Convolution-type operator for series

Question

Suppose $f(z)=\lambda(a_1z+a_2z^2+\cdots)$ is holomorphic in $\{|z|<1\}$ with $\lambda>0$. For each $d\geq 1$ , I am trying to define an operation, $\star_d$ , so that $f(z)\star_d \overline{f(\bar{z})}$ has the form $\lambda^d\sum_{k=d+1}^{\infty}a_k \overline{a}_{k-d}z^k.$ Has anybody come across this before?

Answer 1

By the Cauchy integral formula, $$ f(z) = \sum_{n=0}^\infty a_n z^n = \frac{1}{2\pi i} \oint dw \frac{f(w)}{w-z} = \frac{1}{2\pi i} \sum_{n=0}^\infty \oint dw \frac{f(w)}{w^{n+1}} z^n . $$ If $g(z) = \sum_{n=0}^\infty b_n z^n$, it follows that \begin{align} \sum_{n=1}^\infty a_n b_n z^n &= \frac{1}{(2\pi i)^2} \sum_{n=0}^\infty \oint du \frac{f(u)}{u^{n+1}} \oint dv \frac{g(v)}{v^{n+1}} \, z^n \\ &= -\frac{1}{4\pi^2} \oint du \oint dv \frac{f(u)g(v)}{uv-z} . \end{align} Lets call the above operation $(f\star g)(z)$. If $\lambda=1$ and $d=0$, then your operation is just $(f\star g)(z)$, where $g(z) = \overline{f(\overline{z})}$. For $d>0$, I think it is enough to set $g(z) = z^d \overline{f(\overline{z})}$. I'm not sure what role $\lambda$ plays in the operation, so I guess one can just introduce the $\lambda^d$ factor by hand when $\lambda\ne 1$.