Legendre weighted spectral-Sobolev scale and projection

For $s\ge 0$, define $H_L^s$ as the completion of finite Legendre sums $p={\sum }_{n=0}^M{c}_n{P}_n$ in the coefficient norm

$$\mathrm{\parallel }p{\mathrm{\parallel }}_{H_L^s}^2=\sum _{n=0}^M(1+n(n+1){)}^s\frac{2}{2n+1}\mathrm{\mid }{c}_n{\mathrm{\mid }}^2\mathrm{.}$$

The coefficients are unique because the polynomials have distinct degrees by Legendre polynomials have explicit norms and weighted derivative orthogonality. At $s=0$ this norm is exactly the unweighted $L^2([-1,1])$ norm on finite sums, by the same lemma; its completion is denoted $L_L^2$. Define the degree-$N$ orthogonal projection on finite sums by

$${\mathrm{\Pi }}_{N}u=\sum _{n=0}^N{c}_n{P}_n\mathrm{.}$$

It extends to the coefficient completion by truncation. This explicitly coefficient-defined scale uses the Legendre weight $1$ on $[-1,1]$; no equivalence with an unweighted derivative-based Sobolev norm is asserted here.

For fixed $s\ge 0$, $x\mapsto x^s$ is nondecreasing on $(0,\mathrm{\infty })$: its derivative is $s{x}^{s-1}\ge 0$ by Continuity and derivatives of positive-base real powers and real powers are positive by Real powers for positive bases, with the zero-base positive-exponent convention and The exponential is positive and satisfies $\exp (-x)=1\mathrm{/}\exp (x)$, so the mean-value theorem The mean value theorem, as the case $g(x)=x$ of Cauchy's: for $f$ continuous on $[a,b]$ with $a<b$ and differentiable on $(a,b)$ there is $c\in (a,b)$ with $f(b)-f(a)=f^{\mathrm{\prime }}(c)(b-a)$ applies. Thus the spectral factors $(1+n(n+1){)}^s$ are positive and nondecreasing with $n$; the full weights $w_n$ below need not be nondecreasing.

For $w_n=(1+n(n+1){)}^s\frac{2}{2n+1}>0$, the coefficient map $(c_n)\mapsto (\sqrt{w_n}{c}_n)$ is an isometry onto real ${\mathrm{\ell }}^2$. Thus the coefficient model of $H_L^s$ is exactly $\{(c_n{)}_{n\ge 0}:{\sum }_{n\ge 0}(1+n(n+1){)}^s\frac{2}{2n+1}\mathrm{\mid }{c}_n{\mathrm{\mid }}^2<\mathrm{\infty }\}\mathrm{.}$ Finite-support sequences are dense by truncation, and completeness of ${\mathrm{\ell }}^2$ from The real square-summable sequence space is complete makes this the completion.