An ordinary derivative controls the Legendre tail with a factorial ratio

Fix $k\ge 1$ and put $w_k(x)=(1-x^2{)}^k$. For $n\ge k$, the polynomial $Q_{n,k}:=D^k(w_k{D}^k{P}_n)$ has degree $n$. Integrating by parts $k$ times moves the outer derivative block onto a polynomial $q$ and gives $\int Q_{n,k}q=(-1{)}^k\int w_k{D}^k{P}_n{D}^{k}q=\int P_n{D}^k(w_k{D}^{k}q)$; all boundary terms vanish because derivatives of $w_k$ of orders below $k$ vanish at both endpoints. The final polynomial paired with $P_n$ has degree at most $\mathrm{deg}q<n$, so [F1] makes the integral zero. The leading coefficient of $Q_{n,k}$ is $(-1{)}^k(n+k)!\mathrm{/}(n-k)!$ times the leading coefficient of $P_n$. Orthogonality and the degree statement in [F1] therefore give $Q_{n,k}=(-1{)}^k{\lambda }_{n,k}{P}_n$, where ${\lambda }_{n,k}=(n+k)!\mathrm{/}(n-k)!$.

For $n,m\ge k$, integrate by parts $k$ times and use step 1.1 to obtain ${\int }_{-1}^1{w}_k{P}_n^{(k)}{P}_m^{(k)}=(-1{)}^k{\int }_{-1}^1{P}_n{D}^k(w_k{P}_m^{(k)})={\lambda }_{m,k}{\int }_{-1}^1{P}_n{P}_m\mathrm{.}$ The first identity in [F1] makes this zero for $n\mathrm{\ne }m$ and equal to ${\lambda }_{n,k}2\mathrm{/}(2n+1)$ for $n=m$, proving the repeated derivative orthogonality formula.

Write $c_n=(f,P_n{)}_{L^2}\mathrm{/}\mathrm{\parallel }{P}_n{\mathrm{\parallel }}_2^2$. The absolutely continuous representatives in [F3] justify $k$ integrations by parts; all boundary terms vanish because $w_k$ has endpoint zeros of order $k$. Step 1.1 then gives ${\int }_{-1}^1{w}_k{f}^{(k)}{P}_n^{(k)}={\lambda }_{n,k}{c}_n\mathrm{\parallel }{P}_n{\mathrm{\parallel }}_2^2(n\ge k)\mathrm{.}$ The functions $w_k^{1\mathrm{/}2}{P}_n^{(k)}$ are mutually orthogonal by step 2.1 and have squared norms ${\lambda }_{n,k}\mathrm{\parallel }{P}_n{\mathrm{\parallel }}_2^2$. Applying finite Bessel inequalities to $w_k^{1\mathrm{/}2}{f}^{(k)}$ and then increasing the finite index set yields ${\sum }_{n\ge k}{\lambda }_{n,k}\mathrm{\mid }{c}_n{\mathrm{\mid }}^2\mathrm{\parallel }{P}_n{\mathrm{\parallel }}_2^2\le {\int }_{-1}^1{w}_k\mathrm{\mid }{f}^{(k)}{\mathrm{\mid }}^2\le \mathrm{\parallel }{f}^{(k)}{\mathrm{\parallel }}_2^2\mathrm{.}$

Density [F2] and the best-approximation property of finite orthogonal projections imply ${\mathrm{\Pi }}_{M}f\to f$ in $L^2$: first approximate $f$ by a polynomial and then take $M$ at least its degree. Orthogonality [F1] and Pythagoras give $\mathrm{\parallel }f-{\mathrm{\Pi }}_{N}f{\mathrm{\parallel }}_2^2={\sum }_{n>N}\mathrm{\mid }{c}_n{\mathrm{\mid }}^2\mathrm{\parallel }{P}_n{\mathrm{\parallel }}_2^2$. Since ${\lambda }_{n,k}$ increases for $n\ge k$, step 3.1 bounds this tail by ${\lambda }_{N+1,k}^{-1}\mathrm{\parallel }{f}^{(k)}{\mathrm{\parallel }}_2^2$, which is the stated factorial estimate. When $k=0$, the same finite Pythagoras identity directly gives contraction.