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Willie Wong
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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.

Willie Wong
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