# Hermite-Kakeya Theorem for entire functions

In a question 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

• I'm not sure I understand all the details fully (you want to assume (1), (2), so these are not general entire functions?), but I think such a version follows immediately from the one you quoted. If $f-g$ has a non-real zero, then the same holds for the approximating polynomials when you cut off the products. Jan 3, 2016 at 2:25
• @ChristianRemling. You are right. I only need to prove that $h(z)=f(z)-g(z)$ has only real zeros. Let $f_m(z),g_m(z)$ be the cut off version of the products up to $m$ terms. I think that I might be able to prove that $f'(x)g(x)-g'(x)f(x)>0,x\in\mathbb{R}$. But it might be hard for me to prove that $f'_m(x)g_m(x)-g'_m(x)f_m(x)>0$. Thanks again for the advice.
– mike
Jan 3, 2016 at 2:42
• Where is $r$ in your statement for entire functions? Jan 3, 2016 at 4:32

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.

• Thanks for the solution. What I really need is only the second part: If $f(z)$ and $g(z)$ are expressed by (1) and (2), then $f'(x)g(x)-g'(x)f(x)>0,x\in\mathbb{R}$ is sufficient for $f(z)-g(z)$ to have real zeros only. Is this statement correct?
– mike
Jan 3, 2016 at 7:12