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Fix a smooth function $f:\mathbb{R}\to\mathbb{R}$. Do there exist real numbers $a<b$, an infinite set $S\subset (a, b)$ and an analytic function $g$ defined on $(a-\epsilon, b+\epsilon)$ for some $\epsilon>0$ such that $f|_S=g|_S$?

If $g$ is only required to be defined on $(a, b)$ the question has a positive answer. In fact, we can take any $(a, b)$ we like and set $g=f(a)+\mathrm{sin}(\frac{1}{x-a})$.

We can not require $g$ to be defined on all of $\mathbb{R}$ since we can take $f(x)=\mathrm{exp}(-\frac{1}{|x|})$ for $x\neq 0$ and $f(0)=0$. Then by the pigeonhole principle $S$ must contain infinitely many positive numbers or infinitely many negative numbers; in either case $g$ does not extend to $\mathbb{R}$. Coincidentally, this shows that an arbitrary $(a, b)$ won't do in the original problem.

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    The following link might answer your question: https://web.archive.org/web/20161009194815/mathforum.org/kb/message.jspa?messageID=387148 – Frederik Ravn Klausen Jul 09 '20 at 16:55
  • Do you mean an open interval or an open set? In the former case, if 0 and 1 are limit points of $S$, that may be supposed to be true, we must take $U=(0,1)$. – Fedor Petrov Jul 09 '20 at 19:44
  • I think it is true, and you can choose a and b, and f may be just continuous. In case (a,b) is the whole real line, g can be taken entire. – Pietro Majer Jul 09 '20 at 20:36
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    The latter because any continuous function from R to R can be uniformly approximated by an entire real analytic function. So any $g$ with uniform distance less than 1/2 from $f+\sin x$ coincides with f in an infinite set. The case of a general interval (a,b) follows composing with an analytic diffeo from (a,b) to R. – Pietro Majer Jul 09 '20 at 20:49
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    Following Pietro Majer's comment: approximate your $f$ by a polynomial $P$ so that $|f(x) - P(x)| < 1$ for $x \in (a, b)$, and set $g(x) = P(x) + \sin(\frac{1}{x-a})$. A perhaps more interesting question would ask for $g$ analytic in a larger interval $(a-\epsilon, b+\epsilon)$. – Mateusz Kwaśnicki Jul 09 '20 at 21:00
  • For the mentioned density fact: https://mathoverflow.net/questions/26243/asymptotic-approximation-of-x-alpha-by-entire-functions/26290#26290 – Pietro Majer Jul 09 '20 at 21:00
  • (indeed the case $(a,b)=\mathbb{R}$ is somehow more interesting) – Pietro Majer Jul 09 '20 at 21:05
  • @PietroMajer can't you just take $S=\mathbb{Z}$ in that case? –  Jul 09 '20 at 21:06
  • This question might be useful. If $f(x)$ can be sampled on some infinite $S \subset (a,b)$ such that the Nevanlinna-Pick condition (on a complex disc of appropriate diameter) is satisfied gives you a positive answer, by resorting to complex analytic interpolation. – Igor Khavkine Jul 09 '20 at 22:13
  • I'm not sure I understand your edit. If $f$ is defined as you describe, it has values in $[0,1]$, so $g(x)=\sin(x)$ coincides with $f$ on an infinite set. – Pierre PC Jul 09 '20 at 23:13
  • Also, if you just want some infinite set, $x\mapsto f(a)+sin(\frac1{x-a})$ works just as well, the polynomial $P$ is superfluous. – Pierre PC Jul 09 '20 at 23:15
  • @PierrePC you are correct about the irrelevance of $P$ –  Jul 10 '20 at 03:48
  • @PierrePC but there is no bounded infinite set on which they coincide. I think if two analytic functions coincide on a bounded infinite set they are equal –  Jul 10 '20 at 03:52

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The answer is no, not necessarily.

Let $f$ be any smooth function such that the Taylor series of $f$ about any point $p$ has zero radius of convergence; see this MO answer for an explicit example.

Suppose that $g$ is a smooth function such that for some sequence $(x_n)$ convergent to $p$ (and such that $x_n \ne p$) we have $f(x_n) = g(x_n)$. It is rather easy to see that $f^{(k)}(p) = g^{(k)}(p)$ for every $k \geqslant 0$, and hence the Taylor series of $f$ and $g$ about $p$ coincide. In particular, the Taylor series of $g$ about $p$ has zero radius of convergence, and consequently $g$ is not real-analytic at $p$.

Mateusz Kwaśnicki
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