Let X be a random variable with $\Vert X\Vert_p = E[\vert X\vert^p]^{1/p}<\infty$ for all $p\geq 1$.
Assume for some fixed $r$, and for all $q,s$ satisfying $1\leq q < r < s$,
$$\lim_{p\to\infty} \frac{\Vert X\Vert_p}{p^{1/q}}=0$$
and
$$\limsup_{p\to\infty} \frac{\Vert X\Vert_p}{p^{1/s}}=\infty$$
Does this imply that there is some $K>0$ such that
$$\limsup_{p\to\infty} \frac{\Vert X\Vert_p}{p^{1/r}}=K$$
Best Answer
The claim is false. Here is an explicit counterexample.
Let $X$ be a non-negative random variable with density given by
$$p(x) = \sum_{k=1}^\infty \frac{1}{2^k}C_{3/2+1/2k} \exp\left ( -x^{3/2+1/2k} \right )$$
where
$$C_\alpha = \left ( \int_0^\infty \exp(-x^{\alpha})dx \right )^{-1}.$$
This ensures that the expression above is a genuine density since it is non-negative and integrates to 1.
I will show below that
$$\lim_{p\rightarrow \infty} \frac{\|X\|_p}{p^{1/q}} = 0$$
for all $1 \leq q < 3/2$ and $$\limsup_{p\rightarrow \infty} \frac{\|X\|_p}{p^{1/s}} = +\infty$$ for all $s > 3/2$. This means the value of $r$ must be $r = 3/2$ if it were to exist.
Finally I will show that $$\limsup_{p\rightarrow \infty} \frac{\|X\|_p}{p^{1/(3/2)}} = 0$$ which implies that the $X$ chosen fails to have the property you are looking for.
Intuition
The main trick in this scheme is to observe if $Z_\alpha$ is a random variable pdf $C_\alpha e^{-z^\alpha}$ then $$\|Z_\alpha\|_p \approx p^{1/\alpha}$$ or by raising things to the $p$'th power $$\mathbb{E}[Z_\alpha^p] \approx p^{p/\alpha}$$ From this it follows that $$\mathbb{E}[X^p] = \sum_{k=1}^\infty \frac{1}{2^k} \mathbb{E}[Z_{3/2+1/2k}^p] \approx \sum_{k=1}^\infty \frac{1}{2^k} p^{p/(3/2+1/2k)}.$$ Taking $p$'th roots and dividing by $p^{1/q}$ gives $$\frac{\|X\|_p}{p^{1/q}} \approx \left ( \sum_{k=1}^\infty \frac{1}{2^k} p^{p/(3/2+1/2k)} \right )^{1/p}\frac{1}{p^{1/q}} = \left ( \sum_{k=1}^\infty \frac{1}{2^k} p^{p/(3/2+1/2k) - p/q} \right )^{1/p}$$ Taking the limit as $p \rightarrow \infty$ for different values of $q$ will show the result.
Making this all rigorous is hairy
Step 0: Estimating $C_\alpha$ for $\alpha \in [1,2]$
This step is relatively easy but it's important to check that these constants are not going to cause problems. For $\alpha \in [1,2]$ we have
\begin{align} C_\alpha^{-1} &= \int_0^\infty \exp^{-x^\alpha} dx \\ &\leq 1 + \int_1^\infty \exp^{-x^\alpha} dx \\ &\leq 1 + \int_1^\infty \exp^{-x} dx < 1.5 \end{align} where the first inequality is because $\exp(-x^{\alpha}) \leq \exp(0) = 1$ and the second step is because when $x \geq 1$ it holds that $-x^{\alpha} < -x$ because $\alpha \geq 1$. The last inequality can be numerically verified.
