Limit of the logarithm of the $L^p$ norm over the logarithm of $p$ as $p$ goes to infinity Let $\mu$ be some positive measure on $\mathbb{R}$. For technical reasons, I would like to know if the limit
$$\lim_{p\rightarrow\infty}\frac {\ln \|f\|_{L^p(\mu)}}{\ln p}$$
exists in $[0,\infty]$ for any $f$ (That is, I want the limit to exist, but perhaps not be finite.)
Moreover generally I would like to know if in general,
$$\lim_{p\rightarrow\infty}\frac {\frac{d^k}{dp^k} \ln \|f\|_{L^p(\mu)}}{\frac{d^k}{dp^k} \ln p}$$
exists in $[0,\infty]$ for any $f$ such that $f\in L^p(\mu)$ for all $1\leq p<\infty.$
(Although I only really need it for $k=2.$) Note that these limits are related by L'Hospital's rule.
 A: The limit doesn't always exist. 
For convenience I will use the Lebesgue measure; similar construction can also be done for most $\mu$. 
Let $c_n$ be a sequence of increasing positive integers. Consider the function 
$$ f(x) = \sum_{n = 0}^\infty c_n^2 \chi_{[n, n + (c_n!)^{-1}]}(x)  $$
By the disjoint support we have that 
$$ \|f\|_{L^p}^p = \sum_{n = 0}^\infty \frac{(c_n)^{2p}}{c_n !} $$
Let us choose $c_n$ so that 
$$ c_{n+1} \geq 10 (c_n)^{2n} $$
Lim-sup
If $p = c_N$ for some $N$, we have 
$$ \|f\|_{L^p}^p \geq \frac{(c_N)^{2c_N}}{c_N!} \geq p^p $$
This implies 
$$ \limsup_{p \to \infty} \frac{\ln \|f\|_p}{\ln p} \geq 1$$
Lim-inf
Let $p = (c_N)^{2N}$ for some $n$. 
We have
$$ \sum_{n = 0}^N \frac{(c_n)^{2p}}{c_n!} \leq (c_N)^{2p} \cdot e = e\cdot p^{p/N}$$
On the other hand
$$ \sum_{n = N+1}^\infty \frac{(c_n)^{2p}}{c_n!} \leq \sum_{n = N+1}^\infty \frac{(c_n)^{2p}}{c_n (c_n - 1) \cdots (c_n - 2p + 1) \cdot (c_n - 2p)!} $$
Noting that $2p \leq \frac15 c_{N+1}$ we have 
$$ \leq \sum_{n = N+1}^\infty \left( \frac54\right)^{2p} \cdot \frac{1}{(c_n - 2p)!} \leq \left( \frac54 \right)^{2p} \cdot e $$ 
So 
$$ \|f\|_{L^p} \leq e^{1/p} \left( p^{1/N} + \frac{25}{16}\right) $$
For all sufficiently large $N$, using that $p^{1/N} = (c_N)^2 > 2$ we have
$$ \ln \|f\|_p \leq \frac{1}{p} + \ln 2 + \frac{1}{N} \ln p $$
and hence
$$ \liminf_{p\to \infty} \frac{\ln \|f\|_p}{\ln p} = 0 $$
Remarks
Note that the Lebesgue measure of the support of $f$ is has size less than $e$. So the same construction would work even if your measure is finite. What it really needs is enough sets of arbitrarily small measure. 
To illustrate this final fact, let us consider the special case of the counting measure on $\mathbb{N}$; in other words, let's look at the $\ell_p$ norms on real sequences. (The argument here is a modification of one given by Alexandre Eremenko.) 
By Minkowski's inequality we know that that $\ell_p$ norms are decreasing in $p$. So if we consider the mapping $\psi: [0,1]\ni x \mapsto \ln \|f\|_{\ell_{1/x}}$ for any fixed sequence $f$, we see that $\psi$ is


*

*An increasing function of $x$ (by Minkowski)

*A convex function of $x$ (by Riesz-convexity)


Now the function $y\mapsto e^{-y}$ is convex, and hence the function $\phi: y\mapsto \psi(e^{-y})$ is also convex, as it is the composition of two convex functions, the outer of which is increasing. 
Note that $\phi(\ln p) = \ln \|f\|_{\ell_p}$. 
We can conclude that the desired limit exists in this case since for any convex function $\phi$, the limit $\lim_{x\to\infty} \phi(x) / x$ exists in the sense given in the question (basically since $\phi'$ is increasing). 
A: $\newcommand{\ep}{\epsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\R}{\mathbb{R}}
\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\PP}{\operatorname{\mathsf P}}$ 
In the excellent answer, Willie Wong offered a construction of a measure with a lacunary support set, disproving the conjecture. Let me offer another construction, where the support of the measure is a continuum; however, the construction is still lacunary in the sense that it involves sequences diverging faster than any exponential sequence. 
The main idea is this. Let $p>0$ and $s>0$. Let $\|f\|_p:=\|f\|_{L^p(\mu)}$. Let $f(x)=x$ for all real $x$. Let us start with this trivial lemma: 

Lemma 1. If $\mu(dx)=\frac1s\,e^{-x/s}I\{x>0\}$, where $I$ denotes the indicator, then 
  \begin{equation*}
 \|f\|_p^p=\int_0^\infty x^p \,e^{-x/s}\frac{dx}s=s^p\Ga(p+1). 
\end{equation*}

Then, by Stirling's formula, $\ln\|f\|_p\sim \ln(ps)$ as $s,p\to\infty$. So, if we alternate $s$ between $p$ and $p^2$, $\ln\|f\|_p/\ln p$ will alternate between $2$ and $3$. Now we need to glue pieces of such two alternating sequences of measures, with $s$ alternating between $p$ and $p^2$, to get one measure. The guiding idea in doing this is that for large $p$ the mass $m_p(dx):=x^p \,e^{-x/s}I\{x>0\}\frac{dx}s$ is mostly concentrated near the point $ps$.  
To proceed, we shall need two more simple lemmas, which will be proved at the end of this answer. 

