So I want to show essentially that for a linear functional $\mathcal{L}$ such that $\mathcal{L}[x^n]=\alpha^n$ for $n\geq 0$, that no orthogonal polynomial sequence exists w.r.t $\mathcal{L}$.
Unfortunately due to several reasons my lectures for this module have been suspended for about 3 weeks so we are quite behind hence I was never taught too much about this topic.
I am aware of the theorem:
For $\{P_n\}_{n\geq0}$ a polynomial sequence and $\mathcal{L}$ a linear functional then the following statements are equivalent:
a) $\{P_n\}_{n\geq0}$ is an OPS for $\mathcal{L}$.
b) \begin{cases} \mathcal{L}[\pi(x)P_n(x)]=0, & \text{for any polynomial $\pi(x)$ and any integer $n>deg(\pi(x))$}, \\ \mathcal{L}[\pi(x)P_n(x)]\neq0, & \text{if $deg(\pi(x))=n, n\geq0$.} \end{cases}
c) \begin{cases} \mathcal{L}[x^mP_n(x)]=0, & \text{for any integers $n$ and $m$ such that $n>m$}, \\ \mathcal{L}[x^n(x)P_n(x)]\neq0, & \text{for any $n\geq0$.} \end{cases}
Is it that the linear functional I have does not satisfy one of these properties that would then allow me to show it has no OPS? Or perhaps there is an easier way that might not make use of these relations..
Any hints are very much appreciated.
I believe I have an answer to my question so I will post it for the sake of completeness.
We know that a neccessary and sufficient condition for the existence of an OPS with respect to $\mathcal{L}$ is that $\Delta_n \neq 0$, where $\Delta_n =\det[\mu_{i+j}]_{0\leq i, j\leq n}, n\geq 0$, otherwise known as the Hankel Determinent.
As we have $\mathcal{L}[x^n]=\alpha^n$ we can see that $$\Delta_1=0,$$ hence we conclude that no orthogonal polynomial sequence can exist with respect to $\mathcal{L}$.