For the reverse inequality \begin{align} C_\alpha^{-1} &= \int_0^\infty \exp^{-x^\alpha} dx \\ &\geq \int_1^\infty \exp^{-x^\alpha} dx \\ &\geq \int_1^\infty \exp^{-x^2} dx > 1/10 \end{align} which uses similar ideas and a numerical bound at the end. Ultimately this shows that $$ 2/3 < C_\alpha < 10$$ This is good enough to ensure that the $C_\alpha$ values really do not matter in this probably because they are multiplying by a positive constant that is bounded away from zero and bounded above.
Step 1: Bounds on $\mathbb{E}[Z_\alpha^p]$
Below it is established that for $p \geq 4$ and $\alpha \in (3/2,2]$ that there exist absolute constants $0 < a_1,a_2,c_1,c_2$ with $a_1 < 1 < a_2$ such that
$$c_1(a_1p)^{p/\alpha} \leq \mathbb{E}[Z_\alpha^p] \leq c_2(a_2p)^{p/\alpha}.$$ The arguments below are quite tedious so please skip feel free to skip them.
We can explicitly calculate \begin{align} \mathbb{E}[Z_\alpha^p] &= \int_0^\infty z^p C_\alpha e^{-z^\alpha} dz \\ &= C_\alpha\int_0^\infty (u^{1/\alpha})^p e^{-(u^{1/\alpha})^\alpha} \frac{1}{\alpha} u^{1/\alpha - 1} du \\ &= C_\alpha\frac{1}{\alpha}\int_0^\infty u^{p/\alpha + 1/\alpha - 1} e^{-u} du \\ &= \frac{C_\alpha}{\alpha} \Gamma\left ( \frac{p+1}{\alpha} \right ) \end{align} where we have used the change of variable $u = z^{1/\alpha}$ and the definition of the $\Gamma$ function.
Note here that the leading term satisfies for all $p \geq 1$ and $\alpha \in [3/2,2]$ the bounds $$ \frac{1}{3} < \frac{C_\alpha}{\alpha} < 10$$ which follows from Step 0 and the fact that $\alpha \in (3/2,2]$. This shows that this term is also innocuous.
The real challenge is to estimate $\Gamma\left ( \frac{p+1}{\alpha} \right )$. To start we have \begin{align} \Gamma\left ( \frac{p+1}{\alpha} \right ) &= \left ( \Gamma\left ( \frac{p+1}{\alpha} \right )^{\alpha/p} \right )^{p/\alpha} \end{align} The game is now to control the inner term. For $p \geq 4 \geq 2\alpha$ we have \begin{align} \Gamma\left ( \frac{p+1}{\alpha} \right )^{\alpha/p} > \Gamma\left ( \frac{p}{\alpha} \right )^{\alpha/p} \geq \frac{p}{4\alpha} \geq \frac{p}{8} \end{align} which uses the fact (which I won't prove here) that for $t \geq 2$ $$\Gamma(t)^{1/t} \geq \frac{t}{4}.$$ We also have \begin{align} \Gamma\left ( \frac{p+1}{\alpha} \right )^{\alpha/p} &< \Gamma\left ( \frac{p}{\alpha} + 1\right )^{\alpha/p} \\ &= \Gamma\left ( \frac{p}{\alpha} \right )^{\alpha/p} \left ( \frac{p}{\alpha} \right )^{\alpha/p} \\ &\leq \left ( \frac{p}{\alpha} \right ) \left ( \frac{p}{\alpha} \right )^{\alpha/p} \\ &\leq p ((p)^{1/p})^{\alpha} \alpha^{\alpha / p} \\ &\leq p e^\alpha \alpha^{\alpha/p} \\ &\leq p 4e^2 \end{align} which uses the property of the Gamma function $\Gamma(t+1) = t\Gamma(t)$ and also the inequality valid for $t \geq 2$ (which I also won't prove) $$\Gamma(t)^{1/t} \leq t$$. Combining the arguments above we have shown for $\alpha \in (3/2,2]$ and $p \geq 4$ that $$\frac{1}{3}\left ( \frac{p}{4}\right )^{p/\alpha}\geq \mathbb{E}[Z_\alpha^p] \leq 10\cdot(4e^2p)^{p/\alpha}$$ which completes the step with the specified constants explicitly.
