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The following problem comes from an old complex analysis prelim exam:

Determine all analytic functions $f: H \rightarrow \mathbb{C}$ on the half-plane $H : = \{ z\in \mathbb{C} : \Re(z) > 0 \}$ that satisfy $f(\sqrt{n}) = n$ and $|f^{(n)}(1)| \leq 3$ for all positive integers $n$.

Clearly $f(z) = z^2$ satisfies this, and I wish to show that this is the only example. Note that $f(z) = z^2 + \epsilon \sin(\pi z^2)$ fail to satisfy the derivative bound for any $\epsilon > 0$. Additionally, the derivative bound implies that any such $f$ is analytic and sub-exponential with order 1. I can apply Carlson's theorem to show that $h(z): =f(z) - z^2$ is exactly zero, but this seems like a very heavy hammer to use for a prelim problem.

Any guidance on a more simple proof would be greatly appreciated!

JMill.
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  • $\lvert f^{(n)}(1)\rvert \leqslant C$ for all $n \geqslant 1$ tells you $f$ is entire, of order $\leqslant 1$. The same holds for $f(z) - z^2$. If $h$ weren't identically $0$, its genus would be at least $2$, since it vanishes at $\sqrt{n}$ for all $n \geqslant 1$. But the genus isn't larger than the order. – Daniel Fischer Aug 17 '20 at 19:27

1 Answers1

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Let $g(z)=f(z+1)-(z+1)^2, g(0)=0$; since $|g^{(n)}(0)| \le 3, n \ge 3$ we get that $g$ originally defined on $\Re z >-1$ extends to an entire function that satisfies $|g(z)| \le 3e^{|z|}+|z+1|^2, g(\sqrt n-1)=0, n \ge 1$.

Assume $g$ non-zero and $k \ge 1$ the order of the zero of $g$ at $0$. Then if $M_g(R)= \max_{|z|=R}|g(z)| \le 4e^R, R \ge R_0$ the number $N(R) \ge [R^2]$ of zeroes of $g$ with $|z|\le R$ satisfies (by Jensen theorem):

$\int_0^R \frac{N(t)-k}{t}dt+k \log R+\log |\frac{g^{(k)}(0)}{k!}|=\frac{1}{2\pi}\int_0^{2\pi}\log |g(Re^{it})|dt$

so by easy majorizations using $N(t)-k \ge [t^2]-1 \ge (t/2)^2, t \ge 10$, one gets:

$R^2/8-M \le LHS \le \log 4 + R, R \ge R_0$ for some constant $M$ that incorporates the integral on LHS from $0$ to say $10$ and $\log |\frac{g^{(k)}(0)}{k!}|$, so we get a contradiction for large $R$

Conrad
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