This is a refinement (perhaps a simpler version) of a question I asked here before and couldn't get an answer for.
Fix $\alpha \in (0,1]$ and a small constant $c>0$. For $x \in [0,1]$ and $N\in\mathbb{N}$, we denote by $S(x,N)$ the function that counts the number of pairs of integers $(p,q)$ such that $$|x-\frac{p}{q}|<\frac{c}{q^{\alpha+1}}\qquad(\star)$$
and $1\leq q \leq N$, $gcd(p,q)=1$.
It is known that when $\alpha=1$, all the solutions $(p,q)$ to $(\star)$ are in fact the convergents (i.e. truncation of the continued fraction expansion of $x$). The recursive formula for the denominators of those implies that they grow exponentially (asymptotically), and for all irrationals at least as fast as those for $x$ being the golden ratio (you can consider its fractional part in order to have it in the interval). This, in turn, implies that there exists some constant $C>0$ such that $$||S(x,N)||_{L^{\infty}([0,1])}\leq C \log N.$$

I'm interested in getting an upper bound on this $L^{\infty}$ norm for any $\alpha\in(0,1)$. To be more precise, I wish to show that this bound is sublinear in $N$.
W. Schmidt considered the more general case where the restriction $gcd(p,q)=1$ is dropped (I denote the corresponding counting function by $M(x,N)$). His asymptotic estimate, which holds for almost every $x\in[0,1]$ is $$M(x,N)\leq \sum_{q=1}^{N} \frac{c}{q^{\alpha}}+C(x) (\sum_{q=1}^{N} \frac{c}{q^{\alpha}})^{\frac{1}{2}}.$$ But as far as I can tell, the function $C(x)$ is unbounded, so there is no uniform bound.
One can also try and apply first and second moment estimates on the function $S(x,N)$ (using Chebyshev's inequality). This gives the desired sublinear estimate, but only up to some exceptional subset of $[0,1]$ which turns out to be too big for my purposes.
There seems to be a large gap between the case $\alpha=1$, where everything is known, and smaller values of $\alpha$, which I can't explain at the moment.

-- Addition --

This might be a trivial comment, but it should be emphasized that the trivial bound is $$||S(x,N)||_{L^{\infty}([0,1])}\leq C\cdot N.$$ So I'm really looking for any kind of nontrivial bound.

  • $\begingroup$ I think I have an heuristic argument that for $\tau =(\sqrt{5}-1)/2$ the number of solutions scales as $S(\tau,N) \sim N^{1-\alpha}$. My argument makes use of the fact that you can expand a natural number $q$ in a Fibonacci base ($ q = \sum c_k f_k$, where $f_k$ are Fibonacci numbers and $c_k \in \{0,1\}$ with $c_k c_{k+1}=0$), you then can get estimates for the difference of $\tau q$ to the next integer rather easily. Unfortunately I don’t have the time to write it up properly. And the argument seems to work only for quadratic irrationalities. $\alpha \to 1$ gives correctly $\sim \log N$. $\endgroup$ – Andreas Rüdinger Dec 21 '18 at 23:02

An elementary way of looking at the problem gives the uniform bound you want - the trick is to look at the actual fractions.

Any two non-equivalent fractions with denominator $\le N$ have a difference in absolute value at least $\frac{1}{N^2}$. So the number of primitive pairs $(p,q)$ where $N \le q \le 2N$ satisfying your inequality is:

$$\ll \frac{\frac{1}{N^{1+\alpha}}}{ \frac{1}{N^2}} = N^{1-\alpha}.$$

Putting these dyadic interval bounds together gives

$$S(x, N) \ll N^{1-\alpha}$$

uniformly on $x$ (for $\alpha = 1$ this argument also recovers your $\log N$ bound).


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