Research Problems in Function Theory — Problem 5.60

Definitions and mathematical statement

Let $\mathbb D=\{z:|z|<1\}$. For analytic functions $u(z)=\sum_{k\geq0}u_kz^k$ and $v(z)=\sum_{k\geq0}v_kz^k$, their Hadamard convolution is the coefficientwise product $(u*v)(z)=\sum_{k\geq0}u_kv_kz^k$, which is analytic on $\mathbb D$. For real $\alpha,\beta\geq1$ and $|x|=1$, put \[ K_{\alpha,\beta,x}(z)= \frac{(1+xz)^\alpha}{(1-z)^\beta}. \] Both powers use the analytic logarithms on the disk normalized to have value zero at $z=0$, so the kernel has constant term $1$.

Suppose that $\phi$ is analytic on $\mathbb D$ and that \[ (\phi*K_{\alpha,\beta,x})(z)\ne0 \quad\text{for every }|x|=1\text{ and every }z\in\mathbb D. \] Must the same simultaneous zero-free property hold after lowering the numerator exponent by one, that is, $(\phi*K_{\alpha-1,\beta,x})(z)\ne0$ for every such $x,z$? There is no boundedness, univalence, or coefficient-sign assumption on $\phi$; the hypothesis itself implies $\phi(0)\ne0$.

The implication is required uniformly over all analytic $\phi$ and all real $\beta\geq1$. The source records the implication for integer $\alpha$, so the remaining target is the noninteger range $\alpha>1$. A counterexample must satisfy the first zero-free condition for every boundary parameter $x$, while the lowered kernel fails for at least one parameter and one point of the open disk.

Short English statement

Does a simultaneous zero-free Hadamard-convolution condition survive decreasing a real kernel exponent by one?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2305060 (AMR-022-5060), AMR Open Problem Lists. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Hayman - Research Problems in Function Theory (2018), Problem 5.60. Original source indexed by AMR-022-5060. https://arxiv.org/abs/1809.07200.