Zeros of the Hadamard product of holomorphic functions

Question

Let $A(z) = \sum_{n=0}^{+\infty}a_n z^n$ and $B(z) = \sum_{n=0}^{+\infty}b_n z^n$ be two formal power series with complex coefficients. The Hadamard product of $A$ and $B$ is the formal power series $A\star B$ defined by $$ (A\star B)(z) = \sum_{n=0}^{+\infty}a_n b_n z^n. $$ If $r_A$ (resp. $r_B$) is the radius of convergence of $A$ (resp. $B$), it is clear that the radius of convergence $r_{A\star B}$ of $A\star B$ is greater than or equal to $r_A \cdot r_B.$

Inside the disk $D(0,r_A \cdot r_B)$, what do we know concerning the zeros of $A \star B$ and their relations with the zeros of $A$ and $B$ ?

Note that if $f$ (resp. $g$) is a holomorphic function defined by $A$ (resp. $B$) on the disk $D(0,r_A)$ (resp. $D(0,r_B)$) and if $r \in (0,r_A)$, then \begin{align*} \sum_{n=0}^{+\infty}a_n b_n z^n & = \sum_{n=0}^{+\infty}\left(\frac{1}{2i\pi}\int_{C(0,r)^+}\frac{f(\zeta)}{\zeta^{n+1}}\, d\zeta \right) b_n z^n\\ & = \frac{1}{2i\pi}\int_{C(0,r)^+} f(\zeta) \left(\sum_{n=0}^{+\infty} b_n \left(\frac{z}{\zeta} \right)^n \right) \frac{d\zeta}{\zeta}\\ & = \frac{1}{2i\pi}\int_{C(0,r)^+} f(\zeta)g\!\left(\frac{z}{\zeta}\right) \frac{d\zeta}{\zeta}, \end{align*} for any $z \in D(0,r \cdot r_B)$.

Answer 1

Some cases in point: $A_1 = B_1 = \exp(z)$ entire with no zeros, $A_1 \star B_1 = I_0(2 \sqrt{z})$ with infinitely many.

$$\eqalign{A_2 &= \sum_{n=0}^\infty \frac{z^n}{n!!} = 1 + z e^{z^2/2} + \sqrt{\frac{\pi}{2}} z e^{z^2/2} \text{erf}(z/\sqrt{2})\cr B_2 &= \sum_{n=0}^\infty \frac{n!! z^n}{n!} = e^{z^2/2} + \sqrt{\pi}{2} e^{z^2/2} \text{erf}(z/\sqrt{2})\cr A_2 \star B_2 &= \exp(z)} $$ $A_2$ and $B_2$ are entire and have zeros (presumably infinitely many), $A_2 \star B_2$ has none.

Answer 2

While not sure that it is what the OP is interested in, I would mention that in geometric function theory (univalent functions on the disc, quasi-conformal functions etc) there are a lot of results about convolutions (and implicitly about their zeros) - an early classical example is Grace-Szego theorem for polynomials which asserts that if $P*(1+z)^n, Q*(1+z)^n$ do not vanish in the (open) unit disc, then $P*Q*(1+z)^n$ doesn't vanish in the (open) unit disc either, or more directly but asymmetrically, if $P*(1+z)^n, Q$ do not vanish in the (open) unit disc, neither does $P*Q$ (where $P$ is an arbitrary holomorphic function in the unit disc, but wlog it can be taken as a polynomial of degree at most $n$ in the above, while $Q$ is a polynomial of degree at most $n$ in the second asymmetric version, and arbitrary holomorphic in the first).

There are fairly interesting results about the convolution of univalent (and related classes of) functions too. The book Convolutions in Geometric Functions by S Ruscheweyh, freely available on Research gate as pdf has lots of such, but also I would recommend the books by P. Duren (univalent Functions), T Sheil-Small, Complex Polynomials and references there.

Answer 3

Pemantle states some very general closure properties in "Hyperbolicity and stable polynomials in combinatorics and probability" on p. 33:

If we multiply term by term the coefficient sequences of two real rooted polynomials, at least one of which has nonnegative coefficients, we get another real rooted coefficient sequence. Thus the real root property for polynomials with nonnegative coefficients is closed under Hadamard products.

And, in "Zeros of polynomials and their importance in combinatorics and probability":

Closure properties: the class of univariate polynomials with all real roots is closed under taking the derivative. [Why? The zeros of the derivative interlace the zeros of the polynomial.] It is not only closed under products (obvious) but under coefficient-wise products (Hadamard products – not obvious!).

(Cf. also "Lecture 2: Zeros and coefficients of polynomials in one variable".)