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A function $f(x)$ is defined $\forall x \in \mathbb{Q}$ and $ x \in (0,1)$ as $f(x) = q $ where $x = (p/q)$ in simplest form. Find $\sup_{\tau} \quad \min_{a} \quad f(a) + f(\tau-a)$ where $ a \in \mathbb{Q}$ and $ a \in (0,1)$ and $ \tau \in (\alpha,\beta)$ where $ 0< \alpha < \beta <2 $ and $ \alpha,\beta \in \mathbb{Q} $. $\sup$ and $\min$ are taken over only those $x$ where $f(x)$ is defined. Is the supremum infinity ?


The supremum may or may not depend on $\alpha$ and $\beta$...this has not been ruled out.


I will try something !

Consider $\tau = l/m$ where $l,m \in \mathbb{N}$ in simplest form and $\tau \in \mathbb{Q}\cap(0,2)$.

what is minimum possible value of $q$ where $\tau-a = p/q $ written in simplest form such that $a \in \mathbb{Q} \cap (0,1)$ and $(\tau-a) \in (0,1)$.

As Yuval Filmus suggested in his answer...

one way to choose $a$ is that it has same denominator (in simplest form) as that of $\tau$ and numerator $n$ such that $l-n$ is relatively prime to $m$, (If such a choice is possible under given conditions), then $f(a)+f(\tau-a)$ becomes $2m$.

The question is whether such a choice leading to $f(a)+f(\tau-a) = 2m$ is the minimum possible value ? If not I request you to give a counter example.

If it is, then the supremum (over $\tau$ varying in any arbitrary open interval contained in $(0,2)$) is $\infty$ as we can increase $m$ arbitrarily by varying $\tau$ in any arbitrary open interval $(\alpha,\beta)$.


small change in Question...instead of $0<\alpha<\beta<2$ we have $-1<\alpha<\beta<1$

this change is not made with any intention of bringing out a certain solution.Also its influence on the solution is hoped to be very small.Inconvenience is regretted.

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For questions like this, it is nice to give some indication of what you have tried so far and where you got stuck. – Rahul Nov 28 '10 at 5:17
Even though i have formulated the question...every time i attempt to solve I get lost and tend to move away from the actual question. I find it hard to get hold of anything in it ! – Rajesh Dachiraju Nov 28 '10 at 11:29
I badly need some help atleast to get started somewhere – Rajesh Dachiraju Nov 28 '10 at 11:30
up vote 2 down vote accepted

EDIT: This is a consolidated solution which previously appeared in parts.

Lemma 1. For each natural $M$ and real $\alpha < \beta$ there's a rational number in $(\alpha,\beta)$ whose reduced form has a denominator larger than $M$.

Proof. Choose some power of two $2^n > \max\left(M, \frac{2}{\beta-\alpha}\right)$. Since $2^{-n} < \frac{\beta-\alpha}{2}$, there are at least two fractions of the form $x/2^n$ in $(\alpha,\beta)$. One of them has odd $x$ and so is in reduced form.

Lemma 2. If $p_1/q_1 + p_2/q_2 = P/Q$, all rationals in reduced form, then $(q_1+q_2)^2 \geq 4Q$.

Proof. Taking a common denominator, $\frac{p_1 q_2 + p_2 q_1}{q_1q_2} = \frac{P}{Q}$ and so $Q|q_1q_2$ (since $P/Q$ is reduced). In particular $Q \leq q_1q_2 \leq \left(\frac{q_1+q_2}{2}\right)^2$, where the last inequality is the arithmetic-geometric mean inequality.

Theorem. For every real $M$ and reals $\alpha < \beta$ there is a rational $r \in (\alpha,\beta)$ such that $\min_{a \in \mathbb{Q}} f(a) + f(r-a) > M$.

Proof. Using lemma 1, take a rational $r \in (\alpha,\beta)$ with denominator $Q > M^2/4$. Using lemma 2, we conclude that for every rational $a$, $f(a) + f(r-a) \geq 2\sqrt{Q} = M$.

Corollary. $$\sup_r \min_a f(a) + f(r-a) = \infty.$$

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@Yuval Filmus: ....what is B ? a natural number ? why is q1+q2 being considered here ? – Rajesh Dachiraju Nov 28 '10 at 5:29
@Yuval Filmus: $P = 1$ and $Q = q_1q_2$ ?? please confirm – Rajesh Dachiraju Nov 28 '10 at 5:48
@Yuval Filmus: using the property that when $p,q$ are relatively prime then there exist $x,y \in \mathbb{N}$ such that $px+qy = 1$. – Rajesh Dachiraju Nov 28 '10 at 5:52
@Yuval Filmus: I am not able to make out how it leads me to answer the question. – Rajesh Dachiraju Nov 28 '10 at 5:57
B is a real number, it corresponds to the supremum. – Yuval Filmus Nov 28 '10 at 6:38

For a given $\tau=\frac{1}{q}, q$ prime it seems that $a$ should be (perhaps among other choices) $\frac{\tau}{2}$ with $\sup_{\tau} \quad \min_{a} \quad f(a) + f(\tau-a)=2q$. Can you prove this? If so, then the sup over $\tau$ is infinity as you can find $\tau$ with arbitrarily large denominator.

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the supremum is over $\tau$ varying in any arbitrary open interval contained in (-1,1). It should be proven for every arbitrary open interval contained in (-1,1). Hence the assumption that the numerator of $\tau$ (in simplest form) being equal to $1$ may not cover all possibilities. – Rajesh Dachiraju Nov 28 '10 at 17:50
You are right. I think the same argument can be made for any $\tau=\frac{p}{q}$ with $q$ prime. The frightening thing is for $\frac{1}{n(n+1)}$, where $\frac{1}{n}$ and $\frac{1}{n+1}$ work for a sum of 2n+1. Even that goes to infinity with $\tau$, but only as the square root. – Ross Millikan Nov 28 '10 at 22:00

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