Lemma 2. For every $y\in[0,ps]$ there is some $c\in[1/2,1]$ such that 
  \begin{equation*}
 \int_y^\infty x^p \,e^{-x/s}\frac{dx}s=cs^p\Ga(p+1). 
\end{equation*}
Lemma 3. For every $y\ge2ps$, 
  \begin{equation*}
 \int_y^\infty x^p \,e^{-x/s}\frac{dx}s
 \le\exp\{p\ln(2ps)-y/(2s)\}. 
\end{equation*}

Let 
\begin{equation*}
 \mu(dx)=\sum_{j=1}^\infty\frac{dx}{s_j}\,e^{-x/s_j}I\{x_j<x<x_{j+1}\},
\end{equation*}
where
\begin{equation*}
s_j:=
\begin{cases}
p_j&\text{ if $j$ is odd}, \\
p_j^2&\text{ if $j$ is even}, 
\end{cases}
\qquad
 p_j:=2^{2^j},\qquad x_j:=p_j s_j,  
\end{equation*}
so that $p_{j+1}=p_j^2$. 
Then 
\begin{multline*}
 \|f\|_p^p=\int_\R x^p\mu(dx)=\sum_{j=1}^\infty I_j(p),\\ 
 I_j(p):=\int_{x_j}^{x_{j+1}} \exp\{g_{j,p}(x)\}\frac{dx}{s_j},\\
 g_{j,p}(x):=p\ln x-x/s_j. 
\end{multline*}
In view of Lemma 1 and Stirling's formula, for large odd $k$ one has
\begin{equation*}
 I_k(p_k) \le s_k^{p_k} p_k^{p_k}=p_k^{2p_k} \tag{1}  
\end{equation*}
and, for $j<k$, 
\begin{equation*}
 I_j(p_k)\le s_j^{p_k} p_k^{p_k} \le p_j^{2p_k} p_k^{p_k} 
 \le p_k^{2p_k},
\end{equation*}
whence 
\begin{equation*}
 \sum_{j<k}I_j(p_k) \le k p_k^{2p_k}=p_k^{(2+o(1))p_k}. \tag{2}
\end{equation*}
Next, for large odd $k$ and $j>k$, 
\begin{equation*}
 p_k\ln(2p_ks_j)-p_k\ln(p_ks_k)
 \le p_k\ln s_j\le p_k\ln(p_j^2)\le p_j^{1/2}\ln(p_j^2)\le p_j/4,
\end{equation*}
whence, by Lemma 3, 
\begin{multline*}
 I_j(p_k)\le\exp\{p_k\ln(2p_ks_j)-p_j/2\}\le\exp\{p_k\ln(p_ks_k)-p_j/4\} \\ 
 \le2^{-(j-k)}\exp\{p_k\ln(p_ks_k)\}, 
\end{multline*}
so that 
\begin{equation*}
 \sum_{j>k}I_j(p_k) \le p_k^{2p_k}. \tag{3}
\end{equation*}
Collecting (1), (2), (3), for large odd $k$ we have 
\begin{equation*}
 \ln\|f\|_{p_k}\lesssim 2\ln p_k. 
\end{equation*}
On the other hand, for large even $k$, by Lemma 2 and Stirling's formula,
\begin{equation*}
 \ln\|f\|_{p_k}\ge\frac1{p_k}\,\ln I_k(p_k)= \ln(s_k p_k^{1+o(1)})
 = \ln(p_k^{2} p_k^{1+o(1)})
 \sim3\ln p_k. 
\end{equation*}
So, $\ln\|f\|_p/\ln p$ does not converge as $p\to\infty$. 
In conclusion, let us prove the lemmas. 
Proof of Lemma 1. This is obvious. 
Proof of Lemma 2. Write 
\begin{equation}
 x^p \,e^{-x/s}=e^{g(x)},\quad g(x):=p\ln x-x/s. \tag{4}
\end{equation}
Then $g'(ps)=0$ and $g''(x)$ is increasing in $x>0$. So, $g(ps-u)<g(ps+u)$ for $u\in(0,ps)$. So, for every $y\in[0,ps]$, 
\begin{equation*}
 \int_{-\infty}^\infty x^p \,e^{-x/s}\frac{dx}s
 \ge\int_y^\infty x^p \,e^{-x/s}\frac{dx}s
 \ge\frac12\,\int_{-\infty}^\infty x^p \,e^{-x/s}\frac{dx}s. 
\end{equation*}
To complete the proof of Lemma 2, it remains to refer to Lemma 1. 
Proof of Lemma 3. For $g$ as in (4) and for $x>2ps$, we have $g'(x)=\frac px-\frac1s\le-\frac1{2s}$ and hence 
\begin{equation}
 g(x)\le g(2ps)-\frac1{2s}(x-2ps)=p\ln(2ps)-\frac x{2s}. 
\end{equation}
Now Lemma 3 easily follows. 
