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

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


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.

Tags:

Continuity

Real Analysis

Analytic Continuation

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