Simple zero of complex orthogonal polynomials

Question

Let $\mu$ a positive measure on $\Omega\subset\mathbb{C}$ and we assume that all moment of $\mu$ exists (i.e) for all $i,j\in \mathbb{N}$ $$ \int_\Omega z^i\bar{z}^j d\mu(z)\in\mathbb{C} $$ So on $\mathbb{C}[Z,\bar{Z}]$ we consider the inner product $$ <P,Q>=\int_\Omega P\bar{Q} d\mu(z) $$ and since all polynomials are integrable we can construct a family of orthogonal polynomials $(P_n)_n$ by Gram-Schmidt process so my question is if each of $P_n$ will have simple roots, Or what we can say about zero of $P_n$ ?

Answer 1

First, you have to specify the support. If the support is a finite set with $n$ points then the monic orthogonal polynomials for $\mu$ of degree greater than $n$ are not unique although they are well-defined in $\mathcal{L}^2(\mu)$.

Let us assume for the simplicity that the $\mu$ is a finite Borel measure with infinite and compact support in $\mathbb{C}$. Hence your assumptions hold. If the support lies on $\mathbb{R}$ then the zeros are simple and real. This can be found in many introductory books and articles. See for example "Spectral theory and special functions" by Erik Koelink.

There is also a classical example on the unit circle such that $0$ is the only zero of all orthogonal polynomials. You can find it here (18.33.12): http://dlmf.nist.gov/18.33

In general, zeros may or may not be simple. One more point: On the spacing of the zeros on $\mathbb{R}$, there are fresh results showing that they have "fine structure". You can check "Universality and fine zero spacing on general sets" 2009, Arkiv för Math. by V. Totik.