orthogonal polynomial of the second kind

Question

Let $L: \mathbb{R}[x] \rightarrow \mathbb{R}$ be a positive definite linear functional and let that $\{s_n\}$ be a positive semi-definite sequence such that $L(x^n)= s_n, n\ge 0.$ Given a positive definite sequence, I was able to use the Gram-Schimdt orthogonalization method to construct a sequence of orthogonal polynomials $\{p_n\}$ whose leading coefficient is positive due to the positivity nature of the sequence given. It turns out that this sequence of orthogonal polynomials $\{p_n\}$ satsifies a three term recurrence relation given below \begin{equation} xp_n(x) =b_np_{n+1}(x)+a_np_n(x)+b_{n-1}p_{n-1}(x) , \quad n\ge 0 \end{equation}

We can see the sequence $p_n(x)$ as a solution to the three term recurrence relations stated above. Akhiezer http://www.maths.ed.ac.uk/~aar/papers/akhiezer.pdf as my reference introduced another solution to this three term recurrence relation by defining another solution by \begin{equation} q_n(x)= \displaystyle L\left(\frac{p_n(x)-p_n(y)}{x-y}\right) \end{equation} where the quotient $\frac{p_n(x)-p_n(y)}{x-y}$ is a polynomial in $x$ and $y$ and $q_n(x)$ is a polynomial in variable $x$ and its degree is $n-1$ for any $x,y \in \mathbb{R}$ so that we have $ \displaystyle xq_n(x) =b_nq_{n+1}(x)+a_nq_n(x)+b_{n-1}q_{n-1}(x), n\ge 1$ with $q_0(x)=0$ and $q_1(x)= \frac{1}{b_0}$

My question is how can the linear functional $L$ be defined on rational functions since its domain is $\mathbb{R}[x]$. I have troubles understanding this definition even though I actually confirmed the definition is true by computing $q_1,q_2$.You can please check Akhiezer page 8 for more clarification. Details and explanation will be much appreciated.

Answer 1

For all $P\in\mathbb{R}[X]$, $\frac{P(x)-P(y)}{(x-y)}$ is a polynomial on $X$ and $Y$, to see this let begin by a simple case, $P(x)=1$ \begin{eqnarray} \frac{P(x)-P(y)}{x-y}&=&\frac{1-1}{x-y}\\ &=& 0 \\ &\in & \mathbb{R}[X,Y] \end{eqnarray} Let now $n\in\mathbb{N}$ and $P(x)=x^n$, we will use this lemma

$$ x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dots+x^{n-1-i}y^i+\dots+y^{n-1}) $$

so \begin{eqnarray} \frac{P(x)-P(y)}{x-y}&=&\frac{x^n-y^n}{x-y}\\ &=& (x^{n-1}+x^{n-2}y+\dots+x^{n-1-i}y^i+\dots+y^{n-1}) \\ &\in & \mathbb{R}[X,Y] \end{eqnarray} Now Let $P(x)=\sum_{i=0}^n a_i x^i$ be any polynomial so \begin{eqnarray} \frac{P(x)-P(y)}{x-y}&=&\frac{\sum_{i=0}^n a_i x^i-\sum_{i=0}^n a_i y^i}{x-y}\\ &=&\frac{\sum_{i=0}^n a_i (x^i- y^i)}{x-y} \\ &=& \frac{\sum_{i=0}^n a_i (x-y)Q_i(x,y)}{x-y} \\ &=& \sum_{i=0}^n a_i Q_i(x,y) &\in & \mathbb{R}[X,Y] \end{eqnarray}