Correctly reading this $\lim \inf ()$ and $\lim \sup()$ expression?

Question

I've never thought that I would have difficulties to read such a simple formula, which goes as follows¹:

A well-known unsolved problem in number theory concerns the distribution of $(3/2)^n\pmod1$. The sequence is believed to be uniformly distributed, which is the case for almost all real numbers $\theta^n\pmod1$, but it is not even known to be dense in $[0,1]$. One of the few positive results known for (non-integer) rational $\theta=p/q$ is that of Vijayaraghavan ($1940$), who showed that the set $(p/q)^n\pmod1$ has infinitely many limit points. Vijayaraghavan later remarked that it was striking that one could not even decide whether or not $(3/2)^n\pmod1$ has infinitely limit points in $[0,1/2)$ or in $[1/2,1)$. Both these latter assertions would follow if one could show that $$\limsup_{n\to\infty}\left\{\left(\frac32\right)^n\right\}-\liminf_{n\to\infty}\left\{\left(\frac32\right)^n\right\}>\frac12.$$

I assume surely correctly, that the curly braces mean the fractional part. And $\lim_{n \to \infty} \sup()$ the highest occuring fractional value (or better its limit) and the other the lowest occuring fractional value (or better its limit) . But of course already for small $n$ the left expression approaches $1$ and the right expression approaches $0$, which makes its difference larger than $1/2$.

So obviously I must misread something elementary. Just trying to remove the tomato from my eyes...

¹ Flatto, Lagarias, 1995 "On the range of fractional parts { ξ(p/q) n }"

Answer 1

Given a sequence $a_n$, $\limsup_{n\to\infty}a_n$ is an eventual supremum, not a global supremum of $a_n$. What I mean by that is that it doesn't care about what the early terms do, only what happens as $n\to \infty$.

Slightly more rigorously, if you successively delete entries from the beginning of the sequence $a_n$ (i.e. start at $n = 2$ instead of $n = 1$, then start at $n = 3$ instead of $n = 2$, and so on), then the (global) $\sup a_n$ might stay the same, or it might decrease. Whatever it converges to (be it a real number or $-\infty$), that is $\limsup_{n\to \infty}a_n$

Example: $$ a_n = \frac 1n + (-1)^n\\ \begin{array}{|c|cccccc} \hline n & 1&2&3&4&5&\cdots\\ \hline a_n&0&\frac32&-\frac23&\frac54&-\frac45&\cdots\\\hline \end{array} $$ The largest term in this sequence is $\frac32$, for $n = 2$. So $\sup a_n = \frac32$. However, if we delete the first two terms, then the new largest term is $\frac 54$ for $n = 4$. And so on. The largest term of whatever is left keeps shrinking as we keep deleting terms, and it converges to $1$.

$\liminf_{n\to\infty}a_n$ works the same way, only with $\inf$. Using the example above again, we see that there is no smallest term. But there is a (global) $\inf a_n = -1$. This will stay exactly where it is as we delete terms, as it is the tail end of the sequence which gives it this $\inf$. So $\liminf_{n\to\infty}a_n = -1$.

And be careful labeling $\inf$ and $\sup$ (and their $\lim$ variations) as "min" and "max". It is not the same (although they fill much of the same role). The sequence $a_n$ doesn't have a min, as it never becomes $-1$, but comes ever closer to it, so the $\inf$ is $-1$.

Going back to your original sequence, the whole point of that paragraph is to point out that while it is widely suspected that $\{(3/2)^n\}$ approaches both $1$ and $0$ as $n$ grows, we simply do not know. Maybe all terms are larger than $\frac12$ from some point on. An in that case, the $\liminf$ would be $\frac12$, or even larger.

Answer 2

" But of course already for small n the left expression approaches 1 and the right expression aproaches 0, which makes its difference larger than 1/2."

But we care aobut BIG n!

EDIT AFTER COMMENT:

There is not always a relationship between $\lim \sup$/$\lim \inf$ and global maximums/minimums. You may want to think about it in the following way:

For very big $n$ (meaning, as big of $n$ as we arbitrarely decide to choose), our sequence/function will be oscilating in some range, with its lower boundary being $\lim \inf$ and its upper boundary being $\lim \sup$ Higher values than the $\lim \sup$ may be achieved for smaller $n$

As an example for that, let $f(x):=\frac{1}{x}$ for every $x>=1$ We have a global maximum point at $x=1$ with $\lim \sup$ and $\lim \inf$ (when $n \to \infty$) being both 0