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Recently, some colleagues and I had need to consider the following function defined on a part of the complex plane $$f:(e^{in},e^{im})\quad\mapsto\quad e^{inm},$$ where $n$ and $m$ are non-negative integers. Because these are integer radian angles, it follows that the numbers of the form $e^{in}$ are dense on the unit circle, and we only consider the function $f$ on input of the form $e^{in}$ for $n\in\mathbb{N}$.

Notice that the function $f$ is continuous in each coordinate separately, since if one fixes the first input value at $e^{in}$, then $f(e^{in},z)$ agrees with $z^n$ for $z$ in its domain, and similarly $f(z,e^{im})=z^m$, when $m$ is fixed, and in each case these are continuous unary functions.

But I believe that the function $f$ is likely discontinuous as a binary function on its domain.

Can you prove it?

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  • $\begingroup$ $f(1,1) = 1$, but shouldn't there be points arbitrarily close to $(1,1)$ that map to arbitrarily close to -1? $\endgroup$ – Geoffrey Irving Jun 19 '18 at 23:42
  • $\begingroup$ @GeoffreyIrving That is indeed the kind of thing I suspect is true. But what is your argument? $\endgroup$ – JDH Jun 19 '18 at 23:47
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    $\begingroup$ How do you get uniform continuity? The domain is not compact, since the function is only defined on a countable dense subset of the (square of the) unit circle. $\endgroup$ – JDH Jun 19 '18 at 23:56
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    $\begingroup$ I have a solution now, sent to me by email from Alex Wilkie. I'll have a chance to post it later. $\endgroup$ – JDH Jun 20 '18 at 18:19
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    $\begingroup$ @Robert This function appeared initially in the context of this problem: twitter.com/JDHamkins/status/1006651518534213633?s=19. It doesn't actually work, though, since it is discontinuous. Meanwhile, we have another solution to the problem. $\endgroup$ – JDH Jun 21 '18 at 1:50
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Alex Wilkie sent me the following argument by email, and I post it transcribed here with his permission.

Alex Wilkie's argument

Transcription.

Let $$2\leq N_0 < N_1 <\cdot$$ and $$1\leq k_0 < k_1 <\cdots$$ be strictly increasing sequences of integers, and $$t_0, t_1, \ldots$$ a sequence of real numbers with $$(\ast)\qquad 2\pi k_j-N_j = \frac{t_j}{N_{j-1}}$$ and $$(\ast\ast) \qquad \frac{\pi}{2} \leq |t_j| \leq \pi.$$ Such sequences are easy to construct using Dirichlet's theorem (and, of course, the fact that $\pi$ is irrational).

Then, by $(\ast)$, we have $$e^{iN_j}=e^{i(2\pi k_j-\frac{t_j}{N_{j-1}})}\to 1\quad\text{ as }\quad j\to\infty$$ and so also $$e^{iN_{j-1}}\to 1\quad\text{ as }\quad j\to \infty.$$ Hence, $f(e^{iN_j},e^{iN_{j-1}})\to f(1,1)=1$, if $f$ were continuous.

However, $$f(e^{iN_j},e^{iN_{j-1}})=e^{iN_jN_{j-1}}=e^{i(2\pi k_jN_{j-1}-t_j)}=e^{-it_j},$$ which converges to a point (if it converges at all) in the left half-plane. $\Box$

In fact, only a very weak form of Dirichlet's theorem is needed: just that $|N\pi - M|$ can be made arbitrarily small for integers $M$, $N$.

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