[proofplan] We first derive the three-term recurrence for the orthogonal polynomial sequence, including the exact coefficients needed for telescoping. Then we evaluate that recurrence at two points $x$ and $y$, multiply by the opposite polynomial factor, subtract, and divide by the squared norm. The middle recurrence coefficient cancels, and the two neighbouring terms form a telescoping difference. Summing from $j=0$ to $n$ leaves only the boundary term involving $p_{n+1}$ and $p_n$. [/proofplan]

[step:Derive the three-term recurrence with the correct boundary coefficients] For each $j\geq 0$, the polynomial $t p_j(t)$ has degree $j+1$, so it has a unique expansion in the basis $p_0,\dots,p_{j+1}$. Orthogonality implies that all coefficients of $p_0,\dots,p_{j-2}$ vanish, because if $0\leq m\leq j-2$, then $t p_m(t)$ has degree at most $j-1$ and hence is a linear combination of $p_0,\dots,p_{j-1}$. Therefore \begin{align*} \int_I t p_j(t)p_m(t)w(t)\,d\mathcal{L}^1(t)=\int_I p_j(t)\bigl(t p_m(t)\bigr)w(t)\,d\mathcal{L}^1(t)=0. \end{align*} Thus there are [real numbers](/page/Real%20Numbers) $A_j,B_j,C_j$ such that \begin{align*} t p_j(t)=A_jp_{j+1}(t)+B_jp_j(t)+C_jp_{j-1}(t), \end{align*} where $p_{-1}$ denotes the zero polynomial and hence $C_0p_{-1}=0$. Comparing leading coefficients gives \begin{align*} A_j=\frac{k_j}{k_{j+1}}. \end{align*} For $j\geq 1$, taking the [inner product](/page/Inner%20Product) of the recurrence with $p_{j-1}$ gives \begin{align*} C_jh_{j-1}=\int_I t p_j(t)p_{j-1}(t)w(t)\,d\mathcal{L}^1(t). \end{align*} By symmetry of multiplication by $t$ under this integral, \begin{align*} \int_I t p_j(t)p_{j-1}(t)w(t)\,d\mathcal{L}^1(t)=\int_I p_j(t)\bigl(t p_{j-1}(t)\bigr)w(t)\,d\mathcal{L}^1(t). \end{align*} The coefficient of $p_j$ in the expansion of $t p_{j-1}(t)$ is $k_{j-1}/k_j$, so orthogonality gives \begin{align*} \int_I p_j(t)\bigl(t p_{j-1}(t)\bigr)w(t)\,d\mathcal{L}^1(t)=\frac{k_{j-1}}{k_j}h_j. \end{align*} Hence, for $j\geq 1$, \begin{align*} C_j=\frac{k_{j-1}h_j}{k_jh_{j-1}}. \end{align*} Consequently, for every $j\geq 0$, \begin{align*} t p_j(t)=\frac{k_j}{k_{j+1}}p_{j+1}(t)+B_jp_j(t)+\frac{k_{j-1}h_j}{k_jh_{j-1}}p_{j-1}(t), \end{align*} with the last term interpreted as $0$ when $j=0$. [/step]

[step:Subtract the recurrences at $x$ and $y$ to create a telescoping identity]Fix $x,y\in\mathbb{R}$ with $x\neq y$. For $j\geq 0$, evaluate the recurrence at $t=x$ and multiply by $p_j(y)/h_j$, then evaluate it at $t=y$ and multiply by $p_j(x)/h_j$. Subtracting the second identity from the first cancels the $B_j$ terms and gives \begin{align*} (x-y)\frac{p_j(x)p_j(y)}{h_j}=\frac{k_j}{k_{j+1}h_j}\bigl(p_{j+1}(x)p_j(y)-p_j(x)p_{j+1}(y)\bigr)-\frac{k_{j-1}}{k_jh_{j-1}}\bigl(p_j(x)p_{j-1}(y)-p_{j-1}(x)p_j(y)\bigr). \end{align*} Define $R_{-1}(x,y)=0$, and for $j\geq 0$ define \begin{align*} R_j(x,y)=\frac{k_j}{k_{j+1}h_j}\bigl(p_{j+1}(x)p_j(y)-p_j(x)p_{j+1}(y)\bigr). \end{align*} Then the preceding identity is exactly \begin{align*} (x-y)\frac{p_j(x)p_j(y)}{h_j}=R_j(x,y)-R_{j-1}(x,y). \end{align*}[/step]

