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*