Unique Entire Function with Prescribed Values

Question

Problem. Show there is a unique entire function $f:\mathbb{C}\rightarrow\mathbb{C}$ such that the coefficients of the Taylor expansion about the origin $f(z)=\sum_{n=0}^{\infty}c_{n}z^{n}$ are nonnegative and $f$ satisfies $f(k)=\cosh(\sqrt{k})$ for all positive integers $k$.

The existence of such a function isn't difficult. If we let $\sqrt{z}$ denote the principal branch, then on the positive real line,

$$f(z):=\cosh(\sqrt{z})=\cos(i\sqrt{z})=\sum_{n=0}^{\infty}\dfrac{(-1)^{n}(i\sqrt{z})^{2n}}{(2n)!}=\sum_{n=0}^{\infty}\dfrac{z^{n}}{(2n)!} \tag{1}$$

is analytic. The power series on the right-hand side has infinite radius of convergence, and therefore defines an entire function. Clearly, the Taylor coefficients are nonnegative, and by construction $f(k)=\cosh(\sqrt{k})$ for all positive integers $k$.

My trouble is proving the uniqueness assertion. Clearly, if we have any two entire functions $f_{1}$ and $f_{2}$ satisfying the required hypotheses, then their difference $g:=f_{1}-f_{2}$ has zeros on the positive integers. But I am not able to see much else at this time. In particular, I don't have an idea how to use the hypothesis that the Taylor coefficients are nonnegative to prove uniqueness. If we could show that $g$ vanishes on a set with a limit point, then uniqueness would follow by the identity theorem. Or if we could show that $g$ had a zero somewhere of infinite order, then uniqueness would also follow by the existence of power series expansions.

As this is an old qual problem, I would suggestions for getting on the right track, rather than complete solutions. But any help would be appreciated.

Answer 1

I think the solution to this problem is through the genus $h$ and order $\lambda$ of an entire function and a theorem of Hadamard which relates the two through an inequality. If $f:\mathbb{C}\rightarrow\mathbb{C}$ is an entire function of order $\lambda$ and genus $h$, then $h\leq\lambda\leq h+1$. In particular, if $\lambda$ is non-integral, then $h$ is uniquely determined. See Chapter 5, Section 3.2 of [L.V. Ahlfors, Complex Analysis, 1979] for a proof of this result.

Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be an entire function satisfying the conditions of the problem statement. Since the coefficients in the Taylor expansion $f(z)=\sum_{n}c_{n}z^{n}$ are nonnegative, we see that for every positive integer $k$,

\begin{align*} \left\|f\right\|_{\infty,B_{k}}=\max_{\left|z\right|\leq k}\left|f(z)\right|\leq\sum_{n=0}^{\infty}c_{n}k^{n}=\cosh(\sqrt{k})\leq e^{k^{1/2}} \tag{1} \end{align*}

Now let $\epsilon>0$ be given. If $k\leq r<k+1$, then \begin{align*} \left\|f\right\|_{\infty,B_{r}}=\sum_{n=0}^{\infty}c_{n}r^{n}\leq\sum_{n=0}^{\infty}c_{n}(k+1)^{n}\leq e^{(k+1)^{1/2}}\leq e^{k^{1/2+\epsilon}}\leq e^{r^{1/2+\epsilon}},\tag{2} \end{align*} for all $r>0$ sufficiently large. Thus, $f$ is of order at most $1/2$.

Since the analytic continuation of $\cos(i\sqrt{z})$ constructed in the original question is evidently of order at most $1/2$. If the entire function $g:=f-\cos(i\cdot)$ is not identically zero, then it follows from the triangle inequality and the simple estimate $2\leq e^{r^{\epsilon}}$, for all $r>0$ sufficiently large, that $g$ is of order at most $1/2$. By Hadamard's theorem, $g$ is of genus zero. But the zero set of $g$ includes the positive integers and $\sum_{n}1/n=\infty$, which is a contradiction.

I am curious whether anyone has an elementary solution, in the sense that it does not make use of Hadamard's theorem.

Apologies in advance for any errors. I'll check this again tomorrow tonight when I'm done traveling.

Answer 2

Idea (not sure it will work): If you can show $\frac{f(z)}{\cosh \sqrt z}$ is bounded in the open right half plane, then this function composed with the map $\frac{1+z}{1-z}$ gives a bounded holomorphic function in the unit disc that equals $1$ on a sequence that does not satisfy the Blaschke condition.