Chebyshev weighted spectral Sobolev scale and projection

For $s\ge 0$, define $H_C^s$ to be the completion of finite real Chebyshev sums $p={\sum }_{n=0}^M{c}_n{T}_n$ in the norm

$$\mathrm{\parallel }p{\mathrm{\parallel }}_{H_C^s}^2=\pi \mathrm{\mid }{c}_0{\mathrm{\mid }}^2+\frac{\pi }{2}\sum _{n=1}^M(1+n^2{)}^s\mathrm{\mid }{c}_n{\mathrm{\mid }}^2\mathrm{.}$$

By Chebyshev polynomials are orthogonal for the arcsine weight, the coefficients of a finite sum are unique and this is a norm. At $s=0$, this norm on finite sums is the arcsine-weighted norm from Arcsine-weighted inner product on the Chebyshev interval. This is the spectral Sobolev scale for the named weight and domain; it is defined by Chebyshev coefficients, and no equivalence with an unweighted derivative-based Sobolev norm is asserted.

The weights are positive by Real powers for positive bases, with the zero-base positive-exponent convention and The exponential is positive and satisfies $\exp (-x)=1\mathrm{/}\exp (x)$. For fixed $s\ge 0$, $x\mapsto x^s$ is nondecreasing on $(0,\mathrm{\infty })$: the derivative formula Continuity and derivatives of positive-base real powers gives $s{x}^{s-1}\ge 0$, and 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 $f(b)-f(a)=f^{\mathrm{\prime }}(c)(b-a)$ gives monotonicity. Thus $(1+n^2{)}^s\ge 1$. For $s>0$, the derivative formula also makes $x\mapsto x^{-s}$ nonincreasing on $(0,\mathrm{\infty })$.

The completion is isometric to the weighted coefficient sequence space

$$\{(c_n{)}_{n\ge 0}:c_n\in \mathbb{R}\text{ for every }n,\pi \mathrm{\mid }{c}_0{\mathrm{\mid }}^2+\frac{\pi }{2}\sum _{n\ge 1}(1+n^2{)}^s\mathrm{\mid }{c}_n{\mathrm{\mid }}^2<\mathrm{\infty }\},$$

as follows. Put $w_0=\pi$ and $w_n=\frac{\pi }{2}(1+n^2{)}^s$ for $n\ge 1$. The map $(c_n)\mapsto (\sqrt{w_n}{c}_n)$ is an isometry onto the real square-summable sequence space; it is onto because every $w_n>0$. Finite-support sequences are dense by truncation, and The real square-summable sequence space is complete gives completeness. Therefore the displayed weighted sequence space is exactly the completion. The notation $u={\sum }_{n\ge 0}{c}_n{T}_n$ denotes its coefficient representation and does not assert pointwise convergence of the series. Its norm is the displayed weighted coefficient sum.

At $s=0$, denote this completion by $L_C^2:=H_C^0$ and define

$$\mathrm{\parallel }u{\mathrm{\parallel }}_C^2:=\mathrm{\parallel }u{\mathrm{\parallel }}_{H_C^0}^2=\pi \mathrm{\mid }{c}_0{\mathrm{\mid }}^2+\frac{\pi }{2}\sum _{n\ge 1}\mathrm{\mid }{c}_n{\mathrm{\mid }}^2\mathrm{.}$$

By the orthogonality lemma, this is the unique norm extension from finite sums of the arcsine-weighted norm of Arcsine-weighted inner product on the Chebyshev interval. Since $(1+n^2{)}^s\ge 1$ for $s\ge 0$, every $H_C^s$ embeds continuously in $L_C^2$, so $\mathrm{\parallel }u{\mathrm{\parallel }}_C$ is defined for every $u\in H_C^s$, including when $s=0$.

If an element has a continuous representative given by a uniformly convergent Chebyshev series, its arcsine integral norm agrees with this completion norm: for continuous $h$, the defining pairing gives $\mathrm{\parallel }h{\mathrm{\parallel }}_C\le \sqrt{\pi }\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}$, so uniform convergence implies convergence in the arcsine norm as well.

For $u={\sum }_{n\ge 0}{c}_n{T}_n$ in this coefficient completion, define the degree-$N$ orthogonal projection by

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

The projection is well-defined by coefficient truncation and the positive coefficient norm above.