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 $f$ ?

  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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 $n$.

  • $\begingroup$ Of course, you are right. I have added the condition to avoid your described situation $\endgroup$ – ar.grig Apr 6 '19 at 18:33
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.