Weighted Jacobi endpoint kernel on the reference interval

Apply Leibniz's rule to the $n$th derivative in the definition. Every term differentiates $(1-t{)}^{n+d-1}$ at most $n$ times, so it contains a factor $(1-t{)}^{d-1}$. After division the result is a polynomial. The original factor has degree $2n+d-1$ and leading coefficient $(-1{)}^{n+d-1}$; differentiating $n$ times and dividing by the leading term $(-1{)}^{d-1}{t}^{d-1}$ of $(1-t{)}^{d-1}$ shows that ${\varphi }_n$ has degree $n$ and leading coefficient $\left(\genfrac{}{}{0px}{}{2n+d-1}{n}\right)$. The coefficient of $t^n$ in $t^n(1-t{)}^{n+d-1}$ is $1$, so its $n$th derivative at zero is $n!$ and ${\varphi }_n(0)=(-1{)}^n$.

If $r$ has degree below $n$, multiply the Rodrigues formula by $w_{d}r$ and integrate. Integrating by parts $n$ times is allowed by [A1] and gives ${\int }_0^1{w}_d(t){\varphi }_n(t)r(t)dt=\frac{1}{n!}{\int }_0^1{t}^n(1-t{)}^{n+d-1}{r}^{(n)}(t)dt=0$, because $r^{(n)}=0$. Thus ${\varphi }_n$ is orthogonal to every lower-degree polynomial, and the degree conclusion in step 1.1 makes the family an orthogonal basis of each polynomial space.

For nonnegative integers $a,b$, let $B(a,b)={\int }_0^1{t}^a(1-t{)}^{b}dt$. The case $a=0$ gives $B(0,b)=1\mathrm{/}(b+1)$; for $a\ge 1$, integration by parts with $u=t^a$ and $dv=(1-t{)}^{b}dt$ gives $B(a,b)=aB(a-1,b+1)\mathrm{/}(b+1)$. Iterating yields $B(a,b)=a!b!\mathrm{/}(a+b+1)!$. Applying $n$ integrations by parts to the Rodrigues formula with $r(t)=t^n$ gives ${\int }_0^1{w}_d{\varphi }_n{t}^{n}dt=B(n,n+d-1)=n!(n+d-1)!\mathrm{/}(2n+d)!$. Since $t^n={\varphi }_n\mathrm{/}{\kappa }_n+r$ for some $r$ of degree below $n$, where ${\kappa }_n=\left(\genfrac{}{}{0px}{}{2n+d-1}{n}\right)$, orthogonality gives $\mathrm{\parallel }{\varphi }_n{\mathrm{\parallel }}_{L^2(w_d)}^2={\kappa }_{n}B(n,n+d-1)=1\mathrm{/}(2n+d)$.

Write $q={\sum }_{n=0}^p{c}_n{\varphi }_n$. By the endpoint and norm identities, Cauchy–Schwarz gives $\mathrm{\mid }q(0){\mathrm{\mid }}^2={\mid {\sum }_{n=0}^p{c}_n(-1{)}^n\mid }^2\le ({\sum }_{n=0}^p(2n+d))\left({\sum }_{n=0}^p\frac{\mathrm{\mid }{c}_n{\mathrm{\mid }}^2}{2n+d}\right)=(p+1)(p+d)\mathrm{\parallel }q{\mathrm{\parallel }}_{L^2(w_d)}^2$. Equality is attained by the polynomial $K_p(t,0)={\sum }_{n=0}^p(2n+d)(-1{)}^n{\varphi }_n(t)$: its endpoint value and squared norm both equal ${\sum }_{n=0}^p(2n+d)=(p+1)(p+d)$. Hence the evaluation norm squared is exactly the stated value.