Given $\frac{H(z)}{\prod_{n=1}^\infty \left(1-\left(\frac{z}{\lambda_n}\right)^2\right)} = \phi(z),$ determine if $\phi(z)$ is an entire function.

Question

Let us consider $\lambda_n=n-\frac14$ and define $$\frac{H(z)}{\prod_{n=1}^\infty \left(1-\left(\frac{z}{\lambda_n}\right)^2\right)}=\phi(z) $$ where $H(z)$ is an entire function. I would like to understand if $\phi(z)$ is an entire function. Below my attempt.

Consider the infinite product $\prod_{n=1}^\infty \left(1-\left(\frac{z}{\lambda_n}\right)^2\right)$. This infinite product has zeroes at $z = \pm \lambda_n$, where $\lambda_n = n - \frac{1}{4}$.

Now:

1. If $H(z)$ and the product have the same zeros at $z = \pm \lambda_n$: In this case, both $H(z)$ and the product have zeros at $z = \pm \lambda_n$. When we take the quotient, these zeros cancel out, and $\phi(z)$ remains an entire function.

2. If $H(z)$ has additional zeros compared to the product: In this case, the zeros of the product are canceled out by the corresponding zeros of $H(z)$. The remaining zeros of $H$ become removable singularities, and $\phi(z)$ continues to be entire, am I right?

Answer 1

Let $P(z) = \prod_{n = 1}^\infty \left(1 - \frac{z^2}{\lambda_n^2}\right)$. We have for all $z$ in some ball of center $0$ and radius $R$, $\sum_{n = 1}^\infty \frac{z^2}{\lambda_n^2} \leqslant R^2\sum_{n = 1}^\infty \frac{1}{\lambda_n^2} < +\infty$. We deduce that the product $P$ is uniformly convergent on any compact set when $z$ is not a $\pm\lambda_n$ (in the sens of products i.e. the limit is not zero) thus it defines an entire function on $\mathbb{C}$ whose zeros are exactly the $\pm\lambda_n$ with multiplicity $1$ (because for all $n \neq m$, $\lambda_n \neq \pm\lambda_m$ and the $\lambda_n$ are all non zero).

Therefore, $\phi$ is a meromorphic function on $\mathbb{C}$ by product of two meromorphic functions. It can have poles at $\pm\lambda_n$ if $H$ doesn't vanish at those points. The fact that the zeroes of $P$ all are simple implies that $\phi$ is holomorphic (hence entire) if and only if for all $n$, $H(\pm\lambda_n) = 0$ so the point $1.$ is true.

However, I don't understand your point $2.$, if $H$ has other zeroes (or multiple zeroes at some point of the form $\pm\lambda_n$), then $\phi$ have zeroes at those points too, but it can't add any pole.