Does a function $f$ with the following property exist?

Question

I saw this question yesterday, which asks for fixed $n$, does there exist a continuous function $f: [0, 1] \to \mathbb{R}$ such that $$f\left(\frac k n\right) = {n \choose k}^{-1}$$ The answer is yes, and there are many ways to construct an answer (one can use an interpolating polynomial or simply a set of straight lines, for example). I was wondering if something else could be said if $n$ isn't fixed, namely the following:

Does there exist a continuous function $f: (0, 1) \to \mathbb{R}$ such that for every rational $k / n$ in lowest terms, $$f\left(\frac k n\right) = {n \choose k}^{-1}$$

If so, can one be constructed easily? And how smooth can $f$ be while still satisfying the above property? (I suspect the answer is that there exists an analytic continuation.)

Answer 1

The answer is no. Consider $\alpha \in (0,1)$ any irrational number, and $\frac{p_n}{q_n}$ any sequence of irreducible fractions converging to it.

Then $f(p_n/q_n) \rightarrow f(\alpha)$.

But it’s easy to see that $q_n \rightarrow \infty$ and therefore $\binom{q_n}{p_n}^{-1} \leq q_n^{-1}$ goes to zero.

So $f$ is zero on all irrationals so is identically zero, a contrdiction.

Answer 2

Note that ${2n-1\choose n}\approx{2n\choose n}\approx \frac{4^n}{\sqrt{\pi n}}$ so that $$\lim_{n\to\infty}f(\tfrac n{2n-1})=0\ne f(\tfrac12) $$