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.



Limit of the logarithm of the $L^p$ norm over the logarithm of $p$ as $p$ goes to infinity
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). 


