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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 $p_n0$ 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 $p'_n1$, 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:

  1. Can this assumption be dropped? Does the expression $\int_X |f|^p\log|f| d\mu$ still make sense when $\log|f|\notin L^1$ ?
  2. 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.

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    Yes to the second part of (1), because $|f|^p|\log |f||\lesssim |f|^{p-\epsilon}+|f|^{p+\epsilon} \in L^1$ for $p\in J$. For this reason, I'm sure the answer to the first part of (1) must be *yes* also, but I haven't thought about the details. – 2017-02-10

1 Answers 1

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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

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