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We say that a strictly increasing sequence $x_n$ of reals converges fast to $x$, if for each $k\in\mathbb{N}$ the sequence $n^k\cdot(x_n − x)$ is bounded. It is known that there exists a $C^\infty$-function $f$ such that $f(1/n)=x_n$ and $f(0)=x$.

In which case (sufficient condition on $x_n$) there exists real-analytic function $g$ such that $g(1/n_k)=x_{n_k}$ and $g(0)=x$ for some subsequence $x_{n_k}$ ?

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    $\begingroup$ So $(x_n-x)n^k\to 0$ for all $k$, making all derivatives zero at $t=0$. $\endgroup$
    – Christian Remling
    Commented Apr 8, 2019 at 21:48
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    $\begingroup$ To expand on the comment by Christian Remling, computing divided differences of the values of $g(1/n_k)$ gives sequences that converge to $g^{(p)}(0)$ for any $p$. But by fast convergence of $x_n \to x$ we get $g^{(p)}(0)=0$ for $p>0$. So $g(t)$ cannot be analytic (the covergent sum of its Taylor series $x + 0t + 0t^2 + \cdots$) unless $g(t)$ is constant. $\endgroup$
    – Igor Khavkine
    Commented Apr 8, 2019 at 22:31
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    $\begingroup$ Real-analytic WHERE? $\endgroup$
    – Alexandre Eremenko
    Commented Apr 8, 2019 at 23:58
  • $\begingroup$ Sorry for the question. Obviously, $g$ doesn't exist, because must have all derivatives equal to zero in $t=0$. And it was explained earlier in other question. $\endgroup$
    – ar.grig
    Commented Apr 9, 2019 at 5:20

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