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I have a conjecture which I cannot prove or disprove.

Denote the $i$'th digit of $x$ in binary expansion by $d_i(x)$, where for $i=1$ the MSB is taken. Example: $d_3(110.1)=0$ and $d_4(110.1)=1$ (so it just ignores the dot).

Denote by $p_i$ the $i$'th prime number, i.e. $p_1=2, p_2=3, p_3=5$ and so on.

Let $$h_i^j(x)=\frac {d_i(x)} {p_i^{p_{2j}}}$$ if the $i$'th digit of $x$ is left to the decimal point, and $$h_i^j(x)=\frac {d_i(x)} {p_i^{p_{2j+1}}}$$ otherwise.

Define $f(x):\mathbb R^n\rightarrow\mathbb R$ as:

$$f(x)=\sum_{j=1}^n \sum_{i=1}^\infty h_i^{j+in}(x_j) $$

Conjecture: $f$ is injective and finitely integrable over any finite measure balls on $\mathbb R^n$.

Caution: some explanations might work well for rational numbers only.

Please, if you think if it's incorrect, try to see if you can slightly modify $f$ to make the conjecture correct.


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Shame there is a five tags limit. You also left out [category-theory], [fourier-analysis] and so many others... – Asaf Karagila Jan 4 '13 at 10:33
This was crossposted to MO. In the future, please wait some time before posting your question in multiple fora, and when you do, provide links to the other posts - as you can imagine, it would be frustrating for someone to put time into answering your question here, only to see or hear from you that you'd already gotten the solution elsewhere. – Zev Chonoles Jan 4 '13 at 19:49
Of course, but once it's answered in either locations, I'll immediately post the answer on the second one. I have no intention to abuse (and will not, of course). – Troy McClure Jan 4 '13 at 20:10
up vote 0 down vote accepted

I managed to find an injection as requested, pretty much like the one proposed before. I'll post a link to the full paper when ready

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