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Suppose I have a function $f$, and positive integers $x$ and $y$ such that $x$ is a square, $x \ne y$ and $y \ne \sqrt{x}$.

Now, assume that:

$|\frac{f(x)}{y} - \frac{f(y)}{x}| > 2$

$|\frac{f(\sqrt{x})}{y} - \frac{f(y)}{\sqrt{x}}| > 1$

for all $x, y$ in the domain of $f$.

(Note that $|N|$ is the absolute value of $N$.)

My question is: What properties would this function $f$ necessarily possess? In particular, can $f$ be injective?

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    $\begingroup$ Are you assuming the inequalities for all (x,y)? Your question currently says just one particular pair, which surely can't be what you meant - because then you can't say anything about $f$ except at the three points $\sqrt{x}, x, y$. $\endgroup$
    – Zen Harper
    Commented Mar 2, 2011 at 5:58
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    $\begingroup$ Is $f$ defined just on the positive integers, or on all integers, or on the reals, or what? If the reals, do you insist on continuity? Are your inequalities to hold for one particular pair $x$, $y$, or for all such pairs? Without some clarification, I don't think this is a good question. $\endgroup$
    – Gerry Myerson
    Commented Mar 2, 2011 at 6:00
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    $\begingroup$ It won't look too good if $y=\sqrt(x)$. I think you need to be more clear about what conditions you want. Gerhard "Zero Is Not Always One" Paseman, 2011.03.01 $\endgroup$
    – Gerhard Paseman
    Commented Mar 2, 2011 at 6:02
  • $\begingroup$ @Zen, Gerry: The domain of $f$ is over all positive integers, and yes, I meant that my two inequalities held for all possible pairs $(x, y)$. $\endgroup$
    – Jose Arnaldo Bebita
    Commented Mar 2, 2011 at 6:04
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    $\begingroup$ What's the motivation? $\endgroup$
    – Daniel Litt
    Commented Mar 2, 2011 at 6:51

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I'm assuming the first inequality needs absolute values also; if not then exchanging the roles of $x$ and $y$ forces a number and its opposite to both be greater than two.

In this case, there is such an $f$ which is injective:

Let $f(1)=1$.

Assume $f$ has been defined and satisfies the two properties on $\{1,\ldots,m\}$. To define $f(m+1)$ injectively, still satisfying the two above properties, we need that $f(m+1)$ is not in the range of $f$ as defined so far, and also a finite set of inequalities. ($\lfloor\sqrt{m}\rfloor$ of them if $m+1$ is not a square, and an additional $m$ of them if $m+1$ is a square.) If $f(m+1)$ is defined sufficiently large, all these conditions will be met.

Proceed by induction; the resulting function on $\mathbb{N}$ will be as desired.

You ask "What properties should $f$ posess?" If you're looking to contruct an $f$ which grows as slowly as possible, an inspection of the above construction could give one slowish example.

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  • $\begingroup$ Hi Joe, thanks for providing this construction. I am actually looking at an $f$ which is a bit 'chaotic', but I suspect it's injective over a particular subset of the natural numbers. $\endgroup$
    – Jose Arnaldo Bebita
    Commented Mar 2, 2011 at 8:45
  • $\begingroup$ @Joe, will this same construction work if the inequality $|\frac{f(\sqrt{x})}{y} - \frac{f(y)}{\sqrt{x}}| > 1$ is replaced by $|\frac{f(\sqrt{x})}{y} - \frac{f(y)}{\sqrt{x}}| > \frac{1}{2}$? $\endgroup$
    – Jose Arnaldo Bebita
    Commented Mar 2, 2011 at 9:58
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    $\begingroup$ Do you want the first inequality to hold for all naturals x and y, or only when x is a square? Yes, the same construction will work. If we're defining $f(y)$, then for all sufficiently large choices of $f(y)$, $f(\sqrt{x})/y$ will stay small while $f(y)/\sqrt{x}$ becomes large. When this happens, the inequality will be greater than 1/2. The same happens if we're defining f at a square, for which we must consider it in the role of $x$ as well. $\endgroup$
    – Joe
    Commented Mar 3, 2011 at 5:02

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