In a [question](https://math.stackexchange.com/questions/781461/extensions-of-the-hermite-bielher-and-hermite-kakeya-theorem) asked by Bobby Ocean, the following theorem is cited:


***Hermite-Kakeya Theorem(for polynomials)*** - Given two real-valued polynomials, $f$ and $g$, then $f(x)+g(x) r$ has only real zeros for every $r\in\mathbb{R}$, if and only if, $f$ and $g$ have real interlacing zeros. (see Rahman & Schmeisser, page 197-199).


***Question*** Is there a similar theorem for entire functions as stated below:


***Hermite-Kakeya (for entire functions)*** - Given two *entire functions*, $f$ and $g$, and $f(z)$ and $g(z)$ are real when $z\in\mathbb{R}$, and 


$$f(z)=\prod\_{k=1}^{\infty}\left(1-\frac{z}{\alpha\_k}\right)\tag{1}$$


$$g(z)=\prod\_{k=1}^{\infty}\left(1-\frac{z}{\beta\_k}\right)\tag{2}$$


where $0<\alpha\_1<\alpha\_2<\cdots<\alpha\_k<\cdots<\infty$,$0<\beta\_1<\beta\_2<\cdots<\beta\_k<\cdots<\infty$,


then $f(z)-g(z)$ has only real zeros, if and only if, the zeros of $f$ are strictly interlacing with those of $g$.


For example 
$$f(z)=\cos\sqrt{z}=\prod\_{k=1}^{\infty}\left(1-\frac{z}{((k-1/2)\pi)^2}\right)\tag{3}$$


$$g(z)=\frac{\sin\sqrt{z}}{\sqrt{z}}=\prod\_{k=1}^{\infty}\left(1-\frac{z}{(k\pi)^2}\right)\tag{4}$$


Thanks-
mike



Hermite-Kakeya Theorem for entire functions
Hermite's theorem indeed generalizes to entire functions but your statement for entire functions is incorrect. 
$$2\cos2z-\cos z=4\cos^2z-1-\cos z$$
has all zeros real, but the zeros of $2\cos 2z$ and $\cos z$ do not interlace.
Substitute the square root if you want zeros to be on a ray.


The correct statement: If $f$ and $g$ are real entire functions, and 
$f-ag$ has
only real roots FOR EACH REAL $a$, then
zeros of $f$ and $g$ are interlacent.


Proof. The first statement is equivalent to saying that $f/g$ has imaginary part
of constant sign in the upper half-plane and the opposite sign in the lower
half-plane. The family of such functions in normal in each half-plane.
So it is enough to prove this for polynomials, and for polynomials it is easy.


For the converse to be true, the you need a priori assumptions on your functions,
like the assumption that you make that they are of genus 0, with zeros on a ray.
For a complete discussion of these questions the reference is Levin, Distribution of zeros of entire functions.


EDIT. Under your conditions that the genus is zero and the roots are interlacing, the proof of the converse statement is easy because your $f$ and $g$ are limits of
real polynomials with real zeros. If $f\_n$, $g\_n$ are these polynomials, then
$f\_n/g\_n$ has imaginary part of constant sign in each half-plane, and then you
can pass to the limit. 


