Let $N$ be a non negative integer. Define the sequence of monic orthogonal polynomials $\{P_n(x)\}_{n}$ with respect to the inner product $$ \langle f , g\rangle =\sum^{N}_{k=0}{f(k)g(k)\rho(k)} $$ where $\rho$ is a weight function.
Definition : The $n$-th reproducing kernel is defined by $$ \mathcal{K}_n(x,y) =\sum_{k=0}^{n}{\frac{P_k(x)P_k(y)}{\|P_k\|^2}}. $$ Here, $\|\cdot\|$ stands for the Euclidean norm. We use the following notation for every pair $i,j \ge 0$: $$ \mathcal{K}^{(i,j)}_n (x, y) = \Delta^i_x\big(\Delta^j_y(\mathcal{K}_n(x, y))\big)=\sum_{k=0}^{n}{\frac{\Delta^i(P_k(x))\Delta^j(P_k(y))}{\|P_k\|^2}} $$ with $\Delta f(x)=f(x+1)-f(x)$ and $\Delta^j (f(x))=\Delta (\Delta^{j-1} f(x))$.
Concerning the families of orthogonal polynomials, a Christoffel-Darboux formula is also available
Proposition : For all $n$ , it holds that : if $x\neq y$ $$ \mathcal{K}_n(x,y)=\frac{P_{n+1}(x)P_n(y)-P_{n+1}(y)P_n(x)}{(x-y)\|P_n\|^2} \label{1}\tag{$\ast$} $$
Show that : For all $0\le n\le N$ and $x\neq y$, there exists $\mathcal{A}^{(j)}_n (x, y)$ and $\mathcal{B}^{(j)}_n (x, y)$, such that : $$\mathcal{K}^{(0,j)}_{n-1} (x,y) = \mathcal{A}^{(j)}_n (x, y)P_n(x) +\mathcal{B}^{(j)}_n(x,y)P_{n-1}(x) \label{2}\tag{$\ast\ast$} $$
My effort : After applying the difference operator $\Delta^j$ to \eqref{1} with respect to $y$, we obtain: $$ \mathcal{K}^{(0,j)}_n(x,y)=\frac{1}{\|P_{n-1}\|^2}\bigg[P_n(x)\Delta^j_y\bigg( \frac{P_{n-1}(y)}{x-y}\bigg)- P_{n-1}(x)\Delta^j_y\bigg( \frac{P_{n}(y)}{x-y}\bigg) \bigg] $$ Using an analogue of the Leibniz's rule $$ \Delta^n(f(x)g(x)) =\sum^n_{k=0}{{n\choose k}\Delta^k f(x)\Delta^{n-k}g(x-k)}, $$ and since for any positive integer $k$ we have $$ \Delta^k_y\bigg(\frac{1}{x-y} \bigg)=\frac{k!}{[x-y]_{k+1}} $$ with $ [z]_{k}=z(z-1)...(z-k+1)$ we deduce $$ \begin{array}{lcl} \Delta^j_y\bigg( \frac{P_{n-1}(y)}{x-y}\bigg)&=&\sum_{k=1}^{j}{{j\choose k}\Delta^j_y [P_{n-1}(y)]\frac{(j-k)!}{[x-y-k]_{j-k+1}}}\\ &=&\sum_{k=1}^{j}{\frac{\Delta^j_y [P_{n-1}(y)]}{k!}\frac{j!}{[x-y-k]_{j-k+1}}}\\ \end{array} $$ Since $\frac{[x]_n}{[x]_k}= [x-k]_{n-k}$ if $n\ge k$, we deduce $[x -y- k]_{j+1-k}=\frac{[x-y]_{j+1}}{[x-y]_k}$. Thus, $$ \Delta^j_y\bigg( \frac{P_{n-1}(y)}{x-y}\bigg)=\frac{j!}{[x-y]_{j+1}}\sum_{k=0}^{j}{\frac{\Delta^k P_{n-1}(y)}{k!}[x-y]_k}. $$ in the same way we can show $$ \Delta^j_y\bigg( \frac{P_{n}(y)}{x-y}\bigg)=\frac{j!}{[x-y]_{j+1}}\sum_{k=0}^{j}{\frac{\Delta^k P_{n}(y)}{k!}[x-y]_k}. $$ Therefore, from the above we get \eqref{2}.
Is what I wrote correct? if not please help me