Step 2. Summing things up
This step is a bit easier. We have by the definition of $X$ that \begin{align} \mathbb{E}[X^p] &= \int x^p \left ( \sum_{k=1}^\infty \frac{1}{2^k}C_{3/2+1/2k} \exp\left ( -x^{3/2+1/2k} \right ) \right ) dx \\ &= \sum_{k=1}^\infty \frac{1}{2^k} \int x^pC_{3/2+1/2k} \exp\left ( -x^{3/2+1/2k} \right ) dx \\ &= \sum_{k=1}^\infty \frac{1}{2^k} \mathbb{E}[Z_{3/2+1/2k}^p]. \end{align} Using the upper and lower bounds from the previous step shows for $p \geq 4$ \begin{align} \sum_{k=1}^\infty \frac{1}{2^k} c_1(a_2p)^{p/(3/2+1/2k)} \leq \mathbb{E}[X^p] \leq \sum_{k=1}^\infty \frac{1}{2^k} c_2(a_2p)^{p/(3/2+1/2k)} \end{align}
Step 3: Moments and Inequalities
Using the last step we have \begin{align} \|X\|_p &\leq \left (\sum_{k=1}^\infty \frac{1}{2^k} c_2(a_2p)^{p/(3/2+1/2k)} \right )^{1/p} \\ &\leq \left (\sum_{k=1}^\infty \frac{1}{2^k} c_2(p)^{p/(3/2+1/2k)a_2^p} \right )^{1/p} \\ &= a_2c_2 \left (\sum_{k=1}^\infty \frac{1}{2^k} p^{p/(3/2+1/2k)} \right )^{1/p} \end{align} where we have used that $a_2 > 1$.
Similarly we have \begin{align} \|X\|_p &\geq \left (\sum_{k=1}^\infty \frac{1}{2^k} c_1(a_1p)^{p/(3/2+1/2k)} \right )^{1/p} \\ &\geq \left (\sum_{k=1}^\infty \frac{1}{2^k} c_2(p)^{p/(3/2+1/2k)a_1^{2p}} \right )^{1/p} \\ &= a_1^2c_2 \left (\sum_{k=1}^\infty \frac{1}{2^k} p^{p/(3/2+1/2k)} \right )^{1/p} \end{align} where we have used $c_2 < 1$.
Step 4: Completing the argument
Since $a_1,a_2,c_1,c_2$ are absolute constants they won't effect the convergence to zero or infinity. I'm going to ignore them here and write $\propto$ to mean equality up to these absolute constants.
Now that we have everything in place, let's start concluding the argument. Let $1 \leq q < 3/2$. Let $\epsilon = 3/2 - q > 0$. We have by the last step \begin{align} \frac{\|X\|_p}{p^{1/q}} &\propto \frac{1}{p^{1/q}} \left (\sum_{k=1}^\infty \frac{1}{2^k} p^{p/(3/2+1/2k)} \right )^{1/p} \\ &= \left (\sum_{k=1}^\infty \frac{1}{2^k} p^{p/(3/2+1/2k) - p/q} \right )^{1/p} \\ &\leq \left (\sum_{k=1}^\infty \frac{1}{2^k} p^{p/(3/2) - p/q} \right )^{1/p} \\ &= p^{(q - 3/2)/(3q/2)} = p^{-\epsilon/(3q/2)} \end{align} and taking the limit as $p \rightarrow \infty$ shows that $$\lim_{p\rightarrow \infty}\frac{\|X\|_p}{p^{1/q}} = 0.$$
For the upper bound let $s > 3/2$ and let $k_*$ be large enough that $\frac{3}{2} + \frac{1}{2k} < s$ then again starting with the previous step we have \begin{align} \frac{\|X\|_p}{p^{1/r}} &\propto \frac{1}{p^{1/s}} \left (\sum_{k=1}^\infty \frac{1}{2^k} p^{p/(3/2+1/2k)} \right )^{1/p} \\ &\geq \frac{1}{p^{1/s}} \left ( \frac{1}{2^{k_*}} p^{p/(3/2+1/2k_*)} \right )^{1/p} \\ &= \frac{1}{2^{k_*/p}}p^{1/(3/2+1/2k_*) - 1/s} \end{align} Since $(3/2+1/2k_*) < s$ the exponent in the last expression is strictly positive. Also, $2^{-k_*/p} \rightarrow 2^0 = 1$ so there is no issue there. This shows that
$$\limsup \frac{\|X\|_p}{p^{1/s}} = +\infty.$$
