Ratio of $\ell^p$ seminorms as dimension $n$ increases: $\frac{\|x^{(n)}\|_1}{\|x^{(n)}\|_q} \to \infty$ if $0< q < 1$ and $\|x^{(n)}\|_q \to \infty$?

Question

Setup

Let $x = (x_1, x_2, \dots) \in \mathbb R^{\mathbb N}$ and set $x^{(n)} := (x_1, \dots, x_n)$ for each $n \in \mathbb N$.

For any $p>0$, define the (semi)norm $\|\cdot\|_p$ by $\|x^{(n)}\|_p := \left( \frac 1n \sum_{j=1}^n |x_i|^p \right)^{1/p}$ for all $x^{(n)} \in \mathbb R^n$.

Question

Let $0 < q < 1$ and suppose $\|x^{(n)}\|_q \to \infty$ as $n \to \infty$.

I'm hoping it's true that $$ \frac{\|x^{(n)}\|_1}{\|x^{(n)}\|_q} \to \infty $$ as $n \to \infty$. Is this necessarily the case?

Thoughts

Since $\|\cdot \|_q \leq \| \cdot \|_1$, we know that $\frac{\|x^{(n)}\|_1}{\|x^{(n)}\|_q} \geq 1$ for all $n$. I've tried some other strategies, but haven't managed to get any inequality that would force the ratio to above $1$.

Answer 1

A counterexample:

Pick any $q \in (0, 1)$, let $p:= \frac{1}{q}$ and set $x_j = \log(j)^p$ for all $j$, then using a lower bound from Stirling's formula we get

$$ \|x^{(n)}\|_q = \left( \frac 1n \sum_{j=1}^n \log(j) \right)^p = \left(\frac{\log(n!)}{n}\right)^p \geq \left(\frac{\log\left(n^{n+\frac 12}e^{-n}\right)}{n}\right)^p \geq (\log(n) - 1)^p $$

for all $n > 2$, so $\|x^{(n)}\|_q$ diverges to infinity.

On the other hand we have

$$ \|x^{(n)}\|_1 = \frac 1n \sum_{j=1}^n \log(j)^p \leq \log(n)^p, $$

so for their ratio we can conclude

$$ \frac{\|x^{(n)}\|_1}{\|x^{(n)}\|_q} \leq \frac{\log(n)^p}{(\log(n) - 1)^p} = \left(\frac{1}{1 - \frac{1}{\log(n)}}\right)^p $$

for all $n > 2$ and that last term converges to $1$, so their ratio is bounded.