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Let $0<q<1$ and consider the entire function $f(z)=\displaystyle \sum_{k=0}^\infty q^{k^2}z^k$. For $a>1,$ denote $m_j=f(a^j),\; j=0,1,2,\dots.$

Question: Does there exist an entire function $g(z)=\displaystyle \sum_{k=0}^\infty c_k z^k$ with $c_k>0$ (strictly) and a number $b\neq a$ so that $g(b^j)=m_j$ for all $j=0,1,2,\dots$

If some $c_k$ can be 0, the answer is obvious.

The question arises in my research on Stieltjes classes.

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  • $\begingroup$ I am surely missing something: doesn't $g(z) = f(a^{-1} b z)$ satisfy your requirements? $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jul 16, 2018 at 9:27
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    $\begingroup$ @Mateusz Kwasnicki: Of course not. Just plug $z=b^j$ and see what happens:-) $\endgroup$
    – Alexandre Eremenko
    Commented Jul 16, 2018 at 14:25
  • $\begingroup$ @alexandreeremenko Huh, I must have already entered holiday mode... :-) $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jul 16, 2018 at 14:48
  • $\begingroup$ @Deepti: If I am not missing something obvious again, it seems that the answer is positive when $b = \sqrt{a}$ and $a$ is large enough. However, I have no clue what happens when $a$ is small and/or $b \ne \sqrt{a}$. Does this work for you, or you need arbitrary $a > 1$? $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jul 16, 2018 at 23:23
  • $\begingroup$ @AlexandreEremenko: As a small compensation for making dumb comments, I wrote a (partial) answer. But my mind is still on holidays, so again I expect errors. If you notice any, let me know! $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jul 16, 2018 at 23:25

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The answer is yes, at least in the special case when $a > q^{-4}$ and $b = \sqrt{a}$.

Denote by $h(z)$ the q-Pochhammer symbol with parameter $\tfrac{1}{b}$: $$ h(z) = \prod_{k = 0}^\infty \biggl(1 - \frac{z}{b^k}\biggr) = \biggl(z; \frac{1}{b}\biggr)_{\!\infty} . $$ We claim that for a sufficiently small $\varepsilon > 0$, the function $$ g(z) = f(z^2) - \varepsilon h(z) $$ has the desired properties.

By definition of $g$ and $h$, $$ g(b^j) = f(b^{2 j}) - \varepsilon h(b^j) = f(b^{2 j}) = f(a^j), $$ as desired. It remains to verify that the coefficients $c_k$ of the power series of $g$ are all positive.

Denote by $(-1)^k d_k$ the coefficients of the power series of $h(z)$, that is, $$ h(z) = \sum_{k = 0}^\infty (-1)^k d_k z^k . $$ Clearly, $$ g(z) = f(z^2) - \varepsilon h(z) = \sum_{k = 0}^\infty c_k z^k , $$ where $$ c_k = \begin{cases} \varepsilon d_k & \text{for odd $k$,} \\ q^{k^2}(1 - \varepsilon q^{-k^2} d_k) & \text{for even $k$.} \end{cases} $$ It is known that $$ d_k = \frac{1}{b^{k(k - 1) / 2}} \prod_{j = 1}^k \frac{1}{1 - \tfrac{1}{b^j}} = \frac{1}{b^{k(k - 1) / 2} \bigl(\tfrac{1}{b}; \tfrac{1}{b}\bigr)_k} \, . $$ In particular, $d_k > 0$ for all $k = 0, 1, 2, \ldots$, and so $c_k = \varepsilon d_k > 0$ for odd $k$. In order to prove that $c_k > 0$ for even $k$, we need the following estimate.

Since $b > 1$, we have $1 - \tfrac{1}{b^j} > 1 - \tfrac{1}{b}$ for $j = 1, 2, \ldots, k$, and thus $$ 0 \leqslant d_k \leqslant (1 - \tfrac{1}{b})^{-k} b^{-k (k - 1) / 2} .$$ It follows that $$ q^{-k^2} d_k \leqslant q^{-k} (1 - \tfrac{1}{b})^{-k} (q^2 b)^{-k (k - 1)/2} . $$ Since $q^2 b > 1$, the right-hand side of the above inequality converges to zero as $k \to \infty$. Therefore, if $\varepsilon > 0$ is small enough, we have $\varepsilon q^{-k^2} d_k < 1$ for all $k = 0, 1, 2, \ldots$, and consequently $c_k > 0$ for even $k$, as desired.

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