Strange Functional Equation

A peculiar functional equations question I found:

Find all functions f:Q[1,1]f:\mathbb{Q} \rightarrow [-1,1] such that if xy=1xy=1 or x+y{0,1}x+y\in\{0,1\} for distinct rational xx and yy, then

f(x)f(y)=1f(x) f(y) = -1

Note by Sharky Kesa
3 years, 5 months ago

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Call the required functions as fii=1,2f_i \quad i=1,2.

Then,

fi(x)=fi({x})x∉{12}Zfi(x)=fi(1)xZ+fi(x)=fi(0)xZZ+f1(x)=1x[0,12){12}f1(x)=1x[12,1]f2(x)=1x[0,12){12}f2(x)=1x[12,1]\begin{aligned} f_i(x)&=f_i(\{x\}) \quad \forall x \not\in \left\{-\frac{1}{2}\right\} \cup \mathbb{Z}\\ f_i(x)&=f_i(1) \quad \forall x \in \mathbb{Z}^+\\ f_i(x)&=f_i(0) \quad \forall x \in \mathbb{Z} - \mathbb{Z}^+\\ f_1(x)&= 1 \quad \forall x \in \left [0,\frac{1}{2} \right) \cup \left\{-\frac{1}{2}\right\} \\ f_1(x)&=-1 \quad \forall x \in \left[\frac{1}{2}, 1 \right]\\ f_2(x)&=-1 \quad \forall x \in \left[0,\frac{1}{2} \right) \cup \left\{-\frac{1}{2}\right\} \\ f_2(x)&=1 \quad \forall x \in \left[\frac{1}{2}, 1 \right] \end{aligned}

@Sharky Kesa

Deeparaj Bhat - 3 years, 5 months ago

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@Sharky Kesa What say?

Deeparaj Bhat - 3 years, 5 months ago

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Put x=y=1x=y=1 then, we get (f(1))2=1(f(1))^2=-1.

Hence, no such function ff exists.

@Sharky Kesa

Deeparaj Bhat - 3 years, 5 months ago

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Sorry, I had meant to write for distinct xx and yy.

Sharky Kesa - 3 years, 5 months ago

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Then, we get f(x)=1xQf(x)f(1x)=1xQ{1,1}f(x)f(x)=1xQ{0}f(x)f(1x)=1xQ{12}    f(1+x)=f(x)xQ{0,12}|f(x)|=1 \quad \forall x \in \mathbb{Q}\\f(x)f(\frac{1}{x})=-1 \quad \forall x \in \mathbb{Q}-\{1,-1\}\\f(x)f(-x)=-1 \quad \forall x \in \mathbb{Q}-\{0\}\\f(x)f(1-x)=-1 \quad \forall x \in \mathbb{Q}-\left\{\frac{1}{2}\right\}\\\implies f(1+x)=f(x) \quad \forall x \in \mathbb{Q}-\left\{0,\frac{-1}{2}\right\}

Deeparaj Bhat - 3 years, 5 months ago

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@Deeparaj Bhat This doesn't give us a rule for ff per se but it's a good starting.

Sharky Kesa - 3 years, 5 months ago

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@Sharky Kesa Working on figuring f(1/2)f(-1/2) and on [0,1]...

Deeparaj Bhat - 3 years, 5 months ago

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@Deeparaj Bhat My hint is to consider continued fractions for all rational values.

Sharky Kesa - 3 years, 5 months ago

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@Sharky Kesa Well, I don't know much about continued fractions...

Thanks for the hint anyways.

Deeparaj Bhat - 3 years, 5 months ago

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It should say x+y{0,1}x + y \in \{0,1\}.

Jon Haussmann - 3 years, 5 months ago

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