It's indeed unbounded for every irrational $x$.

Let me identify points of $\mathbb{R}/\mathbb{Z}$ with their representatives on $[0,1)$, and order it by the usual order $<$ of $\mathbb{R}$ applied to the representatives.

Replace $x$ by $y = x/2$, and the question becomes whether for all $m$ there exists $k$ such that the orbit of $0$ in the irrational rotation on $\mathbb{R}/\mathbb{Z}$ by $y$ is in $[0,1/2)$ at least $m$ more times than in $[1/2,1)$, in the first $k$ time steps $y,2y,3y,...,ky$; or that this happens with $[0,1/2)$ and $[1/2,1)$ interchanged.

Since the irrational rotation by $2y$ is topologically transitive, we can find odd $k$ with $ky > 0$ arbitrarily small. For odd $k$ the set $Y_k = \{y,2y,...,ky\}$ has to intersect either $[0,1/2)$ or $[1/2, 1)$ more times than the other. Let's suppose the first case happens for infinitely many $k$. (The other case is symmetric, and one happens by the pigeonhole principle.)

Let now $P_m$ be the statement that there are $ky > 0$ arbitrarily small such that $[0,1/2)$ contains $m$ more elements of $Y_k$ than $[1/2,1)$ does. We have that $P_1$ holds. Observe that if $P_m$ holds and $k$ is as in the definition, then $[0, 1/2)$ also contains $m$ more elements of $Y_{k,a} = \{a+y,a+2y,...,a+ky\}$ than $[1/2,1)$ whenever $a$ is small enough, because $Y_k$ is disjoint from $\{0,1/2\}$.

Let now $m \geq 1$ be any integer such that $P_m$ holds. Let $\epsilon > 0$ be arbitrary and pick $k$ such that $0 < ky < \epsilon/2$ and $[0,1/2)$ contains at least $m$ more elements of $Y_k$ than $[1/2,1)$ does. Let $a$ be as in the previous paragraph.

Using $P_m$ again, take $0 < k'y < \min(\epsilon/2, a)$ such that $[0, 1/2)$ contains $m$ more elements of $Y_{k'} = \{y,2y,...,k'y\}$ than $[1/2,1)$ does. Then $[0,1/2)$ contains $2m$ more elements of $Y_{k'+k}$ than $[1/2,1)$. We have $0 < (k+k')y < \epsilon$, so $P_{2m}$ holds.

Thus $P_m$ holds for all $m$, and the claim follows.

anyirrational $x$. Can someone set me straight? $\endgroup$1more comment