Does a irrational number $x > 1$ exist such that $\{x^n \} \le \frac{1}{2}$ for all positive integers $n$ ?
$x=1+ \sqrt 2$ holds for $n$ odd, but not in even
Does a irrational number $x > 1$ exist such that $\{x^n \} \le \frac{1}{2}$ for all positive integers $n$ ?
$x=1+ \sqrt 2$ holds for $n$ odd, but not in even
The OP asks for an instance of what Dubickas [1] has called a ${\cal Z}$-number: A real number $x>1$ for which there exists a real $\xi\neq 0$ such that $\{\xi x^n\}<1/2$ for every integer $n$.
An example [2] of an irrational ${\cal Z}$-number is $x=\tfrac{1}{2}(7+\sqrt{41})$, when $\{\xi x^n\}$ with $\xi=\tfrac{1}{4}(2+3x)$, converges to $1/4$ for $n\rightarrow\infty$.
The OP asks for an irrational ${\cal Z}$-number with $\xi=1$. It is known [3] that there exists no algebraic non-integer $x>1$ such that $\{x^n\}$ converges to 0. A stronger result (theorem 4 of [2]) indicates that a non-integer ${\cal Z}$-number with $\xi=1$ cannot be a Pisot number. I am not aware of any results for non-algebraic numbers.
For $x \in \mathbb{R}_{\ge 1}$ let $\phi(x) = (\lfloor x^n \rfloor + \frac{1}{2})^{\frac{1}{n}}$, where $n$ is the smallest positive integer such that $(\lfloor x^n \rfloor + \frac{1}{2})^{\frac{1}{n}} < x$; if there is no such $n$, then let $\phi(x) = x$. Clearly $\phi(x) \le x$ for every $x$, and $\phi(x) = x$ if and only if $\{x^n\} \le \frac{1}{2}$ for every $n \ge 1$. So the set $S$ of fixed points of $\phi$ are exactly the special numbers we are looking for. Every positive integer is a fixed point of $\phi$. If $m \in \mathbb{Z}_{\ge 1}$ and $x \in (m,m+1)$, then $\phi(x) \in (m,m+1)$. Note that $\phi(x) \le \phi(y)$ for $1 \le x \le y$. Let $\phi^k$ denote the $k$th iteration of $\phi$ for $k \in \mathbb{Z}_{\ge 0}$, and let $\phi^{\infty}(x) = \lim_{k \to \infty} \phi^k(x)$. It is easy to see that $\phi^{\infty}(x)$ a fixed point of $\phi$.
We have $\phi((\frac{3}{2})^{\frac{1}{k}}) = (\frac{3}{2})^{\frac{1}{k+1}}$ for $k \ge 2$. So $\phi^{\infty}(x) = 1$ for every $x \in [1,2)$. However for larger $x$'s it seems we actually get non-integer fixed points too. E.g., for $x \in [\frac{5}{2},3)$ we seem to get $$\phi^{\infty}(x) \approx 2.488183157043128473812375937641611512972854767026801209322612522.$$ (Iterating $\phi$, one can calculate this numerically to hundreds of digits, the convergence rate seems to be fast (exponential in $n$). To be clear, I do not have a proof for $\phi^{\infty}(x) > 2$.) This number should be $\max(S \cap (2,3))$. Similarly one could study $\max(S \cap (m,m+1))$ for $m \in \mathbb{Z}_{\ge 2}$. I suspect that $S \setminus \{1\}$ is a perfect subset of $\mathbb{R}$ (i.e., it is closed and has no isolated points). If this is true, then $S$ has continuum many elements, so in particular there are irrational and even transcendental numbers in $S$.
In the book Selected Problems in Real analysis page 164 have the proof of following result: For $0\leq p < q\leq1$, there exists a number $x \in \mathbb{R}$ such that $\{x^n\} \in [p,q]$ for all $n \in \mathbb{Z_{>0}}$