Finally we conclude by checking the case where $r=3/2$. In this case we have by the previous step and Jensen's inequality (and the concavity of $x^{1/p}$ for $p \geq 1$) that
\begin{align} 0\leq \frac{\|X\|_p}{p^{1/r}} &\propto \frac{1}{p^{1/r}} \left (\sum_{k=1}^\infty \frac{1}{2^k} p^{p/(3/2+1/2k)} \right )^{1/p} \\ &\leq \frac{1}{p^{1/r}} \sum_{k=1}^\infty \frac{1}{2^k} p^{1/(3/2+1/2k)} \\ &= \sum_{k=1}^\infty \frac{1}{2^k} p^{1/(3/2+1/2k) - 1/r} \end{align} Taking the limit as $p \rightarrow \infty$ gives \begin{align} \lim_{p\rightarrow \infty} \sum_{k=1}^\infty \frac{1}{2^k} p^{1/(3/2+1/2k) - 1/r} &= \sum_{k=1}^\infty \lim_{p\rightarrow \infty} \frac{1}{2^k} p^{1/(3/2+1/2k) - 1/r} \\ &= \sum_{k=1}^\infty 0 = 0 \end{align} The fact that the inner limit converges to zero is because for any $k \geq 1$ it holds that $1/(3/2+1/2k) - 1/(3/2) < 0$. The exchange of the limit and the sum is justified by Tannery's theorem. This establishes that the sequence has a limit and that the limit must be 0 (which implies that the $\limsup$ must also be zero). This completes the counterexample
Debrief
The main things that are important to know are that
a density like $e^{-z^{\alpha}}$ implies moments like $p^{1/\alpha}$.
Having a density which decays like $e^{-z^{r}}$ but not quite that speed is what causes an issue.
This probably works with the density $$\exp(-x^{3/2 + \min(1/x, 1/2)})$$ because this tail gets close to $e^{-x^{3/2}}$ but never quite gets there. The calculations are probably terrible though.
A note on where this idea came from
The main idea is that as $x$ ranges from 0 to $\infty$ each $k$ term in the summation $$\sum_{k=1}^\infty \frac{1}{2^k} C_{3/2+1/2k}\exp(-x^{3/2 + 1/2k})$$ will at some point be essentially the largest term in the summation. It's not something that has a nice formula, but you can plot the terms (best if done in log-scale, and ignoring the $C$ term) and see that as $x$ grows each $k$ term is the maximum in the sum for at least a little while.
What that means is, at least for a small interval $X$ essentially "looks like" something with density $e^{-x^{3/2 + 1/2k}}$ and the moments of this distribution act like $p^{1/(3/2 + 1/2k)}$.
From here I think it's pretty straightforward to see that the behavior above and below $r=3/2$ is what is expected. The only subtle point is what happens at $r=3/2$. This is where the fact that each term is strictly more concentrated than $e^{-x^{3/2}}$ is useful since when we move the limit into the sum each term will tend to zero.
The other fact that is important, but not well known, is that $e^{-x^\alpha}$ has moments like $p^{1/\alpha}$. This has much more well known special cases when $\alpha = 1$ or 2. A good reference for this is Chapter 2 of this book, especially the subsection on "Orlicz norms."
I will say that the choice of $r=3/2$ is arbitrary. The example can be adjusted to use any $r \geq 1$.