# Real analytic function of one variable with given set of values

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 $$f$$ ?

• 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) – Pietro Majer Apr 6 '19 at 13:40
• $f$ must be defined in some open neighborhood of $sup x_n$ – ar.grig Apr 6 '19 at 15:32
• 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 $1$. – Nate Eldredge Apr 6 '19 at 17:24
• @NateEldredge you are right. Question conditions were modified. – ar.grig Apr 6 '19 at 18:35

## 1 Answer

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 $$n$$.

• Of course, you are right. I have added the condition to avoid your described situation – ar.grig Apr 6 '19 at 18:33
• @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.) – Christian Remling Apr 6 '19 at 18:54
• Can you suggest the sufficient condition for existence of $f$? – ar.grig Apr 6 '19 at 19:20