1$, is it true that $\lim_{p\rightarrow 0}\Vert f\Vert_p=0$?


Please enlighten me. Any advice on the questions or the argument I have made is welcome. Thank you in advance.
 A: You do not need $\log|f|\in L_1$.
The same trick you used for the set $|f|\ge1$ works for $|f|<1$ also.
Fix some reals $a<b<c<d$ such that $[a,d]\subset J$ and $p\in (b,c)$. Then 
$$
|\log t| < C \begin{cases}
t^{d-c} & \text{if } t\ge1\\
\frac1{t^{b-a}} & \text{if } t<1\\ \end{cases}
$$
with some constant $C=C(a,b,c,d)$.
For sufficiently large $n$ we have $p_n\in (b,c)$ and
$$
|f(x)|^{p_n'} |\log|f(x)|| \le \begin{Bmatrix}
|f(x)|^c \cdot C|f(x)|^{d-c} & \text{if }|f(x)|\ge1 \\
|f(x)|^b \cdot \frac{C}{|f(x)|^{b-a}} & \text{if }|f(x)|<1 \\
\end{Bmatrix} < C|f(x)|^d+C|f(x)|^a.
$$
So, $C|f(x)|^d+C|f(x)|^a$ is a suitable dominant function.

The second question is interesting only if $f\ne0$ a.e.
Let $X=Z\cup S\cup L$ where $Z=f^{-1}(\{0\})$, $Z=f^{-1}((0,1))$ and $L=f^{-1}([1,\infty))$ are the sets where $f$ is zero, small or large, respectively. Then, applying L'Hospital's rule,
$$
\lim_{p\to+0}\frac{\int_L|f|^p-\int_L1}{p} \stackrel{L'Hosp} = 
\lim_{p\to+0}\frac{\int_L|f|^p\log|f|}{1} \stackrel{Dom.conv} =
\int_L \log|f|
$$
and
$$
\lim_{p\to+0}\frac{\int_S|f|^p-\int_S1}{p} \stackrel{L'Hosp} = 
\lim_{p\to+0}\frac{\int_S|f|^p\log|f|}{1} =
-\lim_{p\to+0}\int_S|f|^p|\log|f|| \stackrel{Mon.conv} =
\int_S \log|f|
$$
so
$$
\lim_{p\to+0}\frac{\int_{S\cup L}|f|^p-\mu(S\cup L)}{p} =
\int_{S\cup L} \log|f| = \int_X \log|f| = -\infty.
$$
Hence, for every $K>0$ there is some $\delta>0$ such that for $0<p<\delta$ we have
$$ \frac{\int_{S\cup L}|f|^p-\mu(S\cup L)}{p} < -K $$
$$ \int_X|f|^p = \int_{S\cup L}|f|^p < \mu(S\cup L) -Kp = 1-Kp < e^{-Kp} $$
$$ \|f\|_p < e^{-K}. $$
