Orthonormal basis in $W^{1,2}([0,1])$

Question

Consider the Hilbertspace $W^{1,2}([0,1])$ (i.e. Sobolev space) with the standard inner product which is defined by: $(f,g) = (f,g)_{L^{2}([0,1])} + (f',g')_{L^{2}([0,1])}$. Here $[0,1]$ is not parametrizing the circle; it is just an interval and the functions on it are not assumed to be periodic. My question is: What is a orthonormal basis of $W^{1,2}([0,1])$? Can one write it down explicitely?

Answer 1

If you change the interval to $[-1,1]$ instead of $[0,1]$, the equivalent inner product would be $$\tag{$*$} (f,g) = (f,g)_{L^2([-1,1])} + \lambda (f',g')_{L^2([-1,1])},$$ where $\lambda = 2$.

One way to get an orthogonal basis is to start with polynomials and apply Gram-Schmidt orthogonalization to them. You get what might be called Sobolev-Legendre polynomials. If you search for Sobolev orthogonal polynomials you find lots of references, of which [MX] seems to be a reasonable review. Digging a bit deeper, one finds [M, Thm.3.3] an explicit recurrence relation for Sobolev-Legendre polynomials $S^\lambda_n(x)$, normalized by $S^\lambda_n(1) = 1$, with respect to the inner product $(*)$:

$$ S^\lambda_n(x) = S^\lambda_{n-2} + a_n (P_n(x) - P_{n-2}(x)) , $$ where $P_n$ are the usual Legendre polynomials and $$ a_n = \sum_{k=0}^{[\frac{n-1}{2}]} \left(\frac{\lambda}{4}\right)^k \frac{1}{(2n)!} \frac{(n+2k-1)!}{(n-2k-1)!} $$

[MX] Francisco Marcellán and Yuan Xu, MR 3360352 On Sobolev orthogonal polynomials, Expo. Math. 33 (2015), no. 3, 308--352.

[M] H. G. Meijer, A short history of orthogonal polynomials in a Sobolev space I. The non-discrete case, Niew Arch. Wisk. 14 (1996), 93--112.

Answer 2

Usual exponents $\exp(2\pi i k x),k\in \mathbb{Z}$, are orthogonal in $W^{1,2}$ (since any two of them are orthogonal in $L^2$ and their derivatives too). They span not the whole $W^{1,2}$, but only a hyperplane of periodic functions (of course trigonometric polynomials are dense in the space of periodic functions even in a stronger sense, say, in $C^1$). We must add a function $h$ orthogonal to all periodic functions, that is, satisfying $\int fh+f'h'=0$ for all periodic $f$. Integrating by parts we reduct this to the conditions $f-f''=0$ and $f'(1)=f'(0)$. This is satisfied by $f(x)=e^{1-x}-e^x$.

Answer 3

Check Computing a family of reproducing kernels for statistical applications, 1996, by Christine Thomas-Agnan