First, we consider, as in elementary statistics, a fixed size (say $n$, but can assume $n=2$ to start with, to eventually relax this) sample $\{x^{m}_1\dots x^{m}_n\}\subset \mathbb{R}.$ Belo, unless otherwise implied $m\to \infty.$ Then it's easy to see that their common variance $V^{(m)}$ satisfies $V^{(m)}\le {R^{(m)}}^2,R^{(m)} = max \{x^{m}_1\dots x^{m}_n\} - min \{x^{(m)}_1\dots x^{(m)}_n\}.$ So it's clear that as $R^{(m)}\to 0, V^{(m)}\to 0$ as well. Indeed it's also true that as $V^{(m)}\to 0, R^{(m)}\to 0,$ without having to use any inequality between $V^{(m)}$ and $R^{(m)}.$
Next, we work with an iid random sample as in probability theory, where by random sample, we mean an iid set of random variables $\{X^{(m)}_1\dots X^{(m)}_n\}: \Omega \to \mathbb{R}, X_i^{(m)} \sim X^{(m)}$ (has the same distribution as $X.$). Now here the variance $V^{(m)}$ of $X^{(m)}$ is a sequence of nonnegatve real numbers whereas the range $R^{(m)}$ is a random variable.
I was wondering whether there's a (probabilistic) inequality that connects $R^{(m)}$ with $V^{(m)}?$ And finally, can we say that $R^{(m)}\to 0$ in distribution/in probability/almost surely/in some $L^p$ norm implies $V^{(m)}:= Var[X^{(m)}]\to 0?$
How about the converse? Does $V^{(m)}\to 0, m\to \infty$ implies $R^{(m)}\to 0$ in distribution/in probability/almost surely/in some $L^p$ norm?
P.S. As hinted through the comments, we can specify that $n$ is fixed, and the thus we can take $n=2.$ But eventually it'd be nice to see the dependence on $n$ as well.
P.P.S. I asked the question on MO just now.