Prove that ##\lim\limits_{x \to \infty} f(x) = 0##

[Meden Agan]

Homework Statement

For $x > 0$, let $f(x)$ be the minimum value of $\left|x-\sqrt{m^2+2 \, n^2}\right|$ for all integers $m, n$. Then, prove that $$\lim_{x \to \infty} f(x)=0.$$

Relevant Equations

Calculus, Number Theory.

Let $f(x)= \min\limits_{m, n \in \mathbb Z} \left|x- \sqrt{m^2+2 \, n^2}\right|$ be the minimum distance between a positive real $x$ and a number of the form $\sqrt{m^2 + 2 n^2}$ with $m, n$ integers.

Let us consider a radius $R$ and let us consider the set $S_R$ of integer points satisfying $m^2 + 2 n^2 \le R^2$.
The ellipse enclosing them has area $A(R) = \dfrac{\pi R^2}{\sqrt 2}$, and number geometry ensures its number of integer points $N(R)$ grows as $\dfrac{\pi}{\sqrt 2} R^2$ plus an error - proportional to $R$; i.e. there exist positive constants $c_1$ and $c_2$ such that $c_1 \, R^2 \le N(R) \le c_2 \, R^2$ for $R$ large enough.

Each pair $(m, n) \in S_R$ determines the distance $r = \sqrt{m^2+2 \, n^2}$ between $0$ and $R$. Two pairs can produce the same $r$, but there are few coincidences with respect to $N(R)$. Thus the number $M(R)$ of different distances satisfies $M(R) \ge c_3 \, R^2$ for a constant $c_3 > 0$. By ordering these distances $0 = r_0 < r_1 < \ldots < r_{M(R)-1} \le R$, the mean space between two consecutive distances is $\dfrac{R}{M(R)} \le \dfrac{1}{c_3 \, R}$.

Now, if we choose a large $x$ and set $R \approx x$, one of the $r_k$s falls within at most $\dfrac{1}{c_3 \, x}$ from $x$. Then, $f(x) \le K/ x$ with $K = 1/c_3$. Since $K/x$ approaches zero when $x \to \infty$, it immediately follows that $$\lim_{x \to \infty} f(x) = 0.$$
The values $\sqrt{m^2+2 \, n^2}$ become so dense on the positive half-line that the minimum deviation from any real number cancels as $x$ increases.

Is that correct?

 

[nuuskur]

A point to consider is why the function is well defined, i.e, why does the minimum exist for $x>0$. (If it doesn't then change minimum to infimum)

Assuming the function is well defined and the limit exists, then the limit must be zero, because we'll just pick a sequence $x_n = \sqrt{1+2n^2}\to\infty$, for example. The minimum is constantly zero along this sequence, so zero is the only limit candidate.
(Side note: a valid proof would also be to show that the limit is zero along any sequence $x_n\to\infty$)

It is a correct approach to show $xf(x)$ is bounded as $x\to\infty$, but I just don't understand what precedes it :/ A more low level explanation for the existence of the constants would be helpful.

 

[Meden Agan]

It is a correct approach to show $xf(x)$ is bounded as $x\to\infty$, but I just don't understand what precedes it :/ A more low level explanation for the existence of the constants would be helpful.

Mhm, could you outline what's unclear step by step? Hope I can provide a detailed clarification.

 

[nuuskur]

its number of integer points $N(R)$ grows as $\dfrac{\pi}{\sqrt 2} R^2$ plus an error - proportional to $R$; i.e. there exist positive constants $c_1$ and $c_2$ such that $c_1 \, R^2 \le N(R) \le c_2 \, R^2$ for $R$ large enough.

Specifically this part.

What I understand is we consider the set of integer points bounded by the ellipse
$\frac{x^2}{R^2} + \frac{y^2}{\left(\frac{R}{\sqrt{2}} \right)^2} = 1$

Are you saying that $N(R)$ is equivalent to $cR^2 + \mbox{Error}(R)$ as $R\to\infty$? Why is that true and why does this imply existence of said constant $c_1,c_2$?