[guided]The purpose of the subtraction is to remove the middle coefficient $B_j$, whose explicit value is irrelevant to the Christoffel-Darboux formula. For a fixed $j\geq 0$, the recurrence at $x$ is \begin{align*} x p_j(x)=\frac{k_j}{k_{j+1}}p_{j+1}(x)+B_jp_j(x)+\frac{k_{j-1}h_j}{k_jh_{j-1}}p_{j-1}(x), \end{align*} with the last term understood as $0$ when $j=0$. Multiplying by $p_j(y)/h_j$ gives an identity whose left-hand side is $x p_j(x)p_j(y)/h_j$. The same recurrence at $y$, multiplied by $p_j(x)/h_j$, has left-hand side $y p_j(y)p_j(x)/h_j$. Subtracting therefore produces the desired factor \begin{align*} (x-y)\frac{p_j(x)p_j(y)}{h_j}. \end{align*} The $B_j$ contribution cancels because it appears as \begin{align*} B_j\frac{p_j(x)p_j(y)}{h_j}-B_j\frac{p_j(y)p_j(x)}{h_j}=0. \end{align*} The remaining upper-neighbour term is \begin{align*} \frac{k_j}{k_{j+1}h_j}\bigl(p_{j+1}(x)p_j(y)-p_j(x)p_{j+1}(y)\bigr), \end{align*} and the lower-neighbour term is \begin{align*} \frac{k_{j-1}}{k_jh_{j-1}}\bigl(p_{j-1}(x)p_j(y)-p_j(x)p_{j-1}(y)\bigr). \end{align*} Writing the lower-neighbour term with the opposite order inside the parentheses changes its sign, so if \begin{align*} R_j(x,y)=\frac{k_j}{k_{j+1}h_j}\bigl(p_{j+1}(x)p_j(y)-p_j(x)p_{j+1}(y)\bigr) \end{align*} and $R_{-1}(x,y)=0$, the whole subtraction becomes the compact difference identity \begin{align*} (x-y)\frac{p_j(x)p_j(y)}{h_j}=R_j(x,y)-R_{j-1}(x,y). \end{align*} This is the exact form needed for telescoping when $j$ is summed from $0$ to $n$.[/guided]

[step:Sum the telescoping identity and divide by $x-y$] Summing \begin{align*} (x-y)\frac{p_j(x)p_j(y)}{h_j}=R_j(x,y)-R_{j-1}(x,y) \end{align*} over $j=0,\dots,n$ gives \begin{align*} (x-y)\sum_{j=0}^{n}\frac{p_j(x)p_j(y)}{h_j}=R_n(x,y)-R_{-1}(x,y). \end{align*} Since $R_{-1}(x,y)=0$, this becomes \begin{align*} (x-y)\sum_{j=0}^{n}\frac{p_j(x)p_j(y)}{h_j}=\frac{k_n}{k_{n+1}h_n}\bigl(p_{n+1}(x)p_n(y)-p_n(x)p_{n+1}(y)\bigr). \end{align*} Because $x\neq y$, division by $x-y$ is valid, and therefore \begin{align*} \sum_{j=0}^{n}\frac{p_j(x)p_j(y)}{h_j}=\frac{k_n}{k_{n+1}h_n}\frac{p_{n+1}(x)p_n(y)-p_n(x)p_{n+1}(y)}{x-y}. \end{align*} This is the Christoffel-Darboux formula. [/step]