# real analytic interpolation

Given monotonic sequence $$x_n$$ of reals. Does there exist a subsequence $$x_{n_k}$$ and real-analytic function $$f$$ on $$(1;+\infty)$$ such that $$f(n_k)=x_{n_k}$$?

• What have you tried? – David G. Stork Dec 23 '19 at 6:24
• @DavidG.Stork: I have tried formal series $$f(s) = a_0 + \sum_{n=1}^\infty a_n\cdot\frac{\prod\limits_{k=0}^{n-1}(s-i_k)}{\prod\limits_{k=0}^{n-1}(i_n-i_k)}$$ where $a_n$ could be determined by induction from equations $f(i_k)=x_{i_k}$. – user2935704 Dec 23 '19 at 8:19

• If $$x_n\to \infty$$, take $$|x_{n_k}|> k^2$$ $$h(z) =\prod_k (1-\frac{z}{x_{n_k}}), \qquad h_m(z)=\frac1{1-\frac{z}{x_{n_m}}}$$ are entire and uniformly bounded by $$\prod_k (1+\frac{|z|}{|x_{n_k}|})$$, take $$g(z)$$ entire growing fast enough on the real axis such that $$|\frac{f(x_{n_m})}{h_m(x_{n_m})g(x_{n_m})} |\le 1/m^2$$ then $$g(z)\sum_m \frac{f(x_{n_m})}{h_m(x_{n_m})g(x_{n_m})} h_m(z)$$ converges locally uniformly thus it is entire
• If $$x_n$$ is bounded, if there exists a function analytic at $$a=\sup x_n$$ agreeing with $$f(x_n)$$ then it is the only one, and it is given by the Taylor series at $$a$$ which is found from the differences of the $$f(x_n)$$.
• sorry, but $h_m(x_{n_m})$ is not defined – user2935704 Dec 23 '19 at 7:50