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

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. $\endgroup$ – Christian Remling Jan 3 '16 at 2:25