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Given two strictly increasing bounded sequences of reals $x_n$ and $y_n$. What is known about existence of real analytic function $f$ with property $f(x_{n_k})=y_{n_k}$ for some subsequence $x_{n_k}$ ?

Edit 1: Of course, $y_n$ can't increase too slow or too fast relative to $x_n$ as it was shown in counterexamples below. So, we can additionally assume the following ($x=\sup x_n,~y=\sup y_n$):

$0<\underline\lim\limits_{n\to\infty}\frac{|x_n-x|}{|y_n-y|},~ \overline\lim\limits_{n\to\infty}\frac{|x_n-x|}{|y_n-y|}<\infty$

or even: $0<\lim\limits_{n\to\infty}\frac{|x_n-x|}{|y_n-y|}<\infty$ if needed.

Edit 2: Clear counterexample is given below. Can one suggest any sufficient conditions on $x_n, y_n$ for existence of $f$ ?

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    $\begingroup$ What requirements on the domain of $f$? Should it include $\sup x_n$? (I guess $f$ locally constant on a disconnected open set is not what you want) $\endgroup$ – Pietro Majer Apr 6 '19 at 13:40
  • $\begingroup$ $f$ must be defined in some open neighborhood of $sup x_n$ $\endgroup$ – ar.grig Apr 6 '19 at 15:32
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    $\begingroup$ How about simply $x_n = 1-1/n$, $y_n = 1-1/\sqrt{n}$? This would seem to ensure that such $f$ cannot even be differentiable at $1$. $\endgroup$ – Nate Eldredge Apr 6 '19 at 17:24
  • $\begingroup$ @NateEldredge you are right. Question conditions were modified. $\endgroup$ – ar.grig Apr 6 '19 at 18:35
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It's not always possible to find such a function, for example take $x_n\to 0-$, $y_n=-e^{-1/x_n^2}$. A holomorphic function $f: D_r(0)\to\mathbb C$, $f\not\equiv 0$, would satisfy $|f(x)|\gtrsim |x|^k$ for some $k\ge 0$, so cannot satisfy $f(x_n)=y_n$ for infinitely many $n$.

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  • $\begingroup$ Of course, you are right. I have added the condition to avoid your described situation $\endgroup$ – ar.grig Apr 6 '19 at 18:33
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    $\begingroup$ @ar.grig: This (your edit) won't help because the same phenomenon can also happen in higher order, say $y_n=x_n-e^{-1/x_n^2}$. (The $x_n,y_n$ have to be consistent with the approximation by Taylor polynomials of arbitrarily high order.) $\endgroup$ – Christian Remling Apr 6 '19 at 18:54
  • $\begingroup$ Can you suggest the sufficient condition for existence of $f$? $\endgroup$ – ar.grig Apr 6 '19 at 19:20

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